Quantitative Aptitude - Arithmetic Ability Questions

Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms.

As speed is same, area of cross sections should be same. 

Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12 

i.e πR2 = 6πR12   

Here R = R and R1 = 12/2 = 6   

⇒R2 = 6×6×6   

R = 14.696   

=> D = X = 14.696 * 2 = 29.3938 cm.

Total number of persons = 9

Out of 9 persons 4 persons can be selected in 9C4 ways =126

1 man,1 woman and 2 children can be selected in 3C1*2C1*4C1 ways =36

So,required probability = 36/126 =2/7

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 

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.

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.