ML Aggarwal Class 6 Maths Solutions Chapter 08 Ratio and Proportion

 

Question 1. Express the following ratios in simplest form:
(i) 20 : 40
(ii) 40 : 20
(iii) 81 : 108
(iv) 98 : 63
Answer: To express a ratio in its simplest form, we divide both terms by their highest common factor (H.C.F.).

(i) 20 : 40
H.C.F. of 20 and 40 is 20.
\( 20 : 40 = \frac{20 \div 20}{40 \div 20} = \frac{1}{2} = 1 : 2 \)

(ii) 40 : 20
H.C.F. of 40 and 20 is 20.
\( 40 : 20 = \frac{40 \div 20}{20 \div 20} = \frac{2}{1} = 2 : 1 \)

(iii) 81 : 108
H.C.F. of 81 and 108 is 27.
\( 81 : 108 = \frac{81 \div 27}{108 \div 27} = \frac{3}{4} = 3 : 4 \)

(iv) 98 : 63
H.C.F. of 98 and 63 is 7.
\( 98 : 63 = \frac{98 \div 7}{63 \div 7} = \frac{14}{9} = 14 : 9 \)
In simple words: A ratio in simplest form means both numbers have no common factor other than 1. Divide both by their H.C.F. to make it as simple as possible.

Exam Tip: Always find the H.C.F. first before reducing. Check your answer by confirming no common factor remains between the simplified terms.

 

Question 2. Fill in the missing numbers in the following equivalent ratios:
(i) \( \frac{14}{21} = \frac{x}{3} = \frac{6}{y} \)
(ii) \( \frac{15}{18} = \frac{...}{6} = \frac{10}{...} = \frac{...}{30} \)
Answer:
(i) \( \frac{14}{21} = \frac{x}{3} = \frac{6}{y} \)

First, reduce \( \frac{14}{21} \) to its simplest form:
\( \frac{14}{21} = \frac{14 \div 7}{21 \div 7} = \frac{2}{3} \)

Now, \( \frac{2}{3} = \frac{x}{3} \implies x = 2 \)

Also, \( \frac{2}{3} = \frac{6}{y} \)

Using cross multiplication: \( 2y = 18 \implies y = 9 \)

Hence, the missing values are 2 and 9.

(ii) \( \frac{15}{18} = \frac{...}{6} = \frac{10}{...} = \frac{...}{30} \)

First, reduce \( \frac{15}{18} \) to its simplest form:
\( \frac{15}{18} = \frac{15 \div 3}{18 \div 3} = \frac{5}{6} \)

Now compare each fraction with \( \frac{5}{6} \):

\( \frac{5}{6} = \frac{...}{6} \implies \) the numerator is 5

\( \frac{5}{6} = \frac{10}{...} \implies \frac{5 \times 2}{6 \times 2} = \frac{10}{12} \)

\( \frac{5}{6} = \frac{...}{30} \implies \frac{5 \times 5}{6 \times 5} = \frac{25}{30} \)

Hence, the missing values are 5, 12, and 25 respectively.
In simple words: Equivalent ratios are different ways of showing the same relationship. Reduce to the simplest form first, then multiply both parts by the same number to get the missing values.

Exam Tip: Cross multiplication helps verify your answer quickly. Always reduce the original fraction first to avoid errors.

 

Question 3. Find the ratio of each of the following in simplest form:
(i) 2.1 m to 1.2 m
(ii) 91 cm to 1.04 m
(iii) 3.5 kg to 250 g
(iv) 60 paise to 4 rupees
(v) 1 minute to 15 seconds
(vi) 15 mm to 2 cm
Answer:
(i) 2.1 m to 1.2 m

Multiply both terms by 10 to remove decimals:
\( 2.1 : 1.2 = 21 : 12 \)

H.C.F. of 21 and 12 is 3.
\( 21 : 12 = \frac{21 \div 3}{12 \div 3} = \frac{7}{4} = 7 : 4 \)

(ii) 91 cm to 1.04 m

Convert m to cm: 1.04 m = 104 cm
\( 91 : 104 \)

H.C.F. of 91 and 104 is 13.
\( 91 : 104 = \frac{91 \div 13}{104 \div 13} = \frac{7}{8} = 7 : 8 \)

(iii) 3.5 kg to 250 g

Convert kg to g: 3.5 kg = 3500 g
\( 3500 : 250 \)

H.C.F. of 3500 and 250 is 250.
\( 3500 : 250 = \frac{3500 \div 250}{250 \div 250} = \frac{14}{1} = 14 : 1 \)

(iv) 60 paise to 4 rupees

Convert rupees to paise: 4 rupees = 400 paise
\( 60 : 400 \)

H.C.F. of 60 and 400 is 20.
\( 60 : 400 = \frac{60 \div 20}{400 \div 20} = \frac{3}{20} = 3 : 20 \)

(v) 1 minute to 15 seconds

Convert minute to seconds: 1 minute = 60 seconds
\( 60 : 15 \)

H.C.F. of 60 and 15 is 15.
\( 60 : 15 = \frac{60 \div 15}{15 \div 15} = \frac{4}{1} = 4 : 1 \)

(vi) 15 mm to 2 cm

Convert cm to mm: 2 cm = 20 mm
\( 15 : 20 \)

H.C.F. of 15 and 20 is 5.
\( 15 : 20 = \frac{15 \div 5}{20 \div 5} = \frac{3}{4} = 3 : 4 \)
In simple words: When units are different, change them to the same unit first. Then divide both numbers by their highest common factor to get the simplest ratio.

Exam Tip: Always convert units to the same measurement before finding the ratio. This is the most common source of errors - double-check your conversions.

 

Question 4. The length and the breadth of a rectangular park are 125 m and 60 m respectively. What is the ratio of the length to the breadth of the park?
Answer: Given:
Length = 125 m
Breadth = 60 m

Ratio of length to breadth = 125 : 60

H.C.F. of 125 and 60 is 5.
\( 125 : 60 = \frac{125 \div 5}{60 \div 5} = \frac{25}{12} = 25 : 12 \)

Hence, the ratio of length to breadth is 25 : 12.
In simple words: To compare two measurements, put them as a ratio and reduce by dividing both by their H.C.F. This shows the proportional relationship between length and breadth.

Exam Tip: State what the ratio represents clearly (length to breadth). Always present your final answer in the simplest form with a colon.

 

Question 5. The population of a village is 4800. If the number of females is 2160, find the ratio of males to that of females.
Answer: Given:
Total population = 4800
Number of females = 2160

Number of males = Total population - Number of females
Number of males = 4800 - 2160 = 2640

Ratio of males to females = 2640 : 2160

H.C.F. of 2640 and 2160 is 240.
\( 2640 : 2160 = \frac{2640 \div 240}{2160 \div 240} = \frac{11}{9} = 11 : 9 \)

Hence, the ratio of males to females is 11 : 9.
In simple words: First, find the number of males by subtracting females from the total. Then express both quantities as a ratio and simplify by dividing by the H.C.F.

Exam Tip: Read the problem carefully to identify what calculation is needed before finding the ratio. Missing this step leads to incorrect ratios.

 

Question 6. In a class, there are 30 boys and 25 girls. Find the ratio of the number of:
(i) boys to that of girls
(ii) girls to that of total number of students
(iii) boys to that of total number of students
Answer: Given:
Number of boys = 30
Number of girls = 25
Total number of students = 30 + 25 = 55

(i) Ratio of boys to girls = 30 : 25

H.C.F. of 30 and 25 is 5.
\( 30 : 25 = \frac{30 \div 5}{25 \div 5} = \frac{6}{5} = 6 : 5 \)

Hence, the ratio of boys to girls is 6 : 5.

(ii) Ratio of girls to total = 25 : 55

H.C.F. of 25 and 55 is 5.
\( 25 : 55 = \frac{25 \div 5}{55 \div 5} = \frac{5}{11} = 5 : 11 \)

Hence, the ratio of girls to total number of students is 5 : 11.

(iii) Ratio of boys to total = 30 : 55

H.C.F. of 30 and 55 is 5.
\( 30 : 55 = \frac{30 \div 5}{55 \div 5} = \frac{6}{11} = 6 : 11 \)

Hence, the ratio of boys to total number of students is 6 : 11.
In simple words: Work out what total you need (either girls and boys separately or with the total), then create the ratio. Divide by the H.C.F. to simplify.

Exam Tip: Pay close attention to what is being compared in each sub-part. Each part requires identifying the correct pair of quantities before simplifying.

