# Mensuration Questions for XAT 2022 – Download PDF

Download Mensuration Questions for XAT PDF â€“ XAT Mensuration questions pdf by Cracku. Top 10 very Important Mensuration Questions for XAT based on asked questions in previous exam papers.

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**Question 1:Â **A circular Ground whose diameter is 35 meters has a 1.4 meters broad garden around it.What is the area of the garden in square meters ?

a)Â 160.16

b)Â 6.16

c)Â 1122.66

d)Â Data Inadequate

e)Â None of these

**Question 2:Â **When the length of the rectangular plot is increased by four times its perimeter becomes 480 meters and area becomes 12800 sq.m. What is its original length(in meters)?

a)Â 160

b)Â 40

c)Â 20

d)Â Cannot be determined

e)Â None of these

**Question 3:Â **Four circles having equal radii are drawn with center at the four corners of a square. Each circle touches the other two adjacent circle. If remaining area of the square is 168 cm, what is the size of the radius of the radius of the circle? (in centimeters)

a)Â 14

b)Â 1.4

c)Â 38

d)Â 21

e)Â 3.5

**Question 4:Â **What would be the cost of building a fence around a square plot with area equal to 361 sq. ft. if the price per foot of building the fence is Rs. 62 ?

a)Â Rs. 4026

b)Â Rs. 4712

c)Â Rs. 3948

d)Â Cannot be determined

e)Â None of these

**Question 5:Â **The sum of the dimensions of a room (i.e. length, breadth and height) is 18 metres and its length, breadth and height are in the ratio of 3 : 2 : 1 respectively. If the room is to be painted at the rate of Rs. 15 per m2, what would be the total cost incurred on painting only the four walls of the room (in Rs.)?

a)Â 3250

b)Â 2445

c)Â 1350

d)Â 2210

e)Â 2940

XAT 2022 Decision Making Course

**Question 6:Â **The sum of the dimensions of a room (i.e. length, breadth and height) is 24 metres and its length, breadth and height are in the ratio of 8: 7 : 5 respectively. If the room is to be painted at the rate of Rs. 12 per m2, what would be the total cost incurred on painting only the four walls of the room (in Rs.)?

a)Â 2592

b)Â 2648

c)Â 2848

d)Â 2120

e)Â 1956

**Question 7:Â **Area of circle is equal to the area of a rectangle having perimeter of 50 cms. and length more than the breadth by 3 cms. What is the diameter of the circle?

a)Â 7 cms.

b)Â 21 cms.

c)Â 28 cms.

d)Â 14 cms.

e)Â None of these

**Question 8:Â **The circumference of a circle is 792 meters. What will be its radius?

a)Â 120 meters

b)Â 133 meters

c)Â 145 meters

d)Â 136 meters

e)Â None of these

**Question 9:Â **Radius of a circular garden is 7 metre more than length of a rectangle whose perimeter is 364 metre and breadth is 84 metre. What will be cost of fencing the garden (only at the circumference), if the cost of fencing is Rs. 8 per metre ?

a)Â Rs.5456

b)Â Rs.6144

c)Â Rs. 5296

d)Â Rs. 5280

e)Â None of these

**Question 10:Â **The perimeter of a square plot is equal to the perimeter of a rectangular plot which is 23 metre long and 19 metre broad. What will be the diagonal of the square plot ?

a)Â $17\sqrt{2}m$

b)Â $21\sqrt{2}m$

c)Â $22\sqrt{2}m$

d)Â $23\sqrt{2}m$

e)Â None of these

**Answers & Solutions:**

**1)Â AnswerÂ (A)**

Radius of circular ground = $r = \frac{35}{2}=17.5$ m and width of broad garden = 1.4 m

=> Radius of outer circleÂ (ground+garden), R = 17.5 + 1.4 = 18.9 m

Area of garden = Area of outer circle – Area of inner circle

= $\pi R^2-\pi r^2 = \pi(R-r)(R+r)$

= $\frac{22}{7}(18.9-17.5)(18.9+17.5)$

= $\frac{22}{7} \times 1.4 \times 36.4$

= $22 \times 0.2 \times 36.4 = 160.16$ $m^2$

=> Ans – (A)

**2)Â AnswerÂ (D)**

Let the length of the plot be $l$ meters and breadth = $b$ meters

New length = $4l$ meters

Perimeter = $2(4l+b)=480$

=> $4l+b=\frac{480}{2}=240$

=> $b=240-4l$ ———(i)

Area = $(4l \times b)=12800$

=> $lb=\frac{12800}{4}=3200$

Substituting value of $b$ from equation (i)

