Top 25 Simple Interest Questions and Answers
What is Interest?
When you borrow money from a bank, there is an amount that you have to pay apart from the amount you borrowed. The extra money you pay is called interest. In simple word Interest is the amount paid by the borrower to the lender.
Principal: The sum lent is called principal.
Simple Interest (SI)= (PxRxT)/100
Where,
P = Principal
R= Rate of interest per annum
T= Time in years
Amount (A)= P + S.I
= P(100+RT)/100
P=(100xI)/(RxT)
R=(100xI)/(PxT)
T=(100xI)/(PxR)
Top 25 Simple Interest Questions and Their Tricky Solutions
In this article we are going to post an
excellent collection of Top 25 Simple Interest questions along with
their tricky solutions asked in various Government examination like
SSC CGL, CHSL, CPO, MTS, JE, GD Constable, Stenographer, IBPS, RRB etc. These
type of questions are repeated every time so you can practice by solving it. Go
through the below Top 25 Simple Interest questions and answers-
Q. (1) What is the Simple Interest on Rs.8000 for 5 years at 10% per annum rate of interest?
(a) 1000
(b) 2000
(c) 3000
(d) 4000
Detail Solution:
Simple Interest =PTR /100
=8000x5x10/100= Rs 4000
Hence, Option (d) is correct.
Q. (2) A sum of money Rs.32000 is lent in two parts, one at the rate of 10% and another at 12%. If the annual interest received is Rs. 3600. What is the amount lent at 12%?
(a) 20000
(b) 38000
(c) 44000
(d) 60000
Detail Solution:
Let X be the money lent at 10% . Then (32000- X) is lent at 12%
Simple Interest on both amounts is equal to Rs.3600
=>Xx1x10/100+ (32000-X)x1x12/100= 3600
=>10X/100 + 32000x 12/100 – 12X/100= 3600
=>2P/100 = 3840-3600
=>2P= 240 x100
=>P =Rs.12000
Thus, Money lent at 12%= (32000-12000)= Rs. 20000.
Hence, Option (a) is correct.
Q. (3) In how many years a sum of Rs. 450 gives Rs. 81 as interest at 4.5% of Simple Interest ?
(a) 5
(b) 4.5
(c) 4
(d) 6
Detail Solution:
T= (100x 81/450x4.5)
= 4 years
Hence, Option (c) is correct.
Q. (4) The difference between the interests received from two scheme on Rs.5000 for two years is Rs.25. The difference between their rates is ?
(a) 0.20%
(b) 0.25%
(c) 0.50%
(d) 1%
Detail Solution:
Let the rates be R1% per annum & R2% per annum.
Then, (5000xR1/100x2)- (5000xR2/100x2) =25
=> 100(R1-R2) = 25
=> R1-R2 =0.25
So, The difference b/w their rates = 0.25%
Hence, Option (b) is correct.
(a) 1800
(b) 1200
(c) 2400
(d) 3000
Detail Solution:
Time=8 years 4 months = 8+4/12 years =25/3years
As We know:
Simple Interest =PTR /100= 2400x25/3x6/100=Rs.1200
Hence, Option (b) is correct.
Q. (6) A certain sum of money amounts to Rs.1008 in 2 years and to Rs.1164 in 3 ½ years. What is the sum and rate of interests?
(a) 10%
(b) 11%
(c) 13%
(d) 12%
Detail Solution:
Simple Interest for (3 ½ - 2)=3/2 years = Rs. (1164-1008)=156
Simple Interest for 2 years = Rs. (156 x 2⁄3 x 2) = Rs.208
Principal =(1008 - 208)= Rs.800
Rate = 208x100/800x2= 13%
Hence, Option (c) is correct.
Q. (7) A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, What is the money lent at 6%?
(a) 800
(b) 900
(c) 1000
(d) 1500
Detail Solution:
Let, the sum lent at 8% be Rs.X and that at 6% be Rs.(1550 - X)
=> [(Px8x1)/100 + (1550-P)x6x1]/100=106
=> 8P + 9300 –6P=10600
=> 2P = 1300
=> P = 650.
Money lent at 8% be= Rs.650 & Money lent at 6% be=(1550 - 650)= Rs. 900.
Hence, Option (b) is correct.
Q. (8) Kamlesh borrowed some money at certain rate of SI for 3 years. If the rate had been 2% higher, He would have to pay Rs. 360 more. What was the amount?
