NCERT Solutions for Class 7 Maths Chapter 7: Comparing Quantities — Complete Guide with Step-by-Step Solutions

Two ratios are in proportion if they are equal: $a : b = c : d$ or $\frac{a}{b} = \frac{c}{d}$.

This can also be written as $a : b :: c : d$ (read as "$a$ is to $b$ as $c$ is to $d$").

Cross-multiplication test: $a : b = c : d$ if and only if $a \times d = b \times c$.

Example: Is $3 : 5 = 9 : 15$?
$3 \times 15 = 45$ and $5 \times 9 = 45$. Since $45 = 45$, yes, they are in proportion.

Problem: Find the ratio of $3$ km to $300$ m.

Solution:
Convert to the same unit: $3$ km $= 3000$ m.

Answer: $10 : 1$.

Problem: A map uses a scale of $1$ cm $: 10$ km. If two cities are $4.5$ cm apart on the map, find the actual distance.

Solution:

Answer: $45$ km.

Problem: Out of $40$ students, $25$ passed. Find the ratio of passed to failed.

Solution:
Passed $= 25$, Failed $= 40 - 25 = 15$.

Answer: $5 : 3$.

Problem: Divide Rs. $600$ in the ratio $2 : 3$.

Solution:
Sum of parts $= 2 + 3 = 5$.

First share $= \frac{2}{5} \times 600 = \text{Rs. } 240$.

Second share $= \frac{3}{5} \times 600 = \text{Rs. } 360$.

Verification: $240 + 360 = 600$. Correct.

Problem: A ribbon is cut into two pieces of lengths $45$ cm and $30$ cm. Find the ratio of the longer piece to the shorter piece.

Solution:

Answer: $3 : 2$.

Problem: Divide Rs. $1200$ among $A$, $B$, $C$ in the ratio $2 : 3 : 5$.

Solution:
Sum of parts $= 2 + 3 + 5 = 10$.

$A = \frac{2}{10} \times 1200 = \text{Rs. } 240$.

$B = \frac{3}{10} \times 1200 = \text{Rs. } 360$.

$C = \frac{5}{10} \times 1200 = \text{Rs. } 600$.

Verification: $240 + 360 + 600 = 1200$. Correct.

Problem: Which is a better deal: $3$ kg for Rs. $45$ or $5$ kg for Rs. $80$?

Solution:
Price per kg in Deal 1: $\frac{45}{3} = \text{Rs. } 15$ per kg.
Price per kg in Deal 2: $\frac{80}{5} = \text{Rs. } 16$ per kg.

Deal 1 is cheaper per kg, so Deal 1 is better.

Problem: Find the ratio of $45$ minutes to $1.5$ hours.

Solution:
Convert to the same unit: $1.5$ hours $= 90$ minutes.

Answer: $1 : 2$.

Problem: Convert $\frac{3}{8}$ to a percentage.

Solution:

Answer: $37.5\%$.

Problem: Convert $0.45$ to a percentage.

Solution:

Answer: $45\%$.

Problem: Find $15\%$ of $250$.

Solution:

Answer: $37.5$.

Problem: A price increased from Rs. $200$ to Rs. $250$. Find the percentage increase.

Solution:
Increase $= 250 - 200 = \text{Rs. } 50$.

Answer: $25\%$ increase.

Problem: The number of students decreased from $80$ to $68$. Find the percentage decrease.

Solution:
Decrease $= 80 - 68 = 12$.

Answer: $15\%$ decrease.

Problem: $30\%$ of a number is $45$. Find the number.

Solution:

Answer: The number is $150$.

Problem: Rahul scored $420$ out of $500$. Find his percentage.

Solution:

Answer: $84\%$.

Problem: A school has $800$ students. If $60\%$ are girls, how many boys are there?

Solution:
Girls $= 60\%$ of $800 = \frac{60}{100} \times 800 = 480$.

Boys $= 800 - 480 = 320$.

Alternatively: Boys $= 40\%$ of $800 = \frac{40}{100} \times 800 = 320$.

Answer: $320$ boys.

Problem: Convert $12.5\%$ to a fraction.

Solution:

Answer: $\frac{1}{8}$.

Problem: A price of Rs. $100$ increases by $10\%$ and then decreases by $10\%$. Is the final price Rs. $100$?

Solution:
After $10\%$ increase: $100 + 10 = \text{Rs. } 110$.
After $10\%$ decrease: $110 - 11 = \text{Rs. } 99$.

The final price is Rs. $99$, NOT Rs. $100$.

