Quantitative Aptitude
AREA MCQs
Areas
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) cm2.
So, Required number of tiles = \(\left(\frac{1517\times902}{41\times41}\right)= 814.\)
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
So, Area = (l x b) = (63 x 40) m2 = 2520 m2.
Let original length = x and original breadth = y.
Original area = xy.
New length = \( \frac{x}{2}\)
New breadth = 3y.
New area = \(\left(\frac{1}{2}x\times3y\right)= \frac{3}{2}xy\)
So, Increase % = \(\left(\frac{1}{2}xy\times\frac{1}{xy}\times100\right)\) % = 50%.
Let breadth = x metres.
Then, length = (x + 20) metres.
Perimeter = \(\left(\frac{5300}{26.50}\right)m.=200m.\)
So, 2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
So, Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
Area to be plastered = [2(l + b) x h] + (l x b)
= {[2(25 + 12) x 6] + (25 x 12)} m2
= (444 + 300) m2
= 744 m2.
So, Cost of plastering = Rs. \(\left(744\times\frac{75}{100}\right)=Rs. 558.\)
 - Perimeter = Distance covered in 8 min. = 12000 x 8 m = 1600 m.
 - 100 cm is read as 102 cm.
A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
Percentage error = 404 x 100 % = 4.04% 100 x 100
 - 2(l + b) = 5 b 1
2l + 2b = 5b
3b = 2l
b = 2 l 3
Then, Area = 216 cm2
l x b = 216
l x 2 l = 216 3
l2 = 324
l = 18 cm
 - Let original length = x metres and original breadth = y metres.
Original area = (ab) m2.
New length = 120 a m = 6 b m. 100 5
New breadth = 120 b m = 6 b m. 100 5
New Area = 6 a x 6 b m2 = 36 ab m2. 5 5 25
The difference between the original area = ab and new-area 36/25 ab is
= (36/25)ab - ab
= ab (36/25 - 1)
= ab (11/25) or (11/25) ab
Increase % = 11 ab x 1 x 100 % = 44%. 25 ab