S.I. & C.I. (19)

1. A tree increases annually by 1⁄5 th of its height. If its height today is 50 cm, what will be the height after 2 years?
1. 64 cm
2. 72 cm
3. 66 cm
4. 84 cm

Explanation :-

Present height = 50 cm ,    Time = 2 years
Rate of increase = (1/5)×100 20%
Height after 2 years = P(1+R/100)^T
=>  50(1+20/100)^50(1+1/5)^50(6/5)^
=>  (50×6×6)/(5×5) 72 cm

2. On a sum of money, the simple interest for 2 years is Rs. 320, while the compound interest is Rs. 340, the rate of interest being the same in both the cases. The rate of interest is:
1. 15%
2. 14.25%
3. 12.5%
4. 10.5%

Explanation :-

Difference between the CI and SI = (R×SI)/(2×100)
Difference between the CI and SI = 340 – 320 = 20
Therefore,  (R×SI)/(2×100) = 20R = (20*2*100) / 320
R = 12.5%

3. A bank offers 10% interest rate compounded annually. A person deposits Rs. 20,000 every year in his account. If he does not withdraw any amount, then how much balance will his account show after four years?
1. Rs. 102102
2. Rs. 102220
3. Rs. 104202
4. Rs. 104222

Explanation :-

Answer:- Amount after 1 years =20000 + 10% of 20000 = 20000+ 2000 = Rs. 22000
Amount after 2 years = 20000 + 22000 + 10% of 22000 = Rs. 46200
Amount after 3 years = 20000 + 46200+ 10% of 46200 = Rs.72820
Amount after 4 years = 20000 +.72820 + 10% of 72820 = Rs.102102

4. A sum of money becomes Rs. 2200 after three years and Rs. 4400 after six years on compound interest. The sum is
1. Rs. 1400
2. Rs. 1100
3. Rs. 1000
4. Rs. 1200

Explanation :-

Let the sum be P and rate of interest be R% per annum.
Amount after 3 years = 2200
P(1+R/100)^2200
P(1+R/100)^2200 ——-— (1)
Amount after 6 years = 4400
P(1+R/100)^4400
P(1+R/100)^4400 ——-— (2)
Dividing (2) by (1)
[P(1+R/100)^6] / [P(1+R/100)]^4400/2200 2
(1+R/100)^2
put this value in equation (1)
× 2200
2200/= Rs.1100

5. What annual payment will discharge a debt of Rs. 1025 due in 2 years at the rate of 5% compound interest?
1. Rs. 560
2. Rs. 560.75
3. Rs. 551.25
4. Rs. 550

Explanation :-

Present Worth = x / (1+R/100)^T
1025 = Present Worth of Rs. x due 1 year hence + Present Worth of Rs.x due 2 year hence
1025 = x/(1+5/100)^+ x/(1+5/100)^2
1025 = x/(105/100 + x/(105/100)^2
1025 = x/(21/20+ x/(21/20)^
1025 = [(x × 20)/21] + [(x × 20 × 20)/(21×21)]
1025 = (20x/21) + (400x/441 )
820x/441 1025
x=(1025 × 441) / 820 = (205×441)/164 Rs. 551.25