Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3
1. Find what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case.
Question (i).
Gardening shears bought for ₹ 250 and sold for ₹ 325
Solution:
C.P. of gardening shears = ₹ 250
S.P. of gardening shears = ₹ 325
Profit = S.P. – C.P.
= ₹ 325 – ₹ 250
= ₹ 75
Profit percentage = \(\left[\frac{\text { Profit }}{\text { Cost price }} \times 100\right] \%\)
= \(\left[\frac{75}{250} \times 100\right] \%\)
= 30%
Question (ii).
A refrigerater bought for ₹ 12,000 and sold at ₹ 13,500
Solution:
C.P. of refrigerator = ₹ 12,000
S.P. of refregerator = ₹ 13,500
Profit = S.P. – C.P.
= ₹ 13,500 – ₹ 12,000
= ₹ 1500
Profit percentage = \(\left(\frac{\text { Profit }}{\text { C.P. }} \times 100\right) \%\)
= \(\left(\frac{1500}{12000} \times 100\right) \%\)
= 12.5%
Question (iii).
A cupboard bought for ₹ 2,500 and old at ₹ 3,000.
Solution:
C.P. of card board = ₹ 2,500
S.P. of card board = ₹ 3,000
Profit = S.P. – C.P.
= ₹ 3000 – ₹ 2500
= ₹ 500
Profit percentage = \(\left(\frac{\text { Profit }}{\text { C.P. }} \times 100\right) \%\)
= \(\left(\frac{500}{2500} \times 100\right) \%\)
= 20%
Question (iv).
A shirt bought for ₹ 250 and sold at ₹ 150
Solution:
C.P. of shirt = ₹ 250
S.P. of shirt a ₹ 150
Since S.P. is less than C.P.
So, there will be a loss
Loss = C.P. – S.P.
= ₹ 250 – ₹ 150
= ₹ 100
Loss percentage = \(\left(\frac{\text { Loss }}{\text { C.P. }} \times 100\right) \%\)
= \(\left(\frac{100}{250} \times 100\right) \%\)
= 40%
2. A shopkeeper buys an article for ₹ 735 and sold it for ₹ 850. Find his profit or loss.
Solution:
C.P. of an article = ₹ 735
S.P. of an article = ₹ 850
Profit = ₹ 850 – ₹ 735
= ₹ 115
3. Kirti bought a saree for ₹ 2500 and sold it for ₹ 2300. Find her loss and loss percent.
Solution:
C.P. of saree = ₹ 2500
S.P. of saree = ₹ 2300
Loss = C.P. – S.P.
= ₹ 2500 – ₹ 2300
= ₹ 200
Loss percentage = \(\left(\frac{\text { Loss }}{\text { C.P. }} \times 100\right) \%\)
= \(\frac {200}{2500}\) × 100
= 8%
4. An article was sold for ₹ 252 with a profit of 5%. What was its cost price ?
Solution:
S.P. of an article = ₹ 252
Profit = 5%
Let C.P. of article = ₹ 100
Profit = 5% of ₹ 100
= ₹ 5
S.P. of article = ₹ 100 + ₹ 5
= ₹ 105
If S.P. is ₹ 105, then C.P. = ₹ 100
If S.P. is ₹ 1 then C.P. = ₹ \(\frac {100}{105}\)
If S.P. is ₹ 252, then C.P. = ₹ \(\frac {100}{105}\) × 252
= ₹ 240
5. Amrit buys a book for ₹ 275 and sells it at a loss of 15%. For how much does she sell it ?
Solution:
C.P. of book = ₹ 275
Loss = 15%
∴ Loss on ₹ 275 = ₹ \(\frac {15}{100}\) × 275
= ₹ 41.25
Thus, S.P. of book = ₹ 275 – ₹ 41.25
= ₹ 233.75
6. Juhi sells a washing machine for ₹ 13500. She losses 20% in the bargain. What was the price at which she bought it ?
Solution:
S.P. of washing machine = ₹ 13500
Let C.P. = ₹ 100
Loss = 20%
S.P. = ₹ 100 – ₹ 20
= ₹ 80
If S.P. of washing machine is ₹ 80 then its cost price = ₹ 100
If S.P. of washing machine is ₹ 1 then its cost price = ₹ \(\frac {100}{80}\)
If S.P. of washing machine is ₹ 13500
then its cost price = ₹ \(\frac {100}{80}\) × 13500
= ₹ 16875
7. Anita takes a loan of ₹ 5000 at 15% per year as rate of interest. Find the interest she has to pay at the end of one year.