Key insight: A percentage increase followed by the same percentage decrease does NOT bring you back to the original value. There is always a net decrease of $\left(\frac{x}{10}\right)^2 \%$ for an $x\%$ change.

Problem: A shopkeeper buys a toy for Rs. $250$ and sells it for Rs. $300$. Find the profit percentage.

Solution:
Profit $= 300 - 250 = \text{Rs. } 50$.

Answer: $20\%$ profit.

Problem: A shirt bought for Rs. $400$ is sold for Rs. $380$. Find the loss percentage.

Solution:
Loss $= 400 - 380 = \text{Rs. } 20$.

Answer: $5\%$ loss.

Problem: CP of an article is Rs. $1500$ and loss is $10\%$. Find the SP.

Solution:
Loss $= \frac{10}{100} \times 1500 = \text{Rs. } 150$.

Answer: Rs. $1350$.

Problem: SP $=$ Rs. $540$ and profit $= 8\%$. Find the CP.

Solution:
SP $= \text{CP} + 8\%$ of CP $= 1.08 \times \text{CP}$.

Answer: Rs. $500$.

Problem: A merchant bought goods for Rs. $2000$. At what price should he sell to make $15\%$ profit?

Solution:

Answer: Rs. $2300$.

Problem: An article bought for Rs. $600$ is sold at a gain of $25\%$. Find the SP. If instead it was sold at a loss of $25\%$, what would the SP be?

Solution:
With gain: SP $= 600 \times 1.25 = \text{Rs. } 750$.

With loss: SP $= 600 \times 0.75 = \text{Rs. } 450$.

Note: The same CP and same percentage give different SPs depending on profit or loss.

Problem: Marked price of a bag is Rs. $800$. A discount of $15\%$ is offered. Find the selling price.

Solution:
Discount $= \frac{15}{100} \times 800 = \text{Rs. } 120$.

Answer: Rs. $680$.

Problem: A man sold a bicycle for Rs. $1540$ making a profit of Rs. $140$. Find the CP and profit percentage.

Solution:

Answer: CP $=$ Rs. $1400$, Profit $= 10\%$.

Problem: Find the simple interest on Rs. $5000$ at $8\%$ per annum for $3$ years.

Solution:

Answer: Rs. $1200$.

Problem: Find the amount to be paid after $2$ years on Rs. $10000$ at $12\%$ per annum.

Solution:

Answer: Rs. $12400$.

Problem: At what rate will Rs. $4000$ earn Rs. $640$ as SI in $2$ years?

Solution:

Answer: $8\%$ per annum.

Problem: In how much time will Rs. $2000$ amount to Rs. $2600$ at $5\%$ per annum?

Solution:
SI $= 2600 - 2000 = \text{Rs. } 600$.

Answer: $6$ years.

Problem: The SI on a certain sum for $3$ years at $10\%$ is Rs. $900$. Find the principal.

Solution:

Answer: Rs. $3000$.

Problem: Find the SI on Rs. $6000$ at $10\%$ per annum for $6$ months.

Solution:
$6$ months $= \frac{6}{12} = 0.5$ years.

Answer: Rs. $300$.

Problem: In how many years will Rs. $5000$ double itself at $10\%$ SI per annum?

Solution:
For the amount to double, SI must equal the principal: SI $= \text{Rs. } 5000$.

Answer: $10$ years.

Shortcut: Time to double at $R\%$ SI $= \frac{100}{R}$ years.

Problem: Which gives more interest: Rs. $5000$ at $12\%$ for $2$ years, or Rs. $6000$ at $10\%$ for $2$ years?

Solution:
Investment 1: SI $= \frac{5000 \times 12 \times 2}{100} = \text{Rs. } 1200$.

Investment 2: SI $= \frac{6000 \times 10 \times 2}{100} = \text{Rs. } 1200$.

Both give the same interest of Rs. $1200$.

Answer: Both investments give equal interest.

Answer 1: $\frac{5}{8} \times 100 = 62.5\%$.

Answer 2: Sum $= 5$. First $= \frac{3}{5} \times 750 = 450$. Second $= \frac{2}{5} \times 750 = 300$.

Answer 3: Profit $= 920 - 800 = 120$. Profit $\% = \frac{120}{800} \times 100 = 15\%$.

Answer 4: SI $= \frac{8000 \times 6 \times 2.5}{100} = \frac{120000}{100} = \text{Rs. } 1200$.

Answer 5: $450 = \frac{3000 \times R \times 3}{100}$. $R = \frac{450 \times 100}{9000} = 5\%$.

Answer 6: Discount $= \frac{20}{100} \times 1200 = 240$. SP $= 1200 - 240 = \text{Rs. } 960$.