(a) 4000
(b) 4800
(c) 7500
(d) 6000
Detail Solution:
Let, Sum= P and Rate = R.
Then, [(Px(R+2)x3)/100] – [(PxRx3)/100]=360.
=> 3PR + 6P - 3PR = 36000
=> 6P=36000
=> P=6000
So, Sum = Rs.6000
Hence, Option (d) is correct.
Q. (9) What is the Simple Interest on Rs.8000 from 7th August 2016 to 31st December 2016 at 5% per annum rate of interest?
(a) 160
(b) 180
(c) 240
(d) 300
Detail Solution:
Number of days from 6th August to 30th December= 146 Days ( we do not count the given day)
Simple Interest =PTR /100= 8000 x(146/365)x5/100= Rs. 160
Hence, Option (a) is correct.
Q. (10) Rs. 800 amounts to Rs. 920 in 3 years. If interest rate is increased by 8%, the what will be amount?
(a) 752
(b) 886
(c) 992
(d) 1012
Detail Solution:
Simple Interest = Rs. (920 - 800) = Rs.120
P=Rs.800, T=3 year.
R=((100x 120)/(800x3) )= 5%.
New Rate =(5 + 3)% = 8%.
New Simple Interest =(800x8x3)/100 = Rs.192.
New amount =Rs.(800+192) = Rs. 992.
Hence, Option (c) is correct.
Q. (11) The Simple Interest on a sum of money is 4 /25th of the principal and the rate of interest is equal to the number of years. What is the rate of interest?
(a) 1
(b) 2
(c) 3
(d) 4
Detail Solution:
Simple Interest is 4 /25th of the principal
=>SI =4/25P
Rate of interest per annum is equal to number of years
=> R =T
SI =PTR /100x4/25
P =PxRxR /100x4/25
=>R2 =400/25
=>R=20/5=4%
Hence, Option (d) is correct.
Q. (12) At what rate percent per annum will a sum of money double in 16 years?
(a) 6.25%
(b) 12.5%
(c) 3.33%
(d) 6.66%
Detail Solution:
Let Principal = P,
Then, Simple Interest = P and T = 16 years
Rate = (100xP⁄Px16)%
= 25/4%
Hence, Option (d) is correct.
Q. (13) What would be the Simple Interest obtained on an amount of Rs 6535 at the rate of 10% after 6 years?
(a) 3600
(b) 3921
(c) 3950
(d) 4250
Detail Solution:
Simple Interest = PTR/100
= 6535 x 6
x 10/100
= Rs. 3921
Hence, Option (b) is correct.
Q. (14) A man took a loan from a bank at the rate of 12% per annum Simple Interest . After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was?
(a) 14500
(b) 15000
(c) 15500
(d) 16500
Detail Solution:
Principal = Rs. (100x 5400/12x3) = Rs. 15000
Hence, Option (b) is correct.
Q. (15) What is the Simple Interest on Rs 2400 for 9 months at 4 paisa per rupee per month?
(a) 612
(b) 718
(c) 864
(d) 930
Detail Solution:
P= Rs. 2400, Time Period T = 9 months =9 /12years = ¾ year
Rate of interest =4 paisa per Rupee per month= 4% per month= 48% per annum
Simple Interest =PTR/100= 2400x¾x48/100= Rs 864
Hence, Option (c) is correct.
Q. (16) What will be the ratio of Simple Interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?
(a) 2:3
(b) 3:4
(c) 4:3
(d) None of these
Detail Solution:
Let the principal be P and rate of interest be R%
So, Required Ratio =(PxRx6⁄100 : PxRx9⁄100)
6PR : 9PR =2:3
Hence, Option (a) is correct.
Q. (17) A sum was put at a Simple Interest at a certain rate of 4 years. Had it been put a 4% higher rate, it would have fetched Rs 720 more. What is the principal?
(a) 2000
(b) 3000
(c) 4000
(d) 6000
Detail Solution:
Let, Rate of interest be R and principal be P.
SI at R rate of interest=>Simple Interest = P x 4 x(R+4)/25
If the rate of interest is 4% more, then SI= P x 4 x(R+4) /25
The difference b/w these two Simple Interest s is Rs.720
P x 4 x(R+4) /25 - P x 4 x(R+4)/25= 720
P x 4 x4 /25= 960
So, P= Rs 6000
Hence, Option (d) is correct.
Q. (18) The Simple Interest on a sum of money is 4/9 of the principal. What is the rate percent and time, if both are numerically equal?
