SI & CI

1. 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 charging an interest of 10%, the effective rate of interest becomes:

1. 13% 
2. 10.25% 
3. 15%
4. 11% 
5. None of these

2. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

1. Rs. 850 
2. Rs. 790
3. Rs. 698 
4. Rs. 800
5. Rs. 600

3. Sum of money becomes Rs. 13,380 after 3 years and Rs. 20,070 after 6 years on compound interest. The sum is:

1. Rs. 9200 
2. Rs. 9000 
3. Rs. 8920 
4. Rs. 9040 
5. Rs. 9500

4. A sum of Rs. 12,000 deposited at compound interest becomes double after 5 years. After 20 years, it will become:

1. Rs. 1,10,000 
2. Rs. 1,30,000
3. Rs. 1,24,000 
4. Rs. 1,92,000
5. Rs. 1,50,000

5. A sum of money placed at compound interest doubles itself in 5 years. It will amount to eight times itself at the same rate of interest in:

1. 7 years 
2. 12 years 
3. 15 years
4. 30 years 
5. 21 years

6. If a sum on compound interest becomes three times in 4 years, then with the same interest rate, the sum will become 27 times in:

1. 11 years 
2. 12 years 
3. 24 years
4. 38 years 
5. 21 years

7. The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

1. 7 
2. 4 
3. 5
4. 8 
5. 7

8. A man borrows Rs. 2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments. How much will each instalment be?

1. Rs.1275 
2. Rs.1383 
3. Rs.1352
4. Rs.1287 
5. Rs.1250

9. What annual payment will discharge a debt of Rs. 1025 due in 2 years at the rate of 5% compound interest?

1. Rs.650 
2. Rs.551.25 
3. Rs.560
4. Rs.660.75 
5. Rs.600

10. A man borrows Rs. 12,500 at 20% compound interest. At the end of every year he pays Rs. 2000 as part repayment. How much does he still owe after three such instalments?

1. Rs.14,000 
2. Rs.13,684
3. Rs.15,600 
4. Rs.14,320
5. None of these

Answers:-

1. 2

2. 3

Sum = A₂T₁- A₁T₂ / T₁ – T₂

Sum= 854x3 – 815x4 / 3-4

Sum = 698

3. (3) 8920

4. (4) Rs.1,92,000

1200x(1+r/100)⁵ = 24000 => (1+r/100)⁵ = 2

Therefore, [(1+r/100)⁵]⁴ = 2⁴ = 16

ð  (1+r/100)²⁰ = 16

ð  P(1+r/100)²⁰ = 16P

ð  P= 12000

Therefore, 12000(1+r/100)²⁰= 12000x16 = 192000

5. (3) 15 years.

P(1+r/100)⁵= 2P => (1+r/100)⁵ = 2

Now, P(1+r/100)ⁿ = 8P

ð  (1+r/100)ⁿ = 8 = 2³ = {(1+r/100)⁵}³

ð  (1+r/100)ⁿ = (1+r/100)¹⁵

ð  n = 15 Years.

6. (2) 12 years.

7. (2)  4

P(1+20/100)ⁿ > 2P

Or  (6/5)ⁿ > 2

Now, putting n=3

We find that (6/5)³ <2

Now, putting n = 4, we find the required criteria is satisfied.

Hence, min n=4.

8. (3) Rs. 1352

Present Worth of Rs. X due n years hence is given by :-

Present worth (PW) = x / (1+r/100)ⁿ.

Now, Let the value of each instalment be Rs.X Then,

(PW of Rs.X due 1 year hence)+ (PW of Rs.X due 2 years hence) = Rs. 2550.

ó [X / (1+4/100)] + [X / (1+4/100)²] = 2550

ó 25X/26 + 625X / 676 = 2550

ó 1275X = 2550x676

ó X = 2x676 = 1352.

9. (2) Rs. 551.25

10. (4) Rs. 14,320

Balance

= Rs.[{12500x (1+20/100)³} – {2000x(1+20/100)² + 2000x(1+20/100) +2000}]

= Rs. [(12500x(6/5)³ – (2000x36/25 + 2000x6/5 + 2000)]

= Rs. [ 21600 – (2880+ 2400 + 2000)]

= Rs. 14320.