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Quantitative aptitude

aswatha (udumalpet) (90 Points)

24 May 2012

If an investment of Rs.1000 and Rs.100 yield an income of Rs.90 and Rs.20 respectively , for earning Rs.50 investment of Rs______ will be required.

a)less than Rs.500

b)over Rs.500

c)Rs.485

d)Rs.486

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2 Replies

intermediate(ipc)course (no) (1460 Points)
Replied 24 May 2012

investment is 500/- i.e. 1100*50/110= 500/-

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Tejaswi Kasturi (student-cpt) (427 Points)
Replied 25 May 2012

The usual way people solve this question is as follows

let x be the part of investment in 1000 and (1-X) be the investment in 100 so total investment is 1000*X + 100*(1-X) = 100 + 900*X now profit is 90*X + (1-X)*20 = 20 + 70*X but this is equal to 50 so 70*X = 50 - 20 = 30 implying X = 3/7 so total investement is 100 + 2700/7 = 3400/7 = 485.71 = 486 the answer would be d.

But this is not the only answer possible answer infact answer can be anything between 1300/3 and 5000/9.

the correct method of solving this problem:

let us assume that total investment is A. now let us put X part of investment in Rs.1000 and (1-X) part of investment in Rs.100

now A*X is the investment in Rs.1000 and (1-X)*A is the investment in Rs.100 Now

for Rs.1000 ----------------------- profit is Rs.90

for Rs.A*X -------------------------- profit is ??? it is 90*A*X/1000 = 9*A*X/100

for Rs.100 -------------------------- profit is Rs.20

for Rs (1-X)*A--------------------------profit is ???? it is (1-X)*A*20/100 now their sum is given to be 50

so (9*A*X+20*(1-X)*A)/100 = 50 implies A(9X +20 - 20X) = 5000 implies

A(20 - 11X) = 5000 ------------(1)

Now to get upper bound and lower bound of A we have to assume that each investment is greater than 0 less than 1000 and 100 respectively(In other words we are saying that you can't put more than 1000 in first investment and more than 100 in second investment and also that you cannot put negative investment (i.e. short selling) ) so

now we get the bounds 0<A*X<1000 , 0<A*(1-X)<100

now from equation 1 we get 11A*X = 20A-5000 so A*X = (20A-5000)/11

now (1-X)*A = A- A*X = A - (20A-5000)/11 = (5000 -9A)/11

now we have 0<(20A-5000)/11<1000 and 0<(5000-9A)11<100

first we will take the second inequality

0<(5000-9A)/11<100 multilply it by -1 and reverse the inequality signs

0>(9A-5000)/11>-100 multiply the inequality by 11

0>(9A-5000)>-1100 add 5000 to all the terms in the inequality

5000>9A>(5000-1100) implies 5000>9A>3900 divide the inequality by 9 we get 5000/9 > A > 1300/3 i.e. 555.5555>A>433.333 The other inequality when solved gives 800>A>250 so the bounds of A are A should be greater than 433.333 and less than 555.5555 A can take any value in between so all the options are correct