 

Question 7. In a year, Reena earns Rs.1,50,000 and saves Rs.50,000. Find the ratio of:
(i) money she earns to the money she saves
(ii) money that she saves to the money she spends
Answer: Given:
Money Reena earns = Rs.1,50,000
Money Reena saves = Rs.50,000

Money Reena spends = Money earned - Money saved
Money Reena spends = Rs.1,50,000 - Rs.50,000 = Rs.1,00,000

(i) Ratio of money earned to money saved = 1,50,000 : 50,000

Simplifying: 150 : 50 = 3 : 1

Hence, the ratio of money earned to money saved is 3 : 1.

(ii) Ratio of money saved to money spent = 50,000 : 1,00,000

Simplifying: 50 : 100 = 1 : 2

Hence, the ratio of money saved to money spent is 1 : 2.
In simple words: Work out how much money is spent by subtracting savings from earnings. Then write the pairs as ratios and reduce to simplest form.

Exam Tip: Remember that money spent = money earned - money saved. Calculate this value first before attempting any ratio.

 

Question 8. The monthly expenses on transport of a student have increased from Rs.350 to Rs.500. Find the ratio of:
(i) increase in expenses to original expenses
(ii) original expenses to increased expenses
(iii) increased expenses to increase in expenses
Answer: Given:
Original expenses = Rs.350
Increased expenses = Rs.500
Increase in expenses = Rs.500 - Rs.350 = Rs.150

(i) Ratio of increase in expenses to original expenses = 150 : 350

H.C.F. of 150 and 350 is 50.
\( 150 : 350 = \frac{150 \div 50}{350 \div 50} = \frac{3}{7} = 3 : 7 \)

Hence, the required ratio is 3 : 7.

(ii) Ratio of original expenses to increased expenses = 350 : 500

H.C.F. of 350 and 500 is 50.
\( 350 : 500 = \frac{350 \div 50}{500 \div 50} = \frac{7}{10} = 7 : 10 \)

Hence, the required ratio is 7 : 10.

(iii) Ratio of increased expenses to increase in expenses = 500 : 150

H.C.F. of 500 and 150 is 50.
\( 500 : 150 = \frac{500 \div 50}{150 \div 50} = \frac{10}{3} = 10 : 3 \)

Hence, the required ratio is 10 : 3.
In simple words: Find the increase amount first by subtracting the old from the new. Then form three different ratios using the three quantities: original, increased, and increase.

Exam Tip: Be careful to distinguish between "increase in expenses," "original expenses," and "increased expenses" - these are three different quantities. Read each part carefully.

 

Question 9. Out of 30 students in a class, 6 like football, 12 like cricket and the remaining like tennis. Find the ratio of:
(i) number of students liking football to number of students liking tennis.
(ii) number of students liking cricket to total number of students.
Answer: Given:
Total students = 30
Students liking football = 6
Students liking cricket = 12
Students liking tennis = 30 - 6 - 12 = 12

(i) Ratio of football lovers to tennis lovers = 6 : 12

H.C.F. of 6 and 12 is 6.
\( 6 : 12 = \frac{6 \div 6}{12 \div 6} = \frac{1}{2} = 1 : 2 \)

Hence, the required ratio is 1 : 2.

(ii) Ratio of cricket lovers to total students = 12 : 30

H.C.F. of 12 and 30 is 6.
\( 12 : 30 = \frac{12 \div 6}{30 \div 6} = \frac{2}{5} = 2 : 5 \)

Hence, the required ratio is 2 : 5.
In simple words: Calculate the number liking tennis by subtracting football and cricket lovers from the total. Then form the required ratios and simplify.

Exam Tip: Always calculate any unknown quantity first. In this case, find tennis lovers before writing ratios.

 

Question 10. Which ratio is greater?
(i) 3 : 4 or 2 : 3
(ii) 11 : 21 or 15 : 28
Answer:
(i) 3 : 4 or 2 : 3

The given ratios are 3 : 4 and 2 : 3, which are equivalent to the fractions \( \frac{3}{4} \) and \( \frac{2}{3} \) respectively.

Find L.C.M. of 4 and 3:
L.C.M. = 2 × 2 × 3 = 12

\( 3 : 4 = \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} \)

\( 2 : 3 = \frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12} \)

Clearly, \( \frac{9}{12} > \frac{8}{12} \)

Hence, (3 : 4) > (2 : 3).

(ii) 11 : 21 or 15 : 28

The given ratios are 11 : 21 and 15 : 28, which are equivalent to the fractions \( \frac{11}{21} \) and \( \frac{15}{28} \) respectively.

Find L.C.M. of 21 and 28:
21 = 3 × 7
28 = 4 × 7
L.C.M. = 3 × 4 × 7 = 84

\( 11 : 21 = \frac{11}{21} = \frac{11 \times 4}{21 \times 4} = \frac{44}{84} \)

\( 15 : 28 = \frac{15}{28} = \frac{15 \times 3}{28 \times 3} = \frac{45}{84} \)

Clearly, \( \frac{44}{84} < \frac{45}{84} \)

Hence, (15 : 28) > (11 : 21).
In simple words: Convert both ratios to fractions. Find a common denominator using the L.C.M., then compare the numerators to see which is larger.

Exam Tip: Converting ratios to fractions with a common denominator is the clearest way to compare them. Always find the L.C.M. of the denominators.

 

Question 11. Divide Rs.560 between Ramu and Munni in the ratio 3 : 2.
Answer: Given: Total money = Rs.560, Ratio = 3 : 2. Find total parts: 3 + 2 = 5 parts. Find the value of 1 part: 560 ÷ 5 = Rs.112. Now calculate shares by multiplying each part with Rs.112: Ramu's share: 3 × Rs.112 = Rs.336. Munni's share: 2 × Rs.112 = Rs.224.
In simple words: Split the 560 rupees into 5 equal parts. Ramu gets 3 parts and Munni gets 2 parts, so Ramu gets Rs.336 and Munni gets Rs.224.

Exam Tip: Always add the ratio numbers to find total parts first, then divide the total amount by this sum to get the value of one part.

 

Question 12. Two people invested Rs.15,000 and Rs.25,000 respectively to start a business. They decided to share the profits in the ratio of their investments. If their profit is Rs.12000, how much does each get?
Answer: Given: First investment = Rs.15,000, Second investment = Rs.25,000. Ratio of investments = 15000 : 25000 = 15 : 25 = 3 : 5. Total profit = Rs.12,000. Find total parts: 3 + 5 = 8 parts. Find the value of 1 part: 12000 ÷ 8 = Rs.1,500. Now calculate shares by multiplying each part with Rs.1,500: First person's share: 3 × Rs.1,500 = Rs.4,500. Second person's share: 5 × Rs.1,500 = Rs.7,500.
In simple words: The two people invested money in the ratio 3 : 5, so they split the profit of Rs.12,000 in the same ratio. The first person gets Rs.4,500 and the second person gets Rs.7,500.

Exam Tip: When sharing profit or loss based on investments, simplify the ratio of investments first before finding total parts.

 

Question 13. The ratio of Ankur's money to Roma's money is 9 : 11. If Ankur has Rs.540, how much money does Roma have?
Answer: Given: Ratio (Ankur : Roma) = 9 : 11, Ankur's money = Rs.540. Since 9 parts represent Ankur's money, 9 parts = Rs.540. Therefore, 1 part = 540 ÷ 9 = Rs.60. Roma's money = 11 parts = 11 × Rs.60 = Rs.660.
In simple words: Ankur's money equals 9 parts and Roma's money equals 11 parts. If 9 parts make Rs.540, then each part is Rs.60, so 11 parts make Rs.660.

Exam Tip: Identify which part of the ratio corresponds to the given amount, then find the value of one part and use it to find other quantities.

 

Question 14. The ratio of weights of tin and zinc in an alloy is 2 : 5. How much zinc is there in 31.5 g of alloy?
Answer: Given: Ratio (Tin : Zinc) = 2 : 5, Total weight of alloy = 31.5 g. Find total parts: 2 + 5 = 7 parts. Find the value of 1 part: 31.5 ÷ 7 = 4.5 g. Weight of zinc = 5 parts = 5 × 4.5 g = 22.5 g.
In simple words: The alloy is made of 7 equal parts - 2 parts tin and 5 parts zinc. Each part weighs 4.5 g, so the zinc weighs 5 × 4.5 = 22.5 g.

Exam Tip: When dealing with alloy or mixture problems, convert the ratio into parts and use these parts to find the required amount.

