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- 11. The difference between the compound interest and simple interest accrued on an amount at the end of three years at the rate of 12% is Rs.381.888. What is the amount?

A.Rs. 9500

B.Rs. 8400

C.Rs.8000

D.Rs.8200

Answer & Explanation

Answer: Option

Explanation:**C.I = P[(1+R/100)**

= P[(1+12/100)

= P[(21952/15625) - 1]

= 6327P/15625

S.I = (P x 3 x 12)/100 = Rs. 9P/25

6327P/15625 - 9P/25 = 381.888

=> 702P/15625 = 381.888

=> P = (381.888 x 15625)/702

= Rs. 8500

^{T}- 1]= P[(1+12/100)

^{3}- 1] = P[(28/25)^{3}- 1]= P[(21952/15625) - 1]

= 6327P/15625

S.I = (P x 3 x 12)/100 = Rs. 9P/25

6327P/15625 - 9P/25 = 381.888

=> 702P/15625 = 381.888

=> P = (381.888 x 15625)/702

= Rs. 8500

- 12. A sum is invested for 3 years compounded at 5%, 10% and 20 % respectively. In three years, if the sum amounts to Rs. 1386, then find the sum.

A. Rs. 1500

B.Rs. 1000

C.Rs. 1200

D.Rs. 1400

Answer & Explanation

Answer: Option B

Explanation:**1386 = P(1+5/100)(1+10/100)(1+20/100)**

1386 = P(21/20)(11/10)(6/5)

P = (1386 × 20 × 10 × 5)/(21 × 11 × 6) = (66 × 20 × 10 × 5)/(11 × 6 = 20 × 10 × 5) = Rs. 1000

i.e., the sum is Rs.1000

1386 = P(21/20)(11/10)(6/5)

P = (1386 × 20 × 10 × 5)/(21 × 11 × 6) = (66 × 20 × 10 × 5)/(11 × 6 = 20 × 10 × 5) = Rs. 1000

i.e., the sum is Rs.1000

- 13. Divide Rs. 3364 between A and B, so that A's Share at the end of 5 years may equal to B's share at the end of 7 years, compound interest being at 5 percent.

A.Rs. 1764 and Rs.1600

B.Rs. 1756 and Rs.1608

C.Rs. 1722 and Rs.1642

D.None of these

Answer & Explanation

Answer: Option A

Explanation:**A's share after 5 years = B's share after 7 years**

(A's present share)(1+5100)

=>(A's present share)/(B's present share) = (1+5/100)

i.e, A's present share : B's present share = 441 : 400

Since the total present amount is Rs.3364, A's share = 3364 × 441/(441 + 400)

= (3364 × 441)/841 = 4 × 441 = Rs. 1764

B's present share = 3364 - 1764 = Rs.1600

(A's present share)(1+5100)

^{5}= (B's present share)(1+5100)^{7}=>(A's present share)/(B's present share) = (1+5/100)

^{7}/(1+5/100)^{5}= (1+5/100)^{(7−5)}= (1+5/100)^{2}=(21/20)^{2}=441/400i.e, A's present share : B's present share = 441 : 400

Since the total present amount is Rs.3364, A's share = 3364 × 441/(441 + 400)

= (3364 × 441)/841 = 4 × 441 = Rs. 1764

B's present share = 3364 - 1764 = Rs.1600

- 14. If a sum on compound interest becomes three times in 4 years, then with the same interest rate, the sum will become 81 times in:

A.12 years

B.18 years

C.16 years

D.14 years

Answer & Explanation

Answer: Option C

Explanation:**The sum P becomes 3P in 4 years on compound interest**

3P = P(1+R/100)

⇒ 3 = (1+R/100)

Let the sum P becomes 81P in n years

81P = P(1+R/100)

⇒ 81 = (1+R/100)

⇒ (3)4 = (1+R/100)

⇒ ((1+R/100)4)

⇒ (1+R/100)

n=16

i.e, the sum will become 81 times in 16 years

3P = P(1+R/100)

^{4}⇒ 3 = (1+R/100)

^{4}Let the sum P becomes 81P in n years

81P = P(1+R/100)

^{n}⇒ 81 = (1+R/100)

^{n}⇒ (3)4 = (1+R/100)

^{n}⇒ ((1+R/100)4)

^{4}= (1+R/100)^{n}⇒ (1+R/100)

^{16}= (1+R/100)^{n}n=16

i.e, the sum will become 81 times in 16 years

- 15. Arun borrowed a certain sum from Manish at a certain rate of simple interest for 2 years. He lent this sum to Sunil at the same rate of interest compounded annually for the same period. At the end of two years, he received Rs. 2400 as compound interest but paid Rs. 2000 only as simple interest. Find the rate of interest.

A.40%

B.30%

C.20%

D.10%

Answer & Explanation

Answer: Option A

Explanation:**Let the sum be x**

Simple interest on x for 2 years = Rs.2000

Simple interest = PRT/100

2000 = x Ã— R Ã— 2100

â‡’ x R = 100000 --- (1)

Compound Interest on x for 2 years = 2400

P(1+R/100)

x (1+R/100)

x (1 +2R/100+R210000)âˆ’x=2400x(2R100+R210000)=2400

2xR/100 + xR2/10000 = 2400--- (2)

Substituting the value of xR from (1) in (2) ,we get

(2Ã—100000)/100 + 100000 Ã— R10000 = 2400

2000+10R = 2400

10R=400

R=40%

Simple interest on x for 2 years = Rs.2000

Simple interest = PRT/100

2000 = x Ã— R Ã— 2100

â‡’ x R = 100000 --- (1)

Compound Interest on x for 2 years = 2400

P(1+R/100)

^{T}âˆ’ P = 2400x (1+R/100)

^{2}âˆ’ x = 2400x (1 +2R/100+R210000)âˆ’x=2400x(2R100+R210000)=2400

2xR/100 + xR2/10000 = 2400--- (2)

Substituting the value of xR from (1) in (2) ,we get

(2Ã—100000)/100 + 100000 Ã— R10000 = 2400

2000+10R = 2400

10R=400

R=40%

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