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Instructions: This Monk test contains JK Bank Area Problems with answer. This will help you in JK bank exam preparation.

This is a FREE online JK Bank Monk Test on topic JK Bank Area Problems with answer. You DO NOT pay money to anyone to attend this test.
Each question carries 1 mark, no negative marks.
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All the best :-).

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:

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:

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?

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:

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

Explanation:

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

Original area = (xy) m2.

 New length = 120 x m = 6 x m. 100 5
 New breadth = 120 y m = 6 y m. 100 5
 New Area = 6 x x 6 y m2 = 36 xy m2. 5 5 25

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

= (36/25)xy – xy

= xy(36/25 – 1)

= xy(11/25) or (11/25)xy

 Increase % = 11 xy x 1 x 100 % = 44%. 25 xy

### #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?

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?

Explanation:

Other side =
 15 2 – 9 2 2 2
ft
=
 225 – 81 4 4
ft
=
 144 4
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:

Explanation:

Let original length = x and original breadth = y.

Decrease in area
 = xy – 80 x x 90 y 100 100
 = xy – 18 xy 25
 = 7 xy. 25
 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?

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:

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?

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. 41 x 41

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

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?

Explanation:

Let original length = x and original breadth = y.

Original area = xy.

 New length = x . 2

 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?

Explanation:

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?

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.