Quantitative Aptitude - Arithmetic Ability Questions

M x T / W = Constant
where, M= Men (no. of men)
T= Time taken
W= Work load
So, here we apply
M1 x T1/ W1 = M2 x T2 / W2
Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?
Note that here, W1 = W2 = 1 road, ie. equal work load.
Clearly, substituting in the above equation we get, M2 = 14 men.

Distance covered by B in 9 sec. = (100/45)*9m = 20m

 A beats B by 20 metres.

If the speed of the faster horse be fs and that of slower horse be ss then

 fs+ss=501=50 

and       50ss-50fs=56

Now, you can go through options.

The speed of slower  horse is 20km/h

Since,   20+30=50

and  5020-5030=56    

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters

Let us Assume that another 'P' liters of water are added to the mixture to make water 25% of the new mixture. So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P)

Thus, (30 + P) = 25% of (150 + P)

Solving, we get P = 10 liters

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235  =>  ab = 35

Now, co-primes with product  35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35)  and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is  (55,77)

Hence , required number = 77