 

Exercise 8.2

 

Question 1. Check whether the given two ratios form a proportion or not:
(i) 4 : 6 and 12 : 18
(ii) 15 : 45 and 40 : 120
(iii) 14 : 4 and 18 : 6
(iv) 12 : 18 and 28 : 12
Answer: In a proportion, product of extremes = product of means.
(i) 4 : 6 and 12 : 18
Product of extremes = 4 × 18 = 72
Product of means = 6 × 12 = 72
Since the two products are equal, these ratios form a proportion. Hence, 4 : 6 and 12 : 18 form a proportion.
(ii) 15 : 45 and 40 : 120
Product of extremes = 15 × 120 = 1800
Product of means = 45 × 40 = 1800
Since the two products are equal, these ratios form a proportion. Hence, 15 : 45 and 40 : 120 form a proportion.
(iii) 14 : 4 and 18 : 6
Product of extremes = 14 × 6 = 84
Product of means = 4 × 18 = 72
Since the products are not equal, these ratios do not form a proportion. Hence, 14 : 4 and 18 : 6 do not form a proportion.
(iv) 12 : 18 and 28 : 12
Product of extremes = 12 × 12 = 144
Product of means = 18 × 28 = 504
Since the products are not equal, these ratios do not form a proportion. Hence, 12 : 18 and 28 : 12 do not form a proportion.
In simple words: Two ratios form a proportion when the product of the first and last numbers equals the product of the middle two numbers.

Exam Tip: Always use the cross-multiplication rule (product of extremes = product of means) to check if two ratios form a proportion.

 

Question 2. Write true (T) or false (F) against each of the following statements:
(i) 16 : 24 = 20 : 30
(ii) 16 : 24 = 30 : 20
(iii) 21 : 6 :: 35 : 10
(iv) 5.2 : 3.9 :: 3 : 4
Answer:
(i) 16 : 24 = 20 : 30
Product of extremes = 16 × 30 = 480
Product of means = 24 × 20 = 480
Since the products are equal, the statement is True.
(ii) 16 : 24 = 30 : 20
Product of extremes = 16 × 20 = 320
Product of means = 24 × 30 = 720
Since the products are not equal, the statement is False.
(iii) 21 : 6 :: 35 : 10
Product of extremes = 21 × 10 = 210
Product of means = 6 × 35 = 210
Since the products are equal, the statement is True.
(iv) 5.2 : 3.9 :: 3 : 4
Product of extremes = 5.2 × 4 = 20.8
Product of means = 3.9 × 3 = 11.7
Since the products are not equal, the statement is False.
In simple words: To check if two ratios form a proportion, multiply the outer numbers and the middle numbers. If both products match, the statement is true.

Exam Tip: Be careful with decimal numbers - multiply accurately to avoid calculation errors.

 

Question 3. Find which of the following are in proportion:
(i) 12, 16, 6, 8
(ii) 2, 3, 4, 5
(iii) 18, 10, 9, 5
(iv) 18, 9, 10, 5
Answer: Four numbers a, b, c, d are in proportion if and only if a × d = b × c.
(i) 12, 16, 6, 8
Product of extremes = 12 × 8 = 96
Product of means = 16 × 6 = 96
Since the products are equal, 12, 16, 6, 8 are in proportion.
(ii) 2, 3, 4, 5
Product of extremes = 2 × 5 = 10
Product of means = 3 × 4 = 12
Since 10 ≠ 12, these numbers are not in proportion.
(iii) 18, 10, 9, 5
Product of extremes = 18 × 5 = 90
Product of means = 10 × 9 = 90
Since the products are equal, 18, 10, 9, 5 are in proportion.
(iv) 18, 9, 10, 5
Product of extremes = 18 × 5 = 90
Product of means = 9 × 10 = 90
Since the products are equal, 18, 9, 10, 5 are in proportion.
In simple words: Four numbers form a proportion when the first number times the last number equals the second number times the third number.

Exam Tip: To save time, simplify each ratio first and check if they are equal, or use cross-multiplication directly.

 

Question 4. Are the following statements true?
(i) 39 kg : 36 kg = 26 men : 24 men
(ii) 45 km : 60 km = 12 hours : 15 hours
(iii) 40 people : 200 people = Rs.1000 : Rs.5000
(iv) 7.5 litres : 15 litres = 15 children : 30 children
Answer:
(i) 39 kg : 36 kg = 26 men : 24 men
39 : 36 = \( \frac{39}{36} = \frac{13}{12} \)
26 : 24 = \( \frac{26}{24} = \frac{13}{12} \)
Both ratios are equal. Hence, the statement is True.
(ii) 45 km : 60 km = 12 hours : 15 hours
45 : 60 = \( \frac{45}{60} = \frac{3}{4} \)
12 : 15 = \( \frac{12}{15} = \frac{4}{5} \)
\( \frac{3}{4} \neq \frac{4}{5} \). The ratios are not equal. Hence, the statement is False.
(iii) 40 people : 200 people = Rs.1000 : Rs.5000
40 : 200 = \( \frac{40}{200} = \frac{1}{5} \)
1000 : 5000 = \( \frac{1000}{5000} = \frac{1}{5} \)
Both ratios are equal. Hence, the statement is True.
(iv) 7.5 litres : 15 litres = 15 children : 30 children
7.5 : 15 = \( \frac{7.5}{15} = \frac{1}{2} \)
15 : 30 = \( \frac{15}{30} = \frac{1}{2} \)
Both ratios are equal. Hence, the statement is True.
In simple words: Reduce each ratio to its simplest form by dividing both parts by their common factors. If the simplified forms match, the statement is true.

Exam Tip: Always simplify ratios before comparing them - this makes it easier to check if they are equal.

 

Question 5. Determine if the following ratios form a proportion. Also write the middle terms and extreme terms when the ratios form a proportion:
(i) 25 cm : 1 m and Rs.40 : Rs.160
(ii) 39 litre : 65 litre and 6 bottles : 10 bottles
(iii) 2 kg : 80 kg and 30 sec : 5 minutes
(iv) 200 g : 2.5 kg and Rs.4 : Rs.50
Answer:
(i) 25 cm : 1 m and Rs.40 : Rs.160
Convert m into cm: 1 m = 100 cm. First ratio = 25 : 100 = 1 : 4. Second ratio = 40 : 160 = 1 : 4. Both ratios are equal, so they form a proportion. Proportion: 25 : 100 :: 40 : 160. Extreme terms: 25 cm and Rs.160; Middle terms: 100 cm and Rs.40.
(ii) 39 litre : 65 litre and 6 bottles : 10 bottles
First ratio = 39 : 65 = 3 : 5. Second ratio = 6 : 10 = 3 : 5. Both ratios are equal, so they form a proportion. Proportion: 39 : 65 :: 6 : 10. Extreme terms: 39 litre and 10 bottles; Middle terms: 65 litre and 6 bottles.
(iii) 2 kg : 80 kg and 30 sec : 5 minutes
Convert 5 minutes into seconds: 5 min = 5 × 60 sec = 300 sec. First ratio = 2 : 80 = 1 : 40. Second ratio = 30 : 300 = 1 : 10. Since 1 : 40 ≠ 1 : 10, the ratios are not equal. Hence, the ratios do not form a proportion.
(iv) 200 g : 2.5 kg and Rs.4 : Rs.50
Convert kg into g: 2.5 kg = 2.5 × 1000 g = 2500 g. First ratio = 200 : 2500 = 2 : 25. Second ratio = 4 : 50 = 2 : 25. Both ratios are equal, so they form a proportion. Proportion: 200 : 2500 :: 4 : 50. Extreme terms: 200 g and Rs.50; Middle terms: 2500 g and Rs.4.
In simple words: First, change all measurements to the same units. Then simplify both ratios. If they match, they form a proportion - the first and last numbers are extremes, and the middle two are middle terms.

Exam Tip: Always convert units to the same standard before comparing ratios to avoid errors.

 

Exercise 8.3

 

Question 1. If the cost of 9 m cloth is Rs.378, find the cost of 4 m cloth.
Answer: Given: Cost of 9 m cloth = Rs.378. Cost of 1 m cloth = \( \frac{378}{9} \) = Rs.42. Cost of 4 m cloth = 4 × Rs.42 = Rs.168.
In simple words: Find the cost of cloth per metre first by dividing the total cost by the total length. Then multiply this cost per metre by 4 to get the cost of 4 m.

Exam Tip: Always find the unit rate first (cost per item, weight per item, distance per litre) and then use it to find the required quantity.

