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The equal amounts of money are deposited in two banks each at 15% per annum for 3.5 years and 5 years respectively. If the difference between their interests is Rs.144, find the each sum?

A.

Rs.3467

B.

Rs.640

C.

Rs.500

D.

None

Answer with explanation

Answer: Option BExplanation

(P*5*15)/100 – (P*3.5*15)/100 = 144

75P/100 – 52.5P/100 = 144

22.5P = 144 * 100

=> P = Rs.640

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Find the compound interest accrued on the principal of Rs. 4000 at the end of 2 years at 10 % per annum.

A.

820 Rs

B.

830 Rs

C.

840 Rs

D.

810 Rs

Answer with explanation

Answer: Option CExplanation

Principal = Rs. 4000, t= 2 years , rate of percent, r = 10 %

Amount = P(1 + r/100)^t = 4000 x (1 + 10/100)^2 = 4000 x (11/10) x (11/10) = 40 x 121 = 4840 Rs.

Amount = Principal + CI => 4000 + CI = 4840 => CI = 840 Rs.

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Simple Interest on a certain sum at the rate of 10% per annum for 6 years and 7 years differs by rs.650/- what is the sum?

A.

Rs 6700

B.

Rs 7000

C.

Rs 6500

D.

Rs 8000

Answer with explanation

Answer: Option CExplanation

Simple interest for one year = 650

PRT/100 = 650

P * 10 * 1/100 = 650

P = 6500

The sum is Rs.6500

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In 10 years, A wants simple interest to be of the amount. So, what should be the rate percent per annum?

A.

5.66%

B.

5%

C.

4%

D.

6%

Answer with explanation

Answer: Option CExplanation

Let sum = x. Then, S.I. = 2x

5 , Time = 10 years.

Rate = (100 * 2x/x * 5 * 10)% = 4%

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The simple interest on a sum of money is 4/9 of the principal. Find the rate percent and time, if both are numerically equal.

A.

6(2/3) years 8 months

B.

4(4/3) years 7 months

C.

7(2/3) years 6 months

D.

None of these

Answer with explanation

Answer: Option AExplanation

Let the sum be Rs P.

Then SI = (4/9) P. Also given R=T.

Since SI = PTR/100 ==>> we find that 4/9P = P x R x R /100

=> R^2 = 100 x 4/9

=> R = 20/3 = 6( 2/3)

Hence the rate = 6 (2/3) % and

time = 6( 2/3) years 8 months.

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A sum at simple interest of 13(1/2)% per annum amounts to Rs. 3080 in 4 years. Find the sum.

A.

Rs. 1550

B.

Rs. 1680

C.

Rs. 2000

D.

Rs. 1850

Answer with explanation

Answer: Option CExplanation

Given:

Amount – Rs. 3080; R = 27/2 %; N = 4 yrs

Let the sum be x.

SI = Amount – Principal

SI = 3080 – x

WKT, SI = PNR/100

3080 – x = (x * 27 * 4)/(100*2)

3080 – x = 27x/50

154000 – 50x = 27x

154000 = 27x + 50x

154000 = 77x

x = 154000/77

x = Rs. 2000

Therefore, sum is Rs. 2000.

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A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6(1/4)% p.a for 2 years. Find his gain in the transaction per year.

A.

Rs. 112.50

B.

Rs. 125

C.

Rs. 225

D.

Rs. 167.50

Answer with explanation

Answer: Option AExplanation

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A certain sum of money amounts to Rs.1008 in 2 years and to Rs.1164 in 3.5 years.find the sum and rate of interest.

A.

12%

B.

13%

C.

15%

D.

20%

Answer with explanation

Answer: Option BExplanation

in 3.5 yrs the amount 1164

in 2 yrs the amount 1008 –

___________________________

the 1.5 yrs interest is 156

the 1 yr interest is 156/1.5 =104

the 2 yrs interest is 208 rs

principal=1008-208= 800

rate of interest=(104/800)*100

=13%

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Ramya gives 50 times the rent per annum to purchase a plot from L.I.C. Find the rate of interest from the amount paid by him.

A.

1%

B.

2%

C.

3%

D.

1.5%

Answer with explanation

Answer: Option BExplanation

Let the annual rent be Rs. x. I = x; P = 50x, n = 1

The rate of interest =100I/Pn =100x/(50x x 1)%= 2%

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4000 was divided into two parts such a way that when first part was invested at 3% and the second at 5%, the whole annual interest from both the investments is Rs.144, how much was put at 3%?

A.

Rs.2500

B.

Rs.5000

C.

Rs.2700

D.

Rs.2800

Answer with explanation

Answer: Option DExplanation

(x*3*1)/100 + [(4000 – x)*5*1]/100 = 144

3x/100 + 200 – 5x/100 = 144

2x/100 = 56 è x = 2800

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What annual installment will discharge a debt of Rs 1092 due in 3 years at 12% simple interest?

A.

325

B.

545

C.

560

D.

550

Answer with explanation

Answer: Option AExplanation

Let each instalment be Rs.x .

Then, `[x + (x ** 12 ** 1)/100] + [ x + (x **12 ** 2)/100] + x =1092`

`( (28x)/25 ) + ( (31x)/25 ) + x = 1092`

`(28x + 31x + 25x) = (1092 ** 25)`

`x = (1092 ** 25)/84= 325`

Each instalment = Rs. 325

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