**1. A sum of money becomes four times in 20 years at simple interest. Find the rate of interest.**

Then, the sum after 20 years is ‘4P’

∴ Simple interest = 4P – P = 3P

Formula for S.I = (PRT / 100)

∴ 3P= (P × R × 20) / 100

R = (3P × 100) / (20 × P) = 15 %

Answer: A

**2. In 4 year, Rs. 6000 amounts to Rs. 8000. In what time at the same rate, will Rs. 525 amounts to Rs. 700?**

S.I = A – P = 8000 – 6000 =2000; R =?

Formula for S.I = (PRT / 100)

2000 = (6000 × R × 4) / 100

∴ R = 25/3 %

Now, we calculate T , for A = 700; P= 525;

S.I = A – P = 700 – 525 = 175

S.I = (525 × 25 / 3 × T) / 100

175 = (525 × 25 × T) / (100 × T)

Hence, T = 4 years

Answer: C

**3. Swathi borrowed some money at the rate of 6% per annum for the first three years, at the rate of 9% per annum for the next five years and at the rate of 13% per annum for the period beyond eight years. If she pays a total interest of Rs.8160 at the end of eleven years, how much money did she borrow?**

Then, according to the question [(P × 6 × 3) / 100] + [(P × 9 × 5) / 100] + [(P × 13 × 3) / 100] = 8160

(18P + 45P + 39P)/100 = 8160

102P / 100 = 8160

∴ P = 8000

Hence, Swathi borrowed Rs. 8000

Answer: A

**4. If a certain sum at compound interest becomes double in 5 year, then in how many years, it will be 16 times at the same rate of interest?**

X becomes 2X in 5 years

2X becomes 4X in 10 years

4X becomes 8X in 15 years

Hence, 8X becomes 16X in 20 years

Answer: C

**5. At simple interest, a sum becomes three times in 20 years. Find the time in which the sum will be double at the same rate of interest?**

∴ S.I = 3P – P = 2P

Formula for S.I = (PRT) / 100

2P = (P × R × 20) / 100

R = 10%

To find Time (T),

∴ S.I = 2P – P = P

P = (P × 10 × T) / 100

Hence, T = 10 yrs.

Answer: B

**6. If the difference between the simple interest and compound interest on some amount at 20% pa for 3 years is 48, then what must be the principal amount?**

P = D × 1003 / R2 (300 + R)

P= 48 × 1003 / 202 (300+20)

P = (48×100×100×100) / [400 (320)]

P= 375

Answer: B

**7. A sum of money invested at compound interest amounts to Rs. 800 in 3 years and Rs.882 in 5 years. What is the rate of interest?**

A= P (1+R/100)T

Hence, P [1 + (R/100)]3 = 800 …..(1)

P [1 + (R/100)]5 = 882 ….(2)

Since, (2)/(1) = P [1 + (R/100)]5 / P [1 + (R/100)]3 = 882/800

[1 + (R/100)]2 = 441/400

[1 + (R/100)] = √(441 / 400) = 21 / 20

∴ R = (1/20) × 100 = 5%

Answer: C

**8. Simple interest for the sum of Rs.1500 is Rs.50 in 4 years and Rs.80 in 8 years. Find the rate of SI?**

According to the question, = [(1500 × R × 8) / 100] - [(1500 × R × 4) / 100] = 80 – 50 (12000R – 6000R) / 100 = 30

6000R / 100 = 30

R = 30/60 = ½ = 0.5%

Answer: A

** 9. Simple interest on an amount after 24 months at the rate of 2% per quarter is 960. What is the amount?**

Hence, P = (S.I × 100) / (R×T)

P = (960 × 100) / (8×2) = 6000

Answer: C

**10. At what rate percent per annum will a sum of money double in 8 years?**

R = (S.I × 100) / (P×T) = (x ×100) / (x × 8) = 12.5%

Answer: C

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