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
Answer & Explanation Answer:- 2 Explanation :- Present height = 50 cm , Time = 2 years 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: Answer:- 3 Explanation :- Difference between the CI and SI = (R×SI)/(2×100) 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? Answer & Explanation Answer:- 1 Explanation :- Answer:- Amount after 1 years =20000 + 10% of 20000 = 20000+ 2000 = Rs. 22000 4. A sum of money becomes Rs. 2200 after three years and Rs. 4400 after six years on compound interest. The sum is Answer & Explanation Answer:- 2 Explanation :- Let the sum be P and rate of interest be R% per annum. 5. What annual payment will discharge a debt of Rs. 1025 due in 2 years at the rate of 5% compound interest? Answer & Explanation Answer:- 3 Explanation :- Present Worth = x / (1+R/100)^T
Rate of increase = (1/5)×100 = 20%
Height after 2 years = P(1+R/100)^T
=> 50(1+20/100)^2 = 50(1+1/5)^2 = 50(6/5)^2
=> (50×6×6)/(5×5) = 72 cm
1. 15%
2. 14.25%
3. 12.5%
4. 10.5%
Difference between the CI and SI = 340 – 320 = 20
Therefore, (R×SI)/(2×100) = 20R = (20*2*100) / 320
R = 12.5%
1. Rs. 102102
2. Rs. 102220
3. Rs. 104202
4. Rs. 104222
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
1. Rs. 1400
2. Rs. 1100
3. Rs. 1000
4. Rs. 1200
Amount after 3 years = 2200
P(1+R/100)^T = 2200
P(1+R/100)^3 = 2200 ——-— (1)
Amount after 6 years = 4400
P(1+R/100)^T = 4400
P(1+R/100)^6 = 4400 ——-— (2)
Dividing (2) by (1)
[P(1+R/100)^6] / [P(1+R/100)]^3 = 4400/2200 = 2
(1+R/100)^3 = 2
put this value in equation (1)
P × 2 = 2200
P = 2200/2 = Rs.1100
1. Rs. 560
2. Rs. 560.75
3. Rs. 551.25
4. Rs. 550
1025 = Present Worth of Rs. x due 1 year hence + Present Worth of Rs.x due 2 year hence
1025 = x/(1+5/100)^1 + x/(1+5/100)^2
1025 = x/(105/100 ) + x/(105/100)^2
1025 = x/(21/20) + x/(21/20)^2
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