case study questions on ratio and proportion

Ratios and Proportion Practice Questions

Ratio and proportion hold a special place in competitive exams. The questions from this topic is a very common occurrence. It is important to have enough practice for this questions to answer them in the exam. Here below are the ratios and proportion practice questions.

Browse more Topics under Ratios And Proportions

Propotion Of Quantities
Proportionals (Third, Fourth and Mean)
Comparison of Ratios
Invertendo, and Alternendo
Componendo, and Dividendo
Componendo-Dividendo
Duplicate Ratios

Ratio and Proportion Practice Questions

Part 1: basic ratio and proportion questions..

Directions: In this section, the questions asked are the basic ratio and proportion questions that can be asked in the exam.

1. Divide Rs. 1870 in three parts such that half of the first part, one-third of the second part and one-sixth of the third part are equal.

A. 270, 840, 1160 B. 341, 243, 245 C. 400, 800, 670 D. None of the above

2. A and B are the two alloys of copper and brass prepared by mixing metals in the proportion of 7:2 and 7:11 respectively. If the equal quantities of two alloys are melted to form a third alloy called C, then the proportion of copper and brass in C will be,

A. 5:9 B. 5:7 C. 7:5 D. 9:5

3. The incomes of X and Y are in the ratio of 3:2 and their expenditures are in the ratio of 5:3. If each of them saves Rs. 1000, then, A’s income can be,

A. Rs. 3000 B. Rs. 4000 C. Rs. 9000 D. Rs. 6000

4. The students in three batches in school is in the ratio of 2:3:5. If 20 students in each batch are increased than the ratio changes to 4:5:7. The total number of students in the three before the increase was,

A. 100 B. 10 C. 90 D. 150

5. Divide the amount of Rs. 500 between P, Q, R, and S such that P and Q together get the thrice as much as R and S together. Q gets four times of what R geta and R gets 1.5 times as much as S. Now the value that Q gets will be

A. 75 B. 125 C. 150 D. 300

6. Rs. 2250 is divided among three friends Ajay, Vijay, and Raj in such a way that 1/6th of Ajay’s share, 1/4th of Vijay’s share and 2/5th of Raj’s share is equal. Find Ajay’s share.

A. Rs. 1080 B. Rs. 720 C. Rs. 450 D. Rs. 1240

7. After an increase of 7 in both numerator as well as the denominator, the fraction changes to 3/4. What was the original fraction?

A. 5/12 B. 7/9 C. 2/5 D. 3/8

8. Divide Rs. 680 among P, Q, and R such that P gets 2/3 of what Q gets and Q gets 1/4 of what R gets. Find the share of R.

A. Rs. 480 B. Rs. 360 C. Rs. 420 D. Rs. 300

1. D. None of the above

3. D. Rs. 6000

5. D. 300

6. A. Rs. 1080

8. A. Rs. 480

Part 2: Proportion Practice Questions

Directions: For this section, the questions related to proportions are asked

1. If a/(b + c) = b/(c + a) = c/(a + b), then each fraction will be equal to,

A. (a + b + c)^2 B. ½ C. ¼ D. 0

2. If a:b = c:d, then the value of a2 + b2/c2 + d2 is,

A. ½ B. (a + b)/(c + d) C. (a – b)/(c – d) D. ab/cd

3. If 6×2 + 6y2 = 13xy, what is the ratio of x to y?

A. 1:4 B. 4:5 C. 3:2 D. 1:2

4. If a, b, c, d are in continued proportion then a – d/b – c>/x. What is the value of x?

A. 2 B. 1 C. 0 D. 3

5. If P varies as R, and Q varies as R, then which of the following is false?

A. (P + Q) varies R B. (P – Q) varies 1/R C. √PQ varies R D. PQ varies R2

6. If a and b are positive integers than √2 always lies between:

A. (a + b)/(a – b) and ab B. a/b and (a + 2b)/(a + b) C. a and b D. ab/(a + b) and (a – b)/ab

2. D. ab/cd

5. B. (P – Q) varies 1/R

6. B. a/b and (a + 2b)/(a + b)

Part 3: Miscellaneous ratio and proportion questions

Directions: In this section, various types of ratio and proportion questions are given.

1. If 4 examiners can examine a certain number of answer books in 8 days by working 5 hours a day, for how many hours a would 2 examiners have to work in order to examine twice the number of answer books in 20 days?

A. 8 B. 6 C. 7½ D. 9

2. Three friends decided to rent a farm for Rs 7000 per year. A outs 110 cows in the farm for 3 months, B puts 110 cows for 6 months and C puts 440 cows for 3 months. Find the total percentage of expenditure that A should pay.

A. 20% B. 16.66% C. 14.28% D. 11.01%

3. At constant temperature, the pressure of a definite mass of a gas is inversely proportional to the volume. If the pressure given is reduced by 20%, find the respective change in the volume.

A. +25% B. -25% C. +16.66% D. -16.66%

4. A group of people row a certain course up the river in 84 minutes; they can row the same course downstream in 9 minutes less than they can row it in the still river. How long would they take to row down with the river?

A. 45 or 23 minutes B. 60 minutes C. 19 minutes D. 63 or 12 minutes

5. If 30 men working 7 hours a day can do a piece of work in 18 days, in how many days will 21 men working 8 hours a day do the same work?

A. 30 days B. 22.5 days C. 24 days D. 45 days

6. If the ratio of the sine of the angles of triangles is given as 1:1:√2, then the ratio of the square of the greatest side to the sum of the squares of the other two sides is

A. 3:4 B. 2:1 C. 1:1 D. 1:2

7. A mixture of milk and water are in the ratio of 5:1. By adding 5 liters of water, the ratio of the milk to water becomes 5:2. The quantity of milk in the mixture is:

A. 16 litres B. 25 litres C. 24 litres D. 22.75 litres

2. C. 14.28%

4. D. 63 or 12 minutes

5. B. 22.5 days

case study questions on ratio and proportion

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could someone please explain the following questions and answers 1. Raju and Sanjay had 35% and 45% rupees more than Ajay respectively. What is the ratio of Raju and Sanjay’s money?

A. 7:9 B. 27:29 C. 37:39 D. 27:39

The correct answer is C.

2. Two men earn a yearly salary in the ratio 10:13. If there spending is in the ratio of 4:5 and the man spending lesser of the two saves Rs. 6000 while the other one saves Rs. 8000, then find the salary of the person who is higher paid.

A. Rs. 12000 B. Rs. 14000 C. Rs. 13000 D. Rs. 11000

3. If the ratio of the ages of Priya and Sunanda is 6:5 at present, and after fifteen years from now, the ratio will be changed to 9:8, then find the Priya’s current age.

A. 22 years B. 30 years C. 34 years D. 38 years

The correct answer is B.

4. P, Q, and R played cricket. P’s runs are to Q’s runs and Q’s runs are to R’s runs at 3:2. All of them scored a total of 342 runs. How many runs did P make?

A. 140 B. 154 C. 168 D. 162

The correct answer is D.

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Class vii math, class 7 ratio and proportion, introduction to ratio, properties of ratio, ratio in the simplest form, comparison of ratios, equivalent ratio, continued proportion.

Ratio and Proportion Test

Ratio and Proportion Worksheet

Answer Sheet

Ratio is a comparison of measures of two or more quantities of the same kind by division. For example, if 'p' and 'q' are two quantities of the same kind and having same units, then the p ⁄ q is called the ratio of 'p' to 'q'.

The ratio of p to q can be denoted as p ⁄ q or p : q.

Here 'p' and 'q' are known as terms of the ratio. The 'p' is known as first term or antecedent and 'q' is known as the second term or consequent.

A ratio does not alter, if it's first and second terms are multiplied or divided by the same non-zero number.

p ⁄ q = ap ⁄ aq (p : q = ap : aq, where a ≠ 0)
p ⁄ q = p÷a ⁄ q÷a (p : q = p ⁄ q : p ⁄ q , where a ≠ 0)

A ratio p : q is said to be in the simplest form if it's first term 'p' and second term 'q' have no common factor other than 1. Steps to convert a ratio into it's simplest form.

Step ‐ 1. Convert both the terms of the ratio into same unit. Step ‐ 2. Find out the HCF of first and second terms. Step ‐ 3. Divide the first and second terms by their HCF

Let's see some examples.

Example 1. Find the ratio 5 m to 75 cm. Solution. First, we must convert both the terms into same unit. 5 m = 500 cm New ratio is 500 : 75. Find HCF of 500 and 75. 500 = 2 × 2 × 5 × 5 × 5 75 = 3 × 5 × 5 HCF of 500 and 75 = 25 500 ⁄ 75 = 500÷25 ⁄ 75÷25 = 20 ⁄ 3

Example 2. Express the 56 : 160 in its simplest form. Solution. First find out the HCF of 56 and 160. 56 = 2 × 2 × 2 × 7 160 = 2 × 2 × 2 × 2 × 2 × 5 So, HCF of 56 and 160 = 2 × 2 × 2 = 8 Now, divide 56 and 160 by their HCF to get its simplest form. 56 : 160 = 56 ⁄ 160 = 56÷8 ⁄ 160÷8 = 7 ⁄ 20 = 7 : 20

Example 3. Divide Rs. 2500 between Julie and Viona in the ration 2 : 3. Solution. Sum of both the terms of the ratio = 2 + 3 = 5 Julie's share would be = 2 ⁄ 5 × 2500 = Rs. 1000 Viona's share would be = 3 ⁄ 5 × 2500 = Rs. 1500 So, Julie and Viona would get Rs. 1000 and Rs. 1500 respectively.

Example 4. Divide 15000 tons of rice among Odisha, west Bengal and Bihar in the ratio of 4 : 5 : 6. Solution. Sum of three terms of the ratio = 4 + 5 + 6 = 15 Odisha's share would be = 4 ⁄ 15 × 15000 = 4000 tons West Bengal's share would be = 5 ⁄ 15 × 15000 = 5000 tons Bihar's share would be = 6 ⁄ 15 × 15000 = 6000 tons So, Odisha, West Bengal and Bihar would get 4000 tons, 5000 tons and 6000 tons respectively.

Example 5. Ratio of the boys and girls in a school is 3 : 5. If there are 120 boys in the school, then find out the girls strength. Solution. Let's assume total students in the school is 'm'. 3 ⁄ 8 × m = 120 => m = 120 × 8 ⁄ 3 => m = 40 × 8 => m = 320 Number of students in the school is 320. Number of girls in the school = 320 − 120 = 200

Example 6. What must be added to each term of the ratio 2 : 3 so that it may become equal to 5 : 9? Solution. Let the number to be added is 'a'. (2 + a) : (3 + a) = 5 : 9 => 2+a ⁄ 3+a = 5 ⁄ 9 => 9 × (2 + a) = 5 × (3 + a) => 18 + 9a = 15 + 5a => 4a = -3 => a = -3 ⁄ 4

To compare two given ratios, we must follow the following steps. Step ‐ 1. Convert both the ratios into fraction in the simplest form. Step ‐ 2. Find the LCM of the denominators of the fractions generated from step 1. Step ‐ 3. Obtain both the fractions and their denominators. Divide the LCM by the denominators of both the fractions. Multiply the individual results with it's corresponding numerator and denominators. Step ‐ 4. Compare the numerators of both the fractions obtained in step 3, fraction having larger numerator will be greater than the other. Let's see some examples.

Example 1. Compare the ratios 5 : 9 and 2 : 3. Solution. Write both the ratios in fraction form. 5 : 9 = 5 ⁄ 9 and 2 : 3 = 2 ⁄ 3 Now, LCM of 9 and 3 is equal to 9. Here, we must make both the fraction's denominator as 9. 5 ⁄ 9 has denominator 9 already, so we do not have to do anything for it. 2 ⁄ 3 = 2×3 ⁄ 3×3 = 6 ⁄ 9 5 ⁄ 9 6 ⁄ 9 Hence 5 ⁄ 9 2 ⁄ 3

Example 2. Compare the ratios 4 : 9 and 5 : 24. Solution. Fraction form of both the ratios are 4 : 9 = 4 ⁄ 9 and 5 : 24 = 5 ⁄ 24 LCM of 8 and 24 is equal to 72 Here, we must make both the fraction's denominator as 72. 4 ⁄ 9 = 4×8 ⁄ 9×8 = 32 ⁄ 72 5 ⁄ 24 = 5×3 ⁄ 24×3 = 15 ⁄ 72 32 ⁄ 72 > 15 ⁄ 72 Hence, 4 ⁄ 9 > 5 ⁄ 24

A ratio obtained by multiplying or dividing the numerator and denominator of a ratio by same number is known as equivalent ratio. Let's see one example to understand it better.

Example 1. Find the equivalent ratio of 3 : 5. Solution. First convert the ratio into fraction. 3 : 5 = 3 ⁄ 5 3 ⁄ 5 = 3×2 ⁄ 5×2 = 3×3 ⁄ 5×3 = 3×4 ⁄ 5×4 and so on. 3 ⁄ 5 = 6 ⁄ 10 = 9 ⁄ 15 = 12 ⁄ 20

Four numbers 'p', 'q', 'r' and 's' are said to be in proportion, if the ratio of 'p' and 'q' is equal to the ratio of 'r' and 's'. p : q = r : s In other words p : q = r : s if and only if ps = qr Let's see some examples.

Example 1. Check if 5 : 15 and 7 : 21 are in proportion. Solution. 5 : 15 = 1 : 3 and 7 : 21 = 1 : 3 Hence, 5 : 15 = 7 : 21 is a proportion.

Example 2. Check if 6 : 9 and 30 : 45 are in proportion. Solution. Let's convert both the ratios into simplest form. 6 : 9 = 2 : 3 and 30 : 45 = 2 : 3 Hence, 6 : 9 = 30 : 45 is a proportion.

Three numbers 'p', 'q', and 'r' are said to be in continued proportion if 'p', 'q', 'q', 'r' in proportion. P : q = q : r If 'p', 'q', 'q', 'r' are in proportion i.e. p : q = q : r p ⁄ q = q ⁄ r => pr = q 2 => q 2 = pr Let's see some examples on proportion.

Example 1. Check if 30, 40, 60, 80 are in proportion. Solution. 30 : 40 = 30 ⁄ 40 = 3 ⁄ 4 60 : 80 = 60 ⁄ 80 = 3 ⁄ 4 Hence, 30 : 40 = 60 : 80 are in proportion.

