Area Questions and Answers updated daily – Aptitude



15 Area Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Area online test. Area Questions with detailed description, explanation will help you to master the topic.

Area Questions

1. 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?



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Correct Ans:18cm
Explanation:


2. 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



SHOW ANSWER
Correct Ans:558
Explanation:
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.
Cost of plastering = Rs.(744 x75/100) = Rs. 558.



3. 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?



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Correct Ans: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.


4. 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?



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Correct Ans:814
Explanation:
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 x 41) cm2.

Required number of tiles =(1517 x 902/41 x 41)= 814.




5. The diagonal of a rectangle is 41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:



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Correct Ans:18cm
Explanation:
l2 + b2 = 41.

Also, lb = 20.

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

(l + b) = 9.

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


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



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Correct Ans:2520m2
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) m2 = 2520 m2.


7. 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?



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Correct Ans:
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.


8. 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?



SHOW ANSWER
Correct Ans:88
Explanation:
Given that area of the field = 680 sq. feet
=> lb = 680 sq. feet

Length(l) = 20 feet

=> 20 × b = 680
b=68020=34 feet


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


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



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Correct Ans:60m
Explanation:
Length of the plot is 20 metres more than its breadth.
Hence, let's take the length as l metres and breadth as (l - 20) metres

Length of the fence = perimeter = 2(length + breadth)= 2[ l + (l - 20) ] = 2(2l - 20) metres
Cost per meter = Rs. 26.50
Total cost = 2(2l - 20) × 26.50

Total cost is given as Rs. 5300
=> 2(2l - 20) × 26.50 = 5300
=> (2l - 20) × 26.50 = 2650
=> (l - 10) × 26.50 = 1325
=> (l - 10) = 1325/26.50 = 50
=> l = 50 + 10 = 60 metres


10.  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?



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Correct Ans:56
Explanation:
Perimeter of a rectangle = 2(l + b)
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 is
very important.

Hence number of poles required = 280⁄5 = 56


11. The length of a room is 5.5 m and width is 3.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 800 per sq. metre.



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Correct Ans:16500
Explanation:
Area = 5.5 × 3.75 sq. metre.
Cost for 1 sq. metre. = Rs. 800

Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500


12. 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?



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Correct Ans:20 m
Explanation:
lb = 460 m2 ------(Equation 1)

Let the breadth = b
Then length, l = b×(100+15)100=115b100------(Equation 2)


From Equation 1 and Equation 2,

115b100×b=460b2=46000115=400⇒b=400−−−√=20 m


13. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet?



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Correct Ans:126 sq.ft
Explanation:
Let l = 9 ft.

Then l + 2b = 37
=> 2b = 37 - l = 37 - 9 = 28
=> b = 28/2 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.


14. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?



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Correct Ans:4.04%
Explanation:
Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100
Then the measured value = 100×(100+2)100=102 (∵ error 2% in excess)


Correct Value of the area of the square = 100 × 100 = 10000

Calculated Value of the area of the square = 102 × 102 = 10404

Error = 10404 - 10000 = 404

Percentage Error = ErrorActual Value×100=40410000×100=4.04%


15. 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



SHOW ANSWER
Correct Ans:rs.558
Explanation:
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.
Cost of plastering = Rs. 744 x75 = Rs. 558.
100




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