### 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

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

1^{st }part=3/4 = (3/4) x 100 %

=3 x 25%

=75%

2^{nd} part =(1/4) x 100%

= 1 x 25

= 25%

**2:3:5**

**Solution :-**

1^{st} part = (2/10) x 100%

=2 x 10%

=20%

2^{nd }part = (3/10) x 100%

=3 x 10

=30%

3^{rd }part = (5/10) x 100 %

=5 x 10

= 50%

**© 1:4**

**Solution:-**

1^{st} part = (1/5) x 100%

= 1 x 20%

2^{nd} part = (4/5) = (4/5) x 100%

= 4 x 20

= 80%

**(d)1:2:5**

**Solution:-**

1^{st }part = 1/8 = (1/8) x 100%

= (100/8)%

= 12.5%

2^{nd} part = (4/5) = (4/5) x 100%

= 4 x 20

= 80%

**(d) 1:2:5**

** Solution:-**

1^{st} part = 1/8 =(1/8) x 100%

=(100/8) %

=12.5%

2^{nd }part =2/8 = (2/8) x 100%

= (200/8)

=25%

3^{rd }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

=5^{5/7}

^{ }

**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

** **

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

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