Example 2. Check if 4, 16, 36 are in continued proportion? Solution. We know that three numbers p, q, r in continued proportion, if p, q, q, r are in proportion. So, in this case 4, 16, 36 will be in continued proportion if 4, 16, 16, 36 are in proportion. We have to prove that 4 ⁄ 16 should be equal to 16 ⁄ 36 . 4 ⁄ 16 = 1 ⁄ 4 and 16 ⁄ 36 = 4 ⁄ 9 It is proved that 4 ⁄ 16 ≠ 16 ⁄ 36 So, 4, 16, 36 are not in continued proportion.

Example 3. The first three terms of a proportion are 2, 5, and 32 respectively. Find the fourth term. Solution. Let's assume the fourth term to be 'y'. 2 ⁄ 5 = 32 ⁄ y => 2y = 32 × 5 => y = 32×5 ⁄ 2 => y = 16 × 5 => y = 80 So, the fourth number is 80.

Example 4. What must be added to the numbers 25, 35, 55, 75 so that they are in proportion? Solution. Let's assume the required number is 'p'. Then 25 + p, 35 + p, 55 + p, 75 + p are in proportion. 25+p ⁄ 35+p = 55+p ⁄ 75+p => (25 + p)(75 + p) = (55 + p)(35 + p) => 25 × 75 + 25p + 75p + p 2 = 55 × 35 + 55p + 35p + p 2 => 1875 + 100p + p 2 = 1925 + 90p + p 2 => 100p − 90p = 1925 − 1875 => 10p = 50 => p = 50 ⁄ 10 => p = 5 So, the number is 5.

Class-7 Ratio and Proportion Test

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Class-7 Ratio and Proportion Worksheet

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Arinjay Academy » Maths » Ratio and Proportion Problems and Solutions for Class 7

Ratio and Proportion Problems and Solutions for Class 7

Ratio and proportion problems and solutions for class 7, deals with various concepts which are as under:-.

Convert Ratio into its simplest form
Ratio of two quantities, by converting them into same units
Equivalent ratios
Find the numbers when their ratio and sum are given
Divide sum of money between two persons when ratio are given
Comparison of ratios
Are the given ratio in proportion
Are the given number are in proportion
Find the value of y when four numbers are in proportion
Finding Ratio A : B : C
Finding Ratio A : B : C if Ratio A : B and B : C given
Converting Ratio into another Ratio
Finding ratio of present ages
Finding ratio of ages of persons after some years
Finding ratio of ages of persons some year ago.
Third proportional value
Mean proportional value
Continued proportion

In order to convert the given ratio to Simplest Form, we should follow the following steps : –

Find the HCF of both the numerator and denominator
Dividing Both numbers by their HCF

The result is the ratio in its simplest form.

Question 1 :

Convert the ratio 66 : 18 in its simplest form

HCF of 66 and 18 is 6

Since, 66 : 18

Hence, the simplest form of 66 : 18 is 11 : 3

Ratio and Proportion Problems and Solutions for Class 7 – Ratio of two quantities, by converting them into same units

Question 2 :

Find the ratio of 48 min to 4 hours

Taking both the quantities in same unit, we have

4 hours = ( 4 x 60 ) = 240 min

The equation now becomes 48 min : 240 min

Hence, the required ratio is 1 : 5

Ratio and Proportion Problems and Solutions for Class 7 – Equivalent ratios

In order to find Equivalent Ratios of any given ratio, we multiply or divide the numerator and denominator of the ratio by the same non zero number.

Question 3 :

Find the Equivalent ratio of 6 : 7 ?

On multiplying or dividing each term of a ratio by the same non zero number, we get a ratio equivalent to the given ratio

Both numerator and denominator of given fraction is multiplied by same non zero number i.e 4

24/28 is an equivalent ratio of 6/7

7/6 is not an equivalent ratio of 6/7. As both 6 and 7 are not mutiply by

same non zero number

25/21 is not an equivalent ratio of 6/7. As both 6 and 7 are not mutiply

by same non zero number

22/21 is not an equivalent ratio of 6/7. As both 6 and 7 are not mutiply

Ratio and Proportion Problems and Solutions for Class 7 – Find the numbers when their ratio and sum are given

Question 4 :

Two numbers are in the ratio 5 : 7 and their sum is 120. Find the numbers?

Let the required number be 5a and 7a

Since the sum of these two numbers is given, we can say that

5a + 7a = 120

So, the first number is 5a = 5 x 10

Second number is 7a = 7 x 10

Hence, two numbers are 50 and 70

Ratio and Proportion Problems and Solutions for Class 7 – Divide sum of money between two persons when ratio are given

Question 5 :

Divide ₹ 2000 between X and Y in the ratio 5 : 3

Total money = ₹ 2000

Given ratio = 5 : 3

Sum of ratio terms = ( 5 + 3 )

Give: 5/8 part of ₹ 2000 to X

Give: 3/8 part of ₹ 2000 to Y

X ‘s share = ₹ ( 2000 x 5/8) = ₹ 1250

Y ‘s share = ₹ ( 2000 x 3/8) = ₹ 750

Ratio and Proportion Problems and Solutions for Class 7 – Divide sum of money among three persons when ratio are given

Question 6 :

Divide ₹ 5000 among X , Y and Z in the ratio 1 : 2 : 7

Total money = ₹ 5000

Given ratio = 1 : 2 : 7

Sum of ratio terms = ( 1 + 2 + 7 )

Share of X = ₹ ( 5000 x 1/10) = ₹ 500

Share of Y = ₹ ( 5000 x 2/10) = ₹ 1000

Share of Z = ₹ ( 5000 x 7/10) = ₹ 3500

Ratio and Proportion Problems and Solutions for Class 7 – Comparison of ratios

To Compare two Ratios, we should follow the following steps : –

Write both the Ratios as Fractions
Convert both the Fractions into Like Fraction:- – Find the L.C.M of denominator of both the Fractions – Make the denominator of each fraction equal to their L.C.M.
In case of Like fractions, the number whose numerator is greater is larger.

Question 7 :

Compare the ratios ( 3 : 4 ) and ( 4 : 3 )

We can write

( 3 : 4 ) = 3/4 and ( 4 : 3 ) = 4/3

Now, let us compare 3/4 and 4/3

LCM of 4 and 3 is 12

Making the denominator of each fraction equal to 12

In case of Like fractions, the number whose numerator is greater is larger. Hence we can say 9/12 < 16/12

That is, 3/4 < 4/3

Hence, ( 3 : 4 ) < ( 4 : 3 )

Ratio and Proportion Examples With Answers – Proportion

When Two Ratios are equal then we say that they are in Proportion and use the symbol “: :” or “=” to equate two ratios.

Four Numbers in Proportion

Let a, b, c, d are four numbers said to be in proportion. then, a : b = c : d or a : b :: c : d here a and d are called the extreme terms or extremes. b and c are called the middle terms or means. When Four numbers are in proportion then, Product of extremes = Product of means. i.e, In proportion a : b :: c : d, (a x d) = (b x c)

Ratio and Proportion Problems and Solutions for Class 7 – Are the given ratio in proportion

Question 8 :

Are the ratios 15 m : 45 m and 30 km : 90 km in proportion?

We have 15 m : 45 m

30 km : 90 km

Since, the ratios 15 m : 45 m and 30 km : 90 km are equal to 1/3. So, they are in proportion.

Ratio and Proportion Problems and Solutions for Class 7 – Are the given number are in proportion

Question 9 :

Are 3, 6, 5, 15 in proportion?

Product of means = Product of extremes

Here, Means are 6 and 5

Extremes are 3 and 15

Product of extremes = 3 x 15 = 45

Product of means = 6 x 5 = 30

Since, Product of extremes ≠ Product of means

Hence, 3 , 6 , 5 , 15 are not in Proportion

Ratio and Proportion Problems and Solutions for Class 7 – Find the value of y when four numbers are in proportion

Question 10 :

If 20 : 12 : : y : 6, find the value of y?

We know that, Product of means = Product of extremes

In the given numbers, we can say that 12 , y are means and 20 , 6 are extremes

12 x y = 20 x 6

Hence, y = 10

Question 11 :

If 45 : y : : y : 5, find the value of y?

Clearly, Product of means = Product of extremes

y x y = 45 x 5

y² = 45 x 5

Hence, y = 15

Ratio and Proportion Problems and Solutions for Class 7 – Finding Ratio A : B : C

Question 12 :

If 3A = 5B = 4C , find A : B : C?

3A = 5B = 4C = k

This implies that 3A = k

Also if 5B = k

Further, if 4C = k

A : B : C = k/3 : k/5 : k/4

LCM of 3 , 5 , 4 is 60

Multiplying each of the ratio by 60/k we get the ratios as

= 20 : 12 : 15

Hence, A : B : C = 20 : 12 : 15

Ratio and Proportion Problems and Solutions for Class 7 – Finding Ratio A : B : C if Ratio A : B and B : C given

Question 13 :

If A : B = 4 : 9 and B : C = 12 : 2, find A : B : C

Given A : B = 4 : 9

and B : C = 12 : 2

To find A : B : C we have to make the value of common term in both the ratios equal

that is, B = 9

For this B : C = 1 : 2/12 ( On dividing each term by 12 )

B : C = ( 9 : 2/12 x 9 ) (On multiplying each term by 9 )

B : C = 9 : 18/12

We get, B : C = 9 : 3/2

Since, A = B = 4 : 9 and B : C = 9 : 3/2

Therefore, A : B : C = 4 : 9 : 3/2

Hence, A : B : C = 8 : 18 : 3 (Multiplying each term by 2 )

Ratio and Proportion Problems and Solutions for Class 7 – Converting Ratio into another Ratio

Question 14 :

What must be added to each term of the ratio 7 : 9 so that the new ratio becomes 7 : 8 ?

Let the required number to be added be ‘ a ‘

then, ( 7 + a ) : ( 9 + a ) = 7 : 8

8 ( 7 + a ) = 7 ( 9 + a )

56 + 8a = 63 + 7a

8a – 7a = 63 – 56

Hence, the required number is 7

Ratio and Proportion Problems and Solutions for Class 7 – Finding ratio of present ages

Question 15 :

Present age of Sanjay is 52 years and the age of his Daughter is 28 years. Find the ratio of present age of Daughter to the present age of Sanjay ?

Present age of Sanjay = 52 years

Present age of Daughter = 28 years

To find the ratio of present age of Daughter to the present age of Sanjay

Hence, the ratio of the present age of Daughter to Sanjay is 7 : 13

Ratio and Proportion Problems and Solutions for Class 7 – Finding ratio of ages of persons after some years

Question 16 :

Present age of Seema is 22 years and the age of her Brother is 26 years. Find the ratio of Seema’s age to her Brother’s age after 10 years.

Present age of Seema = 22 years

After 10 years Seema’s age = 22 + 10 = 32 years

Present age of Brother = 26 years

After 10 years Brother’s age = 26 + 10 = 36 years

Ratio of age of Seema and Brother after 10 years

Taking, HCF of 32 and 36 is 4

Hence, the ratio of Seema’s age to her Brother’s age after 10 years is 8 : 9

Ratio and Proportion Problems and Solutions for Class 7 – Finding ratio of ages of persons some year ago.

Question 17 :

Present age of Ritu is 23 years and the age of her Brother is 18 years. What was the ratio of Ritu’s age to her Brother’s age 3 years ago?

Present age of Ritu = 23 years

Ritu’s age 3 years ago = 23 – 3 = 20 years

Present age of Brother = 18 years

Brother’s age 3 years ago = 18 – 3 = 15 years

Ratio of age of Ritu and Brother 3 years ago

Hence, the ratio of Ritu’s age to Brother’s age is 4 : 3

Ratio and Proportion Problems and Solutions for Class 7 – Third proportional value

Let say a and b be two numbers and c is in third proportion with a and b

a : b = b : c

Question 18 :

Find the third proportion to 6 and 12?

Let, the third proportion to 6 and 12 be a

6 : 12 :: 12 : a

(Product of Extremes= Product of Means)

Here, Extremes are= 6 and a

Means are = 12 and 12

6 x a = 12 x 12

Hence, the value of ‘a’ is 24

Ratio and Proportion Problems and Solutions for Class 7 – Mean proportional value

Let say a and b be two numbers and c is mean proportional between a and b

a : c = c : b

Question 19 :

Find the mean proportional between 10 and 40 ?

Let, the mean proportional between 10 and 40 be ‘ a ‘

10 : a :: a : 40

(Product of extremes = Product of means)

Here, Extremes are 10 and 40

Means are a and a

10 x 40 = a x a

a² = 10 x 40 = 400

Hence, the value of ‘a’ is 20

Ratio and Proportion Problems and Solutions for Class 7 – Continued proportion

Three numbers are said to be in Continued Proportion if the ratio of first and second number is equal to the ratio of second and third number.

If a, b, c are in continued proportion

a : b : : b : c

Question 20 :

If 12, 42, a are in continued proportion, find the value of a?

Given – 12 , 42 , a are in continued proportion.

12 : 42 :: 42 : a

Here, Extremes are 12 and a

Means are 42 and 42

12 x a = 42 x 42

Hence, the value of ‘a’ is 147

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Cbse notes for class 7, worksheets for class 7, 4 thoughts on “ratio and proportion problems and solutions for class 7”.

Amazing!! Helped me a lot in my test !! Thanks a lot

YOU GIVE TO EXCELLENT HELP

Really nice worksheet there so many variety questions helps us to practice a lot .keep it up

Many Thanks Yuvraj

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Important Questions Class 6 Maths Chapter 12 – Ratio and Proportion

Maths is an important subject that is taught in school. Chapter 12 of Class 6 Maths is about ratios and proportions. The ratio is a way of comparison. It is a unitless comparison between measures. On the other hand, when two ratios are equal, they are called proportions.

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The chapter deals with the basic ideas of ratio and proportion. It teaches students how to calculate ratios and the different relations between them. It also gives the idea about proportion. Students must practice questions from this chapter because the concepts are new to them.

Extramarks is a leading company that provides all the study materials related to CBSE and NCERT. Our experts have made the Important Questions Class 6 Maths Chapter 12 so students can practice questions. They collected the questions from different sources such as the textbook exercises, CBSE sample papers , CBSE past years’ question papers and important reference books.

Extramarks provide all the important study materials needed by students. You may register on our official website and download these study materials. You will find the CBSE syllabus, CBSE sample questions, CBSE past years’ question papers, CBSE extra questions, CBSE revision notes, NCERT books, NCERT exemplar, NCERT solutions, NCERT important questions, vital formulas and many more.

