NCERT Solutions: Comparing Quantities

Q1: Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.

(a) Gardening shears bought for ₹250 and sold for ₹325.

• Profit/Loss: ₹325 - ₹250 = ₹75 (Profit)
• Profit per cent: $\frac{75}{250}\times 100=30\%$

(b) A refrigerator bought for ₹12,000 and sold at ₹13,500.

• Profit/Loss: ₹13,500 - ₹12,000 = ₹1,500 (Profit)
• Profit per cent: $\frac{1500}{12000}\times 100=12.5\%\backslash frac\{1500\}\{12000\}\; \backslash times\; 100\; =\; 12.5\backslash \%$

(c) A cupboard bought for ₹2,500 and sold at ₹3,000.

• Profit/Loss: ₹3,000 - ₹2,500 = ₹500 (Profit)
• Profit per cent: $\frac{500}{2500}\times 100=20\%\backslash frac\{500\}\{2500\}\; \backslash times\; 100\; =\; 20\backslash \%$

(d) A skirt bought for ₹250 and sold at ₹150.

• Profit/Loss: ₹150 - ₹250 = -₹100 (Loss)
• Loss per cent: $\frac{100}{250}\times 100=40\%\backslash frac\{100\}\{250\}\; \backslash times\; 100\; =\; 40\backslash \%$

Q2: Convert each part of the ratio to a percentage:

(a) 3: 1

• Total parts = 3 + 1 = 4
• Percentage of first part: (3/4) × 100 = 75%
• Percentage of second part: (1/4) × 100 = 25%

(b) 2 : 3: 5

• Total parts = 2 + 3 + 5 = 10
• Percentage of first part: (2/10) × 100 = 20%
• Percentage of second part: (3/10) × 100 = 30%
• Percentage of third part: (5/10) × 100 = 50%

$\backslash frac\{5\}\{10\}\; \backslash times\; 100\; =\; 50\backslash \%$

(c) 1: 4

• Total parts = 1 + 4 = 5
• Percentage of first part: $\frac{1}{5}\times 100=20\%\backslash frac\{1\}\{5\}\; \backslash times\; 100\; =\; 20\backslash \%$
• Percentage of second part: $\frac{4}{5}\times 100=80\%\backslash frac\{4\}\{5\}\; \backslash times\; 100\; =\; 80\backslash \%$

(d) 1: 2: 5

• Total parts = 1 + 2 + 5 = 8
• Percentage of first part: $\frac{1}{8}\times 100=12.5\%\backslash frac\{1\}\{8\}\; \backslash times\; 100\; =\; 12.5\backslash \%$
• Percentage of second part: $\frac{2}{8}\times 100=25\%\backslash frac\{2\}\{8\}\; \backslash times\; 100\; =\; 25\backslash \%$
• Percentage of third part: $\frac{5}{8}\times 100=62.5\%\backslash frac\{5\}\{8\}\; \backslash times\; 100\; =\; 62.5\backslash \%$

Q3: The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

• Decrease in population: 25,000 - 24,500 = 500
• Percentage decrease:

Q4: Arun bought a car for ₹3,50,000. The next year, the price went up to ₹3,70,000. What was the Percentage of price increase?

• Increase in price: ₹3,70,000 - ₹3,50,000 = ₹20,000
• Percentage increase:

Q5: I buy a T.V. for ₹10,000 and sell it at a profit of 20%. How much money do I get for it?

• Profit: 20% of ₹10,000
• Selling price: ₹10,000 + ₹2,000 = ₹12,000

Q6: Juhi sells a washing machine for ₹13,500. She loses 20% in the bargain. What was the price at which she bought it?

Ans: Loss: 20% of Cost Price (CP)
Selling Price (SP) = ₹13,500
Therefore, SP = CP - 20% of CP

Percentage of carbon:

(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?
Ans: Carbon percentage: 12%
Therefore, 12%  of weight of chalk = 3g 
Let the weight of the chalk be W

Q8: Amina buys a book for ₹275 and sells it at a loss of 15%. How much does she sell it for?

• Loss: 15% of ₹275
  
• Selling price: ₹275 - ₹41.25 = ₹233.75

Q9: Find the amount to be paid at the end of 3 years in each case:

(a) Principal = ₹1,200 at 12% p.a.

• Simple Interest: (1200 × 12 × 3)/100 = ₹432
• Amount: ₹1,200 + ₹432 = ₹1,632

(b) Principal = ₹7,500 at 5% p.a.

• Simple Interest: (7500 × 5 × 3)/100 = ₹1,125
• Amount: ₹7,500 + ₹1,125 = ₹8,625

Q10: What rate gives ₹280 as interest on a sum of ₹56,000 in 2 years?

Ans: Interest (I): ₹280
Principal (P): ₹56,000
Time (T): 2 years
Using the formula for Simple Interest: $I=\frac{P}{\times }$
$\frac{\mathrm{<img\; src="https://cn.edurev.in/ApplicationImages/Temp/1421561\_609b03fa-58b2-4822-8764-2b0d19ad5496\_lg.png"\; style="display:block;margin:5px\; auto;text-align:center"\; loading="lazy"\; data-gumlet="false"\; alt="Exercise\; 7.2">}}{}$

Q11: If Meena gives an interest of ₹45 for one year at 9% rate p.a., what is the sum she has borrowed?
Ans:  Interest (I): ₹45
Rate (R): 9%
Time (T): 1 year
Using the formula for Simple Interest: I = P x R x T/100
$\backslash frac\{P\; \backslash times\; R\; \backslash times\; T\}\{100\}$$=\; ₹500$