Question

Harish deposited ₹ 4500 in an account that earns 3.5% interest compounded annually.what is his account balance after 2 years?​

Answer 1

Answer:

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

Q 2 - A certain sum of money amounts to Rs. 1008 in 2 years and to Rs. 1164 in 31⁄2 years. find the sum and the rate of interest?

A - Rs. 800, Rate 14%

B - Rs. 850, Rate 12%

C - Rs. 800, Rate 13%

D - Rs. 500, Rate 13%

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%

Q 3 - At what rate percent per annum will a sum of money double in 16 years?

A - 5%

B - 3%

C - 4%

D - 61⁄4%

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

Q 11 - what is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum?

A - 120

B - 150

C - 155

D - 650

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

Q 12 - A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a in 5 years. What is the sum?

A - 10000

B - 8500

C - 9000

D - 8925

Answer - B

Explanation

Principal = (100 x 4016.25⁄9 x 5)

(401625⁄45)

= 8925

Answer 2

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• Principle = Rs. 4500
• Rate = 3.5%
• Time = 2 years

_______________________

$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• Amount = ???

______________________

$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

$\sf{\red{\boxed{\bold{Amount= p( 1 + \dfrac{r}{100})^t}}}}$

⠀⠀⠀⠀

: $\sf\implies\:{\bold{ Amount = 4500( 1 + \dfrac{3.5}{100})^2 }}$

⠀⠀⠀⠀

: $\sf\implies\:{\bold{ Amount = 4500( 1 + \dfrac{35}{1000})^2 }}$

⠀⠀⠀⠀

: $\sf\implies\:{\bold{ Amount = 4500( \dfrac{1000+ 35}{1000})^2 }}$

⠀⠀

: $\sf\implies\:{\bold{ Amount = 4500(\dfrac{1035}{1000})^2}}$

⠀⠀⠀⠀

: $\sf\implies\:{\bold{ Amount = 4500\times\dfrac{1035\times 1035}{1000\times 1000} }}$

⠀⠀⠀⠀

: $\sf\implies\:{\bold{ Amount = 45\times\dfrac{1035\times 1035}{10000} }}$

⠀⠀⠀⠀

: $\sf\implies\:{\bold{ Amount = 45\times \dfrac{1071225}{10000} }}$

⠀⠀⠀⠀

: $\sf\implies\:{\bold{ Amount = \dfrac{48205125}{10000} }}$

⠀⠀⠀⠀

: $\sf\implies\:{\bold{ Amount = Rs. 4820.5125}}$

⠀⠀⠀⠀

: $\sf\implies\:{\bold{ Amount = Rs. 4820.51 }}$

⠀⠀⠀⠀

⠀⠀⠀⠀

______________________

$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• Harish's account's balance after 2 years will be Rs. 4820.51