=> $l(240-4l)=3200$

=> $240l-4l^2=3200$

=> $l^2-60l+800$

=> $l=\frac{-(-60) \pm \sqrt{(-60)^2-(4 \times 1 \times 800)}}{2}$

=> $l=\frac{60 \pm \sqrt{3600-3200}}{2} = \frac{60 \pm \sqrt{400}}{2}$

=> $l=\frac{60 \pm 20}{2}$

=> $l=\frac{60+20}{2},\frac{60-20}{2}$

=> $l=\frac{80}{2},\frac{40}{2}$

=> $l=40,20$ meters

=> Ans – (D)

**3)Â AnswerÂ (A)**

Diameter of circle = side of square = $d$ cm

Area of square = $d^2$ sq. cm

Area of 1 quadrant = $\frac{1}{4} \times \pi \times (\frac{d}{2})^2$

= $\pi \times \frac{d^2}{16}$

=> Area of 4 quadrants = $4 \times \pi \times \frac{d^2}{16} = \frac{\pi d^2}{4}$ sq. cm

Area of shaded region = 168

=> $d^2 – \frac{\pi d^2}{4} = 168$

=> $\frac{1}{4} [d^2 (4 – \pi)] = 168$

=> $d^2 = \frac{168 \times 4}{4 – \pi} = \frac{168 \times 4}{4 – \frac{22}{7}}$

=> $d^2 = \frac{168 \times 4 \times 7}{28 – 22} = \frac{168}{6} \times 28$

=> $d = \sqrt{28 \times 28} = 28$ cm

$\therefore$ Radius = $\frac{28}{2} = 14$ cm

**4)Â AnswerÂ (B)**

Let side of square plot = $s$ ft

=> Area = $s^2 = 361$

=> $s = \sqrt{361} = 19$ ft

Perimeter of square plot = $4 s = 4 \times 19 = 76$ ft

Cost of building fence per foot = Rs. 62

$\therefore$ Cost of building fence around the plot = $76 \times 62 = Rs. 4712$

**5)Â AnswerÂ (C)**

Let the dimension of the room be $3x , 2x , x$ metres

Acc. to ques, => $3x + 2x + x = 18$

=> $x = \frac{18}{6} = 3$ metres

Curved surface area of the room = $2 h (l + b)$

= $2 \times x \times (3x + 2x) = 2x \times 5x$

= $10 (x)^2 = 10 \times (3)^2$

= $10 \times 9 = 90 m^2$

$\therefore$ Total cost incurred on painting only the four walls of the roomÂ = $15 \times 90$

= $Rs. 1,350$

**6)Â AnswerÂ (A)**

Let the dimension of the room be $8x , 7x , 5x$ metres

Acc. to ques, => $8x + 7x + 5x = 24$

=> $x = \frac{24}{20} = 1.2$ metres

Curved surface area of the room = $2 h (l + b)$

= $2 \times 5x \times (8x + 7x) = 10x \times 15x$

= $150 (x)^2 = 150 \times (1.2)^2$

= $150 \times 1.44 = 216 m^2$

$\therefore$ Total cost incurred on painting only the four walls of the roomÂ = $12 \times 216$

= $Rs. 2,592$

**7)Â AnswerÂ (D)**

Let the breadth of rectangle be $x$ cm

=> Length of rectangle = $(x + 3)$ cm

Now, perimeter = $2 (x + x + 3) = 50$

=> $2x + 3 = 25$

=> $x = \frac{22}{2} = 11$

Thus, breadth = 11 cm and length = 14 cm

$\because$ Area of circle = Area of rectangle

=> $\pi r^2 = 14 \times 11$

=> $r^2 = \frac{14 \times 11 \times 7}{22}$

=> $r = \sqrt{7 \times 7} = 7$ cm

$\therefore$ Diameter = 2 * 7 = 14 cm

**8)Â AnswerÂ (E)**

Circumference of circle = 792 m

=> $2 \pi r = 792$

=> $r = \frac{792 \times 7}{22 \times 2}$

=> $r = 7 \times 18 = 126$ m

**9)Â AnswerÂ (D)**

Perimeter of rectangle = 364 metre

=> $2 (l + b) = 364$

=> $l + 84 = \frac{364}{2} = 182$

=> $l = 182 – 84 = 98$ metre

$\therefore$ Radius of circular garden = 98 + 7 = 105 metre

Circumference of garden = $2 \pi r$

= $2 \times \frac{22}{7} \times 105$

= $30 \times 22 = 660$ metre

$\therefore$ Cost of fencing = $660 \times 8$

= Rs.Â $5,280$

**10)Â AnswerÂ (B)**

Perimeter of rectangular plot = $2 (l + b)$

= 2Â (23 + 19) = 2 * 42

= 84 metre = Perimeter of square plot

=> Side of square plot = 84/4 = 21 metre

$\therefore$ Diagonal = $\sqrt{2}$ * side

= $21\sqrt{2}$ metre