 

Question 2. The weight of 36 books is 12 kg. What is weight of 75 such books?
Answer: Given: Weight of 36 books = 12 kg. Weight of 1 book = \( \frac{12}{36} \) kg = \( \frac{1}{3} \) kg. Weight of 75 books = 75 × \( \frac{1}{3} \) kg = 25 kg.
In simple words: Find how much one book weighs by dividing the total weight by the number of books. Then multiply this weight per book by 75 to get the total weight.

Exam Tip: Simplify the fraction for the unit rate before multiplying - it makes calculations faster and reduces errors.

 

Question 3. Five pens cost Rs.115. How many pens can you buy in Rs.207?
Answer: Given: Cost of 5 pens = Rs.115. Cost of 1 pen = \( \frac{115}{5} \) = Rs.23. Number of pens that can be bought in Rs.207 = \( \frac{207}{23} \) = 9.
In simple words: First, find the cost of one pen by dividing the total cost by the number of pens. Then divide the amount you have by the cost per pen to find how many pens you can buy.

Exam Tip: Make sure the final answer is a whole number - if not, check your calculations again.

 

Question 4. A car consumes 8 litres of petrol in covering a distance of 100 km. How many kilometres will it travel in 26 litres of petrol?
Answer: Given: Distance covered in 8 litres of petrol = 100 km. Distance covered in 1 litre of petrol = \( \frac{100}{8} \) km = 12.5 km. Distance covered in 26 litres of petrol = 26 × 12.5 km = 325 km.
In simple words: Find how far the car goes on one litre of petrol by dividing the distance by the litres used. Then multiply this distance per litre by 26 litres.

Exam Tip: In mileage or consumption problems, always find the unit rate (distance per litre, consumption per unit) first.

 

Question 5. A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Answer: Given: Diesel needed for 594 km = 108 litres. Diesel needed for 1 km = \( \frac{108}{594} \) litre = \( \frac{2}{11} \) litre. Diesel needed for 1650 km = 1650 × \( \frac{2}{11} \) = \( \frac{3300}{11} \) = 300 litres.
In simple words: Find how much diesel the truck needs per kilometre by dividing the total diesel by the distance. Then multiply this amount per kilometre by 1650 km.

Exam Tip: Always reduce fractions to their simplest form before multiplying to make calculations easier and to avoid rounding errors.

 

Question 6. A transport company charges Rs.5400 to carry 80 quintals of weight. What will it charge to carry 126 quintals of weight (same distance)?
Answer: Given: Charge for carrying 80 quintals = Rs.5400. First, find the cost per quintal by dividing Rs.5400 by 80, which gives Rs.67.50 per quintal. Now multiply this unit rate by 126 to get the total charge: 126 × Rs.67.50 = Rs.8505. Therefore, the company will charge Rs.8505 to carry 126 quintals.
In simple words: Divide the total cost by the weight to find the cost of one quintal. Then multiply that cost by the new weight to get the answer.

Exam Tip: Always find the unit rate (cost per unit) first, then multiply by the new quantity—this method works for all unitary problems.

 

Question 7. 42 metres of cloth is required to make 20 shirts of the same size. How much cloth will be required to make 36 shirts of that size?
Answer: Given: Cloth needed for 20 shirts = 42 m. Calculate cloth per shirt: 42 ÷ 20 = 2.1 m per shirt. For 36 shirts, multiply the unit amount by 36: 36 × 2.1 = 75.6 m. Thus, 75.6 m of cloth is needed to make 36 shirts.
In simple words: Find how much cloth is used for one shirt, then multiply that by the number of shirts you want to make.

Exam Tip: Always express the quantity per single unit first (per shirt, per item), then scale up or down to the required amount.

 

Question 8. Cost of 5 kg of local rice in a village is Rs.107.50. (i) What will be the cost of 8 kg of rice? (ii) What quantity of rice can be purchased in Rs.64.5?
Answer: Given: Cost of 5 kg of rice = Rs.107.50. Cost per kilogram = Rs.107.50 ÷ 5 = Rs.21.50.
(i) For 8 kg: 8 × Rs.21.50 = Rs.172. Therefore, 8 kg of rice costs Rs.172.
(ii) To find how much rice can be bought for Rs.64.5: Rs.64.5 ÷ Rs.21.50 = 3 kg. So, 3 kg of rice can be bought for Rs.64.5.
In simple words: Find the price of one kg. Then either multiply for more kg or divide the money by the price per kg to find the quantity.

Exam Tip: For part (i), multiply unit price by quantity; for part (ii), divide total money by unit price to get quantity.

 

Question 9. 20 tons of iron costs Rs.6,00,000. Find the cost of 560 kg of iron. (1 ton = 1000 kg)
Answer: Given: 20 tons = 20 × 1000 = 20,000 kg. Cost of 20,000 kg = Rs.6,00,000. Cost per kilogram = Rs.6,00,000 ÷ 20,000 = Rs.30. For 560 kg: 560 × Rs.30 = Rs.16,800. Therefore, the cost of 560 kg of iron is Rs.16,800.
In simple words: Convert tons to kilograms, then find the cost per kg. Multiply this rate by the weight you need.

Exam Tip: Always convert units to a common measure first before calculating unit rates, especially when the problem mixes different units.

 

Question 10. Cost of 4 dozen bananas is Rs.180. How many bananas can be purchased for Rs.37.50?
Answer: Given: 4 dozen bananas = 4 × 12 = 48 bananas. Cost of 48 bananas = Rs.180. Cost per banana = Rs.180 ÷ 48 = Rs.3.75. Number of bananas that can be bought for Rs.37.50 = Rs.37.50 ÷ Rs.3.75 = 10 bananas. Therefore, 10 bananas can be purchased for Rs.37.50.
In simple words: First convert dozens to individual bananas. Then find the cost of one banana. Finally, divide the total money by the cost per banana.

Exam Tip: Watch for unit conversions (like dozen to individual items)—missing this step is a common mistake in problems involving grouped quantities.

 

Question 11. Aman purchases 12 pens for Rs.156 and Payush buys 9 pens for Rs.108. Can you say who got the pens cheaper?
Answer: For Aman: Cost of 12 pens = Rs.156. Cost per pen = Rs.156 ÷ 12 = Rs.13. For Payush: Cost of 9 pens = Rs.108. Cost per pen = Rs.108 ÷ 9 = Rs.12. Since Rs.12 is less than Rs.13, Payush got the pens at a cheaper rate. Thus, Payush bought the pens cheaper than Aman.
In simple words: To compare prices, find the cost of one item for each person. The person with the lower unit price got the better deal.

Exam Tip: Always reduce to a unit rate when comparing prices or values—comparing total costs directly can be misleading if quantities differ.

 

Question 12. Rohit made 42 runs in 6 overs and Virat made 63 runs in 7 overs. Who made more runs per over?
Answer: For Rohit: Runs in 6 overs = 42. Runs per over = 42 ÷ 6 = 7 runs per over. For Virat: Runs in 7 overs = 63. Runs per over = 63 ÷ 7 = 9 runs per over. Since 9 is greater than 7, Virat scored at a higher rate. Therefore, Virat made more runs per over than Rohit.
In simple words: Divide total runs by the number of overs to find runs per over. The larger number shows better performance.

Exam Tip: In rate-based comparisons (runs per over, speed per hour), always calculate the rate for each person before comparing, not the totals.

 

Exercise 8.4

 

Question 1. Find the value of: (i) 18% of Rs.450 (ii) 14% of \( 16\frac{2}{3} \) kg (iii) \( 27\frac{3}{4} \)% of Rs.1200 (iv) \( \frac{5}{8} \)% of 600 m (v) \( 6\frac{1}{4} \)% of 1 hour 20 minutes (vi) 0.6% of 5 km
Answer:
(i) 18% of Rs.450 = \( \frac{18}{100} \times 450 = \frac{8100}{100} = \)Rs.81

(ii) First convert \( 16\frac{2}{3} \) kg to an improper fraction: \( 16\frac{2}{3} = \frac{50}{3} \) kg. Then: 14% of \( \frac{50}{3} = \frac{14}{100} \times \frac{50}{3} = \frac{700}{300} = \frac{7}{3} = 2\frac{1}{3} \) kg

(iii) Convert the percentage: \( 27\frac{3}{4}\% = \frac{111}{4}\% \). Then: \( \frac{111}{4}\% \) of Rs.1200 = \( \frac{111}{4 \times 100} \times 1200 = \frac{111 \times 1200}{400} = \)Rs.333

(iv) \( \frac{5}{8}\% \) of 600 m = \( \frac{5}{8 \times 100} \times 600 = \frac{3000}{800} = \)3.75 m

(v) Convert: \( 6\frac{1}{4}\% = \frac{25}{4}\% \) and 1 hour 20 minutes = 80 minutes. Then: \( \frac{25}{4}\% \) of 80 = \( \frac{25}{4 \times 100} \times 80 = \frac{2000}{400} = \)5 minutes

(vi) Convert distance: 5 km = 5000 m. Then: 0.6% of 5000 m = \( \frac{0.6}{100} \times 5000 = \frac{3000}{100} = \)30 m
In simple words: To find a percentage of a quantity, multiply the quantity by the percentage written as a fraction over 100. Always convert mixed numbers and units to standard forms first.

