Quantitative Aptitude - Arithmetic Ability Questions

Converting each of the given fractions into decimal form, we get

2/3 = 0.66, 3/4 = 0.75, 4/5 = 0.8, 5/6 = 0.833 

Since 0.833>0.8>0.75>0.66 

So, 5/6 > 4/5 > 3/4 > 2/3 

Therefore,  Required Difference = 5/6 - 2/3 = 1/6

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.

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

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

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

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.