NCERT Solutions for class 7 maths Comparing Quantities Exercise – 8.3

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%

  1. 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

 

NCERT Solutions for class 7 maths Comparing Quantities – Exercise – 8.3   Q1. Tell what is the profit or loss…