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Also, get access to CBSE Class 6 Maths Important Questions for other chapters too:

Important Questions Class 6 Maths Chapter 12 – With Solutions

The experts of Extramarks have made this question series by taking help from several sources. They have taken help from the textbook exercises, CBSE sample papers, CBSE past years’ question papers, important reference books and NCERT exemplar. They have also solved the questions, and experienced professionals have further checked the answers. Thus, students will find the important questions and solutions if they follow the Important Questions Class 6 Maths Chapter 12. The vital questions are-

Question 1. There are a total of 20 girls and 15 boys in the classroom.

(i) What is the final ratio of the number of girls present to the number of boys?

(ii) What is the final ratio of the number of girls present to the total number of students present in the class?

Number of girls present = 20 girls

Number of boys present = 15 boys

Total number of students present = 20 + 15

(i) Ratio of the number of girls present to the number of boys is = 20 / 15 = 4 / 3

(ii) Ratio of the number of girls present to the total number of students is = 20 / 35 = 4 / 7

Question 2. Out of 30 (thirty) students in a class, six students like football, 12 students like cricket and the remaining students like tennis. Find the ratio of

(i) Number of students that like football to number of students liking tennis.

(ii) Number of students liking cricket to total number of students.

The number of students who like football is = 6

The number of students who like cricket is = 12

The number of students who like tennis is = 30 – 6 – 12

(i) Ratio of the Number of students that like football to the number of students that like tennis

= 6 / 12 = 1 / 2

(ii) Ratio of the Number of students who like cricket to the total number of students who are

Question 3. In a year, Seema earns around ₹ 1,50,000 and saves ₹ 50,000. So, find the ratio of

(i) Money that Seema earns to the money she saves

(ii) Money that she saves to the money she spends.

Money earned by Seema is = ₹ 150000

Money saved by her is = ₹ 50000

The Money spent by her = ₹ 150000 – ₹ 50000 = ₹ 100000

(i) Ratio of money earned to the money saved = 150000 / 50000 = 15 / 5

(ii) Ratio of money saved to the money spent = 50000 / 100000 = 5 / 10

Question 4. Find the ratio of the following statements:

(i) 81 to 108

(ii) 98 to 63

(iii) 33 km to 121 km

(iv) 30 minutes to 45 minutes

(i) 81 / 108 is equal to = (3 × 3 × 3 × 3) / (2 × 2 × 3 × 3 × 3)

(ii) 98 / 63 = (14 × 7) / (9 × 7)

(iii) 33 / 121 = (3 × 11) / (11 × 11)

(iv) 30 / 45 = (2 × 3 × 5) / (3 × 3 × 5)

Question 5. Find the ratio of the following statements:

(i) 30 minutes to 1.5 hours

(ii) 40 cm to 1.5 m

(iii) 55 paise to ₹ 1

(iv) 500 ml to 2 litres

30 min = 30 / 60

= 0.5 hours

Required ratio = (0.5 × 1) / (0.5 × 3)

1.5 m = 150 cm

Required ratio = 40 / 150

₹ 1 = 100 paise

Required ratio becomes = 55 / 100 = (11 × 5) / (20 × 5)

1 litre = 1000 ml

2 litre = 2000 ml

Required ratio becmes = 500 / 2000 = 5 / 20 = 5 / (5 × 4)

Question 6. There is a total of 102 teachers in a school of 3300 no students. Find the final ratio of the number of teachers present to the number of students.

Total number of teachers in a school = 102

The total number of students in a school is = 3300

The ratio of the number of teachers present to the number of students becomes = 102 / 3300

= (2 × 3 × 17) / (2 × 3 × 550)

Question 7. In a college, out of 4320 students present, 2300 students are girls. And the rest are boys. Find the ratio of

(i) Number of girls to the total number of students.

(ii) Number of boys to the number of girls.

(iii) Number of boys to the total number of students.

It is Given

The total number of students is = 4320

Number of girls present = 2300

Number of boys present = 4320 – 2300

(i) Ratio of the Number of girls present to the total number of students present = 2300 / 4320

= (2 × 2 × 5 × 115) divided by (2 × 2 × 5 × 216), which will become,

= 115 / 216

(ii) Ratio of the Number of boys present to the number of girls present = 2020 / 2300

= (2 × 2 × 5 × 101) divided by (2 × 2 × 5 × 115)

= 101 / 115

(iii) Ratio of the Number of boys present to the total number of students present = 2020 / 4320

= (2 × 2 × 5 × 101) divided by (2 × 2 × 5 × 216)

= 101 / 216

Question 8. Out of 1800 students present in a school, 750 students opted for basketball, 800 students opted for cricket, and the remaining students opted for table tennis. If a student can choose only one single game, then find the ratio of

(i) Number of students who opted for the sport basketball to the number of students who chose the sport table tennis.

(ii) Number of students who opted for the sport cricket to the number of students opting for the sport basketball.

(iii) Number of students who opted for the sport basketball to the total number of students.

(i) Ratio of the Number of students who opted for the sport basketball to the number of students that opted for the sport table tennis = 750 / 250 = 3 / 1

(ii) Ratio of the Number of students who opted for the sport cricket to the number of students opting for basketball

= 800 / 750 = 16 / 15

(iii) Ratio of the Number of students who opted for the sport basketball to the total number of students

= 750 / 1800 = 25 / 60 = 5 / 12

Question 9. Divide 20 pens between Sheela and Seema in a ratio of 3: 2.

Terms of 3: 2 = 3 and 2

Sum of these terms = 3 + 2

Hence Sheela will get 3 / 5 of the total pens, and Seema will get 2 / 5 of the total pens.

Number of pens present with Sheela = 3 / 5 × 20

Number of pens having Seema = 2 / 5 × 20

Question 10. A Mother wants to divide ₹ 36 between her two daughters, Shreya and Bhavna, in the ratio of their respective ages. If the age of Shreya is 15 years and the age of Bhavna is 12 years, find how much money Shreya and Bhavna will receive.

The ratio of ages of Shreya and bhavna = 15 / 12

Hence, the mother wants to divide the ₹ 36 in the ratio of 5: 4

Terms of 5: 4 are denoted as 5 and 4

The Sum of these terms will be = 5 + 4

Here Shreya will get 5 / 9 of the total money, and Bhavna will get 4 / 9 of the total money.

The amount Shreya get = 5 / 9 × 36

The amount Bhavna get = 4 / 9 × 36

Therefore Shreya will get ₹ 20, and Bhavna will get ₹ 16

Question 11. The present age of the father is 42 (forty-two) years, and that of his son is 14 years. Hence Find the ratio of

(i) Present age of the father to the present age of the son

(ii) Age of the father to the age of his son, when the son was 12 years old.

(iii) Age of the father to the age of the son after 10 (ten) years.

(iv) Age of the father to the age of his son, when the father was 30 (thirty) years old, will be.

(i) Present age of father = 42 years

The present age of son = is 14 years

Required ratio 42 / 14

(ii) The son was 12 years old two years ago. Hence the age of the father two years ago will be,

= 42 – 2 = 40 years

Required ratio will be = 40 / 12 = (4 × 10) / (4 × 3) = 10 / 3

(iii) After ten years the age of father will be = 42 + 10 = 52 years

After 10 years the age of son becomes = 14 + 10 = 24 years

Required ratio will be = 52 / 24 = (4 × 13) / (4 × 6)

(iv) 12 years ago, age of father was 30 years

At that time, the age of son was = 14 – 12

Required ratio = 30 / 2 = (2 × 15) / 2

Question 12. Determine if the following are in proportion.

(a) 15, 45, 40, 120

(b) 33, 121, 9, 96

(d) 32, 48, 70, 210

(e) 4, 6, 8, 12

(f) 33, 44, 75, 100

15 / 45 = 1 / 3

40 / 120 = 1 / 3

Hence, 15: 45 = 40:120

∴ These are in a proportion

33 / 121 = 3 / 11

9 / 96 = 3 / 32

Hence 33:121 ≠ 9: 96

∴ These are not in a proportion

24 / 28 = 6 / 7

36 / 48 = 3 / 4

Hence, 24: 28 ≠ 36:48

∴ These are not in proportion.

32 / 48 = 2 / 3

70 / 210 = 1 / 3

Hence, 32: 48 ≠ 70: 210

4 / 6 = 2 / 3

8 / 12 = 2 / 3

Hence 4: 6 = 8: 12

33/ 44 = 3/ 4

75 / 100 = 3 / 4

Hence, 33:44 = 75: 100

∴ These are in proportion.

Question 13 . Are the following statements true?

(i) 40 persons : 200 persons = ₹ 15 : ₹ 75

(ii) 7.5 litres : 15 litres = 5 kg : 10 kg

(iii) 99 kg : 45 kg = ₹ 44 : ₹ 20

(iv) 32 m : 64 m = 6 sec : 12 sec

(v) 45 km : 60 km = 12 hours : 15 hours

40 / 200 = 1 / 5

15 / 75 = 1 / 5

Hence, True

7.5 / 15 = 1 / 2

5 / 10 = 1 / 2

99 / 45 = 11 / 5

44 / 20 = 11 / 5

32 / 64 = 1 / 2

6 / 12 = 1 / 2

45 / 60 = 3 / 4

12 / 15 = 4 / 5

Hence, False

Question 14. Determine if the following ratios form a proportion. Also, write the middle terms and the extreme terms where the ratios form a proportion.

(i) 25 cm : 1 m and ₹ 40 : ₹ 160

(ii)39 litres: 65 litres and six bottles : 10 bottles

(iii) 2 kg : 80 kg and 25 g : 625 g

(i) 200 mL : 2.5 litre and ₹ 4 : ₹ 50

25 cm = 25 / 100 m

0.25 / 1 = 1 / 4

40 / 160 = 1 / 4

Yes, these are in a definite proportion.

Middle terms are said to be 1 m, ₹ 40, and Extreme terms are found to be 25 cm, ₹ 160

(ii) 39 litres : 65 litres and 6 bottles : 10 bottles

39 / 65 = 3 /5

6 / 10 = 3 / 5

Yes, these are in definite proportion

Middle terms are said to be 65 litres, 6 bottles, and the Extreme terms are 39 litres, 10 bottles.

2 / 80 = 1 / 40

25 / 625 = 1 / 25

No, these are not in a definite proportion

(iv) 200 mL : 2.5 litre and ₹ 4 : ₹ 50

2.5 litre = 2500 ml

200 / 2500 = 2 / 5

4 / 50 = 2 / 25

Middle terms are said to be 2.5 litres, ₹ 4 and the Extreme terms are 200 ml, ₹ 50

Question 15. Esha earns ₹ 3000 in 10 (ten) days. How much will she earn in 30 (thirty) days?

Money earned by Esha in 10 days = ₹ 3000

Money earned by her in one day = 3000 / 10

So, the total money earned by her in 30 days = 300 × 30

Question 16. If it has rained 276 mm in the previous 3 days, how many cm of rain will fall in the next one full week (7 days)? Assuming that the rain continues to fall in the same rate.

Measure of rain in the previous 3 days = 276 mm

Measure of rain in one single day = 276 / 3

So, a measure of rain in one week which is, 7 days = 92 × 7

Question 17. The Cost of 5 kg of wheat is ₹ 91.50.

(i) What will be the cost of 8 kg of wheat?

(ii) What quantity of wheat can be purchased in ₹ 183?

(i) Cost of 5 kg wheat is = ₹ 91.50.

Cost of 1 kg wheat is = 91.50 / 5

So, cost of 8 kg wheat is = 18.3 × 8

(ii) Wheat purchased in ₹ 91.50 is = 5 kg

Wheat purchased in ₹ 1 is = 5 / 91.50 kg

So, wheat purchased in ₹ 183 is = (5 / 91.50) × 183

Question 18. Shreya pays ₹ 15000 as rent for 3 months. How much does Shreya have to pay for a whole year, given that the rent per month remains the same?

Rent paid by Shreya in 3 months = ₹ 15000

Rent for 1 month is = 15000 / 3

So, rent for 12 months which is 1 year, becomes = 5000 × 12

∴ Rent paid by Shreya in 1 year is ₹ 60,000

Question 19. The cost of 4 dozen bananas is ₹ 180. How many bananas can be purchased for ₹ 90?

Number of bananas that are bought in ₹ 180 = 4 dozens

= 48 bananas

The number of bananas bought in ₹ 1 is = 48 / 180

So, the number of bananas bought in ₹ 90 will be = (48 / 180) × 90

= 24 bananas

∴ 24 bananas can be purchased for ₹ 90

Question 20. The weight of 72 books is 9 kg. So What is the weight of 40 such books?

The weight of 72 books is = 9 kilograms.

The weight of 1 book is = 9 / 72

So, the weight of 40 books becomes = (1 / 8) × 40

∴ The weight of 40 books is 5 kilograms.

Question 21. A truck requires at least 108 litres of Diesel to cover a distance of at least 594 km. So How much Diesel will be required by the same truck to cover a distance of around 1650 km?

The diesel required for 594 km is = 108 litres

Diesel required for 1 km is = 108 / 594

= 2 / 11 litre

So, diesel required for 1650 km is = (2 / 11) × 1650

= 300 litres

∴ Hence Diesel required by the truck to cover a distance of at least 1650 km is 300 litres

Benefits of Solving Important Questions Class 6 Maths Chapter 12

Practice is very important for students. It clears their concepts and generates interest in the subject matter. Many students fear maths because they do not understand the subject matter. Practice is really important to clear their doubts and boost confidence. The experts of Extramarks have made the Important Questions Class 6 Maths Chapter 12 to help students solve questions. There will be multiple benefits to solving the questions. These are-

The experts have collected the questions from sources like textbook exercises, CBSE sample papers, NCERT exemplars and important reference books. They have included a few questions from CBSE past years’ question papers so that students may have an idea regarding questions in exams. Thus, the Maths Class 6 Chapter 12 Important Questions will help students solve questions regularly.
The experts have solved the questions too. They have given a step-by-step solution to each question so that students can easily understand the solution. Thus, the students can follow the answers if they cannot solve a question. They can check their answers with the experts’ answers too. Experienced professionals have further checked the answers to ensure the best quality of content for students. Thus, the Important Questions Class 6 Maths Chapter 12 will guide students thoroughly.
Practice helps students in many ways, such as it helps them to build confidence in the subject. We see many students are afraid of Maths. It is simply because their concepts need to be clarified. Practice helps a lot to clear the doubts. So, the Class 6 Maths Chapter 12 Important Questions will build the habit of solving questions regularly among students. This habit will surely help students to make progress in the subject.

