CBSE Class 7 Mathematics Comparing Quantities Ratio Assignment Set 02

VII - Mathematics Assignments No-02 - Comparing Quantities

Ratios and Proportions

Question 1. The ratio of the income and saving of a family is 8:3. Find the expenditure if the saving is Rs. 1332.
Answer: Let the income be \(8x\) and the saving be \(3x\).
Given saving = Rs. 1332
\(3x = 1332\)
\(x = \frac{1332}{3} = 444\)
Income = \(8 \times 444 = \text{Rs. } 3552\)
Expenditure = Income - Saving
Expenditure = \(3552 - 1332 = \text{Rs. } 2220\)
In simple words: Find total income using the given savings, then subtract savings from income to get the expenditure.

Exam Tip: Always define variables for the ratio terms (like \(8x\) and \(3x\)) before setting up the equation.

Question 2. The ratio of Copper and Zinc in an alloy is 7:5. If the weight of Copper in the alloy is 13.30 kg. Find the weight of Zinc in it.
Answer: Ratio of Copper to Zinc = \(7:5\)
Let Copper = \(7x\) and Zinc = \(5x\)
Given weight of Copper = 13.30 kg
\(7x = 13.30\)
\(x = \frac{13.30}{7} = 1.9 \text{ kg}\)
Weight of Zinc = \(5x = 5 \times 1.9 = 9.5 \text{ kg}\)
In simple words: Use the given weight of copper to find the common multiplier \(x\), then multiply it by 5 to get the weight of zinc.

Exam Tip: Maintain proper decimal places during division to avoid calculation errors.

Question 3. A rectangular field is 6.3 m long and 2.1 m wide. Find the ratio of the dimensions of the field.
Answer: Length = 6.3 m
Breadth = 2.1 m
Ratio of length to breadth = \(\frac{6.3}{2.1}\)
\(= \frac{63}{21} = \frac{3}{1} = 3:1\)
In simple words: Divide the length by the width and simplify the fraction to lowest terms.

Exam Tip: Both quantities must be in the same unit before finding their ratio (here both are already in meters).

Question 4. Which is larger ratio? \(2:3\) OR \(5:7\)
Answer: Convert the ratios into fractions: \(\frac{2}{3}\) and \(\frac{5}{7}\)
To compare, make the denominators equal by taking the LCM of 3 and 7, which is 21.
\(\frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}\)
\(\frac{5}{7} = \frac{5 \times 3}{7 \times 3} = \frac{15}{21}\)
Since \(\frac{15}{21} > \frac{14}{21}\), therefore \(5:7 > 2:3\).
In simple words: Convert both ratios to fractions with a common denominator to easily see which fraction is bigger.

Exam Tip: Cross-multiplication is a quick alternative: compare \(2 \times 7\) (14) with \(3 \times 5\) (15); since 15 is larger, \(5:7\) is larger.

Question 5. Arun divides Rs. 10,000 between his sons Anil and Ajay in the ratio 5:3. Find the share of each of them.
Answer: Total amount = Rs. 10,000
Ratio of division = \(5:3\)
Sum of ratio terms = \(5 + 3 = 8\)
Anil's share = \(\frac{5}{8} \times 10000 = \frac{50000}{8} = \text{Rs. } 6250\)
Ajay's share = \(\frac{3}{8} \times 10000 = \frac{30000}{8} = \text{Rs. } 3750\)
In simple words: Add the ratio parts to find the total parts, then find each person's fractional share of the total amount.

Question 6. Arun walks 15 km in 3 hours and Anil walks 18 km in 6 hours. Find the ratio of their speed.
Answer: Speed of Arun = \(\frac{\text{Distance}}{\text{Time}} = \frac{15 \text{ km}}{3 \text{ hrs}} = 5 \text{ km/hr}\)
Speed of Anil = \(\frac{\text{Distance}}{\text{Time}} = \frac{18 \text{ km}}{6 \text{ hrs}} = 3 \text{ km/hr}\)
Ratio of their speeds = \(5:3\)
In simple words: Calculate the speed of each person first, then form the ratio of their speeds.

Exam Tip: Speed is calculated as distance divided by time. Ensure units are consistent.

Question 7. In CWG, Indian team won 6 out of 8 matches played by them. What is the ratio of
(a) number of matches won to the no. of matches played?
(b) Number of wins to the number of losses?
Answer:
Total matches played = 8
Matches won = 6
Matches lost = \(8 - 6 = 2\)
(a) Ratio of matches won to matches played = \(\frac{6}{8} = \frac{3}{4} = 3:4\)
(b) Ratio of wins to losses = \(\frac{6}{2} = \frac{3}{1} = 3:1\)
In simple words: Find the number of losses by subtraction, then set up the required ratios and simplify them.

Exam Tip: Read sub-parts carefully to ensure the order of terms in the ratio matches the question.

Question 8. The simplest form of the ratio "Rs. 5 to 250 paise"
Answer: Convert Rs. 5 into paise:
1 Rupee = 100 paise
Rs. 5 = \(5 \times 100 = 500 \text{ paise}\)
Ratio = \(\frac{500 \text{ paise}}{250 \text{ paise}}\)
\(= \frac{500}{250} = \frac{2}{1} = 2:1\)
In simple words: Convert both quantities to the same unit (paise) before writing and simplifying the ratio.

Exam Tip: Never find a ratio between two quantities of different units without converting them first.

Question 9. Two numbers are in the ratio 1:2. If 4 is added to each of them, the ratio becomes 4:7. Find the nos.
Answer: Let the two numbers be \(x\) and \(2x\).
According to the question, when 4 is added to each:
\(\frac{x + 4}{2x + 4} = \frac{4}{7}\)
Cross-multiplying:
\(7(x + 4) = 4(2x + 4)\)
\(7x + 28 = 8x + 16\)
\(28 - 16 = 8x - 7x\)
\(x = 12\)
First number = \(x = 12\)
Second number = \(2x = 2 \times 12 = 24\)
In simple words: Assume the numbers as \(x\) and \(2x\), set up the algebraic equation based on the given condition, and solve for \(x\).

Exam Tip: Cross-multiplication is the standard method to solve fraction equations of this type.

Question 10. The ratio of present ages of A and B are 3:2. Five years ago, their ages were in the ratio 14:11. What are their present ages
Answer: Let the present ages of A and B be \(3x\) and \(2x\) respectively.
Five years ago:
Age of A = \(3x - 5\)
Age of B = \(2x - 5\)
Given ratio 5 years ago = \(14:11\)
\(\frac{3x - 5}{2x - 5} = \frac{14}{11}\)
Cross-multiplying:
\(11(3x - 5) = 14(2x - 5)\)
\(33x - 55 = 28x - 70\)
\(33x - 28x = 55 - 70\)
\(5x = 15\)
\(x = 3\)
Present age of A = \(3x = 3 \times 3 = 9 \text{ yrs}\)
Present age of B = \(2x = 2 \times 3 = 6 \text{ yrs}\)
In simple words: Subtract 5 from both present ages to get their past ages, set up the ratio equation, and solve for \(x\).

Exam Tip: For age problems, always define present ages using a variable first, then express past or future ages relative to the present.