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.


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

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

Sol :

I=PRT100

=800×5×2100

=80



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

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

Sol :

I=PRT100

=5000×65×3100
=975


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

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

Sol :

I=PRT100 or =R100IPT

=100×3641300×7
=4


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

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

Sol :

I=PRT100 or  P=100IRT

=100×4405×4

=2200



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

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

Sol :

I=PRT100 or  T=100IPR

=100×4551300×7
=5



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

(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)=100IPT

=100×35550×2

=3.25


Ans : Interest Rate was 3.25%



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

(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=PRT100

or T=100IPR

=100×781.201860×12×100

=3.5

Ans : 3.5 years



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


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

Sol :

I=PRT100

=7300×5×123365100

=123

Ans : 123



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

(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=PRT100
or R=100IPT
=100×80400×4
=5

If rate if interest increased 2% per annum,
New (R1)=(5+2)=7%

Then, I1=P1R1T1100
=400×4×7100=112

Therefore, New amount=(P+1)
=400+112=512



Q10 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

(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=PRT100

38p=PR(254)100

38=R16

R=16×38=6


Ans : Rate of interest is 6%




Q11 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

(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=S.I×100P×T
=x×100x×8=1008
=25.5
Ans : Rate of interest is 12.5%



Q12 | Ex-8H |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |myhelper

(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=PRT100
1=50×T3×100
T=100×350
=6

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