Simple Interest Questions For Class 8
Answers
Answer:
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Step-by-step explanation:
Question 1. Find the simple interest and amount on:
i) Rs.
4500 for
2
\frac{1 }{ 2} years at
7
\frac{2 }{ 3}
\% per annum
ii)
6360 for
6 years
3 months at
8\% per annum
iii) Rs.
19200 for
11 months at
9
\frac{3 }{ 4}
\% per annum
iv)
58400 for
75 days at
6
\frac{1 }{ 2}
\% per annum
Answer:
i) Rs.
4500 for
2
\frac{1 }{ 2} years at
7
\frac{2 }{ 3}
\% per annum
S.I.=
\frac{P \times R \times T }{ 100}
P = Rs. 45000, R = 7
\frac{2 }{ 3}
\% =
\frac{23 }{ 3}
\%
T =2
\frac{1 }{ 2} year
=
\frac{5 }{ 2} year
S.I.=
\frac{45000 \times \frac{23 }{ 3} \times \frac{5 }{ 2 }}{ 100}
= 8625 Rs.
Amount
= P+S.I = 53625 Rs.
ii)
6360 for
6 years
3 months at
8\% per annum
P = 6360 Rs.
R = 8\%
T=6
\frac{1 }{ 4}
=
\frac{25 }{ 4} year
S.I.=
\frac{P \times R \times T }{ 100}
=
\frac{6360 \times 8 \times \frac{25 }{ 4} }{ 100}
= 3180 Rs.
Amount
= 6360+3180=9540 Rs.
iii) Rs.
19200 for
11 months at
9
\frac{3 }{ 4}
\% per annum
P = 19200 Rs. ,
R = 9
\frac{3 }{ 4}
\% = \frac{39 }{ 4}
\%
T= 11 Months
=
\frac{11 }{ 12} years
S.I.
=
\frac{19200 \times \frac{39 }{ 4} \times \frac{11 }{ 12} }{ 100}
= 1716 Rs.
Amount
= P+SI = Rs. 19200+1716 = 20916 Rs.
iv)
58400 for
75 days at
6
\frac{1 }{ 2}
\% per annum
P= 58400 Rs. , latex R= 6 $
\frac{1 }{ 2}
\%=
\frac{13 }{ 2}
\%
T= 75 days
=
\frac{75 }{ 365} years
S.I.=
\frac{58400 \times \frac{13 }{ 2} \times \frac{75 }{ 365} }{ 100}
= 780 Rs.
Amount
= 59180 Rs.
Rishav takes a loan of Rs 10000 from a bank for a period of 1 year. The rate of interest is 10% per annum. Find the interest and the amount he has to the pay at the end of a year.
Namita borrowed Rs 50,000 for 3 years at the rate of 3.5% per annum. Find the interest accumulated at the end of 3 years.
Mohit pays Rs 9000 as an amount on the sum of Rs 7000 that he had borrowed for 2 years. Find the rate of interest.
