NCERT solution class 7 chapter 8 Comparing Quantities exercise 8.3 mathematics

EXERCISE 8.3

---

Introduction - Ex. 8.3, Comparing Quantities - NCERT Class 7th Maths Solutions

OPEN IN YOUTUBE

---

Q 1, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 1
Tell what is the prof‌it or loss in the following transactions. Also f‌ind prof‌it percent or loss percent in each case.
(a) Gardening shears bought for Rs 250 and sold for Rs 325.
Sol :
(a) Cost price = Rs 250
Selling price = Rs 325
Prof‌it = 325 — 250 = Rs 75
$\text{Profit }\%=\dfrac{\text{profit}}{\text{C.P}}\times 100$
$=\dfrac{75}{250}\times 100$
= 30%

---

(b) A refrigerator bought for Rs 12,000 and sold at Rs 13,500.
Sol :
(b) Cost price = Rs 12000
Selling price = Rs 13,500
Prof‌it : 13500 — 12000 = Rs 1500
$\text{Profit }\%=\dfrac{\text{profit}}{\text{C.P}}\times 100$
$=\dfrac{1500}{12000}\times 100$
= 12.5 %

---

(c) A cupboard bought for Rs 2,500 and sold at Rs 3,000.
Sol :
(c) Cost price = Rs 2500
Selling price = Rs 3000
Prof‌it = 3000 — 2500 = Rs 500
$\text{Profit }\%=\dfrac{\text{profit}}{\text{C.P}}\times 100$
$=\dfrac{500}{2500}\times 100$
= 20 %

---

(d) A skirt bought for Rs 250 and sold at Rs 150.
Sol :
(d) Cost price = Rs 250
Selling price = Rs 150
Loss = 250—150=Rs 100
$\text{Loss }\%=\dfrac{\text{Loss}}{\text{C.P}}\times 100$
$=\dfrac{100}{250}\times 100$
= 40 %

---

Q 2, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 2
Convert each part of the ratio to percentage:
(a) 3:1
Sol :
Total parts = 3 + 1 = 4
1^st part $=\dfrac{3}{4}=\dfrac{3}{4}\times 100\% $
= 75 %
2^nd part $=\dfrac{1}{4}=\dfrac{1}{4}\times 100$
= 25 %

---

(b) 2:3:5
Sol :
Total parts = 2 + 3 + 5 = 10
1^st part $=\dfrac{2}{10}=\dfrac{2}{10}\times 100\% $
= 20 %
2^nd part $=\dfrac{3}{10}=\dfrac{3}{10}\times 100$
= 30 %
3^nd part $=\dfrac{5}{10}=\dfrac{5}{10}\times 100$
= 50 %

---

(c) 1:4
Sol :
Total parts = 1 + 4 = 5
1^st part $=\dfrac{1}{5}=\dfrac{1}{5}\times 100\% $
= 20 %
2^nd part $=\dfrac{4}{5}=\dfrac{4}{5}\times 100$
= 80 %

---

(d) 1:2:5
Sol :
Total parts = 1 + 2 + 5 = 8
1^st part $=\dfrac{2}{10}=\dfrac{1}{8}\times 100\% $
= 12.5 %
2^nd part $=\dfrac{3}{10}=\dfrac{2}{8}\times 100$
= 25 %
3^nd part $=\dfrac{5}{10}=\dfrac{5}{8}\times 100$
= 62.5 %

---

Q 3, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 3
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Sol :
Initial population = 25000
Final population = 24500
Decrease = 500
$\text{percentage decrease} = \dfrac{decrease}{original value}\times 100$
$\% decrease=\dfrac{500}{25000}\times 100$
= 2 %

---

Q 4, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 4
Arun bought a car for Rs 3,50,000. The next year, the price went up to Rs 3,70,000. What was the percentage of price increase?
Sol :
Initial price = Rs 350000
Final price = Rs 370000
Increase = Rs 20000
$\text{percentage increase} = \dfrac{increase}{original value}\times 100$
$\% decrease=\dfrac{20000}{350000}\times 100$
$=5\dfrac{5}{7}\%$

