This document contains solutions to 30 multiple choice questions related to interest rate calculations. Various interest formulas are used including simple interest, compound interest, effective interest rate, and others. Principal amounts, interest rates, time periods and compounding frequencies are given. The appropriate formula is selected and calculated to arrive at the answer.

1. 1. Answer: [c]
Solution: P*R*T/100=20000*2*10.100= 4000
2. Answer: [b]
Solution:Using Effective Percentage or Net Percentage Method: a+b+ab/100
10+10+100/100 = 21%
Now find 21% of Principal Amount i.e 21% of 20000=4200 Rs
3. Answer: [b]
Solution: A = P * R * T/100, Put the value of A, P and T to find R
A=2500, P= 2000, T=2 years, Now
A= P * R *T/100=> 2500 = 2000*2*R/100= 37.5%
4. Answer: [b]
Solution: Substituting the options one by one in pascals Triangle, as t=2 value
corresponding to it is 1 2 1,
Put R = 20% in the pascal triangle equation
A= 2000*1+(20% of 2000)*2+(20% of 400)
As A=2880 now above computation should give 2880 then R=20% will be right and on
computing above the values are equal so Answer is 20%
5. Answer: [c]
Solution: a+b+ab/100 as Rate of interest is compounded half yearly then R will halved
i.e r=10% nor 20%
Now, 10+10+100/100 = 21%
Now, 21% of Principal= CI
21/100*400=Rs 84
6. Answer: [a]
Solution: Under Compound Interest, at the end of each year the principal amount
Increases, therefore the interest in nth year will be greater than (n-1)th year, also this
difference increases as the value of n increases
Therefore only when p>q, the given situation arises
7. Answer: [b]
Solution: a+b+ab/100
As rate of interest is computed half yearly Rate of interest het half
Therefore, r=5%
So 5+5+25/100=10.25%
8. Answer: [b]
Solution: Since it issimple interest,interestforeachyearis same.
Sumof interestaccruedonit forthe 6th, 7th and 8th years

2. = 3000 + 3000 + 3000
= Rs.9000
9. Answer: [c]
Solution: A= P(1+r/100)^n
1440=1200(1+R/100)^2
1440/1200=(100+R)^2
100
(12/10)^2=(100+R)^2
100
120-100=R
R=20%
11. Answer: [b]
Solution: Using formula (n-1)*100/T=R so as n=2
2-1)*100/8=25/2
Now T=(n-1)*100/R
T=(4-1)*100/25/2=24
12. Answer: [a]
Solution: First 8 years: 100 to 200. Next 8 years: 200 to 400 at the same rate.
14. Answer: [c]
Solution:A=P+SI
3000=P[1+RT/100]
P=2000
15. Answer: [c]
Using formula A=P(1+R/100)^n
Second year he paid 1440 to clear the debt
Let the amount remaining after first year be p
P(1+(20/100))^1=1440
P=1200
So total amount after first year=1200+1200=2400
Now, p(1+(20/100))^1=2400
P=2000

3. 16. Answer:[C]
SI=(p*r*t)/100
SI for 16 years=(8800*12*16)/100
=Rs.16,896
Sum after 16 years =P+SI
=8800+16,896
=Rs.25,696
17. Ans: [d]
SI for 4 years =(4080-2480)
=1600
Principal amount=(2480-1600)
=Rs.880
18. Ans:[B]
SI for 8 years
(50000x8x2⁄100) + (50000x3x10⁄100) + (50000x2x5⁄100)+
(50000x7x1⁄100)
= 8000+15000+5000+3000
= 31500
Amount paid to repay the loan=50000+31500
=Rs.81500
19 Ans: [a]
P = 80000
R=10%
Amount after 1year = 88000
Let he has repaid X Rs. At the end of 1st year. So the principal for the 2nd year will be
(88000-X).
Now,
1.1(88000-X) = 55000
X =38000

4. 20 Ans: [b]
Applying the rule of alligaton
6% 8%
7.6%
0.4 1.6
So the ratio becomes 1:4.
So Ashok should invest 18000*(1/5) = Rs.3600 in the 1st scheme.
21 Ans: [b]
Interest earned in 2nd year = Rs.1980
Interest earned in 3rd year = Rs.2178
Comparing the two interest we got the value of R = (2178-1980)/1980*100
= 10%
Interest earned in 1st year = 1980-(1980/11)
= Rs. 1800
22 Ans: [d]
CI – SI =P(r/100)2
25.60 = P(8/100)2
P = Rs.4000
23 Ans: [ B]
Let,
Principal = Rs. 100.
Amount = Rs. 200.
Rate = r%
Time = 4 years.
Now,
A = P*[1+(r/100)]n
;
200 = 100*[1+(r/100)]4
;
2 = [1+(r/100)]4
; ........... (i)
If sumbecome 16 times in the time n years,
then,
16 = (1+(r/100))n
;
24
= (1+(r/100))n
; ........ (ii)
Using eqn (i) in (ii), we get;
([1+(r/100)]4
)4
=(1+(r/100))n
;

5. [1+(r/100)]16
=(1+(r/100))n
;
Thus, n = 16 years.
24 Ans: [D]
A=P+SI
A = (P+(P*R*T*)/100)
10640 = ( P+(P*8*5)/100)
10640 = P(1+(8*5)/100)
10640 = P*5/7
P = Rs.7600
25 Ans: [c]
Let each instalment be Rs.x .
1st year = [x + (x * 10 * 2)/100]
2nd year = [ x + (x *10 * 1)/100]
3rd year = x
Then, [x + (x * 10 * 2)/100] + [ x + (x *10 * 1)/100] + x =1815
3x + ( 20x/100 ) + ( 10x/100 ) = 1815
33x =18150
x = 550
Each instalment = Rs. 550
26 Ans: [b]
Suppose he borrowed Rs. X from 1st bank,so money borrowed from the 2nd bank will
be Rs. (55000-X)
According to the question,
(X*8)/100+((55000-X)12)/100 = 4900
Solving the equation we got X = Rs.42500
So Man borrowed Rs. 42500 from 1st bank.
27 Ans: []
Solution:

6. 28 Ans: [c]
Man initially have Rs.25000 and he need the final amount at the rate of 8%.
So final amount that he is supposed to receive is 25000+(25000*(8/100)) = Rs. 27000
For the 1st year SI = CI
So calculating the interest for the 1st year on the given amounts
For Rs.5500,interest earned is Rs.270
For Rs.4000,interest earned is Rs.260
For Rs.3500,interest earned is Rs.192.50
For Rs.7000,interest earned is Rs.595.
So for Rs.20000,interest earned is Rs.1317.50
Total interest to be earned = Rs 2000
Remaining interest to be received from remaining Rs. 5000 = Rs.682.50.
So, The rate of interest for the remaining amount is = (682.50/5000)*100
= 13.55%
29 Ans: [B]
CI – SI =P(r/100)2
126.72 = P(12/100)^2
P = Rs.8800
30 Ans: []