Quantitative Aptitude - Arithmetic Ability Questions

By car 240 km at 60 kmph
Time taken = 240/60 = 4 hr.

By train 240 km at 60 kmph
Time taken = 400/100 = 4 hr.

By bus 240 km at 60 kmph
Time taken = 200/50 = 4 hr.

So total time = 4 + 4 + 4 = 12 hr.
and total speed = 240+400+200 = 840 km
Average speed of the whole journey = 840/12 = 70 kmph.

Let A worked for x days.

x/21 + 4/28 = 1  => x = 18

A worked for 18 days. So, A can complete the remaining work in 18 - 4 = 14 days.

One day work of 6 boys and 8 girls is given as 6b + 8g = 1/10 -------->(I)

One day work of 26 boys and 48 women is given as 26b + 48w = 1/2 -------->(II)

Divide both sides by 2 in (I) and then multiply both sides by 5

Now we get, 15b + 20g = 1/4.

Therefore, 15 boys and 20 girls can do the same work in 4 days.

only (c) is correct since it is divisible by 4.

Let the original number of element  be x then the new no.of elements will be 

            4x5 = K                               

So K must be divisible by 4

Since,      x =  Kx5/4

Let C.P.= Rs. 100.
Then, Profit = Rs. 320,
S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.

Profit = Rs. (420 - 125) = Rs. 295

Required percentage = (295/420) * 100
= 70%(approx)

Total number of elementary events = 10C5

Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is 7C5.

So,required probability =7C5/ 10C5 = 1/12.

Let the present ages of A and B be 'x' and 'y' respectively

From the given data, 

[(x-2) + (y-2)]/2 = 26

=> x+y = 56

But given the age of A, 5 years hence is 40 yrs => present age of A = 40 - 5 = 35 yrs

=> x = 35 => y = 56 - 35 = 21

=> Age of B = 21 yrs

Given B is 5 years younger to C,

=> Age of C = 21 + 5 = 26 yrs

=> Required Difference between ages of A and C = 35 - 26 = 9 yrs.