(a) 6 Years 1 month
(b) 6 Years 5 month
(c) 6 Years 7 month
(d) 6 Years 8 month
Detail Solution:
Let, Sum = Rs. P
Then, Simple Interest = Rs. 4P/9
Let, Rate = R% and time = R years.
Then, (PxRxR)/100=4P/9 or R2 =400/9
=> R = 20/3 = 6 2/3.
=> Rate = 20/3% and Time=20/3 years = 6 years 8 months
Hence, Option (d) is correct.
Q. (19) What annual installment will discharge a debt of Rs. 1092 due in 3 years at 12% Simple Interest ?
(a) 325
(b) 375
(c) 425
(d) 500
Detail Solution:
Let, each Installment be Rs. I
Then, [ I+ {(Ix12x1)/100}] + [I+ {(Ix12x2)/100} ] + I = 1092
=>[((28I/25) + (31I/25) + I] = 1092
=>(28I+31I+25I)=(1092x25)
=> I= (1092x25)/84 = Rs.325.
Each installment = Rs. 325
Hence, Option (a) is correct.
Q. (20) Find the amount which yields Simple Interest of Rs. 4016.25 at the rate of 9% per annum in 5 years.
(a) 8150
(b) 8500
(c) 8925
(d) 9250
Detail Solution:
We have P=(100xI)/(RxT)
Principal = (100x 4016.25)⁄(9 x 5)=(401625/45)= 8925
Hence, Option (c) is correct.
Q. (21) At what rate of interest per annum, will Rs 5500 be obtained as Simple Interest on Rs 25000 for 2 years 9 months?
(a) 9%
(b) 8%
(c) 12%
(d) 15%
Detail Solution:
Given that, P= Rs. 25000 and
Time = 2 years 9 months= 29/12=11/4years
Simple Interest =PTR /100
⇒5500= 25000 x11/4xR /100
R = 8 %
Hence, Option (b) is correct.
Q. (22) A sum of money becomes 4 times in 9 years at Simple Interest. In how many years will it become 8 times of itself at the same rate?
(a) 12
(b) 18
(c) 30
(d) 21
Detail Solution:
Let, Sum = P, Then, Amount= 4P
Simple Interest= 4P-P=3P
To get Simple Interest 3P, it takes 9 years.
To get Simple Interest P, it takes 9/3= 3 years
Now it has to become 8 times
The Simple Interest to be obtained is 7P
To get Simple Interest P, it takes 3 years
Similarly, To get 7P as Simple Interest , it takes 7x3 = 21 years
TRICK:
(x-1)/t1= (y-1)t2
Where, x= nth term, t1 & t2 = time period
(4-1)x9 = (8-1)xt2
3/9 = 7/t2=>t2 = 21 years
Hence, Option (d) is correct.
Q. (23) A sum was put at Simple Interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs. 360 more. What is the sum?
(a) 4000
(b) 5000
(c) 6000
(d) 7000
Detail Solution:
Let, Sum be = P and Rate = R
Then, [Px(R+2)x3⁄100) - (PxRx3⁄100)]=360
3PR + 6P - 3PR =36000
6P=36000
P=6000
Hence, Option (c) is correct.
Q. (24) A certain amount earns Simple Interest of Rs. 1750 after 7 years. Had the interest been 2 % more, how much more interest would it have earned?
(a) 120
(b) 240
(c) 360
(d) Data inadequate
Detail Solution:
We have to know the Simple Interest, principal and time to What is the rate. Since the principal is not given, so the data is inadequate.
Hence, Option (d) is correct.
Q. (25) Divide Rs. 2379 into 3 parts so that their amounts after 2,3 and 4 years respectively may be equal, the rate of interest being 5% per annum at Simple Interest . The first part is?
(a) 759
(b) 792
(c) 818
(d) 828
Detail Solution:
Let the parts be a, b and [2379 - (a + b)]
a + (a x 25⁄100) = b + (b x 35⁄100) = c + (c x 45⁄100)
11a⁄10 = 23b⁄20 = 6c⁄5 = k
a = 10k⁄11
b = 20k⁄23
c = 5k⁄6
But, a + b + c = 2379
10k⁄11 + 20k⁄23 + 5k⁄6 = 2379
1380k + 1320k + 1265k = 2379 x 11 x 23 x 6
k = 2379 x 11 x 23 x 6⁄3965= 3 x 11 x 23 x 6⁄5
a = 828
Hence, Option (d) is correct.
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