Area – Aptitude Questions and Answers | Latest Online Area MCQ Aptitude Test
Area – Aptitude Questions and Answers | Online Area MCQ Aptitude Test Quiz Name Area Category Online Aptitude Test Number of Questions 25 Time 30 Minutes Exam Type MCQ (Multiple Choice Questions) 1. The length to width ratio of a rectangular park is 3:2. If a guy cycling along the park’s perimeter at a pace of... explore below more
Area – Aptitude Questions and Answers | Online Area MCQ Aptitude Test
Quiz Name
Area
Category
Online Aptitude Test
Number of Questions
25
Time
30 Minutes
Exam Type
MCQ (Multiple Choice Questions)
1. The length to width ratio of a rectangular park is 3:2. If a guy cycling along the park’s perimeter at a pace of 12 km/hr completes one circuit in 8 minutes, the park’s area (in square metres) is: a. 15360 b. 153600 c. 30720 d. 307200
Answer: B
Explanation: Perimeter = Distance travelled in eight minutes = 12000/60 x eight metres = 1600 metres.Let’s say the length is 3x metres and the width is 2x metres.2(3x + 2x) = 1600 or x = 160, respectively.The length is 480 metres and the width is 320 metres.Area = 153600 m2 (480 x 320) m2.
2. When measuring the length of a square’s side, an error of 2% is created. The percentage of error in the square’s calculated area is: a. 2% b. 2.02% c. 4% d. 4.04%
Answer: D
Explanation: 100 centimetres is read as 102 centimetres.A1 is equal to (100 x 100) cm2 and A2 is equal to (102 x 102) cm2.[(102)2 – (100)2] = (A2 – A1).(102 + 100) x (102 – 100) 404 cm2 =The percentage error is calculated as 404/100 x 100 x 100 percent = 4.04 percent.
3. A rectangle’s perimeter to breadth ratio is 5 to 1. What is the length of a rectangle with a surface area of 216 square cm? a. 16 cm b. 18 cm c. 24 cm d. Data inadequate
Answer: B
Explanation: 5/1 = 2(l + b)/b,5b = 2l + 2b,2l + 3b,2 l/3 = b.Then the area equals 216 cm2.216 = l x b.216 = l x 2 l/3.324 = l218 cm in length.
4. If each of a rectangle’s sides is increased by 20%, the percentage increase in its area is: a. 40% b. 42% c. 44% d. 46%
Answer: C
Explanation: Let’s say the original length is x metres and the original width is y metres.(xy) m2 was the original area.120/100 x m = 6/5 x m is the new length.New breadth = 120 / 100 y m = 6/5 y m.New Area = 36/25 xy m2 = 6/5 x x 6/5 y m2.The difference between the old area = xy and the new area = 36/25 xy is = (36/25)xy – xy = xy(36/25 – 1) = xy(11/25) or (11/25)xy Increase percent =1/25 xy x 1/xyx 100 percent = 44 percent.
5. A 60-meter-long and 40-meter-wide rectangular park with two concrete crossroads in the middle and a lawn on the remainder of the park. What is the width of the road if the grass is 2109 square metres? a. 2.91 m b. 3 m c. 5.82 m d. None of these
Answer: B
Explanation: The park’s area is equal to (60 x 40) m2 = 2400 m2.The lawn is 2109 m2 in size.Crossroads area = (2400 – 2109) m2 = 291 m2.Allow for a road width of x metres. Then,291 = 60x + 40x – x20 = x2 – 100x + 291,0 = (x – 97)(x – 3).x equals 3.
6. The diagonal of a rectangular closet’s floor is 7 feet. The closet’s shorter side is 4 feet long. a. 5 1/4 b. 13 1/2 c. 27 d. 37
Answer: C
Explanation: 2 – (9/2 ) Other side = root (15/2 ),2 feet Equals root 225/4-81/4 feet = root 144/4 feet = 6 feet.Closet area = (6 x 4.5) sq ft = 27 sq ft.
7. When a towel was bleached, it lost 20% of its length and 10% of its width. The area decreases by the following percentage: a. 10% b. 10.08% c. 20% d. 28%
Answer: D
Explanation: Let’s say the original length is x and the original width is y.Area reduction = xy – 80/100 x x 90/100 y= xy – 18/25 xy = 7/25 xy.Reduced percent = 7/25 xy x 1/xy x 100 percent = 28 percent.
8. A man went across a square lot diagonally. What was the approximate percentage saved by not walking along the edges? a. 20 b. 24 c. 30 d. 33
Answer: C
Explanation: Let the square’s side(ABCD) be x metres.After that, AB + BC = 2x metres.(1.41x) m = AC = 2x.2x metres = (0.59x) m .Savings percentage = 0.59x/2x x 100 percent = 30% (approx.)
