## Area Problems Solved Questions For IBPS, SBI and Other Bank Exams - Page 4

**Question 1**

The dimension of a rectangular plot is 36m x 20m. When the breadth is decreased to 14.4 m, what will be the length of the plot if the area remains same?

a) 28.4m b) 50m c) 52.2m d) 34m

**Answer : ** b) 50m.

Solution :

Dimension of the rectangle plot = 36m x 20m.

Length of the plot l = 36m

Breadth of the plot b = 20m.

Area of the plot = lb = 36 x 20 = 720 m^{2} .

Since breadth is decreased to 14.4 m, the new breadth = 14.4 m

Note that, the area is unchanged.

new length x new breadth = 720 m^{2} .

New length = 720 m^{2} / 14.4 m = 50m.

Hence, the required answer is 50m.

**Question 2**

The perimeter of a rectangle is 208 cm and breadth is 40% less than the length. Find the area of the rectangle.

a) 1585 cm^{2} b) 3505 cm^{2} c) 1595 cm^{2} d) 2535 cm^{2}

**Answer : ** d) 2535 cm^{2}

Solution :

Let L and B be the length and breadth of a rectangle.

Then, perimeter = 2(L + B) units.

Given that, perimeter = 2(L + B) = 208 cm

Therefore, L + B = 104 cm ………(1)

Breadth of the rectangle is 40% less than the length.

i.e., Breadth = L – 40% of L

b = L – 40 L / 100 = 60 L / 100 cm

Sub. B value in (1), we get

L + 60L / 100 = 104 cm

160L / 100 = 104 cm

L = 65 cm

Then, L + B = 65 + B = 104 => B = 39 cm

Now, L = 65 cm and B = 39cm

Area = LB = 65 x 39 cm^{2} .

= 2535 cm^{2}

Hence the answer is option d.

**Question 3**

Find the area of the rectangle whose perimeter is 412 cm and difference between the length and breadth is 46m.

a) 10080 cm^{2} b) 10200 cm^{2} c) 11080 cm^{2} d) 9080 cm^{2}

**Answer : ** a) 10080 cm^{2}.

Solution :

Let L and B be the length and breadth of the rectangle.

Perimeter of the rectangle = 2(L + B) units.

Given that, perimeter = 412 cm

L + B = 206 cm ………(1)

Difference between the length and breadth = L - B = 46 m …….(2)

Adding (1) and (2), we get = 2L = 252 cm

L = 126 cm.

B = 80 cm.

Area = LB = 126 x 80 = 10080 cm^{2}.

**Question 4**

The length of a rectangular plot is 40% more than its breadth. It cost Rs.12288 to build a wall around the plot at Rs.128 per meter. Find the area of the plot.

a) 560 m^{2} b) 840 m^{2} c) 390 m^{2} d) 440 m^{2}

**Answer : ** a) 560 m^{2}.

Solution :

Let L and B be the length and breadth of the rectangle.

Length is 40% more than the breadth. i.e., L = B + 40% of B

L = B + 40L / 100 = 140B / 100 cm.

Total cost to build a wall around the plot is Rs. 12288 at Rs.128 per meter.

Circumference of the plot = Rs. 12288 / Rs.128 = 96 meters.

i.e., Perimeter = 2(L + B) = 96 m.

Therefore, L + B = 48 m ……1)

Substituting the values of L and B in above eqn, we get,

L + B = 140B / 100 + b = 240B / 100 = 48

B = 4800 / 240 = 20 m

L = 48 – 20 = 28 m.

Area = LB = 20 x 28 = 560 m^{2}.