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1424.The diagonal of the floor of a rectangular closet is 7$ \dfrac{1}{2} $ feet. The shorter side of the closet is 4$ \dfrac{1}{2} $ feet. What is the area of the closet in square feet?
5 $ \frac{1}{4} $
13 $ \frac{1}{2} $
27
37
Explanation:
Other side = $ \sqrt{\left(\dfrac{15}{2} \right)^2-\left(\dfrac{9}{2} \right)^2} $ ft
= $ \sqrt{\dfrac{225}{4} -\dfrac{81}{4} } $ ft
= $ \sqrt{\dfrac{144}{4} } $ ft
= 6 ft.

$\therefore$ Area of closet = $\left(6 \times 4.5\right)$ sq. ft = 27 sq. ft.

1425.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?
40
50
None of these
Data inadequate
Explanation:

Let breadth = x metres.

Then, length = (x + 20) metres.

Perimeter =$ \left(\dfrac{5300}{26.50} \right) $m = 200 m.

$\therefore$ 2[(x + 20) + x] = 200

$\Rightarrow$ 2x + 20 = 100

$\Rightarrow$ 2x = 80

$\Rightarrow$ x = 40.

Hence, length = x + 20 = 60 m.

1426.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?
2.91 m
3 m
5.82 m
None of these
Explanation:

Area of the park = (60 x 40) m2 = 2400 m2.

Area of the lawn = 2109 m2.

$\therefore$ Area of the crossroads = (2400 - 2109) m2 = 291 m2.

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

60x + 40x - x2 = 291

$\Rightarrow$ x2 - 100x + 291 = 0

$\Rightarrow$ (x - 97)(x - 3) = 0

$\Rightarrow$ x = 3.

1427.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?
814
820
840
844
Explanation:

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

Area of each tile = $(41 \times 41)$ cm2.

$\therefore$ Required number of tiles =$ \left(\dfrac{1517 \times 902}{41 \times 41} \right) $= 814.
1428.The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?
14 metres
20 metres
18 metres
12 metres
Explanation:

$lb=460$ m2...(Equation 1)

Let the breadth = b

Then length, $l=b \times \dfrac{(100 + 15)}{100} = \dfrac{115b}{100}$ ...(Equation 2)

From Equation 1 and Equation 2,

$\dfrac{115b}{100} \times b = 460$

$b^2 = \dfrac{46000}{115} = 400$

$\Rightarrow b = \sqrt{400} = 20\text{ m}$

1429.The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
40%
42%
44%
46%
Explanation:

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

Original area = $(xy)$ m2.

New length =$ \left(\dfrac{120}{100} x\right) $m=$ \left(\dfrac{6}{5} x\right) $m.
New breadth =$ \left(\dfrac{120}{100} y\right) $m=$ \left(\dfrac{6}{5} y\right) $m.
New Area =$ \left(\dfrac{6}{5} x \times\dfrac{6}{5} y\right) $m2=$ \left(\dfrac{36}{25} xy\right) $m2.

The difference between the original area = $xy$ and new-area 36/25 $xy$ is

= (36/25)$x$$y$ - $x$$y$

= $x$$y$(36/25 - 1)

= $x$$y$(11/25) or (11/25)$x$$y$

$\therefore$ Increase % =$ \left(\dfrac{11}{25} xy \times\dfrac{1}{25} \times 100\right) $%= 44%.
1430.The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
1520 m2
2420 m2
2480 m2
2520 m2
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.

$\therefore Area = (l \times b) = (63 \times 40)$ m2 = 2520 m2.

1431.A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?
400
365
385
315
Explanation:

Let the areas of the parts be x hectares and (700 - x) hectares

Difference of the areas of the two parts = x - (700 - x) = 2x - 700

one-fifth of the average of the two areas = $\dfrac{1}{5}\dfrac{[\text{x}+(700-\text{x})]}{2}$

$=\dfrac{1}{5} \times \dfrac{700}{2}=\dfrac{350}{5}=70$

Given that difference of the areas of the two parts = one-fifth of the average of the two areas

=> 2x - 700 = 70

=> 2x = 770

$\Rightarrow x = \dfrac{770}{2}= 385$

Hence, area of smaller part = (700 - x) = (700 – 385) = 315 hectares.

1432.The diagonal of a rectangle is $ \sqrt{41} $ cm and its area is 20 sq. cm. The perimeter of the rectangle must be:
9 cm
18 cm
20 cm
41 cm
Explanation:

$ \sqrt{l^2 + b^2} $ = $ \sqrt{41} $ .