9. A rectangle has a diagonal of 41 cm and a surface size of 20 square centimetres. The rectangle’s perimeter must be: a. 9 cm b. 18 cm c. 20 cm d. 41 cm
Answer: B
Explanation: 41 = l2 + b2.In addition, lb = 20.2lb = 41 + 40 = 81 (l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81,(9 + l + b).2(l + b) = 18 cm is the perimeter.
10. How many square tiles are needed to pave the floor of a room that is 15 m 17 cm long and 9 m 2 cm wide? a. 814 b. 820 c. 840 d. 844
Answer: A
Explanation: H.C.F. of 1517 cm and 902 cm = 41 cm for the biggest tile.Each tile has a surface area of (41 x 41) cm2.1517 x 902/ 41 x 41 = 814 is the required number of tiles.
11. The distance between the length and width of a rectangle is 23 metres. The area of a circle with a perimeter of 206 metres is: a. 1520 m2 b. 2420 m2 c. 2480 m2 d. 2520 m2
Answer: D
Explanation: (l – b) = 23 and 2(l + b) = 206 or (l + b) = 103 are the results.We obtain l = 63 and b = 40 when we solve the two equations.(63 x 40) m2 = 2520 m2 Area = (l x b) = (63 x 40) m2.
12. A rectangle’s length is halved while its width is quadrupled. What is the area change as a percentage? a. 25% increase b. 50% increase c. 50% decrease d. 75% decrease
Answer: B
Explanation: Let’s say the original length is x and the original width is y.xy is the original area.x/2 is the new length.3 years is the new breadth.3 xy = x/2 x 3y/2 = new areaIncrease in percentage = 1/2 xy x 1/xy x 100 percent = 50%
13. A rectangular plot’s length is 20 metres longer than its width. What is the length of the property in metres if the cost of fencing the plot is Rs. 5300 at a rate of Rs. 26.50 per metre? a. 40 b. 50 c. 120 d. None of these
Answer: D
Explanation: Let’s say the width is x metres.The length is then equal to (x + 20) metres.5300 m / 26.50 m = 200 m Perimeter.200 = 2[(x + 20) + x].100 = 2x + 20 80 = 2x.x equals 40.As a result, length = x + 20 = 60 metres.
14. A rectangular field will be walled on three sides, leaving a 20-foot-long side unfenced. How many feet of fence will be required if the field measures 680 square feet? a. 34 b. 40 c. 68 d. 88
Answer: D
Explanation: We have the following measurements: l = 20 ft and lb = 680 sq. ft.As a result, b = 34 ft.Fencing length = (l + 2b) = (20 + 68) feet = 88 feet.
15. A tank is 25 metres long, 12 metres wide, and 6 metres deep. At 75 paise per sq. m, the cost of plastering its walls and bottom is: a. Rs. 456 b. Rs. 458 c. Rs. 558 d. Rs. 568
Answer: C
Explanation: [2(l + b) x h] + (l x b) = [2(25 + 12) x 6] + (25 x 12) Area to be plastered (444 + 300) m2, 744 m2.Plastering costs Rs. 744 x 75/100 = Rs. 558.
16. 1. The playground has a total size of 1600 m2. What is the size of the perimeter? I. It’s a square playground to the nth degree. II. At a rate of Rs. 20 per metre, erecting a fence around the playground costs Rs. 3200. a. I alone sufficient while II alone not sufficient to answer b. II alone sufficient while I alone not sufficient to answer c. Either I or II alone sufficient to answer d. Both I and II are not sufficient to answer
Answer: C
Explanation: 1600 m2 in size.I. Side = 1600 m = 40 m; II. Side = 1600 m = 40 m; III. Side = 1600 m As a result, the perimeter is (40 x 4) m = 160 m.I am the only one who can provide an answer.Perimeter = Total cost = 3200/20 m = 160 m Cost per metre II by itself provides the solution.
17. A rectangle’s area is equal to a right-angles triangle’s area. What is the rectangle’s length? I. The triangle’s base measures 40 cm. II. The triangle’s height is 50 cm. a. I alone sufficient while II alone not sufficient to answer b. II alone sufficient while I alone not sufficient to answer c. Either I or II alone sufficient to answer d. Both I and II are not sufficient to answer
Answer: D
Explanation: Assume that the area of a rectangle equals the area of a right-angles triangle.1/2 x B x H = l x b.B = 40 cm, according to I.H = 50 cm, according to II. To find l, we also require b, which isn’t supplied.Data alone isn’t enough to make a decision.
