Quantitative Aptitude - Arithmetic Ability Questions

Total hours worked = 8 x 3 + 6 x 2 = 36

Total earned = 324. Hourly wage = 324 / 36 = 9

After 10 days : 150 men had food for 35 days.

Suppose 125 men had food for x days.

Now, Less men, More days (Indirect Proportion)

∴ 125 : 150 :: 35 : x    ⇔   125 x x = 150 x 35

 ⇒x=150×35125=42

He has gain = 15 - 12 = 3,
Gain% = (3/12) x 100 = (100/4) = 25.
He has 25% gain.

 Vowels are A I A I O, 

 C    A   S    T   I    G   A    T   I    O   N

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

n(E)=5P52!*2!(A,I are 2 times)*6P62!(T is 2 times)=21600

n(S)=11!2!*2!*2!(A, I are 2 times)=4989600

216004989600=0.043

To solve a pair of simultaneous equations such as those given we can add or subtract them.

Adding we get 4x + 4y = 20

Therefore 2x + 2y = 10

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) = 8 /14=4/7.

Let x liter be the per day filling and v litr be the capacity of the reservoir, then 

      90x + v = 40000 * 90     -----(1)

      60x + v= 32000 * 60     ------(2)

solving eq.(1) and (2) , we get

      x = 56000

Hence , 56000 liters per day can be used without the failure of supply.

Let the length of the stationary train Y be LY
Given that length of train X, LX = 300 m
Let the speed of Train X be V.
Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.
=> 300/V = 25 ---> ( 1 )
(300 + LY) / V = 60 ---> ( 2 )
From (1) V = 300/25 = 12 m/sec.
From (2) (300 + LY)/12 = 60
=> 300 + LY = 60 (12) = 720
=> LY = 720 - 300 = 420 m
Length of the stationary train = 420 m