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Q.1 15 envelopes cost 15.5. How many envelopes can be bought for 186

Q.2 What is the value of x in the given proportion

15: 18 :: x : 6

Product of means = Product of extremes

18 × x = 15 × 6

or, x = 90/18

Q.3 The ratio of 2.5 m to 10 cm is:

Q.4 Are 36, 48, 42 and 56 in proportion

Ratio of 36 to 48 = 36 48 = 3 : 4 Ratio of 42 to 56 = 42 56 = 3 : 4 Since , 36 : 48 = 42 : 56 36 , 48 , 42 and 56 are in proportion .

Q.5 Find the ratio of 2 mm to 1 m.

To find the ratio, first we have to convert the two quantities into same units.

1 m = 1000 mm

Required ratio = 2 : 1000 = 1 : 500.

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Important questions class 6 maths, chapter 1 - knowing our numbers.

Chapter 2 - Whole Numbers

Chapter 3 - playing with numbers, chapter 4 - basic geometrical ideas, chapter 5 - understanding elementary shapes, chapter 6 - integers, chapter 7 - fractions, chapter 8 - decimals, chapter 9 - data handling, chapter 10 - mensuration, chapter 11 - algebra, chapter 13 - symmetry, chapter 14 - practical geometry, faqs (frequently asked questions), 1. what are the main concepts of class 6 maths chapter 12.

Class 6 Maths Chapter 12 is about ratios and proportions. It helps students to build the concept of ratios, how to calculate the ratio, concepts of proportion, how to calculate proportion etc. It is a relatively new concept for students, who must practice questions from the chapter. Our experts have made the Important Questions Class 6 Maths Chapter 12 for the benefit of students. They can take help from it for their better preparation.

2. How can the Important Questions Class 6 Maths Chapter 12 help students?

Extramarks is a leading organisation that provides all the study materials related to CBSE and NCERT. The experts of Extramarks have made the Important Questions Class 6 Maths Chapter 12 to help students solve questions regularly. They have collected the questions from several sources such as the CBSE sample papers, textbook exercises, CBSE past years’ question papers and important reference books. They solved the questions too. Thus, the Important Questions Class 6 Maths Chapter 12 will help the students to clear their doubts, boost their confidence and improve their preparation for exams.

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Ratio and Proportion class 6 extra questions with answers

Ratio and proportion class 6

In our daily life, many times we compare two quantities of the same type. The comparison by division is the Ratio. when we compared the two quantities in terms of ‘how many times’. This comparison is known as the Ratio. We denote ratio using symbol ‘:’ .

Example : Cost of a pen is Rs. 10 and cost of a pencil is Rs. 2. How many times the cost of a pen that of a pencil? Obviously it is five times.

$\frac{10}{2}= \frac{5}{1}= 5: 1$

If two ratios are equal, we say that they are in proportion and use the symbol ‘::’ or ‘=’ to equate the two ratios.

Ratio and proportion class 6 extra questions with answers

1) Raj purchased 3 pens for Rs. 15 and Anu purchased 10 pens for Rs. 50. Whose pens are more expensive? Solution : Ratio of number of pens purchased by Raj to the number of pens purchased by Anu = 3 : 10. Ratio of their costs = 15 : 50 = 3 : 10 Both the ratios 3 : 10 and 15 : 50 are equal. Therefore, the pens were purchased for the same price by both.

$\mathrm{\frac{2x}{x}} =\frac{2}{1}= 2 : 1$

3) Express the ratio 150 : 400 in its simplest form . Solution : To express the ratio 150:400 in its simplest form, you need to find the greatest common factor (GCF) of the two numbers and then divide both numbers by that factor. The GCF of 150 and 400 is 50. Dividing both numbers by 50.

$\frac{150}{400}=\frac{150\div 50}{400\div 50}=\frac{3}{8}= 3 : 8$

So, the simplified form of the ratio 150:400 is 3:8 .

4) Find the ratio of 200grams to 4 kg . Solution : To find the ratio of 200 grams to 4 kilograms, you need to make sure that the units are the same. Since 1 kilogram is equal to 1000 grams, you can express 4 kilograms as 4000 grams. The ratio of 200 grams to 4000 grams is 200 : 4000 . Now, simplify the ratio by finding the greatest common factor (GCF) of 200 and 4000, which is 200. Divide both numbers by the GCF.

$\frac{200}{4000}=\frac{200\div 200}{4000\div 200}=\frac{1}{20}$

So, the ratio of 200 grams to 4 kilograms is 1:20.

$\frac{480}{340}$

10) What must be subtracted from each term of the ratio 3 : 2. So, that the ratio becomes 2 : 5 ?

$\frac{3-x}{7-x}=\frac{2}{5}$

Therefore, the number of girls is 20 .

12) A recipe calls for a ratio of 2 cups of flour to 3 cups of sugar. If you want to make half the recipe, how much flour should you use? Solution: Since you want to make half the recipe, you need to find half of each part of the ratio. Half of 2 cups is 1 cup. Therefore, you should use 1 cup of flour.

$\frac{x}{6}=\frac{5}{8}$

Therefore the length of the rectangle is 3.75 units.

Ratio sums for class 6 (unsolved with answers )

(1) If p : q = 3 : 4 and q : r = 8 : 9. Find p: q: r .

(2) 3 : 5 :: 60 : x, find the value of x.

3) If A : B = 6 : 7 and B : C = 8 : 9, then find A : C .

4) What is ratio in between 7 months and 7 yr?

5)Divide Rs. 4000 among A , B and C so that their shares may be in the ratio of 5 : 7 : 8 .

6)The ratio of two numbers is 3 : 8 and their difference is 116. What is the largest number?

7)What must be added to each term of the ratio 49 : 68, so that it becomes 3 : 4?

8) The sum of the squares of three numbers is 116 and their ratio is 2 : 3 : 4. Find the numbers.

9) A sum of money is to be distributed between Ajay and Sanjay in the proportion of 7 : 11, respectively. Sanjay gets Rs.6000 more than Ajay. How much did Ajay get?

10) The ratio of copper and zinc is 11: 6. How much zinc is there in 850 kg of brass?

$\frac{5x-2y}{3x+2y}$

12)A picture is 60 cm wide and 1.8 m long. Find the ratio of its width to its perimeter.

13)Neelam’s annual income is Rs. 288000. Her annual savings amount to Rs. 36000. Find the ratio of her savings to her expenditure.

14) On a shelf, books with green cover and that with brown cover are in the ratio 2 : 3. If there are 18 books with green cover, then find the number of books with brown cover.

15) If a bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds, then the ratio of the distance travelled by them in one hour is

$\frac{25}{31}$

12) 1 : 8 13) 1 : 7 14) 27 15) 5 : 8

Ratio and proportion extra questions for class 6 pdf

Ratio and Proportion Class 6 Maths Chapter 12 Extra Questions

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Practice Questions for Class 6 Mathematics Chapter 12

Ch-12 Ratio and Proportion

Kirti = Rs 40, Kamal = Rs 20
Kirti = R s 60, Kamal = Rs 30
Kirti = Rs 20, Kamal = Rs 40
Kirti = Rs 30, Kamal = Rs 60
None of these

Match the following:

Fill In The Blanks.

If 4, a, a, 36 are in proportion then a is equal to ____.
32 m : 64 m : : ____.
5 : 4 = ____.
If a = 2b then a : b = ____.

State whether the following statements are True or False:

A ratio equivalent to 3 :7 is 9 : 21.
The ratio 35 : 84 in simplest form is 7 : 12.
A ratio can be equal to 1.
5 :2 = 2 : 5.

Find X in the proportion X : 6 = 25 : 5

The weight of 25 copies is 5 kg. Find the weight of 30 such copies?

Are the following statement true? 45km : 60km = 12 hours : 15 hours.

Write True or False against the following statement :8 : 9 : : 24 : 27.

Are the following statement true? 7.5litre : 15litre = 5kg : 10kg.

Write True or False against the following statement: 5.2 : 3.9 : : 3 : 4.

If 2A = 3B = 4C, find A : B : C

Kirti = Rs 20, Kamal = Rs 40 Explanation: ratio = 1 : 2 ∴ the total parts is 1+2=3so kriti’s share = {tex}{1 \over 3} \times 60 = {{60} \over 3} = {{60 \div 3} \over {3 \div 3}} = {{20} \over 1} = 20{/tex} Rskamal’s share = {tex}{2 \over 3} \times 60 = {{120} \over 3} = {{120 \div 3} \over {3 \div 3}} = {{40} \over 1} = 40{/tex} Rs
a:b = b:c Explanation: Three terms(a ,b and c) are said to be in proportion if the ratio of the first and the second is equal to the ratio of the second and third. a:b=b:c
it is 1:2 Explanation: Re 1 ⇒ 100 paisa ratio = 50 paisa : Re 1= 50:100 = {tex}{{50} \over {100}} = {{50 \div 50} \over {100 \div 50}} = {1 \over 2} = 1:2{/tex}
it is 2:3 Explanation: 30:45 {tex}{{30} \over {45}} = {{30 \div 15} \over {45 \div 15}} = {2 \over 3} = 2:3{/tex}
(c) Rs 16000 Explanation: sol : given ratio of ravi and rani 2:3 , total part is 2+3=5 and given profit is 40000ravi share in profit= {tex}{2 \over 5} \times 40000 = {{80000} \over 5} = Rs16000{/tex}
{tex} \to {/tex} r
{tex} \to {/tex} p
{tex} \to {/tex} q
{tex} \to {/tex} s
6secs : 12sec
X:6 = 25:5 {tex} \Rightarrow \frac{X}{6} = \frac{{25}}{5}{/tex} {tex} \Rightarrow \frac{X}{6} = \frac{5}{1}{/tex} (Dividing {tex}\frac{{25}}{5}{/tex} by 5) {tex} \Rightarrow X = 5 \times 6 = 30{/tex} {tex} \Rightarrow X = 30{/tex}
It is given that Weight of 25 copies = 5 kg ∴ Weight of 1 copy = 5 ÷25 = 1/5 kg ∴ Weight of 30 copies = 30 {tex} \times {/tex} 1/5= 6 kg
45km : 60km = {tex}\frac{45}{60}=\frac{45 \div 15}{60 \div 15}{/tex} [ {tex}\therefore{/tex} H.C.F.(45, 60) = 15] {tex}=\frac{3}{4}{/tex} = 3 : 4 12 hours : 15 hours {tex}=\frac{12}{15}=\frac{12 \div 3}{15 \div 3}{/tex} [ {tex}\therefore{/tex} H.C.F.(12, 15) = 3] {tex}=\frac{4}{5}{/tex} = 4 : 5 Since, the two ratios are not equal, therefore the given statement is false(F).
24 : 27 {tex}=\frac{24}{27}=\frac{24 \div 3}{27 \div 3}{/tex} [ {tex}\therefore{/tex} H.C.F.(24, 27) = 3] {tex}=\frac{8}{9}{/tex} = 8 : 9 {tex}\because{/tex} 8 : 9 = 24 : 27 {tex}\therefore{/tex} 8 : 9 : : 24 : 27 is true(T).
7.5 litre : 15 litre {tex}=\frac{7.5}{15}=\frac{7.5 \times 10}{15 \times 10}=\frac{75}{150}=\frac{75 \div 75}{150 \div 75}{/tex} [ {tex}\therefore{/tex} H.C.F.(75, 150) = 75] {tex}=\frac{1}{2}{/tex} = 1 : 2 5kg : 10kg {tex}=\frac{5}{10}=\frac{5 \div 5}{10 \div 5}{/tex} [ {tex}\therefore{/tex} H.C.F.(5, 10) = 5] {tex}=\frac{1}{2}{/tex} = 1 : 2 Since, the two ratios are equal, therefore, the given statement is true (T).
5.2 : 3.9 {tex}=\frac{5.2}{3.9}=\frac{5.2 \times 10}{3.9 \times 10}=\frac{52}{39}=\frac{52 \div 13}{39 \div 13}{/tex} [ {tex}\therefore{/tex} H.C.F.(52, 39) = 13] {tex}=\frac{4}{3}{/tex} = 4 : 3 {tex}\therefore{/tex} 4 : 3 {tex} \neq{/tex} 3 : 4 {tex}\therefore{/tex} 5.2 : 3.9 : : 3 : 4 is false (F).
Let 2A = 3B = 4C = x So, {tex}A = \frac{x}{2},B = \frac{x}{3},C = \frac{x}{4}{/tex} The L.C.M of 2, 3 and 4 is 12 Therefore, A:B:C {tex} = \frac{x}{2} \times 12:\frac{x}{3} \times 12:\frac{x}{4} \times 12 = 6x:4x:3x = 6:4:3{/tex} Therefore A:B:C = 6:4:3.

Chapter Wise Extra Questions for Class 6 Mathematics

Knowing our Numbers
Whole Numbers
Playing with Numbers
Basic Geometrical Ideas
Understanding Elementary Shapes
Data handling
Mensuration
Ratio and Proportion
Practical Geometry

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Unit 4: Proportional relationships and percentages

Lesson 2: ratios and rates with fractions.

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6th Class Mathematics Ratio and Proportion Question Bank

Done ratio, proportion & unitary method total questions - 40.

question_answer 1) In the word 'BEAUTIFUL', the ratio of the number of consonants to the number of vowels is

A) \[4:5\] done clear

B) \[4:6\] done clear

C) \[4:7\] done clear

D) \[4:8\] done clear

question_answer 2) Product of mean\[=?\]

A) Product of extremes done clear

B) Product of two numbers done clear

C) Product of median done clear

D) Product of ratio done clear

question_answer 3) The cost of 5 Mental ability books is Rs. 90. What will be the cost of 8 such books?

A) Rs. 124 done clear

B) Rs. 144 done clear

C) Rs. 140 done clear

D) Rs. 145 done clear

question_answer 4) If \[a:b::c:d,\]the correct statement is

A) \[ab=cd\] done clear

B) \[bc=ac\] done clear

C) \[ad=bc\] done clear

D) \[bc=da\] done clear

question_answer 5) The value of \[x\] in the proportion, \[9:5::\]\[36:x-3\]

A) 38 done clear

B) 23 done clear

C) 18 done clear

D) 28 done clear

question_answer 6) If \[x:y=3:1\] then find\[({{x}^{3}}-{{y}^{3}}):({{x}^{3}}+{{y}^{3}})\]

A) \[13:14\] done clear

B) \[14:13\] done clear

C) \[10:11\] done clear

D) \[11:10\] done clear

question_answer 7) First, second, third terms of a proportion are respectively 20, 18 and 40. Find its fourth term.