Exam Tip: Convert mixed number percentages to improper fractions and convert different units to a single standard unit before multiplying—this prevents errors.

 

Question 2. In a class of 60 students, 45% are girls. Find the number of boys in the class.
Answer: Given: Total students = 60. Percentage of girls = 45%. Number of girls = 45% of 60 = \( \frac{45}{100} \times 60 = \frac{2700}{100} = \)27. Number of boys = Total students - Number of girls = 60 - 27 = 33. Therefore, there are 33 boys in the class.
In simple words: Find how many girls there are using the percentage. Then subtract from the total to find the number of boys.

Exam Tip: Remember that percentages must add up to 100%, so if one group is 45%, the other group is 55% (or 100% - 45%)—you can use either method.

 

Question 3. Mr. Malkani saves 22% of his salary every month. If his salary is Rs.12750 per month, what is his expenditure?
Answer: Given: Monthly salary = Rs.12750. Percentage saved = 22%. Savings = 22% of Rs.12750 = \( \frac{22}{100} \times 12750 = \frac{280500}{100} = \)Rs.2805. Expenditure = Salary - Savings = Rs.12750 - Rs.2805 = Rs.9945. Therefore, Mr. Malkani's monthly expenditure is Rs.9945.
In simple words: Calculate how much is saved by finding the percentage. The rest of the salary (salary minus savings) is what he spends.

Exam Tip: Salary = Savings + Expenditure. If you know savings, subtract from total salary to find expenditure—this is a key relationship to remember.

 

Question 4. On a rainy day, 94% of the students were present in a school. If the number of students absent on that day was 174, find the total strength of the school.
Answer: Given: Percentage present = 94%. Percentage absent = 100% - 94% = 6%. Number absent = 174. Let total strength = x. Then: 6% of x = 174, which means \( \frac{6}{100} \times x = 174 \)

\[ x = \frac{174 \times 100}{6} = \frac{17400}{6} = 2900 \]

Therefore, the total strength of the school is 2900 students.
In simple words: If 94% were present, then 6% were absent. Use the absent number to find the whole total by setting up an equation.

Exam Tip: When a percentage is given and you know the actual count in that percentage, always find the whole by: (actual count ÷ percentage as decimal) = whole.

 

Exercise 8.5

 

Question 1. The speed of a car is \( 105\frac{1}{5} \) km/h, find the distance covered by it in \( 3\frac{3}{5} \) hours.
Answer: Given: Speed of car = \( 105\frac{1}{5} = \frac{526}{5} \) km/h. Time = \( 3\frac{3}{5} = \frac{18}{5} \) hours. Distance = Speed × Time = \( \frac{526}{5} \times \frac{18}{5} = \frac{9468}{25} = 378\frac{18}{25} \) km. Therefore, the car covers \( 378\frac{18}{25} \) km in \( 3\frac{3}{5} \) hours.
In simple words: Convert mixed numbers to improper fractions. Multiply speed by time to find distance. Convert the result back to a mixed number if needed.

Exam Tip: Always convert mixed numbers to improper fractions before multiplying—this avoids calculation errors and keeps the work neat.

 

Question 2. If the speed of a car is 50.4 km/h, find the distance covered in 3.6 hours.
Answer: Given: Speed of car = 50.4 km/h. Time = 3.6 hours. Distance = Speed × Time = 50.4 × 3.6. Calculate: \( 50.4 \times 3.6 = \frac{504}{100} \times \frac{36}{100} \times 100 = \frac{504 \times 36}{100} = \frac{18144}{100} = \)181.44 km. Therefore, the car covers 181.44 km in 3.6 hours.
In simple words: Multiply the speed (in km per hour) by the time (in hours) to get the total distance traveled.

Exam Tip: For decimal multiplication, count total decimal places in both numbers and place the decimal in the answer accordingly.

 

Question 3. If a car covers a distance of 201.25 km in 3.5 hours, find the speed of the car.
Answer: Given: Distance = 201.25 km. Time = 3.5 hours. Speed = \( \frac{\text{Distance}}{\text{Time}} = \frac{201.25}{3.5} = \frac{201.25 \times 10}{3.5 \times 10} = \frac{2012.5}{35} = \)57.5 km/h. Therefore, the speed of the car is 57.5 km/h.
In simple words: To find speed, divide the total distance by the time taken. The answer tells you how many kilometers traveled per hour.

Exam Tip: When dividing decimals, multiply both numerator and denominator by 10 (or 100) to remove decimals, making the division easier.

 

Question 4. A bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds, then find the ratio of the distances travelled by them in one hour.
Answer: Given: Bus travels 160 km in 4 hours. Distance covered by bus in 1 hour = \( \frac{160}{4} = \)40 km. Train travels 320 km in 5 hours. Distance covered by train in 1 hour = \( \frac{320}{5} = \)64 km. Ratio of distances in one hour (bus : train) = 40 : 64. Find H.C.F. of 40 and 64, which is 8. Simplify: \( \frac{40 \div 8}{64 \div 8} = \frac{5}{8} \) = 5 : 8. Therefore, the ratio of distances travelled by them in one hour is 5 : 8.
In simple words: Find how far each vehicle travels in one hour (by dividing total distance by hours). Then write the ratio of these one-hour distances in simplest form.

Exam Tip: Always reduce ratios to simplest form by dividing both terms by their H.C.F.—examiners expect the final answer in lowest terms.

 

Objective Type Questions - Mental Maths

 

Question 1. Fill in the blanks: (i) In the ratio 3 : 5, the first term is ... and second term is ... (ii) In a ratio, the first term is also called ... and second term is also called ... (iii) If two terms of a ratio have no common factor (except 1), then the ratio is said to be in ... (iv) To simplify a ratio, we divide the two terms by their ... (v) The simplest form of the ratio 8 : 12 is ... (vi) 90 cm : 1.5 m = ... (vii) When two ratios are equal, they are said to be in ... (viii) 4.5% of Rs.40 is equal to ...
Answer:
(i) In the ratio 3 : 5, the first term is 3 and the second term is 5.

(ii) In a ratio, the first term is called the antecedent and the second term is called the consequent.

(iii) If two terms of a ratio share no common factor except 1, the ratio is in simplest form (or lowest terms).

(iv) To simplify a ratio, divide both terms by their H.C.F. (Highest Common Factor).

(v) The simplest form of 8 : 12 is 2 : 3. (Reason: H.C.F. of 8 and 12 is 4, so \( \frac{8}{4} : \frac{12}{4} = 2 : 3 \))

(vi) 90 cm : 1.5 m = 3 : 5. (Reason: Convert 1.5 m to 150 cm, so 90 : 150 = 3 : 5)

(vii) When two ratios are equal, they are said to be in proportion.

(viii) 4.5% of Rs.40 = Rs.1.80. (Reason: \( \frac{4.5}{100} \times 40 = \frac{180}{100} = \)Rs.1.80)
In simple words: The first number in a ratio is the antecedent; the second is the consequent. To simplify, find the largest number that divides both, and divide. Equal ratios form a proportion.

Exam Tip: Always convert units to the same measure before writing a ratio (cm to cm, or m to m)—mixing units in a ratio is a common error.

 

Question 2. State whether the following statements are true (T) or false (F): (i) Ratio exists only between two quantities of the same kind. (ii) Ratio has no units. (iii) Ratio a : b is same as the ratio b : a. (iv) If we multiply or divide both terms of a ratio by the same non-zero number, then the ratio remains the same. (v) The ratio a : b is said to be in simplest form if HCF of a and b is 1.
Answer:
(i) True. A ratio is a comparison of two quantities that must be of the same type and expressed in the same units.

(ii) True. A ratio is a dimensionless (unitless) pure number because units cancel when quantities of the same kind are compared.

(iii) False. The ratio a : b is different from b : a because the order matters—the first term and second term have different meanings and values.

(iv) True. If both terms are multiplied or divided by the same non-zero number, the value of the ratio does not change—it remains equivalent (e.g., 2 : 3 = 4 : 6 = 6 : 9).

