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

Exercise 8H

Question

Find the missing value.

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(1) I if P = Rs. 800, R = 5% p.a., T = 2 years.

Sol :

$I=\frac{\text{PRT}}{100}$

$=\frac{800 \times 5 \times 2}{100}$

=80

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(2) I if P = Rs. 5000, R = 6.5% p.a. T = 3 years.

Sol :

$I=\frac{\text{PRT}}{100}$

$=\frac{5000 \times 65\times 3}{100}$

=975

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(3) R is I = Rs. 364 P = Rs. 1300, T = 7 years.

Sol :

$I=\frac{\text{PRT}}{100}$ or $=R\frac{100I}{PT}$

$=\frac{100 \times 364}{1300 \times 7}$

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(4) P if I = Rs. 440, R = 5%, T = 4 years

Sol :

$I=\frac{\text{PRT}}{100}$ or $P=\frac{100I}{RT}$

$=\frac{100 \times 440}{5 \times 4}$

=2200

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(5) T if I = Rs. 455, R = 7%, P = Rs. 1300

Sol :

$I=\frac{\text{PRT}}{100}$ or $T=\frac{100I}{PR}$

$=\frac{100 \times 455}{1300 \times 7}$

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(6) After 2 years, the credit balance in a saving account earning simple interest was Rs. 585.75. The original amount was Rs. 550. What was the interest rate?

Sol :

Simple Interest(585.75-550)=35.75

Interest Rate (R)$=\frac{100I}{PT}$

$=\frac{100 \times 35}{550 \times 2}$

=3.25

Ans : Interest Rate was 3.25%

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(7) In what time will Rs. 1860 amount to Rs. 2641.20 at simple interest of 12% per annum.

Sol :

SI=2641.20-1860=781-20

$I=\frac{PRT}{100}$

or $T=\frac{100I}{PR}$

$=\frac{100\times 781.20}{1860\times 12 \times 100}$

=3.5

Ans : 3.5 years

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(8) Find the simple interest on Rs. 7300 from 10 May 2017 to 10 September 2017 at 5% per annum.

Sol :

$I=\frac{PRT}{100}$

$=\frac{7300 \times 5\times \frac{123}{365}}{100}$

=123

Ans : 123

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(9) A sum of Rs. 400 amounts to Rs. 480 in 4 years. What will it amount ti, if the rate of interest is increased by 2%?

Sol :

I=480-400=80

$I=\frac{PRT}{100}$
or $R=\frac{100I}{PT}$
$=\frac{100 \times 80}{400 \times 4}$
=5

If rate if interest increased 2% per annum,

New (R1)=(5+2)=7%

Then, $I_{1}=\frac{P_{1}R_{1}T_{1}}{100}$
$=\frac{400 \times 4 \times 7}{100}=112$

Therefore, New amount=(P+1)

=400+112=512

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(10) On a certain sum, the simple interest at the end of 6-1/4 years becomes 3/8 of the sum. Find the rate of interest.

Sol :

$I=\frac{PRT}{100}$

$\frac{3}{8}p=\frac{PR\left(\frac{25}{4}\right)}{100}$

$\frac{3}{8}=\frac{R}{16}$

$R=\frac{16\times 3}{8}=6$

Ans : Rate of interest is 6%

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(11) A sum doubles itself in 8 years at simple interest. Find the rate of interest per annum.

Sol :

Let , the sum of money be x

After 8 years amount becomes =2x

S.I=2x-x=x

∴Rate per annum$=\frac{S.I \times 100}{P\times T}$
$=\frac{x\times 100}{x\times 8}=\frac{100}{8}$
=25.5

Ans : Rate of interest is 12.5%

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(12) In how many years will the simple interest on a sum of money be equal to the principal at the rate of 16-2/3% per annum?

Sol :
Let, Principle is P and Interest is also P then $P=\frac{PRT}{100}$
$1=\frac{50 \times T}{3\times 100}$
$T=\frac{100\times 3}{50}$
=6

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

Exercise 8H

Question

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