Quantitative Aptitude - Arithmetic Ability Questions

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.

In a simultaneous throw of two dice, n(S) = 6 x 6 = 36

Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

 P(E)=n(E)n(S)=636=16

We have to rearrange the equation to make R the subject.

Start by cross multiplying by (r + R); V (r + R) = 12R

Multiply out the bracket Vr + VR = 12R

A                   B                 C

3x                 4x                5x 

(3x+x)          2x               (5x+x)         (B Gives 1/4 to A and 1/4 to C) 

=4x              2x                 6x 

(4x+x)          2x                 5x              ( C Gives 1/6 to A) 

=5x              2x                 5x

Therefore, 5 : 2 : 5

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.

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.

Let x and y be the two parts of 96.

x/7 = y/9 => x:y = 7:9

=> The smallest part is = 7/16 x 96 = 42.

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.