18. What is the triangle’s height in metres? I. The triangle’s area is 20 times its base. II. The triangle’s circumference is equal to the perimeter of a square with a side of 10 cm. a. I alone sufficient while II alone not sufficient to answer b. II alone sufficient while I alone not sufficient to answer c. Either I or II alone sufficient to answer d. Both I and II are not sufficient to answer
Answer: A
Explanation: I. A = 20 x B 1/2x B x H = 20 x B H = 40; II. A = 20 x B 1/2x B x H = 20 x B H = 40; III. A = 20 x B 1/2x B x H =I am the only one who can provide an answer.According to II, the triangle’s circumference is 40 cm.This does not reveal the triangle’s height.
19. How much would it cost to paint the inside walls of a room at a rate of Rs. 20 per square foot? I. The floor’s circumference is 44 feet. II. The room’s longest wall is 12 feet tall. a. I alone sufficient while II alone not sufficient to answer b. II alone sufficient while I alone not sufficient to answer c. Either I or II alone sufficient to answer d. Both I and II are necessary to answer
Answer: D
Explanation: 2R = 44, according to I.H = 12 is the result of II.2RH = 2A = 2RH = 2RH = 2RH = (44 x 12).The price of the artwork is Rs (44 x 12 x 20).As a result, the solution is given by combining I and II.
20. What is the hall’s square footage? I.Flooring costs Rs. 2.50 per square metre in materials. II.The cost of labour to lay the hall flooring is Rs. 3500. III.The total cost of the hall’s flooring is Rs. 14,500. a. I and II only b. II and III only c. All I, II and III d. Any two of the three
Answer: C
Explanation: I. Material cost per m2 = Rs. 2.50,2. The cost of labour is Rs. 3500.The total cost of the project is Rs. 14,500.Let’s say the area is A square metres. (14500 – 3500) = Rs. 11,000 in material costs,A = 11000 x 2 /5 = 4400 m2. 5A/2 = 11000 A = 11000 x 2 /5 = 4400 m2.As a result, all of I, II, and III are required to obtain the solution.
21. How big is a right-angled triangle’s area? I.The triangle’s perimeter is 30 cm. II. The triangle’s base and height have a 5:12 ratio. III.The size of the triangle is the same as the area of a 10 cm long rectangle. a. I and II only b. II and III only c. I and III only d. III, and either I or II only
Answer: A
Explanation: From II, height = 5:12 as a basic .Let’s say the base is 5x and the height is 12x.The hypotenuse is thus equal to (5x)2 + (12x)2 = 13x.The triangle’s circumference is 30 cm, according to I.30 x = 5x + 12x + 13x = 5x + 12x + 13x = 5x + 12x + 13x = 5x + 12.As a result, the base is 5x = 5 cm, and the height is 12x = 12 cm.30 cm2 area = 1/2 x 5 x 12 cm2.As a result, the solution is given by combining I and II.Because the rectangle’s width is not specified, III is clearly redundant.
22. What is the rectangular field’s area? I. The field’s circumference is 110 metres. II. The length exceeds the breadth by 5 metres. III. The length and width ratios are 6:5 correspondingly. a. I and II only b. Any two of the three c. All I, II and III d. I, and either II or III only
Answer: B
Explanation: 2(l + b) = 110 l + b = 55 I. 2(l + b) = 110 I. 2(l + b) = 110 I. 2(l +),II. l = (b + 5) l – b l – b l – b l – b l – b l,l/b = 6 /5 5l – 6b = 0 III. l/b = 6 /5 5l – 6b = 0 These are three l and b equations. We may solve them in pairs.The solution can be found in any two of the three options.
23. What is the area of the rectangle given? I. The rectangle’s perimeter is 60 cm. II. The rectangle’s width is 12 cm. III. The total length of two adjacent sides is 30 cm. a. I only b. II only c. I and II only d. II and either I or III
Answer: D
Explanation: We can calculate the length and width of the rectangle using I and II, and hence the area.As a result, III is superfluous.Also, using II and III, we can calculate the length and width, and hence the area.As a result, I am no longer needed.
24. At Rs. 5 per m2, how much would it cost to paint the two adjacent walls of a hall that has no windows or doors? I. The hall is 24 square metres in size. II. The hall’s width, length, and height are in the proportions of 4 : 6 : 5. III. One wall is 30 square metres in size. a. I only b. II only c. III only d. All I, II and III are required.
Answer: C
Explanation: Let l = 4x, b = 6x, and h = 5x be the values from II.The hall’s size is then (24×2) m2.The hall is 24 m2 in size, according to I.We obtain 24×2 = 24 x = 1 by multiplying II and I.l = 4 metres, b = 6 metres, and h = 5 metre.As a result, the area of two neighbouring walls = [(l x h) + (b x h)] m2 may be calculated, as well as the cost of painting two nearby walls.As a result, III is superfluous.
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