Quantitative Aptitude - Arithmetic Ability Questions

Given 2000 ---- 54 days

The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days.

i.e, 2000 ----- 39 days == m ---- 20 days

=> m x 20 = 2000 x 39

m = 3900

So total men for 20 days is 3900

=> 2000 old and 1900 new reinforcement.

Hence, reinforcement = 1900.

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.

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

Number of shares = 44558.25= 540.

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

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

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

Angle of triangle = 180 × 3/12

= 45 deg

Other angles of the triangle are = 60 deg, 75 deg

Angle of quadrilateral

= 360 - (45 + 60 + 75)

= 180 deg

Therefore, the four angles of the quadrilateral are 45, 60, 75 & 180 degrees.

Hence, the sum of largest and smallest angle of the quadrilateral are 180 + 45 = 225 deg.

We know that perimeter of rectangle = 2(l+b) = 2(8+7)= 30cm
Given perimeter of square is twice the perimeter of rectangle = 2(30) = 60cm
Therefore, side of the square is = 1/4 x 60 = 15cm
Circumference of the required semicircle = πr + 2r = 227× 152 + 2×152 = 23.57 + 15 = 38.57cm.