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# Arithmetic Aptitude :: Simple Interest

Home > Arithmetic Aptitude > Simple Interest > General Questions

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11. A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate percent.

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

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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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13. 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?

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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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Tags:  Infosys

14. 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.

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

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Answer: Option A

Explanation:

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

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16. 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?

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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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Tags:  IBM

17. 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

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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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Tags:  Cognizant

18. 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?

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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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Tags:  Cognizant

19.

A certain sum of money becomes Rs.750 in 2 years and becomes Rs.873 in 3.5 years.Find the sum and rate of interest.

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Answer: Option D

Explanation:

Lets assume principle as P, rate as R and Interest as SI

P + SI for 3.5 yrs = Rs 873 --------- (1)

P + SI for 2 yrs = Rs 756 -----------(2)

By solvig eq. (1) & (2) SI for 1.5 yrs = Rs 117

Therefore, SI for 2 yrs = Rs (117/1.5) * 2= Rs 156

So, P = 756 - 156 = Rs 600 R = (100 x 156) / ( 600 x 2 ) = 13% per annum.

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20. 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?

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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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