- 1. A person took some amount with some interest for 2 years, but increase the interest for 1%, he paid Rs.120/- extra, then how much amount he took?

A.Rs.5500/-

B.Rs.6000/-

C.Rs.4000/-

D.Rs.7000/-

Answer & Explanation

Answer: Option B

Explanation:2 years = Rs.120/-

year = 120/2

Rate of Interest = 1%

100/1% × 120/2 = Rs.6000/-

P = Rs.6000/-

- 2. The rate of Interest on a sum of money is 2% p.a. for the first 3 years, 4% p.a. for the next 4 years, and 5% for the period beyond 7 years. If the S.I, Occured on the sum for the total period of 8 years is Rs. 540/-, the sum is

A.2,200

B.2,000

C.2,100

D.2,250

Answer & Explanation

Answer: Option B

Explanation:**I**

I

I

3P/50 + 4P/25 + P/20 = 540

The L.C.M of 50, 25, 20 = 100

(6P + 16P + 5P)/100 = 540

27P/100 = 540

27P = 54000

P = 54000/27

P = 2000

_{1}= (P x 3 x 2)/100 = 3P/50I

_{2}= (P x 4 x 4)/100 = 4P/25I

_{3}= (P x 1 x 5)/100 = P/203P/50 + 4P/25 + P/20 = 540

The L.C.M of 50, 25, 20 = 100

(6P + 16P + 5P)/100 = 540

27P/100 = 540

27P = 54000

P = 54000/27

P = 2000

- 3. A sum of 1200 lent at S.I at 10% will become twice in how many years?

A.12 years

B.14 years

C.10 years

D.8 years

Answer & Explanation

Answer: Option C

Explanation:**To become twice means S.I should be as much as initial sum i.e, 1200**

(1200 x 10 x T)/100 = 1200

T = 10 years

(1200 x 10 x T)/100 = 1200

T = 10 years

- 4. At what rate p.a a sum of Rs. 2400 will become Rs. 3600/- in 5 years.

A.10%

B.15%

C.20%

D.12%

Answer & Explanation

Answer: Option A

Explanation:**principal P = 2400/-**

**Rate of interest R = ?**

**Time T = 5 years**

**Accumulated Amount A = Rs.10400/-**

**A = P + I**

**A = P + PTR/100**

**A = P(1+ TR/100)**

2400[ 1 + (5 x R)/100] = 3600

2[(20 + R)/20] = 3

20 + R = 30

R = 10%

2400[ 1 + (5 x R)/100] = 3600

2[(20 + R)/20] = 3

20 + R = 30

R = 10%

- 5. A sum when lent at S.I 3% per annum amount to Rs.840/- after 4 years

A.

B.

C.

D.

Answer & Explanation

Answer: Option

Explanation:**P = ?**

T = 4 years

R = 3%

S.I + P = 840

P(1 + (TR/100) = S.I + P

P[1 + (4 x 3)/100] = 840

P[(25 + 3)/25] = 840

P = 840 x (25/28)

P = 750

T = 4 years

R = 3%

S.I + P = 840

P(1 + (TR/100) = S.I + P

P[1 + (4 x 3)/100] = 840

P[(25 + 3)/25] = 840

P = 840 x (25/28)

P = 750

**To whom this Simple Interest Questions and Answers section is beneficial?**

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