A) 36 done clear

B) 34 done clear

C) 35 done clear

D) 37 done clear

question_answer 8) The ratio of 2 liters to \[600\text{ }ml\] is

A) \[1:300\] done clear

B) \[1:30\] done clear

C) \[3:10\] done clear

D) \[10:3\] done clear

question_answer 9) A comparison by __________ is called ratio.

A) Division done clear

B) Multiplication done clear

C) Addition done clear

D) Subtraction done clear

question_answer 10) Which of the following is not equivalent ratio to compare 5 pens to 6 pencils?

A) \[15:18\] done clear

B) \[8:10\] done clear

C) \[20:24\] done clear

D) \[25:30\] done clear

question_answer 11) If \[a:b=3:5\]and \[b:c=6:7,\]find \[a:b:c\]

A) \[18:30:35\] done clear

B) \[35:30:18~\] done clear

C) \[15:36:35\] done clear

D) \[4:5:6\] done clear

question_answer 12) Find the Mean proportional between \[\frac{1}{4}\] and \[\frac{1}{25}\]

A) \[\frac{1}{10}\] done clear

B) \[\frac{1}{20}\] done clear

C) \[\frac{1}{30}\] done clear

D) \[\frac{1}{40}\] done clear

question_answer 13) 26 Cartons of 15 pens each cost Rs. 2145. Find the cost of 40 Cartons of 18 pens.

A) Rs. 3960 done clear

B) Rs. 3690 done clear

C) Rs. 3450 done clear

D) Rs. 3567 done clear

question_answer 14) If 30% of \[A=0.25\] of \[B=\frac{1}{5}\]of C, then find\[A:B:C=\]?

A) \[12:15:10\] done clear

B) \[10:12:15\] done clear

C) \[10:15:12\] done clear

D) \[15:12:10\] done clear

question_answer 15) A bus consumes 28 litres of petrol in covering 2100 km. How much petrol will be needed to cover a distance of 3600 Km?

A) 48 litres done clear

B) 42 litres done clear

C) 40 litres done clear

D) 46 litres done clear

question_answer 16) If the sum of A, B and C is 98. If \[A:B=\frac{2}{3}\]and \[B:C=\frac{5}{8}\]then\[B=?\]

A) 15 done clear

B) 20 done clear

C) 30 done clear

D) 32 done clear

question_answer 17) In a Global Pratibha School trip, Bus 1 has 4 teachers and 14 students, Bus 2 has 3 teachers and 7 students. Bus 3 has 4 teachers and 28 students. Which of the buses has the least teacher- student ratio?

A) Bus 1 done clear

B) Bus 2 done clear

C) Bus 3 done clear

D) Cannot say done clear

question_answer 18) The mean proportion of \[a\] and \[b\] is 10 and the value of \[a\] is four times the value of\[b\]. The value of \[a+b(a>0,\,\,b>0)\] is

A) 20 done clear

B) 25 done clear

C) 101 done clear

D) 29 done clear

question_answer 19) If \[A:B=\frac{1}{2}:\frac{3}{8},\] \[B:C=\frac{1}{3}:\frac{5}{9}\]and \[C:D=\frac{5}{6}:\frac{3}{4}\]then find\[A:B:C:D=\]?

A) \[6:4:8:10\] done clear

B) \[6:8:9:10\] done clear

C) \[8:6:10:9\] done clear

D) \[4:6:8:10\] done clear

question_answer 20) The ratio of two numbers is\[2:3\]. If 3 is added to both the numbers then ratio is \[3:4,\]find the sum of two numbers.

A) 10 done clear

B) 15 done clear

C) 20 done clear

D) 25 done clear

question_answer 21) If\[x=\frac{1}{3}y\] and \[y=\frac{1}{2} z\] then \[x:y:z=\]?

A) \[3:2:1\] done clear

B) \[1:2:3\] done clear

C) \[1:3:6\] done clear

D) \[2:4:6\] done clear

question_answer 22) In 5 minutes, Rohit can prepare 12 letters for mailing. At the same rate, how long will it take him to prepare 192 letters.

A) 80 min done clear

B) 40 min done clear

C) 20 min done clear

D) 90 min done clear

question_answer 23) A bag contains Rs. 187 in the form of 1 Rs. 50 paise and 10 paise coins in the ratio \[3:4:5\]. Find the number of 10 paise coins.

A) 170 done clear

B) 160 done clear

C) 136 done clear

D) 102 done clear

question_answer 24) Seats for mathematics, Physics and Chemistry in a school are in the ratio \[5:7:8.\] There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?

A) \[2:3:4\] done clear

B) \[6:7:8\] done clear

C) \[6:8:9\] done clear

D) None of these done clear

question_answer 25) Twelve persons can make 360 toys in 8 days. If three persons join them, then how many toys can be made in 8 days.

A) 370 done clear

B) 369 done clear

C) 450 done clear

D) 539 done clear

question_answer 26) If \[\frac{a}{b}=\frac{c}{d}=\frac{e}{f},\]then \[\frac{6a+9c+2c}{6b+9d+2f}=\]?

A) \[\frac{a}{b}\] done clear

B) \[\frac{c}{d}\] done clear

C) \[\frac{e}{f}\] done clear

D) All of these done clear

question_answer 27) If \[(3x+4y):\] \[(3x-4y)=\] \[(3c+8d):\] \[(3x-8d)\] then which of the following is true.

A) \[xd=yc\] done clear

B) \[2xd=yc\] done clear

C) \[2xd-cd\] done clear

D) \[xy-cd\] done clear

question_answer 28) A pipe can fill an empty tank in 12 minutes and another pipe can fill it in 24 min. If both the pipes are kept open simultaneously, then in how many minutes will the tank get filled?

A) 8 min done clear

B) 9 min done clear

C) 10 min done clear

D) 15 min done clear

question_answer 29) Find the value of\[x:y,\frac{8x-10y}{3x+4y}=\frac{4}{5}\]

A) \[55:29\] done clear

B) \[36:17\] done clear

C) \[33:14\] done clear

D) \[16:11\] done clear

question_answer 30) If the cost of 15 mangoes is Rs. 180, then what is the cost of 25 mangoes (in Rs.)?

A) 220 done clear

B) 360 done clear

C) 200 done clear

D) 300 done clear

question_answer 31) 20 men can finish the construction of a wall in 18 days. How many men are added to finish the work in half a day?

A) 700 done clear

B) 800 done clear

C) 900 done clear

D) 720 done clear

question_answer 32) If 30% of \[(A+B)=60%\] of \[(A-B),\] then find\[A:B\].

A) \[4:2\] done clear

B) \[3:1\] done clear

C) \[1:3\] done clear

D) \[3:2\] done clear

question_answer 33) If \[x:9=y:7=z:4,\]then find\[\frac{x+y+z}{z}=?\]

A) 4 done clear

B) 6 done clear

C) 7 done clear

D) 5 done clear

question_answer 34) If \[\frac{x}{y}=\frac{y}{z}=\frac{z}{R},\] then \[\frac{{{y}^{3}}+{{z}^{3}}+{{R}^{3}}}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}\]

A) \[\frac{x}{y}\] done clear

B) \[\frac{y}{z}\] done clear

C) \[\frac{z}{R}\] done clear

D) \[\frac{R}{x}\] done clear

question_answer 35) A man completes \[\frac{5}{8}\] of a job in 20 days. At this rate, how many more days will it take him to finish the job?

A) 12 done clear

B) 16 done clear

C) 17 done clear

D) 18 done clear

question_answer 36) Six numbers R, O, H, I, T, S are such that R\[RO=1\], \[OH=\frac{1}{2},\] \[HI=6,\] \[IT=2\]and \[TS=\frac{1}{2}.\]What is the value of (RI: OT : HS) = ?

A) \[4:3:27\] done clear

B) \[6:1:9\] done clear

C) \[8:9:9\] done clear

D) \[72:1:9\] done clear

A) Only I done clear

B) only II done clear

C) both I and II done clear

D) Neither 1 nor li done clear

question_answer 38) The ratio of the students in schools \[x,\,\,y\]and \[z\] is \[5:4:7.\]If the number of students hi the schools are increased by 20%, 25% and 20% respectively. What would be the new ratio of the students in schools x, y and z?

A) \[5:5:7\] done clear

B) \[30:25:42~\] done clear

C) \[30:20:49\] done clear

D) None of these done clear

question_answer 39) 21 binders of GOF can bind 1400 books in 15 days. How many binders will be required to bind 800 books in 20 days?

A) 7 done clear

B) 9 done clear

C) 12 done clear

D) 14 done clear

question_answer 40) The petrol tank of an automobile can hold f, litres. If a litres was removed when the tank was full, what part of the full tank was removed?

A) \[\ell -a\] done clear

B) \[\frac{\ell }{a}\] done clear

C) \[\frac{a}{\ell }\] done clear

D) \[\frac{\ell -a}{a}\] done clear

Study Package

Questions - Ratio, Proportion & Unitary Method

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RD Sharma Solutions
Chapter 9 Ratio Proportion And Unitary Method

RD Sharma Solutions for Class 6 Maths Chapter 9: Ratio, Proportion and Unitary Method

The solutions created by the faculty at BYJU’S are very helpful for the students to score well in the exam. Gaining knowledge about the various concepts which are explained in Mathematics is very important as a few topics are continued in higher classes as well. The solutions are mainly created with the aim of helping the students improve their problem-solving and logical thinking abilities.

By downloading PDFs of solutions, students obtain a clear idea about the concepts covered in each exercise. The solutions contain explanations in an interactive manner to make the subject interesting for the students. RD Sharma Solutions for Class 6 Maths Chapter 9 Ratio, Proportion and Unitary Method PDF is provided here.

RD Sharma Solutions Class 6 Maths Chapter 1 Knowing Our Numbers
RD Sharma Solutions Class 6 Maths Chapter 2 Playing with Numbers
RD Sharma Solutions Class 6 Maths Chapter 3 Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 4 Operations on Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers
RD Sharma Solutions Class 6 Maths Chapter 6 Fractions
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals
RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra
RD Sharma Solutions Class 6 Maths Chapter 9 Ratio, Proportion and Unitary Method
RD Sharma Solutions Class 6 Maths Chapter 10 Basic Geometrical Concepts
RD Sharma Solutions Class 6 Maths Chapter 11 Angles
RD Sharma Solutions Class 6 Maths Chapter 12 Triangles
RD Sharma Solutions Class 6 Maths Chapter 13 Quadrilaterals
RD Sharma Solutions Class 6 Maths Chapter 14 Circles
RD Sharma Solutions Class 6 Maths Chapter 15 Pair of Lines and Transversal
RD Sharma Solutions Class 6 Maths Chapter 16 Understanding Three-Dimensional Shapes
RD Sharma Solutions Class 6 Maths Chapter 17 Symmetry
RD Sharma Solutions Class 6 Maths Chapter 18 Basic Geometrical Tools
RD Sharma Solutions Class 6 Maths Chapter 19 Geometrical Constructions
RD Sharma Solutions Class 6 Maths Chapter 20 Mensuration
RD Sharma Solutions Class 6 Maths Chapter 21 Data Handling – I (Presentation of Data)
RD Sharma Solutions Class 6 Maths Chapter 22 Data Handling – II (Pictographs)
RD Sharma Solutions Class 6 Maths Chapter 23 Data Handling – III (Bar Graphs)
Exercise 9.1 Chapter 9 Ratio, Proportion and Unitary Method
Exercise 9.2 Chapter 9 Ratio, Proportion and Unitary Method
Exercise 9.3 Chapter 9 Ratio, Proportion and Unitary Method
Exercise 9.4 Chapter 9 Ratio, Proportion and Unitary Method
Objective Type Questions Chapter 9 Ratio, Proportion and Unitary Method

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rd sharma class 6 maths solutions chapter 9 ex 1 1

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Access answers to Maths RD Sharma Solutions for Class 6 Chapter 9: Ratio, Proportion and Unitary Method

Exercise 9.1 page: 9.5.

1. Express each of the following in the language of ratios:

(i) In a class, the number of girls in the merit list of the board examination is two times that of boys.

(ii) The number of students passing mathematics test is 2/3 of the number that appeared.

(i) Ratio of the number of girls to that of boys in the merit list is 2: 1.

(ii) Ratio of the number of students passing a mathematics test to that of total students appearing in the test is 2: 3.

2. Express the following ratios in language of daily life:

(i) The ratio of the number of bad pencils to that of good pencils produced in a factory is 1: 9.

(ii) In India, the ratio of the number of villages to that of cities is about 2000: 1.

(i) The number of bad pencils produced in a factory is 1/9 of the number of good pencils produced in the factory.

(ii) The number of villages is 2000 times that of cities in India.

