How long will it take for an amount Tk 450 to yield Tk 81 as interest at 4.5% per annum of simple interest?

Q 1 - Adam borrowed some money at the rate of 6% p.a. for the first two years, at the rate of 9% p.a. for the next three years, and at the rate of 14% p.a. for the period beyond five years. If he pays a total interest of Rs. 11,400 at the end of nine years, how much money did he borrow?

A - 12,000

B - 13,000

C - 14,500

D - 12,500

Answer - A

Explanation

Let the sum borrowed be Z. Then, (Zx6x2⁄100) + (Zx9x3⁄100) + (Zx14x4⁄100) = 11400 Therefore, 3Z⁄25 + 27Z⁄100 + 14Z⁄25) = 11400 ? 95Z⁄100 = 11400 Z = (11400x100⁄95) = 12000

Answer - C

Explanation

S.I. for 11⁄2years = Rs. (1164-1008) =156 S.I. for 2 years = Rs. (156 x 2⁄3 x 2) = Rs. 208 Principal = Rs. (1008 - 208) = Rs. 800 Now, P = 800, T = 2, and S.I. = 208 Rate = (100x208⁄800x2)% = 13%

Answer - D

Explanation

Let Principal = P, Then, S.I. = P and T = 16 years Rate = (100xP⁄Px16)% = 6(1⁄4)

Q 4 - The simple interest on a certain sum of money for 21⁄2 years at 12% per annum is Rs. 40 less than the simple interest on the same sum for 31⁄2 years at 10% per annum. Find the sum?

A - 400

B - 800

C - 1600

D - 500

Answer - B

Explanation

Let the sum be Z then, (Zx10x7⁄100x2) - (Zx12x5⁄100x2) =40 7Z⁄20 - 3Z⁄10 = 40 Z = 40 x 20 The sum is Rs. 800

Q 5 - 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. Find the sum?

A - 10000

B - 6000

C - 15000

D - 6500

Answer - B

Explanation

Let the sum be = P and original rate = R. Then, (Px(R+2)x3⁄100) - (PxRx3⁄100) = 360 3PR + 6P - 3PR =36000 6P = 36000 P=6000

Q 6 - 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 - 112

B - 112.50

C - 110

D - 100

Answer - B

Explanation

Gain in 2 years = Rs. [(5000x25x⁄42⁄100) - (500x4x2⁄100)] = Rs. (625 - 400) = Rs. 225 Gain in 1 year = Rs. (225⁄2) = Rs. 112.50

Q 7 - How much time for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

A - 4

B - 5

C - 4.5

D - 6

Answer - A

Explanation

Time = (100 x 81⁄450 x 4.5) = 4 years

Q 8 - A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?

A - 6

B - 6.5

C - 5

D - 6.75

Answer - A

Explanation

S.I. = Rs. (15500 - 12500) = Rs. 3000 Rate = (100 x 3000⁄12500 x 4) = 6%

Q 9 - Reema took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of loan period, what was the rate of interest?

A - 10

B - 5

C - 6

D - 7

Answer - C

Explanation

Let Rate = R% and time also R years. Then, (1200 x R x R⁄100) = 432 = 12R2 = 432 R2 = 36 R=6

Q 10 - A man took a loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed b him was?

A - 10000

B - 15000

C - 15500

D - 6500

Answer - B

Explanation

Principal = Rs. (100 x 5400⁄12 x 3) = Rs. 15000

Answer - A

Explanation

Let the present worth be Rs. z then, S.I. = Rs. (132 - z) therefore (zx5x2⁄100) = 132 - z 10z = 13200 - 100z 110z = 13200 z = 120

Answer - B

Explanation

Principal = (100 x 4016.25⁄9 x 5) (401625⁄45) = 8925

Q 13 - Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, What amount will Rs. 800 become in 3 years?

A - 10000

B - 1025

C - 15500

D - 6500

Answer - B

Explanation

S.I. = (956 - 800) = 156 Rate = (100 x 156⁄800 x 3) New Rate = (61⁄2 + 4) = 10 1⁄2 New S.I. = Rs. (800 x 21 x⁄2 3⁄100) =252 therefore New Amount = Rs. (800 + 252) = 1025

Answer - B

Explanation

We need to know the S.I., principal and time to find the rate. Since the principal is not given, so the data is inadequate.

Q 15 - In how many years, Rs. 150 will produce the simple interest @ 8% as Rs. 800 produce in 3 years @ 4(1⁄2?

A - 6

B - 8

C - 9

D - 8

Answer - C

Explanation

P = Rs. 800 R = 4 1⁄2 = (9⁄2 T = 3 years S.I. = Rs. (800 x 9 x ⁄2 3⁄100) = 108 < Now, P = Rs 150, S.I. = Rs. 108, R = 8% Time = (100 x 108⁄150 x 8) = 9 years

Q 16 - A sum invested at 5% simple interest per annum grows to Rs. 504 in 4 years. The same amount at 10% simple interest per annum in 21⁄2 years will grow to?

A - 420

B - 525

C - 450

D - 500

Answer - B

Explanation

Let the sum be Rs. z. Then, S.I. = Rs. (504 - z) therefore (z x 5 x 4⁄100) = 504 - z 20z = 50400 - 100z 120z = 50400 z = 420 Now P = 420, R = 10%, T = 5⁄2 S.I. = (420 x 10 x 5⁄100 x 2) = 105 Amount = Rs (420 + 105) = 525

Q 17 - 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 - 1:4

C - 1:3

D - none

Answer - A

Explanation

Let the principal be P and rate of interest be R% therefore Required Ratio =(PxRx6⁄100⁄PxRx9⁄100) 6PR⁄9PR =2:3

Q 18 - Nitin borrowed some money at 6% for the first three years, 9% for the next 5 years and 13% for the period beyond 8 years. If the total interest paid by him at the end of eleven years is Rs 8160, how much money did he borrow?

A - 8000

B - 10000

C - 12000

D - Data inadequate

Answer - A

Explanation

Let the sum be Z. Then, (Zx6x3⁄100) + (Zx9x5⁄100) + (Zx13x3⁄100) = 8160 = 18Z + 45Z + 39Z = (8160 x 100) 102Z = 816000 Z = 8000

Q 19 - An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is changing an interest of 10%, the effective rate of interest becomes?

A - 12

B - 15

C - 10

D - 10.25

Answer - D

Explanation

Let the sum be Rs. 100. Then, S.I. for first 6 months = Rs. (100 x 10 x 1⁄100 x 2) = Rs 5. S.I. for last 6 months = Rs. (105 x 10 x 1⁄100 x 2) = Rs 5.25 So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25 Therefore Effective rate = (110.25 - 100) = 10.25%

Answer - C

Explanation

S.I. for 1 year = Rs. (854 - 815) = 39 S.I. for 3 years = Rs. (39 x 3) = 117 Therefore Principal = 815 - 117 = 698

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