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Total Questions : 2556 | Page 7 of 256 pages
Question 61.

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

  1.    814
  2.    820
  3.    840
  4.    844
 Discuss Question
Answer: Option A. -> 814

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 =   \(\left(\frac{1517\times902}{41\times41}\right)= 814.\)

Question 62.

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

  1.    1520 m2
  2.    2420 m2
  3.    2480 m2
  4.    2520 m2
 Discuss Question
Answer: Option D. -> 2520 m2

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.

Question 63.

The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

  1.    25% increase
  2.    50% increase
  3.    50% decrease
  4.    75% decrease
 Discuss Question
Answer: Option B. -> 50% increase

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%.

Question 64.

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

  1.    40
  2.    50
  3.    120
  4.    Data inadequate
  5.    None of these
 Discuss Question
Answer: Option C. -> 120

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.

Question 65.

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

  1.    34
  2.    40
  3.    68
  4.    88
 Discuss Question
Answer: Option D. -> 88

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.

Question 66.

 tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

  1.    Rs. 456
  2.    Rs. 458
  3.    Rs. 558
  4.    Rs. 568
 Discuss Question
Answer: Option C. -> Rs. 558

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. \(\left(744\times\frac{75}{100}\right)=Rs. 558.\)

Question 67.

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

  1.    15360
  2.    153600
  3.    30720
  4.    307200
  5.    None of these
 Discuss Question
Answer: Option B. -> 153600
 -   Perimeter = Distance covered in 8 min. =   12000 x 8 m = 1600 m.  
Question 68.

An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

  1.    2%
  2.    2.02%
  3.    4%
  4.    4.04%
  5.    None of these
 Discuss Question
Answer: Option D. -> 4.04%
 -  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
Question 69.

The ratio between the perimeter and the breadth of a rectangle is 5:1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

  1.    16 cm
  2.    18 cm
  3.    24 cm
  4.    Data inadequate
  5.    None of these
 Discuss Question
Answer: Option B. -> 18 cm
 -   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
Question 70.

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

  1.    40%
  2.    42%
  3.    44%
  4.    46%
  5.    None of these
 Discuss Question
Answer: Option C. -> 44%
 -  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

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