Solution:
Here, Principal (P) = ₹ 5000
Rate (R) = 15% per year
Time (T) = 1 year
Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{5000 \times 15 \times 1}{100}\)
= ₹ 750
8. Find the amount to be paid at the end of 3 years in each case :
Question (i).
Principal = ₹ 1200 at 12% p.a.
Solution:
P = ₹ 1200, R = 12% p.a.
T = 3 years
Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{1200 \times 12 \times 3}{100}\)
= ₹432
Amount = Principal + Simple Interest
= ₹ 1200 + ₹ 432
= ₹ 1632
Question (ii).
Principal = ₹ 7500 at 5% p.a.
Solution:
P = ₹ 7500, R = 5% p.a. T = 3 years
SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹\(\frac{7500 \times 5 \times 3}{100}\)
= ₹ 1125
Amount = P + S.I. = ₹ 7500 + ₹ 1125
= ₹ 8625
9. Find the time when simple interest on ₹ 2500 at 6% p.a. is ₹ 450
Solution:
P = ₹ 2500, R = 6% p.a. Time = T = ?,
S.I. = ₹ 450
T = \(\frac{\mathrm{SI} \times 100}{\mathrm{P} \times \mathrm{R}}\)
T = \(\frac{450 \times 100}{2500 \times 6}\)
= 3 years
10. Find the rate of interest when simple interest on ₹ 1560 in 3 years is ₹ 585.
Solution:
Principal (P) = ₹ 1560
Time (T) = 3 years
S.I. = ₹ 585
R = \(\frac{\text { SI } \times 100}{\mathrm{P} \times \mathrm{T}}\)
= \(\frac{585 \times 100}{1560 \times 3}\)
= \(\frac {125}{10}\)
= 12.5
Thus Rate of interest is 12.5% p.a.
11. If Nakul gives an interest of ₹ 45 for, one year at 9% rate p.a. what is the sum he borrowed ?
Solution:
Here S.I.= ₹ 45, R = 9% p.a.
T = 1 year, P = ?
P = \(\frac{\text { S.I. } \times 100}{\mathrm{R} \times \mathrm{T}}\)
= \(\frac{45 \times 100}{9 \times 1}\)
= ₹ 500
12. If ₹ 14,000 is invested at 4% per annum simple interest, how long will it take for the amount to reach ₹ 16240 ?
Solution:
P = ₹ 14,000
R = 4% p.a.
T = ?
A = ₹ 16240
S.I. = A – P
= ₹ 16240 – ₹ 14,000
= ₹ 2240
T = \(\frac{\text { SI } \times 100}{\mathrm{R} \times \mathrm{P}}\)
= \(\frac{2240 \times 100}{14000 \times 4}\)
= 4 years
13. Multiple Choice Questions :
Question (i).
If a man buys an article for ₹ 80 and sells it for ₹ 100, then gain percentage is
(a) 20%
(b) 25%
(c) 40%
(d) 125%
Answer:
(b) 25%
Question (ii).
If a man buys an article for ₹ 120 and sells it for ₹ 100, then his loss percentage is
(a) 10%
(b) 20%
(c) 25%
(d) 16\(\frac {2}{3}\)%
Answer:
(d) 16\(\frac {2}{3}\)%
Question (iii).
The salary of a man is ₹ 24000 per month. If he gets an increase of 25% in the salary, then the new salary per month is
(a) ₹ 2,500
(b) ₹ 28,000
(c) ₹ 30,000
(d) ₹ 36,000
Answer:
(c) ₹ 30,000
Question (iv).
On selling an article for ₹ 100, Renu gains ₹ 20 Her gain percentage is
(a) 25%
(b) 20%
(c) 15%
(d) 40%
Answer:
(a) 25%
Question (v).
The simple interest on ₹ 6000 at 8% p.a. for one year is
(a) ₹ 600
(b) ₹ 480
(c) ₹ 400
(d) ₹ 240
Answer:
(b) ₹ 480
Question (vi).
If Rohini borrows ₹ 4800 at 5% p.a. simple interest, then the amount she has to return at the end of 2 years is.
(a) ₹ 480
(b) ₹ 5040
(c) ₹ 5280
(d) ₹ 5600
Answer:
(c) ₹ 5280