Also, lb = 20.

(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81

$\Rightarrow$ (l + b) = 9.

$\therefore$ Perimeter = 2(l + b) = 18 cm.

1433.A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed?
30
44
56
60
Explanation:

Perimeter of a rectangle = 2(l + b)

where l is the length and b is the breadth of the rectangle

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres

Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which isvery important.

Hence number of poles required = $\dfrac{280}{5}$ = 56

1434.A 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:
Rs. 456
Rs. 458
Rs. 558
Rs. 568
Explanation:
Area to be plastered= $[2\left(l + b\right) \times h] $+ $\left(l \times b\right)$

= ${[2\left(25 + 12\right) \times 6] + \left(25 \times 12\right)}$ m2

= (444 + 300) m2

= 744 m2.
$\therefore$ Cost of plastering = Rs.$ \left(744 \times\dfrac{75}{100} \right) $= Rs. 558.
1436.A rectangular field has 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?
95
92
88
82
Explanation:

Given that area of the field = 680 sq. feet

=> lb = 680 sq. feet

Length(l) = 20 feet

=> 20 × b = 680

=> b $= \dfrac{680}{20} = 34$ feet

Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet

1437.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:
2%
2.02%
4%
4.04%
Explanation:

100 cm is read as 102 cm.

$\therefore$ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.

(A2 - A1) = [(102)2 - (100)2]

= (102 + 100) x (102 - 100)

= 404 cm2.

$\therefore$ Percentage error =$ \left(\dfrac{404}{100 \times 100} \times 100\right) $%= 4.04%
1438.The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field.
37500 m2
30500 m2
32500 m2
40000 m2
Explanation:

Given that breadth of a rectangular field is 60% of its length

$\Rightarrow b = \dfrac{60l}{100} = \dfrac{3l}{5}$

perimeter of the field = 800 m

=> 2 (l + b) = 800

$\Rightarrow 2\left(l + \dfrac{3l}{5} \right)=800$

$\Rightarrow l + \dfrac{3l}{5} = 400$

$\Rightarrow \dfrac{8l}{5} = 400$

$\Rightarrow \dfrac{l}{5} = 50$

$\Rightarrow l = 5 \times 50 = 250\text{ m}$

$\text{b = }\dfrac{3l}{5} = \dfrac{3 \times 250}{5} = 3 \times 50 = 150\text{ m}$

Area = $\text{lb = }250 \times 150 = 37500\text{ m}^2$

1439.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?
34
40
68
88
Explanation:

We have: l = 20 ft and lb = 680 sq. ft.

So, b = 34 ft.

$\therefore$ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

1440.If each side of a square is increased by 25%, find the percentage change in its area?
65.25
56.25
65
56
Explanation:

let each side of the square be a , then area = a x a

New side = 125a / 100 = 5a / 4

New area =$\left(5a \times 5a\right)$ / $\left(4 \times 4\right)$ = (25a²/16) 

increased area== (25a²/16) - a²

Increase %= [(9a²/16 ) x (1/a² ) x 100]% = 56.25%

8243.Find the area of a square of side 35 m.
1220m2
1225m2
1250m2
1200m2
Explanation:

Area of a square = length × length

= 35 × 35 sq. m.

= 1225 m2

8245.Find the area of a rectangle of length 24 mm and breadth 8 mm.?
180m2
185m2
190m2
192m2
Explanation:

Length of the rectangle = 24 mm.

Breadth of the rectangle = 8 mm.

Area of the rectangle = l × b

= 24 × 8m2.

= 192m2.

8247.The diameter of a circle is 8 centimeters. What is the area?
51.12cm
52.42cm
54.40cm
50.24cm
Explanation:

8 cm = 2 × radius

8 cm × 2 = radius

radius = 4 cm

Area= $\pi$ × r2

= $\pi$ × 42

= $\pi$ × (4 × 4)

Area = 50.24 cm

8252.Length of a side of a square field is 275 m. What will be cost of levelling the field at a rate of 10 cent per square metre?
$ 7564.2
$ 7568.50
$ 7542.62
$ 7562.50
Explanation:

Length of the square field = 275 m

Area of the square field = side × side

= 275 m × 275 m

= 75625 m2

Cost of levelling the field for 1 sq m = 10 cent.

Now, we have to find the cost for entire area ie 75625 sq m.

Cost of levelling the field = 75625 × 10 cent

= 756250 cent

= $ 7562.50

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