3. Express each of the following ratios in its simplest form:

(ii) 324: 144

(iii) 85: 391

(iv) 186: 403

It can be written as 60/72

We know that the HCF of 60 and 72 is 12

By dividing the term by 12 we get

(60/72) × (12/12) = 5/6

So we get 60: 72 = 5: 6

It can be written as 324/144

We know that the HCF of 324 and 144 is 36

By dividing the term by 36 we get

(324/144) × (36/36) = 9/4

So we get 324: 144 = 9: 4

It can be written as 85/391

We know that the HCF of 85 and 391 is 17

By dividing the term by 17 we get

(85/391) × (17/17) = 5/23

So we get 85: 391 = 5: 23

It can be written as 186/403

We know that the HCF of 186 and 403 is 31

By dividing the term by 31 we get

(186/403) × (31/31) = 6/13

So we get 186: 403 = 6: 13

4. Find the ratio of the following in the simplest form:

(i) 75 paise to Rs 3

(ii) 35 minutes to 45 minutes

(iii) 8 kg to 400 gm

(iv) 48 minutes to 1 hour

(v) 2 metres to 35 cm

(vi) 35 minutes to 45 seconds

(vii) 2 dozen to 3 scores

(viii) 3 weeks to 3 days

(ix) 48 min to 2 hours 40 min

(x) 3 m 5 cm to 35 cm

It can be written as

75 paise to Rs 3 = 75 paise: Rs 3

We know that 1 Rs = 100 paise

75 paise to Rs 3 = 75 paise: 300 paise

Dividing the two terms by HCF 75

75 paise to Rs 3 = 1: 4

35 minutes to 45 minutes = 35 minutes: 45 minutes

Dividing the two terms by HCF 5

35 minutes to 45 minutes = 7: 9

8 kg to 400 gm = 8 kg: 400 gm

We know that 1 kg = 1000 gm

8 kg to 400 gm = 8000 gm: 400 gm

Dividing the two terms by HCF 400

8 kg to 400 gm = 20: 1

48 minutes to 1 hour = 48 minutes: 1 hour

We know that 1 hour = 60 minutes

48 minutes to 1 hour = 48 minutes: 60 minutes

Dividing the two terms by HCF 12

48 minutes to 1 hour = 4: 5

2 metres to 35 cm = 2 metres: 35 cm

We know that 1 m = 100 cm

2 metres to 35 cm = 200 cm: 35 cm

2 metres to 35 cm = 40: 7

35 minutes to 45 seconds = 35 minutes: 45 seconds

We know that 1 minute = 60 seconds

35 minutes to 45 seconds = 2100 seconds: 45 seconds

Dividing the two terms by HCF 15

35 minutes to 45 seconds = 140: 3

2 dozen to 3 scores = 2 dozen: 3 scores

We know that 1 dozen = 12 score = 20

2 dozen to 3 scores = 24: 60

2 dozen to 3 scores = 2: 5

3 weeks to 3 days = 3 weeks: 3 days

We know that 1 week = 7 days

3 weeks to 3 days = 21 days: 3 days

Dividing the two terms by HCF 3

3 weeks to 3 days = 7: 1

48 min to 2 hours 40 min = 48 min: 2 hours 40 min

48 min to 2 hours 40 min = 48 min: 160 min

Dividing the two terms by HCF 16

48 min to 2 hours 40 min = 3: 10

3 m 5 cm to 35 cm = 3 m 5 cm: 35 cm

3 m 5 cm to 35 cm = 305 cm: 35 cm

3 m 5 cm to 35 cm = 61: 7

5. Find the ratio of

(i) 3.2 metres to 56 metres

(ii) 10 metres to 25 cm

(iii) 25 paise to Rs 60

(iv) 10 litres to 0.25 litre

3.2 metres to 56 metres = 3.2 metres: 56 metres

Dividing the two terms by HCF 1.6

3.2 metres to 56 metres = 2: 35

10 metres to 25 cm = 10 m: 25 cm

10 metres to 25 cm = 1000 cm: 25 cm

Dividing the two terms by HCF 25

10 metres to 25 cm = 40: 1

25 paise to Rs 60 = 25 paise: Rs 60

25 paise to Rs 60 = 25 paise: 6000 paise

25 paise to Rs 60 = 1: 240

10 litres to 0.25 litre = 10 litres: 0.25 litre

Dividing the two terms by HCF 0.25

10 litres to 0.25 litre = 40: 1

6. The number of boys and girls in a school are 1168 and 1095 respectively. Express the ratio of the number of boys to that of the girls in the simplest form.

No. of boys = 1168

No. of girls = 1095

So the ratio of the number of boys to that of the girls = 1168: 1095

Dividing the two terms by HCF 73

Ratio of number of boys to that of the girls = 16: 15

Hence, the ratio of the number of boys to that of girls in simplest form is 16: 15.

7. Avinash works as a lecturer and earns Rs 12000 per month. His wife who is a doctor earns Rs 15000 per month. Find the following ratios:

(i) Avinash’s income to the income of his wife.

(ii) Avinash’s income to their total income.

Avinash salary earned per month = Rs 12000

Avinash wife salary per month = Rs 15000

(i) Avinash’s income to the income of his wife = 12000/15000 = 4: 5

(ii) Avinash’s income to their total income = 12000/ (12000 + 15000) = 4: 9

8. Of the 72 persons working in an office, 28 are men and the remaining are women. Find the ratio of the number of:

(i) men to that of women,

(ii) men to the total number of persons

(iii) persons to that of women.

No. of persons working in an office = 72

No. of men = 28

So the number of women = 72 – 28 = 44

(i) men to that of women = 28: 44

Multiplying and dividing the equation by HCF 4

Men to that of women = (28/44) × (4/4) = 7: 11

(ii) men to the total number of persons = 28: 72

Men to the total number of persons = (28/72) × (4/4) = 7: 18

(iii) persons to that of women = 72: 44

Persons to that of women = (72/44) × (4/4) = 18: 11

9. The length of a steel tape for measurements of buildings is 10 m and its width is 2.4 cm. What is the ratio of its length to width?

It is given that

Length of a steel tape = 10 m

Width of steel tape = 2.4 cm

So the ratio of its length to width = 10 m/ 2.4 cm

Ratio of its length to width = 1000 cm/ 2.4 cm

Dividing the two terms by HCF 0.8 cm

Ratio of its length to width = 1250: 3

Hence, the ratio of its length to width is 1250: 3.

10. An office opens at 9 am and closes at 5 pm with a lunch interval of 30 minutes. What is the ratio of lunch interval to the total period in office?

Duration of office = 9 am to 5 pm = 8 hours

Lunch interval = 30 minutes

So the ratio of lunch interval to the period in office = 30 minutes/8 hours

Ratio of lunch interval to the period in office = 30/ (8 × 60) = 30/480

Dividing the two terms by HCF 30

Ratio of lunch interval to the period in office = (30/480) × (30/30) = 1: 16

Hence, the ratio of lunch interval to the total period in office is 1: 16.

11. A bullock-cart travels 24 km in 3 hours and a train travels 120 km in 2 hours. Find the ratio of their speeds.

Distance travelled by bullock-cart = 24 km in 3 hours

Distance travelled by train = 120 km in 2 hours

Distance travelled by bullock-cart = 24 km/ 3 = 8 km

Distance travelled by train = 120 km/2 = 60 km

So the ratio of their speeds = 8/60

Dividing the two terms by HCF 4

Ratio of their speeds = (8/60) × (4/4) = 2:15

Hence, the ratio of their speeds is 2: 15.

12. Margarette works in a factory and earns Rs 955 per month. She saves Rs 185 per month from her earnings. Find the ratio of:

(i) her savings to her income

(ii) her income to her expenditure

(iii) her savings to her expenditure.

Margarette monthly income = Rs 955

Margarette monthly savings = Rs 185

Margarette expenditure = 955 – 185 = Rs 770

(i) her savings to her income = 185/955

Her savings to her income = (185/955) × (5/5) = 37: 191

(ii) her income to her expenditure = 955/770 = 191: 154

(iii) her savings to her expenditure = 185/770 = 37: 154

Exercise 9.2 page: 9.9

1. Which ratio is larger in the following pairs?

(i) 3: 4 or 9: 16

(ii) 15: 16 or 24: 25

(iii) 4: 7 or 5: 8

(iv) 9: 20 or 8: 13

(v) 1: 2 or 13: 27

3: 4 = 3/4 and 9: 16 = 9/16

LCM of 4 and 16 is 16

Multiplying both numerator and denominator of the term ¾ by 4 to make the denominator 16

3/4 = (3/4) × (4/4) = 12/16 and 9/16

We know that 12 > 9

So we get 12/16 > 9/16

We can write it as

3/4 > 9/16

Hence, 3: 4 > 9: 16.

15: 16 = 15/16 and 24: 25 = 24/25

LCM of 16 and 25 is 400

Multiplying both the terms by relevant numbers to make denominator as 400

15/16 = (15/16) × (25/25) = 375/400 and 24/25 = (24/25) × (16/16) = 384/400

We know that 384 > 375

So we get 384/400 > 375/400

We can write it as 24/25 > 15/16

Hence, 24: 25 > 15: 16.

4: 7 = 4/7 and 5: 8 = 5/8

LCM of 7 and 8 is 56

Multiplying both the terms by relevant numbers to make denominator as 56

4/7 = (4/7) × (8/8) = 32/56 and 5/8 = (5/8) × (7/7) = 35/56

We know that 35 > 32

So we get 35/56 > 32/56

We can write it as 5/8 > 4/7

Hence, 5: 8 > 4: 7.

9: 20 = 9/20 and 8: 13 = 8/13

LCM of 20 and 13 is 260

Multiplying both the terms by relevant numbers to make denominator as 260

9/20 = (9/20) × (13/13) = 117/260 and 8/13 = (8/13) × (20/20) = 160/260

We know that 160 > 117

So we get 160/260 > 117/260

We can write it as 8/13 > 9/20

Hence, 8: 13 > 9: 20.

1: 2 = 1/2 and 13: 27 = 13/27

LCM of 2 and 27 is 54

Multiplying both the terms by relevant numbers to make denominator as 54

1/2 = (1/2) × (27/27) = 27/54 and 13/27 = (13/27) × (2/2) = 26/54

We know that 27 > 26

So we get 27/54 > 26/54

We can write it as 1/2 > 13/27

Hence, 1: 2 > 13: 27.

2. Give two equivalent ratios of 6: 8.

The given ratio = 6: 8

It can be written as = 6/8

Dividing the fraction by 2 we get

6/8 = (6/8) ÷ (2/2) = 3/4

Hence, 3: 4 is an equivalent ratio of 6: 8

Multiply the fraction by 2 we get

6/8 = (6/8) × (2/2) = 12/16

Hence, 12: 16 is an equivalent ratio of 6: 8

So, 3: 4 and 12: 16 are the equivalent ratios of 6: 8.

3. Fill in the following blanks:

12/20 = ☐/5 = 9/☐

We know that LCM of 20 and 5 is 20

It can be written as 20/4 = 5

Dividing the fraction by 4

12/20 = (12/20) × (4/4) = 3/5

So the first number is 3 and the ratio is 3/5.

In the same way,

Consider 2/3 + 3/5 = 9/☐

We know that 9/3 = 3

Multiply the fraction by 3

3/5 = (3/5) × (3/3) = 9/15

So the second number is 15 and the ratio is 9/15.

Exercise 9.3 page: 9.14

1. Which of the following statements are true?

(i) 16: 24 = 20: 30

(ii) 21: 6 = 35: 10

(iii) 12: 18 = 28: 12

(iv) 51: 58 = 85: 102

(v) 40 men: 200 men = Rs 5: Rs 25

(vi) 99 kg: 45 kg = Rs 44: Rs 20

16/24 = 20/30

Dividing 16/24 by 4/4 and 20/30 by 5/5

(16/24) ÷ (4/4) = (20/30) ÷ (5/5)

On further calculation

Hence, 16: 24 = 20: 30 is true.

21/6 = 35/10

Dividing 21/6 by 3/3 and 35/10 by 5/5

(21/6) ÷ (3/3) = (35/10) ÷ (5/5)

Hence, 21: 6 = 35: 10 is true.

12/18 = 28/12

6/9 ≠ 14/6

Hence, 12: 18 = 28: 12 is false.

51/58 = 85/102

51/58 ≠ 5/6

Hence, 51: 58 = 85: 102 is false.

40/200 = 5/25

We get 40/200 = 1/5 and 5/25 = 1/5

Hence, 40 men: 200 men = Rs 5: Rs 25 is true.

99/45 = 44/20

Dividing the fraction by 9

(99/45) ÷ (9/9) = (44/20) ÷ (9/9)

11/5 = 11/5

Hence, 99 kg: 45 kg = Rs 44: Rs 20 is true.

2. Find which of the following are in proportion:

(i) 8, 16, 6, 12

(ii) 6, 2, 4, 3

(iii) 150, 250, 200, 300

We know that

8: 16 = 8/16 = 1/2

6: 12 = 6/12 = 1/2

So we get 8/16 = 6/12

Therefore, 8, 16, 6, 12 are in proportion.

6: 2 = 6/2 = 3/1

So we get 3/1 ≠ 4/3

Therefore, 6, 2, 4, 3 are not in proportion.

150: 250 = 150/250 = 3/5

200: 300 = 200/300 = 4/6 = 2/3

So we get 3/5 ≠ 2/3

Therefore, 150, 250, 200, 300 are not in proportion.

3. Find x in the following proportions:

(i) x: 6 = 55: 11

(ii) 18: x = 27: 3

(iii) 7: 14 = 15: x

(iv) 16: 18 = x: 96

x/6 = 55/11

x = 5 (6) = 30

18/x = 27/3

x = 18/9 = 2

7/14 = 15/x

x = 15 (2) = 30

16/18 = x/96

x = 8/9 (96) = 256/3

4. Set up all proportions from the numbers 9, 150, 105, 1750.

The proportions from the numbers are

9: 150 = 3: 50

9: 105 = 3: 35

150: 9 = 50: 3

150: 105 = 10: 7

150: 1750 = 3: 35

105: 9 = 35: 3

105: 150 = 7: 10

105: 1750 = 3: 50

1750: 150 = 35: 3

1750: 105 = 50: 3

Hence, the proportions that are formed are

9: 150 :: 105: 1750

150: 9 :: 1750: 105

1750: 150 :: 105: 9

9: 105 :: 150: 1750

5. Find the other three proportions involving terms of each of the following:

(i) 45: 30 = 24: 16

(ii) 12: 18 = 14: 21

(i) 45: 30 = 24: 16 can be written as 3: 2 in simplest form

So the other three proportions involving terms are

45: 24 = 30: 16 can be written as 15: 8 in simplest form

30: 45 = 16: 24 can be written as 2: 3 in simplest form

16: 30 = 24: 45 can be written as 8: 15 in simplest form

(ii) 12: 18 = 14: 21 can be written as 2: 3 in simplest form

12: 14 = 18: 21 can be written as 6: 7 in simplest form

21: 18 = 14: 12 can be written as 7: 6 in simplest form

18: 12 = 21: 14 can be written as 3: 2 in simplest form

6. If 4, x, 9 are in continued proportion, find the value of x.

We know that 4, x, 9 are in continued proportion

4: x :: x: 9

x 2 = 9 (4) = 36

7. If in a proportion, the first, second and fourth terms are 32, 112 and 217 respectively, find the third term.

It is given that in a proportion the first, second and fourth terms are 32, 112 and 217

Consider x as the third term

32: 112 :: x: 217

32/112 = x/217

x = 32/112 (217) = 62

8. Show that the following numbers are in continued proportion:

(i) 36, 90, 225

(ii) 48, 60, 75

(iii) 16, 84, 441

Consider the fraction 36/90

By dividing the fraction by 18

36/90 = 2/5

Consider the fraction 90/225

By dividing the fraction by 45

90/225 = 2/5

Hence, 36: 90 :: 90: 225.