(v) True. A ratio a : b is in its simplest (or reduced) form when the highest common factor of a and b equals 1, meaning they share no common divisors.
In simple words: A ratio compares two amounts of the same kind. It has no units. Order matters—3 : 5 is not the same as 5 : 3. Multiplying or dividing both numbers by the same amount keeps the ratio equal. Simplest form means no number divides both terms.

Exam Tip: Remember that 3 : 5 and 5 : 3 are completely different—order is critical in ratios. This is a frequent true/false trap in exams.

 

Question 3. A ratio equivalent to 5 : 7 is
(a) 10 : 21
(b) 15 : 14
(c) 20 : 28
(d) 25 : 49
Answer: (c) 20 : 28
In simple words: When you multiply both parts of a ratio by the same number, you get a ratio that is the same as the original. Here, 5 : 7 becomes 20 : 28 when both parts are multiplied by 4.

Exam Tip: To check if two ratios are equivalent, try multiplying or dividing both terms by the same number - if the result matches one of the options, that is your answer.

 

Question 4. The ratio 384 : 480 in the simplest form is
(a) 2 : 5
(b) 3 : 5
(c) 5 : 4
(d) 4 : 5
Answer: (d) 4 : 5
In simple words: Find the largest number that divides both 384 and 480 evenly. That number is 96. When you divide both parts by 96, you get 4 : 5.

Exam Tip: Always find the H.C.F. (Highest Common Factor) of both terms to reduce a ratio to its simplest form.

 

Question 5. The ratio of 20 minutes to 1 hour is
(a) 20 : 1
(b) 1 : 3
(c) 1 : 4
(d) 2 : 5
Answer: (b) 1 : 3
In simple words: First, convert 1 hour to minutes: 1 hour = 60 minutes. Then write the ratio as 20 : 60. Simplify it by dividing both by 20 to get 1 : 3.

Exam Tip: When comparing quantities with different units, always convert them to the same unit first before forming the ratio.

 

Question 6. The ratio of 150 g to 2 kg is
(a) 75 : 1
(b) 40 : 3
(c) 3 : 40
(d) 3 : 200
Answer: (c) 3 : 40
In simple words: Change 2 kg to grams: 2 kg = 2000 g. Now write the ratio as 150 : 2000. The H.C.F. of 150 and 2000 is 50. Divide both by 50 to get 3 : 40.

Exam Tip: Remember to convert all units to the same measurement before simplifying - grams to grams, or kilograms to kilograms.

 

Question 7. In a class of 40 students, 25 students play cricket and the remaining play tennis. The ratio of number of students playing crickets to the number of students playing tennis is
(a) 5 : 8
(b) 5 : 3
(c) 3 : 5
(d) 8 : 5
Answer: (b) 5 : 3
In simple words: 25 students play cricket. The rest play tennis: 40 - 25 = 15 students. The ratio is 25 : 15. Simplify by dividing both by 5 to get 5 : 3.

Exam Tip: First, find the numbers you need (cricket players and tennis players), then form the ratio in the order asked, and reduce it to simplest form.

 

Question 8. Two numbers are in the ratio 3 : 5. If the sum of numbers is 144, then the smaller number is
(a) 54
(b) 72
(c) 90
(d) 48
Answer: (a) 54
In simple words: If two numbers are in the ratio 3 : 5, imagine them as 3 parts and 5 parts. Together they make 3 + 5 = 8 parts, which equal 144. So each part = 144 ÷ 8 = 18. The smaller number is 3 parts = 3 × 18 = 54.

Exam Tip: In ratio problems, always add the parts of the ratio first, then divide the total by this sum to find the value of one part.

 

Question 9. The ratio of number of girls to the number of boys in a class is 5 : 4. If there are 25 girls in the class, then the number of boys in the class is
(a) 15
(b) 20
(c) 30
(d) 40
Answer: (b) 20
In simple words: Girls and boys are in the ratio 5 : 4. There are 25 girls, which equals 5 parts. So each part = 25 ÷ 5 = 5. Boys are 4 parts = 4 × 5 = 20.

Exam Tip: Match the known quantity to its part in the ratio first, then find the value of one part, and finally calculate the unknown quantity.

 

Question 10. The ratio of the number of sides of a square and the number of edges of a cube is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 2 : 3
Answer: (b) 1 : 3
In simple words: A square has 4 sides. A cube has 12 edges. The ratio is 4 : 12. Reduce this by dividing both by 4 to get 1 : 3.

Exam Tip: Know the basic properties of shapes - a square has 4 sides, a cube has 12 edges - these appear often in ratio problems.

 

Question 11. On a shelf, the books with green cover and that with brown cover are in the ratio 2 : 3. If there are 18 books with green cover, then the number of books with brown cover is
(a) 12
(b) 24
(c) 27
(d) 36
Answer: (c) 27
In simple words: Green and brown books are in the ratio 2 : 3. There are 18 green books, which is 2 parts. So each part = 18 ÷ 2 = 9. Brown books are 3 parts = 3 × 9 = 27.

Exam Tip: Always match the given information to the correct part of the ratio before calculating the unknown value.

 

Question 12. In a box, the ratio of the number of red marbles to that of blue marbles is 4 : 7. Which of the following could be the total number of marbles in the box?
(a) 14
(b) 21
(c) 22
(d) 28
Answer: (c) 22
In simple words: Red and blue marbles are in the ratio 4 : 7. The total parts are 4 + 7 = 11. The total number of marbles must be a multiple of 11 (meaning 11 times some whole number). Among the choices, 22 = 2 × 11, so 22 is the answer.

Exam Tip: When asked about the total in a ratio problem, the total must always be a multiple of the sum of all parts in the ratio.

 

Question 13. If a, b, c and d are in proportion, then
(a) ab = cd
(b) ad = bc
(c) ac = bd
(d) none of these
Answer: (b) ad = bc
In simple words: When four numbers are in proportion, it means a : b = c : d. This gives us a useful rule: multiply the outer terms (called extremes) and the inner terms (called means). These two products are always equal: a × d = b × c.

Exam Tip: Remember the fundamental property of proportion: the product of extremes equals the product of means - this is used in almost every proportion problem.

 

Question 14. If the weight of 5 bags of rice is 272 kg, then the weight of 1 bag of rice is
(a) 50.4 kg
(b) 54.4 kg
(c) 54.004 kg
(d) 54.04 kg
Answer: (b) 54.4 kg
In simple words: Five bags together weigh 272 kg. To find the weight of one bag, divide 272 by 5. The answer is 54.4 kg.

Exam Tip: In unitary method problems, divide the total by the number of items to find the value per item.

 

Question 15. If 7 pencils cost Rs.35, then the cost of one dozen pencils is
(a) Rs.60
(b) Rs.70
(c) Rs.30
(d) Rs.5
Answer: (a) Rs.60
In simple words: First, find the cost of 1 pencil by dividing: 35 ÷ 7 = Rs.5 per pencil. One dozen = 12 pencils. So 12 pencils cost 12 × 5 = Rs.60.

Exam Tip: Break the problem into two steps - first find the unit cost, then multiply by the quantity needed.

 

Question 16. The ratio 2 : 3 expressed as percentage is
(a) 40%
(b) 60%
(c) \( 66\frac{2}{3}\% \)
(d) \( 33\frac{1}{3}\% \)
Answer: (c) \( 66\frac{2}{3}\% \)
In simple words: The ratio 2 : 3 means 2 out of (2 + 3) = 2 out of 5 parts. As a fraction, this is 2/5. Convert to percentage: (2/5) × 100 = 40%. Wait - that is the percentage of the first part. The question asks for the ratio itself as a percentage, which means expressing 2/3 (the first to the second) as a percentage: (2/3) × 100 = 66⅔%.

Exam Tip: When converting a ratio to a percentage, divide the first term by the second term, then multiply by 100.

 

Question 17. 0.025 when expressed as percentage is
(a) 250%
(b) 25%
(c) 4%
(d) 2.5%
Answer: (d) 2.5%
In simple words: To change a decimal to a percentage, multiply by 100. So 0.025 × 100 = 2.5%.

Exam Tip: Always multiply a decimal by 100 to convert it to a percentage - this moves the decimal point two places to the right.

 

Question 18. In a class, 45% of the students are girls. If there are 18 girls in the class, then the total number of students in the class is
(a) 44
(b) 40
(c) 36
(d) 30
Answer: (b) 40
In simple words: Let the total number of students be x. We know that 45% of x = 18. Write this as (45/100) × x = 18. Solve for x: x = (18 × 100) ÷ 45 = 1800 ÷ 45 = 40.

