Quantitative Aptitude - Arithmetic Ability Questions

Do not assume that AB and C are on a straight line. Make a diagram with A and B marked 5 miles apart. Draw a circle centered on B, with radius 6. C could be anywhere on this circle. The minimum distance will be 1, and maximum 11

Number of shares = 44558.25= 540.

Face value = Rs. (540 x 10) = Rs. 5400.

Annual income = Rs.12100*5400 = Rs. 648.

1/10 + 1/15 = 1/6 x 3 = 1/2

1 - 1/2 = 1/2

1/10 + 1/15 - 1/x = 1/2

x = 8.57 min

T.D = (B.G * 100) / (Rate * Time)

(24*100) / (10 * 2)

= 120.

P.W = (100 *T.D) / (Rate * Time)

= (100 * 120) /(10 * 2)

= 600

Given 4men, 12 women and 20 children work for  2 days.

Workdone for 2 days by 4men, 12 women and 20 children = 4 x 26 x 12 + 12 x 28 x 18 + 20 x 218 x 10 = 12

Therefore, remaining work = 1 - 12 = 12

To complete the same work by only men in 1 day,

We know that M1 x D1 = M2 x D2

Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1

12 x 6 = M x 1

=> M = 12 x 6 = 72

=> But the remaining work = 1/2

Men required => 1/2 x 72 = 36

Only men required to Complete the remaining work in 1 day = 36.

Let the numbers be x, x + 2 and X + 4.

Then, x(x + 2)(x +4) / 8 = 720 => x(x + 2)(x +4) = 5760.

√x  *  √(x + 2) * √(x +4) = √x(x +2)(x + 4 ) = √5760  =  24√10.

Let the length of the train be L metres

Speed of first train = 360018xL m/hour

Speed of secxond train = 360012xL m/hour

When running in opposite directions, relative speed = 200 L + 300 L m/hour

Distance to be covered = L + L = 2L metre

Time taken = 2L500Lx3600 sec

                 =14.4 sec

Let the original price of rice be Rs. x

Let rice a man can buy for Rs. 500 at rs. x/kg be = R kgs

From given data, for Rs. 500

x ---- R kgs

(x + 25x/100) ---- (R - 5) kgs

=> Rx = (R - 5)(x + 25x/100)

=> Rx = (R - 5)(125x/100)

=> 100Rx = 125Rx - 625x

=> R = 25

So at rate of Rs. x/kg, man get 25 kgs for Rs. 500

=> x = 500/25 = Rs. 20