Quantitative Aptitude - Arithmetic Ability Questions

Given total number of cows = k
Now, 1st son share = k/2
2nd son share = k/4
3rd son share = k/5
4th son share = 9

(k) + (k/4) + (k/5) + 9 = k
=> k - (19k/20) = 9
=> (20k-19k)/20 = 9
=> k = 180.

From the given data,

The part of honey in the first mixture = 1/4

The part of honey in the second mixture = 3/4

Let the part of honey in the third mixture = x

Then,

1/4             3/4

           x

(3/4)-x     x-(1/4) 

Given from mixtures 1 & 2 the ratio of mixture taken out is 2 : 3

=> 34-xx-14=23

=> Solving we get the part of honey in the third mixture as 11/20

=> the remaining part of the mixture is water = 9/20

Hence, the ratio of the mixture of honey and water in the third mixture is 11 : 9 .

Let the first number be x and the second number be y.

Given

5% of x = (25% of y) + y

=> 5x/100 = 5y/4

=> x = 25y ......(1)

and also given that, x - y = 96

=> x = 96 + y .....(2)

From (1)&(2)

25y = 96 + y

=> 24y = 96

=> y = 4

From (1)

x = 96 + 4 = 100

Hence, the numbers are 100 & 4.

Let the length be 'l' and breadth be 'b'.

b = l × 3/4__________(a)

2(l+b) = 1050

l+b = 525___________(b)

From equations (a) and (b),

l = 300m, b = 225 m

Area = l × b

= 300 × 225

= 67500 sq.m.

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7

Smallest number of 6 digits is 100000

on dividing 100000 by 111 we get 100 as remainder 

 Number to be added = (111 -100) = 11

 Required Number  = 100011

Days remaining 124 – 64 = 60 days

Remaining work = 1 - 2/3 = 1/3

Let men required men for working remaining days be 'm'

So men required = (120 x 64)/2 = (m x 60)/1 => m = 64

Men discharge = 120 – 64 = 56 men.