Instructions: This Monk test contains JK Bank Area Problems with answer. This will help you in JK bank exam preparation.
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Directions to Solve
In the following the questions choose the word which best expresses the meaning of the given word.
Results
#1. The 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:
Answer: Option B
Explanation:
| Perimeter = Distance covered in 8 min. = |  |
12000 | x 8 |  m = 1600 m. |
| 60 |
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 x 320) m2 = 153600 m2.
#2. 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:
Answer: Option D
Explanation:
100 cm is read as 102 cm.
A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 – A1) = [(102)2 – (100)2]
= (102 100) x (102 – 100)
= 404 cm2.
| Percentage error = |
|
404 | x 100 | % |
= 4.04% |
#3. 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?
Answer: Option B
Explanation:
| 2(l b) | = | 5 |
| b | 1 |
2l 2b = 5b
3b = 2l
| b = | 2 | l |
| 3 |
Then, Area = 216 cm2
l x b = 216
| l x |
2 | l | = 216 |
| 3 |
l2 = 324
l = 18 cm.
#4. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
#5. 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?
Answer: Option B
Explanation:
Area of the park = (60 x 40) m2 = 2400 m2.
Area of the lawn = 2109 m2.
Area of the crossroads = (2400 – 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x 40x – x2 = 291
x2 – 100x 291 = 0
(x – 97)(x – 3) = 0
x = 3.
#6. The diagonal of the floor of a rectangular closet is 7 1/2 feet. The shorter side of the closet is 4 1/2 feet. What is the area of the closet in square feet?
Answer: Option C
Explanation:
| Other side | = |
|
ft | ||||||||||||
| = |
|
ft | |||||||||||||
| = |
|
ft | |||||||||||||
| = | 6 ft. |
Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.
#7. 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:
Answer: Option D
Explanation:
Let original length = x and original breadth = y.
| Decrease in area |
|
||||||||||
|
|||||||||||
|
| Decrease % = |
|
7 | xy x | 1 | x 100 | % |
= 28%. |
| 25 | xy |
#8. A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?
Answer: Option C
Explanation:
Let the side of the square(ABCD) be x metres.
Then, AB BC = 2x metres.
AC = 2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.
| Saving % = | |
0.59x | x 100 | % |
= 30% (approx.) |
| 2x |
#9. The diagonal of a rectangle is 41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:
Answer: Option B
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.
#10. 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?
Answer: Option A
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 | |
= 814. |
|
#11. The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
Answer: Option D
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.
#12. The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
Answer: Option B
Explanation:
Let original length = x and original breadth = y.
Original area = xy.
| New length = | x | . |
| 2 |
New breadth = 3y.
| New area = | |
x | x 3y | |
= | 3 | xy. |
| 2 | 2 |
| Increase % = |
|
1 | xy x | 1 | x 100 | % |
= 50%. |
| 2 | xy |
#13. 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?
Answer: Option E
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.
#14. 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?
Answer: Option D
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.
#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:
Answer: Option C
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 x | 75 | |
= Rs. 558. |
| 100 |
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