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Simple Interest Questions

Home > Quantitative Aptitude > Simple Interest > General Questions
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A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate percent.

ASum = Rs. 4400 and Rate=16 2/3 %.

BSum = Rs. 5400 and Rate=14 2/3 %.

CSum = Rs. 5400 and Rate=16 2/3 %.

DNone of these

Answer: Option C

Explanation:

S.I. on Rs. 7350 for 1 year = Rs. (8575-7350) = Rs. 1225.
Therefore, Rate = (100*1225 / 7350*1) % = 16 2/3 %.
Let the sum be Rs. X. then, x[1 + (50/3*100)]^2 = 7350.
=> x * 7/6 * 7/6 = 7350.
=> x = [7350 * 36/49] = 5400.
So, Sum = Rs. 5400.

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In how much time would the simple interest on a certain sum be 0.125 times the principal at 10% per annum?

A2 3/4 years

B1 1/4 years

C2 1/4 years.

D3 1/4 years.

Answer: Option B

Explanation:

Lets assume sum = x.
Then, S.I. = 0.125x = 1/8 *x, GivenR = 10%
Formula: Time = (100 * S.I)/(P * R)
Time = (100 *1/8* x) / (x * 10) = (100 *x)/(x*8*10)
= 5/4 = 1 1/4 years.

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There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

ARs. 3420

BRs. 3120

CRs. 3972

DRs. 3240

Answer: Option C

Explanation:

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
R = (100 x 60)/(100x6) = 10% p.a.
Now, P = Rs. 12000
T = 3 years and R = 10% p.a.
So, C.I. = Rs.[12000 * [(1 + 10/100)^3- 1]]
= Rs. 12000 * 331/1000 = Rs. 3972

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A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 1/2 years. Find the sum and rate of interests.

A700,13%

B700,15%

C800,13%

D800,15%

Answer: Option C

Explanation:

S.I. for 1 and 1/2 years = Rs.(1164-1008) = Rs.156.
SI for 2 years = Rs.(156*(2/3)*2)=Rs.208.
Principal = Rs. (1008 - 208) = Rs. 800.
Now, P = 800, T = 2 and S.l. = 208.
Rate =(100 * 208)/(800 * 2)% = 13%

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What sum of money will accumulate to Rs.5300 at 8% rate of interest in 9 months?

A5000

B5400

C4500

D4000

Answer: Option A

Explanation:

Let assume money = p.
so,106p/100=5300
p=5000

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A car that has an original value of $52,500 depricates$10000 in the first year and there original cost per year. What is its value after 8 years?

A$6,400 B$18,700

C$8,900 D$13,100

E$15,300 Answer: Option D Explanation: Given 1st year depreciation amount =$10000
next 7 year's depreciation amount = 7 * 8% original cost price
= 7*8/100 * 52500 = $29400 Total depreciation amount = ($10000 + $29400) =$39400
Value after 8 years = $(52500-39400) =$13100

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simple interest on an amount at 4% per annum for 13 months is more than the simple interest on the same sum for 8 months at 6% per annum by Rs. 40. what is the principle amount

A3600

B12000

C4800

D24000

Answer: Option B

Explanation:

Lets assume principle ammount = Rs. x
[(P*4*13/100*12) - (p*6*8/100*12)] = 40
=> p=12000

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Given that the interest is only earned on principal, if an investment of Rs.1000.00 amount to Rs.1440.00 in two years, then what is the rate of interest earned?

A20%

B21%

C22%

D23%

Answer: Option A

Explanation:

Solution: I= (P*R*T)/100  INTEREST( I)= 1440-1000= 440

440=(1000*R*2)/100

R=22%

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As per scheme a car loan of Rs. 4 lakh at 12.5 p.c.p. rate of simple interest can be borrowed on a repayment term of lump sum amount at the end of 3 years.
As per scheme (b), the amount can be repaid at the end of 2 years, but compound interest (compounded annually) would be charged at the same rate. What would be the difference in amount of interest between the two schemes?

ARs. 43,125

BRs 43,750

CRs. 41,025

DData inaequate

ENone of these

Answer: Option B

Explanation:

Scheme ,
p=400000
r=12.5%
n=3
SI = 400000*3*12.5/100
SI = 150000
Total Sum = 550000
Scheme (b)
Amount = 400000[1 + 12.5/100]^2
=> 400000*[9/8 * 9/8]
Amount = 506250
Difference is 550000 - 506250 = Rs. 43750

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The difference between simple interest and compound interest on a certain deposit for two years at 8 p.c.p. is Rs. 8. What is the amount of deposit

ARs. 2250

BRs. 1250

CRs. 1550

DRs. 2000

ENone of these

Answer: Option B

Explanation:

Given Intrest rate r = 8%
Deposited for year n = 2 years
Formula: SI = p*n*r/100
=> p*2*8/100 = 4p/25
CI = Amount - p
Amount = p(1 + r/100)^n
=> p(1 + 8/100)^2
CI = p(27/25)^2 - p
p(27/25)^2- p - 4p/25 = 8
p[27*27 - 625 - 100]/625 = 8
4p = 5000 => p = Rs. 1250

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