S.chand publication New Learning Composite mathematics solution of class 7 Chapter 8 Percentage and Its Applications Exercise 8D

 Exercise 8D

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Q1 | Ex-8D |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 1

Increase:

(a) Rs. 50 by 20%

Sol :

$=\frac{20}{100}\times 50$

=10

∴New increased amount=50+10=60

(b) Rs. 700 by 8%

Sol :

$=\frac{8}{100}\times 700$

=56

∴New increased amount=700+56=756

(c) 600 g by 18%

Sol :

$=\frac{18}{100}\times 600$

=108

∴New increased amount=600+108=708g

(d) 4L by 15%

Sol :

$=\frac{15}{100}\times 4$

=0.6

∴New increased weight=(4+0.6)L=4.6 L

(e) 8 km by 19%

Sol :

$=\frac{19}{100}\times 8$

=1.52

∴New increased distance=(8+1.52)=9.52 km

(f) 5 kg by 23%

Sol :

$=\frac{23}{100}\times 5$

=1.15

∴New increased weight=(5+1.15)=6.15 kg

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Q2 | Ex-8D |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 2

Decrease:

(a) Rs. 170 by 10%

Sol :

$=\frac{10}{100}\times 170$

=17

∴New decreased amount=(170+17)=153

(b) Rs. 2500 by 18%

Sol :

$=\frac{18}{100}\times 2500$

=450

∴New decreased amount=(2500-450)=2050

(c) 1 ha by 28%

Sol :

$=\frac{28}{100}\times 1$

=0.28

∴New decreased amount=(1-0.28)=0.72 ha

(d) 12 tonnes by $24\frac{2}{3}$% tonnes

Sol :

$=24\frac{2}{3}\% \text{ of } 12$

$=\frac{74}{300}\times 12$=2.96

∴New decreased weight=(12-2.96)=9.04

(e) 24 m by 22%

Sol :

$=\frac{22}{100}\times 24$

=5.28

∴New decreased unit=(24-5.28)=18.72 m

(f) 480 g by 20%

Sol :

$\frac{20}{100}\times 480$

=96

∴New decreased weight=480-96=384 g

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Q3 | Ex-8D |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 3

The entrance fee at a science museum is Rs. 50. Senior citizens recieve a 20% discount. How much fee does a senior citizen pay?

Sol :

Discount for senior citizen$=2\left(50\times \frac{20}{100}\right)$=10

∴Fees for senior citizen=50-10=40

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Q4 | Ex-8D |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 4

What is the final price of a Rs. 4500 leather jacket that is on sale for 40% off?

Sol :

Discount price$=\left(\frac{40}{100}\times 4500\right)$

=1800

∴Final price of leather jacket=4500-1800=2700

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Q5 | Ex-8D |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 5

Increase Rs. 2000 by 10% and then decrease the new amount by 10%.

Sol :

∴Price increases$=\left(\frac{10}{100}\times 2000\right)$=200

∴New increased amount=(2000+200)=2200

New price decreases$=\left(\frac{2110}{100}\times 2200\right)$=220

∴New decreased amount=2200-220=1980

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Q6 | Ex-8D |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 6

Decrease Rs. 500 by 15% and then increase the result by 15%.

Sol :

∴Price decreases$=\left(\frac{15}{100}\times 500\right)$=75

∴Decreased price=500-75=425

New increased price$=\left(\frac{15}{100}\times 425\right)$=63.75

∴New increased price=425+63.75=488.75

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Q7 | Ex-8D |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 7

Find the per cent increase for the following.

(a) original price = Rs. 800, new price = Rs. 872

Sol :

Increased price=872-800=72

∴Percentage of new price$=\left(\frac{72}{800}\times 100\right)$=9%

(b) old income = Rs. 1,20,000, new income = Rs. 1,50,000

Sol :

∴Increased income=150000-120000=30000

∴Percentage of new income$=\left(\frac{30000}{120000}\times 100\right)$=25%

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Q8 | Ex-8D |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 8

Find the percent decrease for the following.

(a) From 500 to 100

Sol :

500-100=400

∴Decreased percentage$=\left(\frac{400}{500}\times 100\right)$=80%

(b) Original price = Rs. 300, new price = Rs. 270

Sol :

Decreased price=300-270=30

∴Decreased percentage$=\left(\frac{30}{300}\times 100\right)$=10% 

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Q9 | Ex-8D |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

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Question 9

A property which was valued at Rs. 40,000 five years ago is now valued at Rs. 60,000. What is the percentage increase in value?

Sol :

Decreased value=60000-40000=20000

∴Percentage of increased value$=\left(\frac{20000}{40000}\times 100\right)$=50%