Quantitative Aptitude - Arithmetic Ability Questions

From the given data,

=> (2 M + 3W) 8 = (3M + 2W)7

=> 16M + 24W = 21M + 14 W

=> 10W = 5M

=> 2W = M

=> 14W × ? = 7W × 8

? = 4 days

Let 'd' be the distance and 's' be the speed and 't' be the time
d=sxt
45 mins = 3/4 hr and 48 mins = 4/5 hr
As distance is same in both cases;
s(3/4) = (s-5)(4/5)
3s/4 = (4s-20)/5
15s = 16s-80
s = 80 km.
=> d = 80 x 3/4 = 60 kms.

Amount of milk left after 3 operations

401-4403=40*910*910*910=29.16 litres

Let the mother's present age be x years.

Then, the person's present age = 2/5 x years.

(2x/5 + 8) = 1/2 (x + 8)

2(2x + 40) = 5(x + 8) => x = 40.

$$\begin{gathered} 23x35x34x?=3174 \\ 3174x10/3 \\ = 10580 \end{gathered}$$

The CP of profitable cow  = 9900/1.1 = 9000

and profit = Rs. 900

The  CP of loss yielding cow = 9900/0.8 = 12375

and loss = Rs. 2475

so, the net loss = 2475 - 900 = 1575

m+nm-n=7xx ⇒mn=4x3x  

Again   mn=12x2 

and         mn =60x  

so, 60x=12x2 ⇒x=5    

=>  m= 20  and    n= 15

Hence,    1m:1n=120:115=3:4

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216−125=91

The required probability = 91/256