---

Q 5, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 5
I buy a TV. for Rs 10,000 and sell it at a prof‌it of 20%. How much money do I get for it?
Sol :
Cost price = Rs 10000
Prof‌it = 20% of 10000
$\text{Profit }\% = \dfrac{\text{profit}}{\text{C.P}}\times 100$
$20 = \dfrac{\text{profit}}{10000}\times 100$
$\text{Profit}=\dfrac{20}{100}\times 10000$
= 2000
Selling price = Cost price + Prof‌it
S.P = 10000 + 2000 = Rs 12,000

---

Q 6, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 6
Juhi sells a washing machine for Rs 13, 500. She loses 20% in the bargain. What was the price at which she bought it?
Sol :
Selling price = Rs 13500
Loss % = 20%
Let the cost price be x.
∴ Loss = 20% of x
$=\dfrac{20}{100}\times x$
Cost price — Loss = Selling price
$x - \dfrac{20}{100}\times x = 13500$
$x - \dfrac{1}{5}\times x = 13500$
$\dfrac{4}{5}x=13500$
$x=13500\dfrac{5}{4}$
= 16875
Therefore, she bought it for Rs 16875

---

Q 7, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 7
(i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.
Sol :
(i) Ratio of calcium, carbon, and oxygen = 10 : 3 : 12
As 10 + 3 + 12 = 25 ,
Therefore , the percentage of the carbon $=\dfrac{3}{25}\times 100$
= 12 %
(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
Sol :
(ii) Let the weight of the stick be x g.
12 % of x = 3
$\dfrac{12}{100}\times x = 3$
$x=3\times \dfrac{100}{12}$
= 25 g

---

Q 8, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 8
Amina buys a book for Rs 275 and sells it at a loss of 1 5%. How much does she sell it for?
Sol :
Cost price = Rs 275
Loss % = 15 %
Loss = 15% of 275
$=\dfrac{15}{100}\times 275$
Cost price — Loss = Selling price
$275 - \dfrac{15}{100}\times 275 = \text{Selling Price}$
$275 - \dfrac{4125}{100}= \text{Selling Price}$
275 - 41.25 = Selling Price
Selling Price = 233.75

---

Q 9, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 9
Find the amount to be paid at the end of 3 years in each case:
(a) Principal = Rs 1,200 at 12% pa.
Sol :
(a) Principal (P) = Rs 1200
Rate (R) =12 % p.a.
Time (T) = 3 years
$\text{S.I}=\dfrac{P\times R \times T}{100}$
$=\dfrac{1200\times 12 \times 3}{100}$
= 432
Amount = P + S.I.
= 1200 + 432
= Rs 1632
(b) Principal = Rs 7,500 at 5% pa.
Sol :
(b) P = Rs 7500
R = 5 % p.a
T = 3 years
$\text{S.I}=\dfrac{P\times R \times T}{100}$
$=\dfrac{7500\times 5 \times 3}{100}$
= Rs 1125
Amount = 3500 +1125
= Rs 8625

---

Q 10, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 10
What rate gives Rs 280 as interest on a sum of Rs 56,000 in 2 years?
Sol :
$\text{S.I}=\dfrac{P\times R \times T}{100}$
$280=\dfrac{56000\times R \times 2}{100}$
$R=\dfrac{280}{560\times 2}=\dfrac{1}{4}$
R = 0.25
Therefore, 0.25% gives Rs 280 as interest on the given sum.

---

Q 11, Ex 8.3 - Comparing Quantities - Chapter 8 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

QUESTION 11
If Meena gives an interest of Rs 45 for one year at 9% rate p.a. . What is the sum she has borrowed?
Sol :
$\text{S.I}=\dfrac{P\times R \times T}{100}$
$\text{45}=\dfrac{P\times 9 \times 1}{100}$
$\text{P}=\dfrac{45 \times 100}{9}$
= Rs 500
Therefore, she borrowed Rs 500

---