Consider the fraction 48/60

By dividing the fraction by 12

48/60 = 4/5

Consider the fraction 60/75

By dividing the fraction by 15

60/75 = 4/5

Hence, 48: 60 :: 60: 75.

Consider the fraction 16/84

By dividing the fraction by 4

16/84 = 4/21

Consider the fraction 84/441

By dividing the fraction by 21

84/441 = 4/21

Hence, 16: 84 :: 84: 441.

9. The ratio of the length of a school ground to its width is 5: 2. Find its length if the width is 40 metres.

Ratio of length of a school ground to its width = 5: 2

Width of the school ground = 40 m

So the length of the school ground = 5/2 (40) = 100 m

Hence, the length of the school ground is 100 m.

10. The ratio of the sale of eggs on a Sunday to that of the whole week of a grocery shop was 2: 9. If the total sale of eggs in the same week was Rs 360, find the sale of eggs on Sunday.

Ratio of the sale of eggs on a Sunday to that of the whole week of a grocery shop = 2: 9

We know that the sale of eggs in a week is Rs 9 and on Sunday is Rs 2

If eggs of Rs 1 is sold in a week, the cost of eggs on Sunday = Rs 2/9

If the total sale of eggs in the same week was Rs 360, the sale of eggs on Sunday = 2/9 (360) = Rs 80

Hence, the sale of eggs on Sunday is Rs 80.

11. The ratio of copper and zinc in an alloy is 9: 7. If the weight of zinc in the alloy is 9.8 kg, find the weight of copper in the alloy.

Ratio of copper and zinc in an alloy = 9: 7

If the weight of zinc is 7 kg then the weight of copper is 9 kg

If the weight of zinc is 1 kg then the weight of copper = 9/7 kg

So if the weight of zinc is 9.8 kg then the weight of copper = 9/7 (9.8) = 12.6 kg

Hence, the weight of copper in the alloy is 12.6 kg.

12. The ratio of the income to the expenditure of a family is 7: 6. Find the savings if the income is Rs 1400.

Ratio of the income to the expenditure of a family = 7: 6

We know that saving = total income – expenditure

Ratio of saving to the income = [7 – 6]: 7 = 1: 7

It is given that income = Rs 1400

So the saving of the family = 1/7 (1400) = Rs 200

Hence, the saving of the family is Rs 200.

13. The ratio of story books in a library to other books is 1: 7. The total number of story books is 800. Find the total number of books in the library.

Ratio of story books in a library to other books = 1: 7

Consider the ratio as x

So the number of story books = x

Number of other books = 7x

Total number of story books = 800

Number of other books = 7 × 800 = 5600

Total number of books = 5600 + 800 = 6400

Hence, the total number of books in the library is 6400.

Exercise 9.4 PAGE: 9.18

1. The price of 3 metres of cloth is Rs 79.50. Find the price of 15 metres of such cloth.

Price of 3 m of cloth = Rs 79.50

Price of 1 m of cloth = 79.50/3 = Rs 26.5

So the price of 15 m of cloth = 26.5 (15) = Rs 397.50

Hence, the price of 15 m of such cloth is Rs 397.50.

2. The cost of 17 chairs is Rs 9605. Find the number of chairs that can be purchased in Rs 56500.

No. of chairs purchased for Rs 9605 = 17

No. of chairs purchased for Rs 1 = 17/9605

So the number of chairs purchased for Rs 56500 = 17/9605 (56500) = 100

Hence, 100 chairs can be purchased in Rs 56500.

3. Three ferryloads are needed to carry 150 people across a river. How many people will be carried on 4 ferryloads?

No. of people required to carry 3 ferryloads = 150

No. of people required to carry 1 ferryload = 150/3 = 50

So the number of people required to carry 4 ferryloads = 4 (50) = 200

Hence, 200 people are required to carry 4 ferryloads.

4. If 9 kg of rice costs Rs 120.60, what will 50 kg of such a quality of rice cost?

Cost of 9 kg rice = Rs 120.60

Cost of 1 kg rice = 120.60/9 = Rs 13.4

So the cost of 50 kg rice = 13.4 (50) = Rs 670

Hence, 50 kg of such a quality of rice costs Rs 670.

5. A train runs 200 kilometres in 5 hours. How many kilometres does it run in 7 hours?

Distance travelled by train in 5 hours = 200 km

Distance travelled by train in 1 hour = 200/5 = 40 km

So the distance travelled by train in 7 hours = 40 (7) = 280 km

Hence, the train runs 280 km in 7 hours.

6. 10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?

10 boys can dig a pitch in 12 hours

We know that the time taken by one boy = 10 (12) = 120 hours

So the time taken by 8 boys to dig the pitch = 120/8 = 15 hours

Hence, 8 boys will take 15 hours to dig the pitch.

7. A man can work 8 hours daily and finishes a work in 12 days. If he works 6 hours daily, in how many days will the same work be finished?

A man can work 8 hours daily and finishes a work in 12 days

If he works for one hour, then the time taken to finish the work = 8 × 12 = 96 days

If he works 6 hours daily, the days required to finish the work = 96/6 = 16 days

Hence, the man requires 16 days to finish the same work.

8. Fifteen post cards cost Rs 2.25. What will be the cost of 36 post cards? How many postcards can be bought in Rs 45?

Cost of fifteen post cards = Rs 2.25

Cost of one post card = Rs 2.25/15

So the cost of 36 post cards = 2.25/15 (36) = Rs 5.40

No. of postcards that can be purchased in Rs 1 = 15/2.25

Number of postcards that can be bought in Rs 45 = 15/2.25 (45) = 300

Hence, the cost of 36 post cards is Rs 5.40 and 300 post cards can be bought in Rs 45.

9. A rail journey of 75 km costs Rs 215. How much will a journey of 120 km cost?

Cost of rail journey of 75 km = Rs 215

Cost of rail journey of 1 km = Rs 215/75

So the cost of rail journey of 120 km = 215/75 (120) = Rs 344

Hence, the cost of rail journey of 120 km is Rs 344.

10. If the sales tax on a purchase worth Rs 60 is Rs 4.20. What will be the sales tax on the purchase worth Rs 150?

Sales tax on a purchase worth Rs 60 = Rs 4.20

Sales tax on a purchase worth Rs 1 = Rs 4.20/60

So the sales tax on the purchase worth Rs 150 = 4.20/60 (150) = Rs 10.50

Hence, the sales tax on the purchase worth Rs 150 is Rs 10.50.

11. The cost of 17 chairs is Rs 19210. Find the number of such chairs that can be purchased in Rs 113000?

No. of chairs purchased in Rs 19210 = 17

No. of chairs purchased in Rs 1 = 17/19210

So the number of chairs that can be purchased in Rs 113000 = 17/19210 (113000) = 100

Hence, 100 chairs can be purchased in Rs 113000.

12. A car travels 165 km in 3 hours

(i) How long will it take to travel 440 km?

(ii) How far will it travel in 7 hours?

Distance travelled by car = 165 km in 3 hours

So the speed of car = Distance/ time = 165/3 = 55 km per hour

(i) Time taken to travel 440 km = 440/55 = 8 hours

(ii) Distance covered in 7 hours = 55 (7) = 385 km

13. 2 dozens of oranges cost Rs 60. Find the cost of 120 similar oranges?

Cost of 2 dozens of oranges = Rs 60

Cost of 1 orange = Rs 60/24

So the cost of 120 similar oranges = 60/24 (120) = Rs 300

Hence, the cost of 120 similar oranges is Rs 300.

14. A family of 4 members consumes 6 kg of sugar in a month. What will be the monthly consumption of sugar, if the number of family members becomes 6?

Amount of sugar used by a 4 members family = 6 kg

Amount of sugar used by 1 member = 6/4 kg

So the sugar consumed by 6 members of a family = 6/4 (6) = 9kg

Hence, 9 kg is the monthly consumption of sugar, if the number of family members becomes 6.

15. The weight of 45 folding chairs is 18 kg. How many such chairs can be loaded on a truck having a capacity of carrying 4000 kg load?

No. of folding chairs weighing 18 kg = 45

No. of folding chairs weighing 1 kg = 45/18

So the number of folding chairs weighing 4000 kg = 45/18 (4000) = 10000

Hence, 10000 chairs can be loaded on a truck having a capacity of carrying 4000 kg load.

Objective Type Questions page: 9.19

Mark the correct alternative in each of the following:

1. A ratio equivalent of 2 : 3 is (a) 4 : 3 (b) 2 : 6 (c) 6 : 9 (d) 10 : 9

The option (c) is correct answer.

We know that 6: 9 when divided by 3 we get 2: 3.

2. The angles of a triangle are in the ratio 1 : 2 : 3. The measure of the largest angle is (a) 30° (b) 60° (c) 90° (d) 120°

We know that the sum of all the angles = 180°

So the largest angle = 3/ (1 + 2 + 3) × 180

Largest angle = 3/6 × 180 = 90°

3. The sides of a triangle are in the ratio 2 : 3 : 5. If its perimeter is 100 cm, the length of its smallest side is (a) 2 cm (b) 20 cm (c) 3 cm (d) 5 cm

The option (b) is correct answer.

We know that the length of smallest side = 100 × 2/ (2 + 3 + 5) = 200/10 = 20 cm

4. Two numbers are in the ratio 7 : 9. If the sum of the numbers is 112, then the larger number is (a) 63 (b) 42 (c) 49 (d) 72

The option (a) is correct answer.

Consider x as the largest number

7x + 9x = 112

x = 112/16 = 7

7x = 7 × 7 = 49

9x = 9 × 7 = 63

Hence, the largest number is 63.

5. Two ratio 384 : 480 in its simplest form is (a) 3 : 5 (b) 5 : 4 (c) 4 : 5 (d) 2 : 5

384: 480 can be written as

384/480 = 4/5 when divided by 96

6. If A , B , C, divide Rs 1200 in the ratio 2 : 3 : 5, then B’s share is (a) Rs 240 (b) Rs 600 (c) Rs 380 (d) Rs 360

The option (d) is correct answer.

So B’s share = 1200 × 3/ (2 + 3 + 5)

B’s share = 1200 × 3/10 = Rs 360

7. If a bus travels 126 km in 3 hours and a train travels 315 km in 5 hours, then the ratio of their speeds is (a) 2 : 5 (b) 2 : 3 (c) 5 : 2 (d) 25 : 6

We know that speed = distance/time

So the speed of bus = 126/3 = 42 km/h

Speed of train = 315/5 = 63 km/h

So the ratio of their speeds = 42: 63 = 2: 3

8. The ratio of male and female employees in a multinational company is 5 : 3. If there are 115 male employees in the company, then the number of female employees is (a) 96 (b) 52 (c) 69 (d) 66

Consider x as the number of female employees

5/3 = 115/x

By cross multiplication

5x = 115 × 3 = 345

By division

x = 345/5 = 69

9. Length and width of a field are in the ratio 5 : 3. If the width of the field is 42 m, then its length is (a) 50 m (b) 70 m (c) 80 m (d) 100 m

It is given that length and width of a field = 5: 3

Consider x m as the length

Width of the filed = 42 m

So the length can be written as

3x = 42 × 5 = 210

x = 210/3 = 70

10. If 57 : x = 51 : 85, then the value of x is (a) 95 (b) 76 (c) 114 (d) None of these

57/x = 51/85

57 × 85/51 = x

11. The ratio of boys and girls in a school is 12 : 5. If there are 840 girls in the school, then the number of boys is (a) 1190 (b) 2380 (c) 2856 (d) 2142

The options are not correct.

Consider x as the number of boys

Ratio of boys and girls = 12: 5

12/5 = x/840

x = 12/5 × 840 = 2016

12. If 4, a , a , 36 are in proportion, then a = (a) 24 (b) 12 (c) 3 (d) 24

It is given that 4, a, a, 36 are in proportion

We can write it as 4 : a :: a : 36

4 × 36 = a × a

13. If 5 : 4 : : 30 : x, then the value of x is (a) 24 (b) 12 (c) 3/2 (d) 6

x = 30 × 4/5 = 24

14. If a , b , c , d are in proportion, then (a) ab = cd (b) ac = bd (c) ad = bc (d) None of these

It is given that a, b, c, d are in proportion

We can write it as a : b :: c : d

15. If a, b, c, are in proportion, then (a) a 2 = bc (b) b 2 = ac (c) c 2 = ab (d) None of these

It is given that a, b, c are in proportion

a : b :: b : c

16. If the cost of 5 bars of a soap is Rs. 30, then the cost of one dozen bars is (a) Rs 60 (b) Rs 120 (c) Rs 72 (d) Rs 140

Consider Rs x as the cost of one dozen bars

30/5 = x/12

x = 30/5 × 12 = Rs 72

17. 12 men can finish a piece of work in 25 days. The number of days in which the same piece of work can be done by 20 men, is (a) 10 days (b) 12 days (c) 15 days (d) 14 days

Consider x days required by 20 men to do the same work

20/12 = 25/x

x = 12 × 25/20 = 15 days

18. If the cost of 25 packets of 12 pencils each is Rs 750, then the cost of 30 packets of 8 pencils each is

(d) None of these

Cost of 300 pencils = Rs 750

So consider Rs x as the cost of 240 pencils

750: 300 :: x: 240

Cost of 240 pencils = 750/300 × 240 = Rs 600

19. If a, b, c are in proportion, then (a) a : b : : b : c (b) a : b : : c : a (c) a : b : : c : b (d) a : c : : b : c

We know that a, b, c are in proportion

So we get a: b :: b: c

It can be written as ac = b 2

20. The first, second and fourth terms of a proportion are 16, 24 and 54 respectively. The third term is (a) 32 (b) 48 (c) 28 (d) 36

16: 24 = x: 54

16/24 = x/54

x = 16/24 × 54

Chapter 9 Ratio, Proportion and Unitary Method has 4 exercises which help students solve problems using various methods. The concepts which are explained in RD Sharma Solutions Chapter 9 are as follows:

Unitary Method

Chapter Brief of RD Sharma Solutions Class 6 Maths Chapter 9 – Ratio, Proportion and Unitary Method

RD Sharma Solutions created by subject-matter experts are according to the current CBSE syllabus and guidelines. The PDF of solutions can be downloaded and referred to while solving the chapter-wise problems of the RD Sharma textbook. By using the PDF of solutions, students can obtain a better understanding of concepts covered in the chapter.

Ratio and Proportion are used in our daily lives and have various important applications in grocery shopping, recipes and cooking, planning vacation trips with families, etc. Students can download the PDF and use it to learn more about the concepts which are explained in this chapter.

