Last Updated on Mar 14, 2023
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Quiz Description :
Name : Area quiz : Maths / Aptitude practice test
Subject : Mathematics / Aptitude
Topic : Area
Questions: 15 mcq
Time Allowed : 30 Minutes
Important for : SSC CGL, CHSSL, MTS, UPSC Prelims, CAT, GATE, Job interviews ( TCS, WIPRO, IBM, Accenture, BARK, NICL, BHEL, BEL, HCL etc), Bank exams like IBPS Clerk, PO / MT, RRB, SBI PO / Clerk, RBI and for High school students.
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| Pos. | Name | Entered on | Points | Result |
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Question 1 of 15
1. Question
1 pointsThe length of a room is 5.5 m and Width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs. 800 per sq. Meter.
Correct
Area of the floor = (5.5 × 3.75) m2 = 20.625 m2.
Cost of paving = Rs. (800 × 20.625) = Rs 16500.Incorrect
Area of the floor = (5.5 × 3.75) m2 = 20.625 m2.
Cost of paving = Rs. (800 × 20.625) = Rs 16500.Unattempted
Area of the floor = (5.5 × 3.75) m2 = 20.625 m2.
Cost of paving = Rs. (800 × 20.625) = Rs 16500. -
Question 2 of 15
2. Question
1 pointsThe length of a rectangle is 18 cm and its breadth is 10 cm. When the length is increased to 25 cm, what will be the breadth of the rectangle if the area remains the same?
Correct
Let the breadth be b. Then, 25 × b = 18 × 10 < = = > b = (180 / 25) cm = 7.2 cm.
Incorrect
Let the breadth be b. Then, 25 × b = 18 × 10 < = = > b = (180 / 25) cm = 7.2 cm.
Unattempted
Let the breadth be b. Then, 25 × b = 18 × 10 < = = > b = (180 / 25) cm = 7.2 cm.
-
Question 3 of 15
3. Question
1 pointsA rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart, how many poles will be needed?
Correct
Perimeter of the plot = 2. (90+50) = 280 m.
Number of poles = (280/5) = 56 m.Incorrect
Perimeter of the plot = 2. (90+50) = 280 m.
Number of poles = (280/5) = 56 m.Unattempted
Perimeter of the plot = 2. (90+50) = 280 m.
Number of poles = (280/5) = 56 m. -
Question 4 of 15
4. Question
1 pointsThe length of a rectangular plot is 60% more than its breadth. If the difference between the length and the breadth of that rectangle is 24 cm, what is the area of that rectangle?
Correct
Let breadth = x cm. Then length = (160/100 x) cm 8/5 x cm.
So, 8/5 x-x = 24 <==> 3/5 x = (120/3) = 40.
Length = 64 cm, Breadth = 40 cm.
Area = (64 × 40) cm2 = 2560 cm2.Incorrect
Let breadth = x cm. Then length = (160/100 x) cm 8/5 x cm.
So, 8/5 x-x = 24 <==> 3/5 x = (120/3) = 40.
Length = 64 cm, Breadth = 40 cm.
Area = (64 × 40) cm2 = 2560 cm2.Unattempted
Let breadth = x cm. Then length = (160/100 x) cm 8/5 x cm.
So, 8/5 x-x = 24 <==> 3/5 x = (120/3) = 40.
Length = 64 cm, Breadth = 40 cm.
Area = (64 × 40) cm2 = 2560 cm2. -
Question 5 of 15
5. Question
1 pointsA 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, then What is the area of the parking space in square feet?
Correct
Clearly, we have: l = 9 and l+ 2b = 37 or b = 14.
Area = (l × b) = (9 × 14) sq. ft = 126 sq. ft.Incorrect
Clearly, we have: l = 9 and l+ 2b = 37 or b = 14.
Area = (l × b) = (9 × 14) sq. ft = 126 sq. ft.Unattempted
Clearly, we have: l = 9 and l+ 2b = 37 or b = 14.