Exam Tip: In percentage problems, set up an equation using "percentage × total = part", then solve for the unknown by reversing the operations.

 

Question 19. Statement I: When two quantities of the same or different kind are compared by division, it is called a ratio. Statement II: The ratio of 6 hours to 24 hours is 1 : 4.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (b) Statement I is false but statement II is true.
In simple words: Statement I has an error - a ratio compares two quantities of the SAME KIND only, not different kinds. Statement II is correct: 6 hours : 24 hours = 6 : 24 = 1 : 4 when simplified.

Exam Tip: Remember that a ratio must always compare quantities of the same kind - they must have the same units.

 

Question 20. Statement I: The terms of a ratio can be divided by a common non-zero number. Statement II: The first term of a ratio is called the antecedent.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (c) Both Statement I and statement II are true.
In simple words: Statement I is correct - dividing both terms by the same non-zero number gives an equivalent ratio (just like multiplying does). Statement II is also correct - in a ratio a : b, the first term a is called the antecedent, and the second term b is called the consequent.

Exam Tip: Know the vocabulary of ratios: the first term is the antecedent, the second is the consequent, and ratios can be made equivalent by multiplying or dividing both terms by the same non-zero number.

 

Question 21. Statement I: Sugar can be purchased at Rs.40/kg and tea at Rs.300/kg. The ratio of the price of sugar to the price of tea is 2 : 15. Statement II: When you write a ratio, the order in which you write the terms is important.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (c) Both Statement I and statement II are true.
In simple words: Statement I is correct: the ratio 40 : 300 reduces to 2 : 15 when both are divided by 20. Statement II is also correct: order matters because 2 : 15 is not the same as 15 : 2 (unless the two numbers are equal).

Exam Tip: The order of a ratio is critical - always write the quantity mentioned first as the first term and the quantity mentioned second as the second term.

 

Question 22. Statement I: When two ratios are equal, they are said to be in proportion. Statement II: If the numbers a, b, c, d are in proportion, then ab = cd.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (a) Statement I is true but statement II is false.
In simple words: Statement I is correct - proportion means two ratios are equal. Statement II is wrong - it should be ad = bc (the product of the outer terms equals the product of the inner terms), not ab = cd.

Exam Tip: The key property of proportion is: if a : b = c : d, then a × d = b × c. This is sometimes called "cross multiplication" and is one of the most important formulas in ratio and proportion.

 

Question 23. Statement I: At school, recess lasts for 30 minutes, and the Maths period is 40 minutes long. The ratio of Maths period to recess time is 4 : 3. Statement II: A ratio has no units.
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both Statement I and statement II are true.
(d) Both Statement I and statement II are false.
Answer: (c) Both Statement I and statement II are true.
In simple words: Statement I is correct: 40 : 30 simplifies to 4 : 3. Statement II is also correct - once you form a ratio of two quantities with the same units, the ratio itself has no units (the units cancel out).

Exam Tip: A key feature of ratios is that they are unit-free - even though the original quantities may have units like minutes or kilograms, the ratio itself is just numbers.

 

Question 24. Statement I: Raman got 80, 70, 90 marks out of 100 each in Physics, Chemistry and Maths. His total percentage is 80%. Statement II: \( \frac{2}{5} = 40\% \)
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (3) Both Statement I and statement II are true
In simple words: When you add Raman's three marks (80 + 70 + 90 = 240) and divide by the total possible marks (300), you get 80 out of 100, which is 80%. The fraction 2/5, when changed to a percentage, also gives 40%. Both statements are correct.

Exam Tip: Always calculate the total marks obtained and total possible marks carefully before finding the percentage. Double-check fraction-to-percentage conversions by multiplying by 100.

 

Check Your Progress

 

Question 1. From the adjoining figure, find the ratio of:
(i) Number of triangles to the number of circles inside the rectangle.
(ii) Number of squares to the number of all the figures inside the rectangle.
(iii) Number of circles to the number of remaining figures inside the rectangle.
Answer: By counting the shapes in the figure, we get: Number of triangles = 3, Number of circles = 2, Number of squares = 2, Total number of figures = 3 + 2 + 2 = 7.
(i) The ratio of triangles to circles = 3 : 2
(ii) The ratio of squares to all figures = 2 : 7
(iii) Remaining figures (excluding circles) = 3 + 2 = 5. The ratio of circles to remaining figures = 2 : 5
In simple words: Count each type of shape, then write them as ratios using colons. For example, 3 triangles to 2 circles is written as 3 : 2.

Exam Tip: When counting shapes from a figure, recount carefully to make sure you have the correct total. Write your final ratios in simplest form.

 

Question 2. The length of a pencil is 16 cm and its diameter is 6 mm. What is the ratio of the diameter of the pencil to that of its length?
Answer: Given: Length of pencil = 16 cm, Diameter of pencil = 6 mm. First, convert the units so both are the same. Change 16 cm to mm: 16 cm = 16 × 10 = 160 mm. Now form the ratio: diameter to length = 6 : 160. To simplify, find the H.C.F. of 6 and 160, which is 2. Divide both numbers by 2: 6 ÷ 2 = 3 and 160 ÷ 2 = 80. The simplified ratio is 3 : 80.
In simple words: Always convert both measurements to the same unit before writing the ratio. Then reduce the numbers by dividing by their H.C.F.

Exam Tip: Remember to convert units before forming ratios. A common mistake is forgetting to simplify the ratio, so always find the H.C.F. and reduce to lowest terms.

 

Question 3. The length and the breadth of a steel tape are 10 m and 2.4 cm respectively. Find the ratio of the length to the breadth.
Answer: Given: Length of steel tape = 10 m, Breadth of steel tape = 2.4 cm. Convert m to cm: 10 m = 10 × 100 = 1000 cm. The ratio of length to breadth = 1000 : 2.4. To remove the decimal, multiply both terms by 10: (1000 × 10) : (2.4 × 10) = 10000 : 24. Find the H.C.F. of 10000 and 24, which is 8. Divide both by 8: 10000 ÷ 8 = 1250 and 24 ÷ 8 = 3. The simplified ratio is 1250 : 3.
In simple words: Convert to the same unit first, then remove any decimals by multiplying both sides by 10 or 100. Finally, reduce using the H.C.F.

Exam Tip: When ratios contain decimals, multiply both terms by a suitable power of 10 to remove the decimal point. This step must come before you simplify using the H.C.F.

 

Question 4. A certain club has 100 members, out of which 25 play tennis, 28 play badminton, 12 play chess and the rest do not play any game. Find the ratio of number of members who play:
(i) badminton to the number of those who play chess.
(ii) badminton to the number of those who do not play any game.
(iii) tennis to the number of those who do not play any game.
(iv) tennis to the number of those who play either badminton or chess.
Answer: Given: Total members = 100, Tennis players = 25, Badminton players = 28, Chess players = 12. Members who do not play any game = 100 - 25 - 28 - 12 = 35.
(i) Ratio of badminton to chess = 28 : 12. H.C.F. is 4. Simplified: 28 ÷ 4 = 7, 12 ÷ 4 = 3. Ratio = 7 : 3.
(ii) Ratio of badminton to non-players = 28 : 35. H.C.F. is 7. Simplified: 28 ÷ 7 = 4, 35 ÷ 7 = 5. Ratio = 4 : 5.
(iii) Ratio of tennis to non-players = 25 : 35. H.C.F. is 5. Simplified: 25 ÷ 5 = 5, 35 ÷ 5 = 7. Ratio = 5 : 7.
(iv) Players of badminton or chess = 28 + 12 = 40. Ratio of tennis to (badminton or chess) = 25 : 40. H.C.F. is 5. Simplified: 25 ÷ 5 = 5, 40 ÷ 5 = 8. Ratio = 5 : 8.
In simple words: Identify the two groups you need to compare, find the H.C.F., and divide both numbers to get the simplest ratio.

Exam Tip: Always simplify ratios to their lowest terms. For multi-part questions, tackle each part separately to avoid mixing up which groups to compare.

 

Question 5. Do the ratios 15 cm to 3 m and 25 seconds to 3 minutes form a proportion?
Answer: Given: First ratio is 15 cm to 3 m. Convert m to cm: 3 m = 3 × 100 = 300 cm. First ratio = 15 : 300 = 1 : 20. Second ratio is 25 seconds to 3 minutes. Convert minutes to seconds: 3 min = 3 × 60 = 180 seconds. Second ratio = 25 : 180 = 5 : 36. To check if they form a proportion, cross-multiply. Product of extremes = 1 × 36 = 36. Product of means = 20 × 5 = 100. Since 36 ≠ 100, the ratios are not equal. Therefore, the given ratios do not form a proportion.
In simple words: Convert both ratios to the same units first. Then multiply the first and last numbers, and also multiply the middle two numbers. If both products are equal, they form a proportion; if not, they don't.