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Ratio and Proportion Questions & Word Problems | GMAT GRE Maths

Whether you are using units from the Metric system (as we do in this post) or US measurement system (the GMAT being an American test), the concepts don’t change.

Introduction to Ratios

Ratio is the quantitative relation between two amounts showing the number of times one value contains or is contained within the other.

( Reference : Oxford dictionary )

Notation : Ratio of two values a and b is written as a:b or a/b or a to b.

For instance, the ratio of number of boys in a class to the number of girls is 2:3. Here, 2 and 3 are not taken as the exact count of the students but a multiple of them, which means the number of boys can be 2 or 4 or 6…etc and the number of girls is 3 or 6 or 9… etc. It also means that in every five students, there are two boys and three girls.

Question : In a certain room, there are 28 women and 21 men. What is the ratio of men to women? What is the ratio of women to the total number of people?

Men : women = 21 : 28 = 3:4

Women : total number of people = 28 : 49 = 4 : 7

Question : In a group, the ratio of doctors to lawyers is 5:4. If the total number of people in the group is 72, what is the number of lawyers in the group?

Let the number of doctors be 5x and the number of lawyers be 4x.

Then 5x+4x = 72 → x=8.

So the number of lawyers in the group is 4*8 = 32.

Question : In a bag, there are a certain number of toy-blocks with alphabets A, B, C and D written on them. The ratio of blocks A:B:C:D is in the ratio 4:7:3:1. If the number of ‘A’ blocks is 50 more than the number of ‘C’ blocks, what is the number of ‘B’ blocks?

Let the number of the blocks A,B,C,D be 4x, 7x, 3x and 1x respectively

4x = 3x + 50 → x = 50.

So the number of ‘B’ blocks is 7*50 = 350.

If the ratio of chocolates to ice-cream cones in a box is 5:8 and the number of chocolates is 30, find the number of ice-cream cones.

Let the number of chocolates be 5x and the number of ice-cream cones be 8x.

5x = 30 → x = 6.

Therefore, number of ice-cream cones in the box = 8*6 = 48.

Introduction to Proportion

A lot of questions on ratio are solved by using proportion.

Definition & Notation

A proportion is a comparison of two ratios. If a : b = c : d, then a, b, c, d are said to be in proportion and written as a:b :: c:d or a/b = c/d.

a, d are called the extremes and b, c are called the means.

For a proportion a:b = c:d, product of means = product of extremes → b*c = a*d.

Let us take a look at some examples:

In a mixture of 45 litres, the ratio of sugar solution to salt solution is 1:2. What is the amount of sugar solution to be added if the ratio has to be 2:1?

Number of litres of sugar solution in the mixture = (1/(1+2)) *45 = 15 litres.

So, 45-15 = 30 litres of salt solution is present in it.

Let the quantity of sugar solution to be added be x litres.

Setting up the proportion,

sugar solution / salt solution = (15+x)/30 = 2/1 → x = 45.

Therefore, 45 litres of sugar solution has to be added to bring it to the ratio 2:1.

A certain recipe calls for 3kgs of sugar for every 6 kgs of flour. If 60kgs of this sweet has to be prepared, how much sugar is required?

Let the quantity of sugar required be x kgs.

3 kgs of sugar added to 6 kgs of flour constitutes a total of 9 kgs of sweet.

3 kgs of sugar is present in 9 kgs of sweet. We need to find the quantity of sugar required for 60 kgs of sweet. So the proportion looks like this.

3/9 = x/60 → x=20.

Therefore, 20 kgs of sugar is required for 60 kgs of sweet.

Problems on Mixtures / Blends

If a 60 ml of water contains 12% of chlorine, how much water must be added in order to create a 8% chlorine solution?

Let x ml of chlorine be present in water.

Then, 12/100 = x/60 → x = 7.2 ml

Therefore, 7.2 ml is present in 60 ml of water.

In order for this 7.2 ml to constitute 8% of the solution, we need to add extra water. Let this be y ml.

Then, 8/100 = 7.2/y → y = 90 ml.

So in order to get a 8% chlorine solution, we need to add 90-60 = 30 ml of water.

There is a 20 litres of a solution which has 20% of bleach. Extra bleach is added to it to make it to 50% bleach solution. How much water has to be added further to bring it back to 20% bleach solution?

This question has 3 parts.

In the first part, there is 20% of bleach in 20 L of solution → 4 L of bleach in 16 L of water = 20 L of solution. Let’s note the details in a table for better clarity and understanding.

In the second part, Extra bleach is added to bring it to 50% of total solution. Let the amount of bleach added be x litres.

Then, (4+x)/(20+x) = 50/100 → x = 12 L of bleach is added.

Now, there is 4+12 = 16 L of bleach in 16 L of water in a total of 32 L of solution.

Now, to bring the bleach percentage back to 20%, extra water is added and the amount of bleach remains the same. Let this extra amount of water be y litres.

16 L of bleach constitutes 20% of the solution →

16/(32+y) = 20/100 → y = 48.

Therefore, 48 litres of water has to be added to the solution if bleach has to be 20% of the whole solution.

1 kg of cashews costs Rs. 100 and 1 kg of walnuts costs Rs. 120. If a mixture of cashews and walnuts is sold at Rs. 105 per kg,then what fraction of the total mixture are walnuts?

For this type of problems, first step is to determine how much each of the items is above or below the target.

Our target price is Rs. 105. Cashews price is Rs. 5 below the target price and walnuts price is Rs. 15 above the target price.

So, for each kg of cashews added, let’s consider it as ‘-5’ and for each kg of walnuts added, let’s consider it as ‘+15’. These two have to be added in such a way that they cancel out each other. Adding ‘-5’ thrice gives a ‘15’ and adding ‘+15’ once results in cancellation of the terms.

This means that adding 3 kgs of cashews and 1 kg of walnuts gives a mixture that can be sold at Rs. 105 per kg.

So, 3 kgs of cashews present for every 1 kg of walnuts. The ratio of cashews to walnuts is 3:1. Fraction of walnuts in the mixture = 1/(3+1) = 1/4 of the total mixture

Practice Questions in Ratio and Proportion

Problem 1: Click here

On a certain map, 1 cm = 12 km actual distance. If two places are 96 km apart, what is their distance on map?

A. 10 cm B. 12 cm C. 96 cm D. 8 cm

Answer 1: Click here

Explanation :

1cm/12 km = x cm/100 km → x = 8 cm

Problem 2: Click here

A person types 360 words in 4 minutes. How much time does he take to type 900 words? A. 15 B. 90 C. 10 D. 9

Answer 2: Click here

Explanation

4/360 = x/900 → x=10

12 thoughts on “Ratio and Proportion Questions & Word Problems | GMAT GRE Maths”

Hey can you please emphasize on the working of the cashews and walnuts question?

I’m stuck on how we’re assuming this: So, for each kg of cashews added, let’s consider it as ‘-5’ and for each kg of walnuts added, let’s consider it as ‘+15’. These two have to be added in such a way that they cancel out each other. Adding ‘-5’ thrice gives a ‘15’ and adding ‘+15’ once results in cancellation of the terms.

Why do they have to be added to cancel out eaach other?

Would appreciate the help.

I think if they equal zero, then is the only time a ratio can be found. If it isn’t equal to zero you won’t get a ratio.

assume the following Lets the total Qty of the Mixture is 10 KG Lets assume the Qty of Cashew is X Lets assume the Qty of Walnut is 10-X As we know that the purchase qty and sold qty can be equal for instance 10 KG now put this into equation 100(X) + 120(10-X) = 10*105 ” Total purchase = total Sales” X = 7.5 , hence walnut = 2.5 now 2.5/10=25%

given 1 KG of cashews cost 100 and walnuts was 120 given their mixture costs 105 rupees this 105 rupees of mixture cost is for just 1 kg includes X grams of cashews and Y grams of walnuts so, X+Y=1 KG or X+Y=1000 grams——1 1 KG of cashews cost is 100 so ,1000 grams=100 rupees. Hence 1 gram=1/10 rupees same way 1 gram of walnuts costs 12/100 rupees so, X/10 +Y*(12/100)=105——-2 solve equation 1 and 2 . Quantity of X and Y will be 750 grams and 250 grams so their ratio is 3:1 walnuts to the total ratio will be 1/4

Hello I have this peoblem The ratio of the area of the dining room to the family room is 2 to 3. After remodeling the family room is now 1/2 as large as it used to be and has 60 Sq less than the dining room. How many Sq feet is the isning room?

Hi Let the area of the dining room be d and the family room be f. d/2 = b f/3 = k d= 2k f=3k Total area = 5k 5k=3k +60(1/2*3k +1/2*3k + 60) 2k= 60 k = 30 Area of family = 3*30 /2 = 45 Area of dining = 105 45+105 = 150 and total area remains the same

Hi Caterina,

I had a look at Erica’s answer, and unfortunately it doesn’t add up. However, I will stick to the same format she used:

dining room = d family room = f

d : f = 2 : 3

Lets make that ratio easier to handle: d : f = 4 : 6

Now what does 4 : 6 mean? It means that the dining room represents 4 parts whilst the family room represents 6 parts. A part can be any size in a ratio, what matters is the proportion that the ratio describes.

Lets say 1 part = k, therefore d : f = 4k : 6k

Now if the family room halves in size, the ratio becomes 4k : 3k because the family room used to be 6 parts but is now 3 parts.

It is now also 60sq ft less than the dining room, so ‘size of f’ – ‘size of d’ = 60 sq ft therefore 4k – 3k = 60 sq ft i.e 1k = 60sq ft

Remember that the dining room is 4k so its size is:

4k = 4 x 60 sq ft = 240 sq ft

For every 2 boy students there are 3 girl students and for every one teacher there are 10 students whereas for every 4 male teachers there are 5 female teachers. Which of the following is the ratio of number of boy students to the number of male teachers?

For 1 teacher, there are 10 students which contains 4 boys, and 6 girls(because ratio of boys to girls is 2:3)

Therefore ratio of boys to teacher is 4:1

Now for 9 (4 male+5 female) teachers, there are total 90 students which has 36(90*0.4) boys

This means for a group of 36 boys, there are 4 male teachers.

Hence the ratio of boys to male teacher is 36:4 or 9: 1

Three similar lamps use 4 liters of oil in 80 hours. How much oil will 6 lamps of the same kind use in 40 hours?

Worker*Rate*Time=Object

Worker(lamp)=3 Time=80 hours Object = 4 liters

3R80=4. R=4/240 or R=1/60

Worker(lamp)=6 Time=40 hours Object= x

6R40=X (Rate is 1/60 from previous). 6(1/60)40=x (240/60)=x and x=4

Object or Liters is 4 liters.

a farmer has a total yield of 42,000 bu of corn from a 350-acre farmer what total yield should he expect from a similar 560-acre farm?

Demographic, temporal, and spatial analysis of human rabid animal bite cases in Mymensingh District, Bangladesh

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Objective This study aimed to analyze the demographic, temporal, and spatial characteristics of rabid animal bite (RAB) cases in humans across 12 upazilas within Mymensingh district of Bangladesh. Methods Retrospective hospital-based data from RAB cases for 2022 and 2023 were collected from S.K Hospital. The dataset included information on victim demographics, bite details, vaccination information, and Rabies Immune Globulin (RIG) administration. Additionally, monthly case counts from 2016 to 2023 were sourced and analyzed to identify trends. Descriptive statistics and time series analysis using the seasonal decomposition technique were conducted. The risk maps for rabid animal bites in 2022 and 2023 were generated using a standardized incidence rate ratio (SIRR) approach. Findings An almost two-fold increase in the proportion of category 3 bites receiving Rabies Immune Globulin (RIG) from 3.6% in 2022 to 6.5% in 2023 was noted. Only 9.7% of bite cases in 2022 and 16.9% in 2023 received the vaccine promptly after the incident. However, the majority received vaccines within the first 24 hours after being bitten. Moreover, significant seasonal patterns and year-wise increasing trends in RAB cases were observed. Males and individuals <10 years old had a higher risk of being bitten. Dogs (48.2% in 2022) and cats (52.6% in 2023) were identified as the primary animals responsible for the bites. Notably, the legs were the body part most frequently bitten. The bites risk map identified four high risk upazilas. Conclusion There is a significant gap in ensuring timely vaccination delivery. Study results also suggest other potential improvements in healthcare practices or treatment protocols. Increasing RAB cases highlights the need to enhance surveillance and control measures. Targeted awareness campaigns and preventive measures tailored to high-risk groups − including males, children <10 years old, dogs and cats − are imperative. Coordinated efforts among healthcare professionals, policymakers, and community stakeholders are crucial to effectively mitigate the incidence of RAB cases, safeguarding public health and eradicate dog mediated rabies by 2030 in the region.

Competing Interest Statement

The authors have declared no competing interest.

Funding Statement

The author(s) received no specific funding for this work.

Author Declarations

I confirm all relevant ethical guidelines have been followed, and any necessary IRB and/or ethics committee approvals have been obtained.

The details of the IRB/oversight body that provided approval or exemption for the research described are given below:

The Ethical Review Committee of Mymensingh Medical College waived ethical approval for this work.

I confirm that all necessary patient/participant consent has been obtained and the appropriate institutional forms have been archived, and that any patient/participant/sample identifiers included were not known to anyone (e.g., hospital staff, patients or participants themselves) outside the research group so cannot be used to identify individuals.

I understand that all clinical trials and any other prospective interventional studies must be registered with an ICMJE-approved registry, such as ClinicalTrials.gov. I confirm that any such study reported in the manuscript has been registered and the trial registration ID is provided (note: if posting a prospective study registered retrospectively, please provide a statement in the trial ID field explaining why the study was not registered in advance).

I have followed all appropriate research reporting guidelines, such as any relevant EQUATOR Network research reporting checklist(s) and other pertinent material, if applicable.

Data Availability

All data generated from this study are presented within the manuscript.

View the discussion thread.

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Demographic, temporal, and spatial analysis of human rabid animal bite
Objective This study aimed to analyze the demographic, temporal, and spatial characteristics of rabid animal bite (RAB) cases in humans across 12 upazilas within Mymensingh district of Bangladesh. Methods Retrospective hospital-based data from RAB cases for 2022 and 2023 were collected from S.K Hospital. The dataset included information on victim demographics, bite details, vaccination ...

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Ratio and Proportion Questions & Word Problems | GMAT GRE Maths

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Demographic, temporal, and spatial analysis of human rabid animal bite cases in Mymensingh District, Bangladesh

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