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 Area
Aptitude Topics
 Height and Distance
 Simple Interest
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 Volume and Surface Area
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 Problems on H.C.F and L.C.M
 Simplification
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 Chain Rule
 Boat and Streams
 Logarithm
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 True Discount
 Odd Man Out and Series
 Time and Distance
 Time and Work
 Compound Interest
 Partnership
 Problems on Ages
 Clock
 Area
 Permutation and Combination
 Problems on Numbers
 Decimal Fraction
 Square Root and Cube Root
 Ratio and Proportion
 Pipes and Cistern
 Alligation or Mixture
 Races and Games
 Probability
 Banker's Discount
Area
Answer: Option B
Explanation:
Perimeter = Distance covered in 8 min. =  12000  x 8  m = 1600 m.  
60 
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) m^{2} = 153600 m^{2}.
Answer: Option D
Explanation:
100 cm is read as 102 cm.
A_{1} = (100 x 100) cm^{2} and A_{2} (102 x 102) cm^{2}.
(A_{2}  A_{1}) = [(102)^{2}  (100)^{2}]
= (102 + 100) x (102  100)
= 404 cm^{2}.
Percentage error =  404  x 100  %  = 4.04%  
100 x 100 
Answer: Option B
Explanation:
2(l + b)  =  5 
b  1 
2l + 2b = 5b
3b = 2l
b =  2  l 
3 
Then, Area = 216 cm^{2}
l x b = 216
l x  2  l  = 216 
3 
l^{2} = 324
l = 18 cm.
Answer: Option C
Explanation:
Let original length = x metres and original breadth = y metres.
Original area = (xy) m^{2}.
New length =  120  x  m  =  6  x  m.  
100  5 
New breadth =  120  y  m  =  6  y  m.  
100  5 
New Area =  6  x x  6  y  m^{2}  =  36  xy  m^{2}.  
5  5  25 
The difference between the original area = xy and newarea 36/25 xy is
= (36/25)xy  xy
= xy(36/25  1)
= xy(11/25) or (11/25)xy
Increase % =  11  xy x  1  x 100  %  = 44%.  
25  xy 
Answer: Option B
Explanation:
Area of the park = (60 x 40) m^{2} = 2400 m^{2}.
Area of the lawn = 2109 m^{2}.
Area of the crossroads = (2400  2109) m^{2} = 291 m^{2}.
Let the width of the road be x metres. Then,
60x + 40x  x^{2} = 291
x^{2}  100x + 291 = 0
(x  97)(x  3) = 0
x = 3.
Answer: Option C
Explanation:
Other side  =√[ 

]ft  
=√[ 

]ft  
=√[ 

]ft  
=  6 ft. 
Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.
Answer: Option D
Explanation:
Let original length = x and original breadth = y.
Decrease in area 





Decrease % =  7  xy x  1  x 100  %  = 28%.  
25  xy 
Answer: Option C
Explanation:
Let the side of the square(ABCD) be x metres.
Then, AB + BC = 2x metres.
AC = √2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.
Saving % =  0.59x  x 100  %  = 30% (approx.)  
2x 
Answer: Option B
Explanation:
√(l^{2} + b^{2} )= √41.
Also, lb = 20.
(l + b)^{2} = (l^{2} + b^{2}) + 2lb = 41 + 40 = 81
(l + b) = 9.
Perimeter = 2(l + b) = 18 cm.
Answer: Option A
Explanation:
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) cm^{2}.
Required number of tiles =  1517 x 902  = 814.  
41 x 41 
Answer: Option D
Explanation:
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.
Area = (l x b) = (63 x 40) m^{2} = 2520 m^{2}.
Answer: Option B
Explanation:
Let original length = x and original breadth = y.
Original area = xy.
New length =  x  . 
2 
New breadth = 3y.
New area =  x  x 3y  =  3  xy.  
2  2 
Increase % =  1  xy x  1  x 100  %  = 50%.  
2  xy 
Answer: Option E
Explanation:
Let breadth = x metres.
Then, length = (x + 20) metres.
Perimeter =  5300  m = 200 m.  
26.50 
2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.
Answer: Option D
Explanation:
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
Answer: Option C
Explanation:
Area to be plastered  = [2(l + b) x h] + (l x b) 
= {[2(25 + 12) x 6] + (25 x 12)} m^{2}  
= (444 + 300) m^{2}  
= 744 m^{2}. 
Cost of plastering = Rs.  744 x  75  = Rs. 558.  
100 