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Q:

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?

A) 22	B) 44
C) 66	D) 88

Answer & Explanation Answer: D) 88

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.

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Q:

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?

A) 40	B) 20
C) 30	D) none of these

Answer & Explanation Answer: D) none of these

Explanation:

Let breadth = x metres.
Then, length = (x + 20) metres.

Perimeter =5300/23.50

2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

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Q:

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

A) 2520	B) 2420
C) 2320	D) 2620

Answer & Explanation Answer: A) 2520

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) = 2520 sq.m

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Q:

What is the least number of squares tiles required to pave the floor of a room 15.17 m long and 9.02 m broad?

A) 814	B) 714
C) 614	D) 713

Answer & Explanation Answer: A) 814

Explanation:

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = $(41 \times 41) c m^{2}$

Required number of tiles = $\frac{1517 \times 902}{41 \times 41}$ = 814

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Q:

The diagonal of a rectangle is $\sqrt{41}$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

A) 18	B) 28
C) 38	D) 48

Answer & Explanation Answer: A) 18

Explanation:

$\sqrt{(l^{2} + b^{2})} = \sqrt{41}$ (or) $l^{2} + b^{2} = 41$

Also, $l b = 20$

${(l + b)}^{2} = l^{2 +} b^{2} + 2 l b$ = 41 + 40 = 81

(l + b) = 9.

Perimeter = 2(l + b) = 18 cm.

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Q:

A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:

A) 18%	B) 28%
C) 38%	D) 48%

Answer & Explanation Answer: B) 28%

Explanation:

Let original length = x and original breadth = y.

Decrease in area = xy - $(\frac{80}{100} x \times \frac{90}{100} y)$ = (7/25)xy

Decrease % = (7/25)xy * (1/xy) * 100 = 28%

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Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A) 1	B) 2
C) 3	D) 4

Answer & Explanation Answer: C) 3

Explanation:

Area of the park = (60 x 40) = 2400 $m^{2}$

Area of the lawn = 2109 $m^{2}$

Area of the crossroads = (2400 - 2109) = 291 $m^{2}$

Let the width of the road be x metres. Then,

$60 x + 40 x - X^{2} = 291$

$x^{2} - 100 x + 291 = 0$

(x - 97)(x - 3) = 0
x = 3.

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Q:

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

A) 22%	B) 33%
C) 44%	D) 55%

Answer & Explanation Answer: C) 44%

Explanation:

Let original length = x metres and original breadth = y metres.

Original area = xy sq.m

Increased length = $\frac{120}{100}$ and Increased breadth = $\frac{120}{100}$

New area = $\frac{120}{100} x * \frac{120}{100} y = \frac{36}{25} x y m^{2}$

The difference between the Original area and New area is:

$\frac{36}{25} x y - x y$

$\frac{11}{25} x y$ Increase % = $(\frac{\frac{11}{25} x y}{x y}) * 100$ = 44%

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