Exam Tip: Always convert units before checking for proportion. The cross-product method (extremes and means) is reliable for comparing two ratios quickly.

 

Question 6. Divide Rs.500 among Suresh and Awanti in the ratio 3 : 7.
Answer: Given: Total money = Rs.500, Ratio = 3 : 7. Find total parts: 3 + 7 = 10 parts. Find the value of 1 part: 500 ÷ 10 = Rs.50. Calculate each share: Suresh's share = 3 × Rs.50 = Rs.150, Awanti's share = 7 × Rs.50 = Rs.350. Therefore, Suresh gets Rs.150 and Awanti gets Rs.350.
In simple words: Add the ratio numbers to find the total parts. Divide the total amount by the total parts to find the value of one part. Then multiply each ratio number by that value.

Exam Tip: Always verify your answer by adding the two shares - they should equal the original total. This quick check catches arithmetic errors.

 

Question 7. The ratio of the number of girls to that of boys in a school is 9 : 11. If the number of boys in the school is 2035, find:
(i) the number of girls in the school.
(ii) the number of students in the school.
Answer: Given: Ratio (Girls : Boys) = 9 : 11, Number of boys = 2035. Since 11 parts represent the boys, 11 parts = 2035. Find 1 part: 1 part = 2035 ÷ 11 = 185.
(i) Number of girls = 9 parts = 9 × 185 = 1665. Therefore, the number of girls is 1665.
(ii) Total students = Girls + Boys = 1665 + 2035 = 3700. Therefore, the total number of students in the school is 3700.
In simple words: Find the value of one part by dividing the known quantity by its corresponding ratio number. Then multiply each ratio number by that value to find the other quantities.

Exam Tip: When one quantity in a ratio is given, divide it by its ratio number to find the value of one part. This value then helps you calculate all other quantities.

 

Question 8. The ratio of income to expenditure of a family is 7 : 6. Find the savings if the income of family is Rs.42,000.
Answer: Given: Ratio (Income : Expenditure) = 7 : 6, Income = Rs.42,000. Since 7 parts represent the income, 7 parts = Rs.42,000. Find 1 part: 1 part = 42000 ÷ 7 = Rs.6,000. Expenditure = 6 parts = 6 × Rs.6,000 = Rs.36,000. Savings = Income - Expenditure = Rs.42,000 - Rs.36,000 = Rs.6,000. Therefore, the savings of the family is Rs.6,000.
In simple words: Find the value of one part by dividing the known amount by its ratio number. Then find the expenditure using its ratio number. Subtract expenditure from income to get savings.

Exam Tip: Remember the relationship: Savings = Income - Expenditure. Always find the value of one part first when one quantity in a ratio is known.

 

Question 9. An employee earns Rs.72,000 in 3 months.
(i) How much does he earn in 7 months?
(ii) In how many months will he earn Rs.3,60,000?
Answer: Given: Earnings in 3 months = Rs.72,000. Find monthly earnings: Earnings in 1 month = 72000 ÷ 3 = Rs.24,000.
(i) Earnings in 7 months = 7 × Rs.24,000 = Rs.1,68,000. Therefore, the employee earns Rs.1,68,000 in 7 months.
(ii) To earn Rs.3,60,000: Number of months = 3,60,000 ÷ 24,000 = 15 months. Therefore, the employee will earn Rs.3,60,000 in 15 months.
In simple words: Find how much is earned in one month by dividing the total by the number of months. Then use this monthly amount to find earnings for any other number of months or the time needed to earn a target amount.

Exam Tip: Always find the unit rate first (earnings per month). Once you have this, you can solve for any time period or target amount quickly.

 

Question 10. A train travels 110 km in 2 hours and a car travels 245 km in 3\( \frac{1}{2} \) hours. What is the ratio of the speed of the train to that of the car?
Answer: Given: Distance travelled by train = 110 km in 2 hours, Distance travelled by car = 245 km in 3\( \frac{1}{2} \) = \( \frac{7}{2} \) hours. Speed of train = 110 ÷ 2 = 55 km/h. Speed of car = 245 ÷ \( \frac{7}{2} \) = 245 × \( \frac{2}{7} \) = \( \frac{490}{7} \) = 70 km/h. Ratio (Speed of train : Speed of car) = 55 : 70. Find H.C.F. of 55 and 70, which is 5. Simplify: 55 ÷ 5 = 11, 70 ÷ 5 = 14. The ratio is 11 : 14. Therefore, the ratio of the speed of the train to that of the car is 11 : 14.
In simple words: Calculate the speed of each by dividing distance by time. Then form a ratio of the two speeds and simplify using the H.C.F.

Exam Tip: When time is given as a mixed number, convert it to an improper fraction before dividing. Speed = Distance ÷ Time, so be careful with the order.

 

Question 11. What is the ratio of the number of prime numbers to the composite numbers from the numbers 1 to 45?
Answer: Prime numbers between 1 and 45 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43. Number of prime numbers = 14. Note that 1 is neither prime nor composite. Total numbers from 1 to 45 = 45. Composite numbers = 45 - 14 - 1 = 30. Ratio (Primes : Composites) = 14 : 30. Find H.C.F. of 14 and 30, which is 2. Simplify: 14 ÷ 2 = 7, 30 ÷ 2 = 15. The ratio is 7 : 15. Therefore, the required ratio is 7 : 15.
In simple words: List all prime numbers in the given range and count them. Subtract this count from the total, but also subtract 1 (since 1 is neither prime nor composite). This gives the composite numbers. Then form and simplify the ratio.

Exam Tip: Remember that 1 is a special case - it is neither prime nor composite. Don't forget to exclude it when calculating composite numbers.

 

Question 12. Divide Rs.6000 among Irfan, Nagma and Ishan in the ratio 3 : 5 : 7.
Answer: Given: Total money = Rs.6000, Ratio = 3 : 5 : 7. Find total parts: 3 + 5 + 7 = 15 parts. Find the value of 1 part: 6000 ÷ 15 = Rs.400. Calculate each share: Irfan's share = 3 × Rs.400 = Rs.1200, Nagma's share = 5 × Rs.400 = Rs.2000, Ishan's share = 7 × Rs.400 = Rs.2800. Therefore, Irfan gets Rs.1200, Nagma gets Rs.2000 and Ishan gets Rs.2800.
In simple words: Add all the ratio numbers to get the total parts. Divide the total amount by the number of parts to find the value of one part. Multiply this value by each person's ratio number.

Exam Tip: Verify your answer by adding all three shares - they should equal the original total. This is a quick way to catch mistakes.

 

Question 13. Sapna weighs 54 kg on earth and 9 kg on moon. If a monkey weighs 3.5 kg on moon, then how much will it weigh on the earth?
Answer: Given: Sapna's weight on earth = 54 kg, Sapna's weight on moon = 9 kg. Find the ratio of earth weight to moon weight: 54 : 9 = 6 : 1. This means weight on earth = 6 × weight on moon. Monkey's weight on moon = 3.5 kg. Monkey's weight on earth = 6 × 3.5 = 21 kg. Therefore, the monkey will weigh 21 kg on the earth.
In simple words: Find the relationship between earth weight and moon weight using Sapna's data. Then apply this same relationship to find the monkey's earth weight.

Exam Tip: This is a direct proportion problem. Once you find the constant ratio (earth weight ÷ moon weight), use it to solve for unknown weights.

 

Question 14. If 5 men can do a certain construction work in 14 days, then how long will 7 men take to complete the same construction work?
Answer: Given: 5 men complete the work in 14 days. This is an inverse proportion problem - when the number of workers increases, the time needed decreases. Calculate total work: Total work = 5 men × 14 days = 70 man-days. Let 7 men take x days to complete the same work. Then, 7 × x = 70. Solve for x: x = 70 ÷ 7 = 10 days. Therefore, 7 men will take 10 days to complete the same construction work.
In simple words: Calculate the total amount of work in man-days (men multiplied by days). When more workers are available, divide this total by the new number of workers to find the new time needed.

Exam Tip: Inverse proportion means as one quantity increases, the other decreases. The product of the two quantities (men × days) always remains constant for the same work.

This page contains only navigation elements, footer content, and a watermark. There is no educational content to digitize.