Quantitative Aptitude - Arithmetic Ability Questions

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 side of the square plot be 'a' ft.
Given area of the plot (a x a) = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 x 58 = Rs. 3944.

Let us say x boys and x girls joined the group.

(64 + x)/(40 + x) = 4/3

192 + 3x = 160 + 4x => x = 32

Number of members in the group = 64 + x + 40 + x

= 104 + 2x = 168.

Let the numbers be x and y. Then, 2x + 3y = 39...(i)and 3x + 2y = 30 ...(ii)

On solving (i) and (ii), we get :  x = 6 and y = 9.

larger number = 9

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.

when folded along the breadth
we have 2(l/2 +b) = 34 or l+2b = 34...........(1)
when folded along the length, we have 2(l+b/2)=38 or 2l+b =38.....(2)
from 1 &2 we get l=14 and b=10
Area of the paper = 14*10 = 140 sq cm

Amount = Rs.8000*1+51002=Rs.8820

surface area = [2(10*4 + 4*3 + 10*3)]sq.cm = 164 sq.cm