NCERT Solutions for class 7 maths Comparing Quantities – Exercise – 8.3
Q1. Tell what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case.
- Gardening shears bought for Rs 250 and sold for Rs 325
Solution:-
Cost price of gardening shears = Rs 250
Selling price of gardening shears = Rs 325
Profit = (SP) – (CP)
= Rs 325 – Rs 250
= Rs 75
Profit% = { ( Profit /CP) x 100}
= { (75/250) x 100}
= (7500/250)
= (750/25)
= 30%
- A refrigerator bought for Rs 12000 and sold at Rs 13500.
Solution:-
Cost price of refrigerator = Rs 12000
Selling price of refrigerator = Rs 13500
Profit = (SP) – (CP)
= (13500) – (12000)
= 1500
Profit% = {(1500/12000) x 100}
= {150000/12000}
= 150/12
= 12.5%
© A cupboard bought for Rs. 2500 and sold at Rs.3000
Solution:-
From the question , it is given that cost price of cupboard=Rs.2500
Selling price of cupboard =Rs.3000
Profit =(SP)-(CP)
=Rs.(3000-2500)
=Rs.(500)
Profit % ={(500/2500) x 100}
={50000/2500}
=500/25
=20%
- A skirt bought for Rs. 250 and sold at Rs.150.
Solution:-
Loss =(CP)-(SP)
=Rs.(250-150)
=Rs.100
Loss %={(100/250) x 100}
={10000/250}
=40%
Q2. Convert each part of ratio to percentage ;
- 3:1
Solution:-
We have to find total parts by adding the given ratio = 3+1=4
1st part=3/4 = (3/4) x 100 %
=3 x 25%
=75%
2nd part =(1/4) x 100%
= 1 x 25
= 25%
- 2:3:5
Solution :-
1st part = (2/10) x 100%
=2 x 10%
=20%
2nd part = (3/10) x 100%
=3 x 10
=30%
3rd part = (5/10) x 100 %
=5 x 10
= 50%
© 1:4
Solution:-
1st part = (1/5) x 100%
= 1 x 20%
2nd part = (4/5) = (4/5) x 100%
= 4 x 20
= 80%
(d)1:2:5
Solution:-
1st part = 1/8 = (1/8) x 100%
= (100/8)%
= 12.5%
2nd part = (4/5) = (4/5) x 100%
= 4 x 20
= 80%
(d) 1:2:5
Solution:-
1st part = 1/8 =(1/8) x 100%
=(100/8) %
=12.5%
2nd part =2/8 = (2/8) x 100%
= (200/8)
=25%
3rd part = 5/8 = (5/8) x 100%
=(500/8)
=62.5%
Q3. The population of a city decreased from 25000 to 24500. find the percentage decrease .
Solution :-
Initial population of the city = 25000
Final population of the city = 24500
Population decrease = Initial
Population-final population
=25000-24500
=500
Then,
=(500/25000) x 100
=(50000/25000)
=50/25
=2%
Q4. Arun bought a car for Rs. 350000 . The next year ,the price went up to Rs. 370000. What was the percentage of price increase ?
Solution :-
Arun bought a car for =Rs.350000
The price of the car in the next year , went up to =Rs.370000
Then increase in price of car =Rs.370000-350000=20000
=(20000/350000) x 100
=(2/35) x 100
=200/35
=40/7
=55/7
Q5. I buy a T.V. for Rs. 10000 and sell it at a profit of 20% . how much money do I get for it ?
Solution :-
Cost price of thee T.V. = Rs.10000
Percentage of profit = 20%
Profit =(20/100) x 10000
=Rs. 2000
=10000+2000
=Rs.12000
Q6. Juhi sells a washing machine for Rs.13500 .she loses 20% in the bargain .what was the price at which she bought it ?
Solution :-
Selling price of washing machine = Rs.13500
Percentage of loss =20%
CP=Rs {(100/(100-loss%)) x SP}
={(100/(100-20)) x 13500}
={(100/80) x 13500}
={1350000/80}
={135000/8}
=Rs.16875
Q7. i. Chalk contains calcium , carbon and oxygen in the ratio 10:3:12. find the percentage of carbon in chalk .
Solution;-
The ratio of calcium ,carbon and oxygen in chalk = 10:3:12
So , the total part = 10+3+12=25
In the total part amount of carbon =3/25
Then,
Percentage of carbon =(3/25) x 100
=3 x 4
=12%
- If a stick of chalk, carbon is 3g, what is the weight of the chalk stick ?
Solution:-
12% of x=3
(12/100) x (x) = 3
X=3 x (100/12)
X=1 x (100/4)
X=25g
Q8. Amina buys a book for Rs. 275 and sells it at a loss of 15% . How much does she sell it for ?
Solution:-
Cost price book =Rs.275
Percentage of loss =15%
SP={(100-Loss%)/100) x CP )}
={((100-15)/100) x 275)}
={(85/100) x 275 }
=23375 /100
=Rs.233.75
Q9. Find the amount to be paid at the end of 3 year in each case:
- Principle =Rs. 1200at 12% p.a.
SI=(PxRxT)/100
=(1200 x 12 x 3)/100
=(12 x 12 x 3)/1
=Rs.432
Amount =(principal +SI )
=(1200+432)
=Rs.1632
- Principal =Rs.7500 at 5 % p.a.
Solution:-
=(7500 x 5 x 3)/100
=(75 x 5 x 3)/1
=Rs.8625
Q10. What rate gives Rs.280 as interest on a sum of Rs.56000 in 2 years?
Solution:-
R=(100 x SI) / (P x T )
=(100 x 280) / (56000 x 2)
=(1 x 28) / (56 x 2)
=(1 x 14) / (56 x 1)
=(1 x 1) / (4 x 1)
=(1/4)
=0.25%
Q11. If Meena gives an interest of Rs 45 for one year at 9% rate p.a What is the sum she has borrowed?
Solution:-
SI = (P X R X T )
45 = (P X 9 X 1)/100
= 5 X 100
= Rs 500
Hence, she borrowed Rs 500
Use of Is, Am , Are English Gramm