In this post we will discuss questions from simple interest chapter. There are two chapters as far as compounding is concerned.
1. Simple Interest
2. Compound interest
Out of these two, simple interest is relatively less complex and take less time to solve.
Here we are discussing important questions of simple interest that are repeatedly asked in the competition exams. All the questions are solved step by step with simple explanation so that student with weak math background can easily grasp the concept
Simple Interest Questions
(01) What will be the simplest interest earned on an amount of Rs. 16,800 in 9 months at the rate of 6![]()
.25 % p.a.?
Given –
Principal (P) = Rs. 16, 800
Rate of interest (R) = 6.25 % p.a
Time (T) = 9 months = 9/12 = 3/4 years.
Simple Interest (SI) = (P * R * T)/100
= (16800 *25/4 * 3/4)/100
= Rs. 787.50
(02) A person borrows Rs. 5000 for 2 years at 4% p.a simple interest. He immediately lends it to another person at 6![]()
.25 % p.a for 2 years. Find his gain in the transaction per year?
Given;
P = Rs. 5000,
T = 2 years,
R = 4% p.a
Therefore,
SI = (P*R*T)/100 = (5000*4*2)/100
= Rs. 400 —-eq(1)
Second Case, Given;
P = Rs. 5000,
T= 2 years,
R = 6.25 %
Therefore, SI =(5000*6.25*2)/100 =
Rs. 625 —-eq(2)
From eq(1) and eq(2)
Thus, Overall gain for 2 years = Rs. 625 – Rs. 400 = Rs. 225
Therefore, Gain in the transaction per year = = Rs. 112.50
(03) Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
Given:
Principal (P) = Rs. 1200
Let rate of interest be x% p.a
Then, Time = x years (As per question)
We know that:
SI = (P * R * T)/100
432 = (1200 * x * x)/100
Solving the above equation we get
x^2 = 36
Hence x=> 6
So 6% is our rate of interest
(04) A man took a loan from a bank at the rate of 12% p.a. simple interest. After 3 years, he had to pay Rs. 5400 interest only for the period. Find the principal amount borrowed by the man?
Let the Principal amount be Rs. x
We know that:
SI = (P * R * T)/100
5400 = (x * 12 * 3)/100
x × 12 × 3 = 5400 × 100
x = Rs. 15,000
Hence the man borrowed 15,000 Rs
(05) Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs. 800 become in 3 years?
Given:–
Principal = Rs. 800,
Time = 3 years,
Amount = Rs. 956
SI = Amount – Principal = Rs. (956-800) = Rs. 156
We know that-
SI = (P*R*T)/100
156 = (800*R*3)/100
R= (156*100) / (800*3)
R = 6.5%
As per Question;
New Rate of interest = 6.5 + 4 = 10.5%
New Principal = Rs. 800, Time = 3 years
Again, SI = (800 * 10.5 * 3)/100
SI = 252
Therefore, total Amount = Rs. (800 + 252) = Rs. 1052
(06) The simple interest on a certain sum of money at the rate of 5% p.a. for 8 years is Rs. 840. At what rate of interest the same amount of interest can be received on the same sum after 5 years?
Given:
SI = Rs. 840,
R = 5% p.a.,
T= 8 years
We know that:
SI = (P * R * T)/100
840 = (P * 5 * 8)/100
P=(840 * 100)/ (5 * 8)
P= Rs. 2100
So the principal amount is Rs 2100
Again,
SI= (P * R * T)/100
840= (2100 * R * 5 )/100
R = 8% p.a.
Hence 8% is the rate of interest
(07) Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next 5 years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of 11 years is Rs. 8160, how much money did he borrow?
Given:
Total SI = Rs. 8160
Rate of Interests: R1 = 6 % p.a. for T1 = 3 years
R2 = 9% p.a. for T2 = 5 years
R3 = 13% p.a. for T3= 3 years ( As the total duration = 11 years)
Total SI = (P*R1*T1/100) + (P*R2*T2/100)+(P*R3*T3/100)
8160 = {(P*6*3)/100} + {(P*9*5)/100} + {(P*13*3)/100}
8160 *100 = 18P + 45P + 39P
on Solving we get
P= Rs. 8000
Hence Nitin Borrowed 8000 Rs
(08) An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, what is the effective rate of interest?
Let the sum be Rs. 100
As the time duration taken is 6 months, we will take T= 1/2
Then, SI for first 6 months =(100*10*1)/(100*2) = Rs. 5
Now this rs. 5 interest is added into the principal;
so new principal is = 100+5 = 105
SI for last 6 months =(105*10*1)/(100*2) = Rs. 5.25
Amount at the end of one year = Rs. (100 + 5 + 5.25) = Rs. 110.25
Therefore, effective rate = (110.25 – 100) = 10.25%
(09) A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. Find the principal amount
SI for 1 year = Rs. (854 – 815) = Rs. 39
Therefore, SI for 3 years = 39 × 3 = Rs. 117
Thus, Principal amount = Rs. (815 – 117) = Rs. 698
(10) At what rate percent of simple interest will a sum of money double itself in 12 years?
Let the Principal amount be P.
Amount after 12 years = 2P
SI = Amount – Principal = 2P – P = P —-eq(1)
We know that:
SI = (P * R * T)/100
From eq(1), we know that SI = P
P = (P * R * 12)/100
P × R × 12 = 100 × P
R = 100/12
R = 8
.33% is the answer