Area = (l × b) = (9 × 14) sq. ft = 126 sq. ft. -
Question 6 of 15
6. Question
1 pointsThe difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
Correct
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 × b) = (63 × 40) m2 = 2520 m2.Incorrect
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 × b) = (63 × 40) m2 = 2520 m2.Unattempted
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 × b) = (63 × 40) m2 = 2520 m2. -
Question 7 of 15
7. Question
1 pointsThe length of a rectangular plot is 20 meters 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?
Correct
Let breadth = x meters. Then, length = (x+20) metres.
Perimeter = (5300/26.50) m = 200 m.
2 [(x+20) + x] = 200 <==> 2x + 20 = 100 <== > 2x + 80 <==> x = 40.
Hence, length = x+20 = 60m.Incorrect
Let breadth = x meters. Then, length = (x+20) metres.
Perimeter = (5300/26.50) m = 200 m.
2 [(x+20) + x] = 200 <==> 2x + 20 = 100 <== > 2x + 80 <==> x = 40.
Hence, length = x+20 = 60m.Unattempted
Let breadth = x meters. Then, length = (x+20) metres.
Perimeter = (5300/26.50) m = 200 m.
2 [(x+20) + x] = 200 <==> 2x + 20 = 100 <== > 2x + 80 <==> x = 40.
Hence, length = x+20 = 60m. -
Question 8 of 15
8. Question
1 pointsThe breadth of a rectangular field is 60 % of its length. If the perimeter of the field is 800 m. What is the area of the field?
Correct
Let length = x metres. Then, breadth = (60x/100) metres = (3x/5) metres.
Perimeter = [2(x+3x/5)] m = (16x/5) m.
16x/5 = 800 <==> x = (4000/16) = 250.
So, length = 250 m; breadth = 150 m.
Area = (250 × 150) m2 = 37500 sq m.Incorrect
Let length = x metres. Then, breadth = (60x/100) metres = (3x/5) metres.
Perimeter = [2(x+3x/5)] m = (16x/5) m.
16x/5 = 800 <==> x = (4000/16) = 250.
So, length = 250 m; breadth = 150 m.
Area = (250 × 150) m2 = 37500 sq m.Unattempted
Let length = x metres. Then, breadth = (60x/100) metres = (3x/5) metres.
Perimeter = [2(x+3x/5)] m = (16x/5) m.
16x/5 = 800 <==> x = (4000/16) = 250.
So, length = 250 m; breadth = 150 m.
Area = (250 × 150) m2 = 37500 sq m. -
Question 9 of 15
9. Question
1 pointsThe ratio between the length and the perimeter of a rectangular plot is 1: 3. What is the ratio between the length and breadth of the plot?
Correct
l/2 (l +b) =1/3 => 3l = 2l+2b => l = 2b => l /b = 2/1 = 2: 1.
Incorrect
l/2 (l +b) =1/3 => 3l = 2l+2b => l = 2b => l /b = 2/1 = 2: 1.
Unattempted
l/2 (l +b) =1/3 => 3l = 2l+2b => l = 2b => l /b = 2/1 = 2: 1.
-
Question 10 of 15
10. Question
1 pointsThe 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 :
Correct
Perimeter = Distance covered in 8 min. = (12000/60×8) m =1600 m.
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 × 320) m2 = 153600 m2.Incorrect
Perimeter = Distance covered in 8 min. = (12000/60×8) m =1600 m.
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 × 320) m2 = 153600 m2.Unattempted
Perimeter = Distance covered in 8 min. = (12000/60×8) m =1600 m.
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 × 320) m2 = 153600 m2. -
Question 11 of 15
11. Question
1 pointsThe length of a rectangular hall is 5m more than its breadth. The area of the hall is 750 sq m. The length of the hall is:
Correct
Let breadth = x metres. Then, length = (x+5) metres.
Then, x (x+5) = 750 <==> x2+5x-750 = 0 <==> (x+30) (x-25) = 0 <==> (x+30) (x-25) = 0 <==> x = 25.
Length = (x + 5) = 30 m.Incorrect
Let breadth = x metres. Then, length = (x+5) metres.
Then, x (x+5) = 750 <==> x2+5x-750 = 0 <==> (x+30) (x-25) = 0 <==> (x+30) (x-25) = 0 <==> x = 25.
