Aptitude and Reasoning Questions

Let first part is 'K' and second part is 'L'
then K + L = 50000---------eq1
Total profit = profit on x + profit on y
7000 = (K x 10 x 1)/100 + (L x 20 x 1)/100
70000 = K + 2L-----------------------------------eq2
70000 = 50000 + L
so L = 20000 then K = 50000 - 20000 = 30000
first part on 10% is = Rs.30000

a^3 + b^3 = (a+ b) * (a^2 + b^2 - ab)

[ (10.3)^3 + (1)^3 ] / [(10.3)^2  + (1)^2 - (1 * 10.3)]

 => a + b = 10.3 + 1 = 11.3 

In the real numbers, zero does not have a reciprocal because no real number multiplied by 0 produces 1 (the product of any number with zero is zero).

Hence, 0 doesn't have any multiplicative inverse.

At first glance it might seem that this problem cannot be solved because we do not have enough
information. It can be solved as long as you double whatever amount you start with. If we start with
$100, then P = $100 and FV = $200.

FV=P(1+r/n)^nt

Given initial investments ratio = 2 : 3 : 4

At the end of 6 months, A invested an amount such that his total capital became equal to B's initial capital investment

i.e, upto 6 months A's investment is 2 and after 6 months his invstment is 3 = B's investment

Now, Ratio of investment for one year

=> A : B : C = (2×6 + 3×6) : (3×12) : (4×12)

= 30 : 36 : 48

= 5 : 6 : 8

But given B's profit = 3000

=> 6 ratio = 3000

For total => 19 ratio = Rs. 9500.

1st supporter recieve 440 calls
2nd supporter recieve 360 calls
3rd supporter recieve 300 calls

So total calls = 1100 calls ;
Calls this month= 1500
So remaining calls to be distributed is 400

So Now Ratio 1st:2nd:3rd ==> 440:360:300
=> 22:18:15

Now No. of More Calls 2nd supporter will get => [18/(22+18+15)] x 400
=> (18/55) x 400
=> 131 Calls
So 131 more Calls than last month.

Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F of 551 and 1073 = 29;
First number = 551/29 = 19
Third number = 1073/29 = 37.
Required sum = 19 + 29 + 37 = 85.