Answer: D Rs 10 + (Interest on Rs 10 for 11 months) = Rs 11 + (Interest on Re 1 for (10+9 +8 +...+1)) months Or, Rs 10 + interest on Rs 10 for 11 months = Rs 11 + interest of Re 1 for 55 months Or, Rs 10 + interest on Re 1 for 110 months = Rs 11 + interest of Re 1 for 55 months Or, Re 1 = Interest on Re 1 for 55 months 1 = (1*R*55)/(12*100) => R = 240/11

Q. No. 21:

Mr. Jeevan wanted to give some amount of money to his two children, so that although today they may not be using it, in the future the money would be of use to them. He divides a sum of Rs.18,750/- between his two sons of age 10 years and 13 years respectively in such a way that each would receive the same amount at 3% p.a. Compound interest when he attains the age of 30 years.What would be the original share of the younger son?

Answer: A Let the amount given to younger son be Rs. x and the amount given to older son beRs.(18750 – x). The younger son turns 30, after 20 years and the older turns 30 after 17years. As each of them will receive the same amount, we must have: x(1 + 3/100)^{20} =(18750-x)(1+ 3/100)^{17} => x= Rs 8959.80 is the share of younger son.

Q. No. 22:

Abhishek invests a principal in a simple interest based scheme, such that the rate of interest increases in a geometric progression every year with the common ratio 2. If the amount becomes four times in 4 years, then in how many years it will become 52 times?

Answer: B The amount becomes 4 times in 4 yaesr => interest = 3 times the principal P(r+2r+4r+8r) = 3P => r= 20%. Here, 52 times means interest = 51P => 51 P = p/100 * [20+40+80+............2^{n-1}*20] => n= 8 years

Q. No. 23:

Mantu deposited Rs 1000 at the start of the year 2008 in a bank offering interest 10% compounded annually. However, he used to withdraw a fixed amount at the end of the year. When Mantu withdrew that fixed amount at the end of year 2010, he discovered that no money is let in his deposit. Find that fixed amount?

Answer: B Let the fixed amount be Rs r amount at the end of 2008 = Rs [(1000*1.1) - r] Amount at the end of 2009 = Rs [(1000*1.1 - r)*1.1 - r], because interest would be calculated upon the remaining amount. Amount at the end of 2010 was zero Thus,{ [(1000*1.1 - r)*1.1 - r]*1.1 -r} =0 => 1000 = r/1.1^{3} +r/^{2} +r/1.1 => 1331 = 3.31 r => r= Rs 402.11

Q. No. 24:

At simple interest Rs 5000 becomes Rs 6000 in four years. Now, if the rate of simple interest is doubled, then in how much time from now this amount become Rs 10,000?

Answer: B Let the original rate of interest = r% Then, (6000-5000) = (5000*4*r)/100 => r = 5% New rate of SI = 2r = 10% New principal = Rs 6000 Final amount = Rs 10,000 Time required = (10000-6000)/(6000*10) * 100 = 20/3 year = 80 months.