Length = (x + 5) = 30 m.Unattempted
Let breadth = x metres. Then, length = (x+5) metres.
Then, x (x+5) = 750 <==> x2+5x-750 = 0 <==> (x+30) (x-25) = 0 <==> (x+30) (x-25) = 0 <==> x = 25.
Length = (x + 5) = 30 m. -
Question 12 of 15
12. Question
1 pointsThe area of a rectangle is 460 square metres. If the length is 15% more than the breadth .What is the breadth of the rectangular field?
Correct
Let breadth = x meters. Then, length = (115x/100) meters.
X × 115x/100 = 460 <==> x2 = (46000/115) = 400 <==> x = 20.Incorrect
Let breadth = x meters. Then, length = (115x/100) meters.
X × 115x/100 = 460 <==> x2 = (46000/115) = 400 <==> x = 20.Unattempted
Let breadth = x meters. Then, length = (115x/100) meters.
X × 115x/100 = 460 <==> x2 = (46000/115) = 400 <==> x = 20. -
Question 13 of 15
13. Question
1 pointsA 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?
Correct
: We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft.
Length of fencing = (l + 2b) = (20+ 68) ft = 88 ft.Incorrect
: We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft.
Length of fencing = (l + 2b) = (20+ 68) ft = 88 ft.Unattempted
: We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft.
Length of fencing = (l + 2b) = (20+ 68) ft = 88 ft. -
Question 14 of 15
14. Question
1 pointsThe ratio between the perimeter and the breadth of a rectangle is 5: 1. If the area of the area of the rectangle is 216 sq. Cm, What is length of the rectangle?
Correct
2 (l+ b)/b = 5/1 => 2l+ 2b = 5b => 3b = 21 => b = 2/3i.
Then, Area = 216 cm2 => l × b = 216 => I × 2/3i = 216 => i2 = 324 => I = 18 cm.Incorrect
2 (l+ b)/b = 5/1 => 2l+ 2b = 5b => 3b = 21 => b = 2/3i.
Then, Area = 216 cm2 => l × b = 216 => I × 2/3i = 216 => i2 = 324 => I = 18 cm.Unattempted
2 (l+ b)/b = 5/1 => 2l+ 2b = 5b => 3b = 21 => b = 2/3i.
Then, Area = 216 cm2 => l × b = 216 => I × 2/3i = 216 => i2 = 324 => I = 18 cm. -
Question 15 of 15
15. Question
1 pointsA farmer wishes to start a 100 sq. m rectangular, vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is:
Correct
We have: 2b+l = 30 => I = 30- 2b.
Area = 100 m2 => l × b = 100 => b (30-2b) = 100 => b2-15b + 50 = 0
=> (b- 10) ( b- 5 ) = 0 => b = 10 or b = 5.
When b = 10, l = and when b = 5, =20.
Since the garden is rectangular, so its dimension is 20 m × 5 m.Incorrect
We have: 2b+l = 30 => I = 30- 2b.
Area = 100 m2 => l × b = 100 => b (30-2b) = 100 => b2-15b + 50 = 0
=> (b- 10) ( b- 5 ) = 0 => b = 10 or b = 5.
When b = 10, l = and when b = 5, =20.
Since the garden is rectangular, so its dimension is 20 m × 5 m.Unattempted
We have: 2b+l = 30 => I = 30- 2b.
Area = 100 m2 => l × b = 100 => b (30-2b) = 100 => b2-15b + 50 = 0
=> (b- 10) ( b- 5 ) = 0 => b = 10 or b = 5.
When b = 10, l = and when b = 5, =20.
Since the garden is rectangular, so its dimension is 20 m × 5 m.
i am preparing for bank exam(RRB)..i need practice test and also answers
Hello Sirlaka, You can check answers and explanations of questions after the test by “View Questions” Button.
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Sir,
You haven’t given the value of length in ques no. 2
Hello Debjit Ghosh, Thanks for your comment, It was a publishing error and we have updated that question. The value of length in that question is 20 m. These questions are in random mode so may be in next attempt it will come at different position.