MENSURATION QUESTIONS SSC CGL SET 3

Here is another set of MENSURATION QUESTIONS SSC CGL. The solutions of the Mensuration questions SSC CGL are given at the end of the page. You can practice more mensuration Questions SSC CGL and more Maths practice sets Chapter wise study study material here.

MENSURATION QUESTIONS SSC CGL

MENSURATION QUESTIONS SSC CGL Set 3

 
Check Mensuration formulas before solving questions
 
1. Length of the floor of a rectangular auditorium is 6 m more than the radius of a circle with a circumference of 572 m. The perimeter of the floor of the rectangular auditorium is 356 m. What will be the cost of flooring the auditorium (only the floor of the auditorium), if the cost of flooring is ₹ 12/m2?
A.₹87,954
B.₹91,236
C.₹94,284
D.₹75,490

2. The side of a square is 2 cms less than the length of a rectangle and the breadth of the rectangle is 5 cms less than the side of the square. The area of the square is 324 sq. cms. What is the area of the rectangle?
A.250 sq. cms
B.260 sq. cms
C.254 sq. cms
D.258 sq. cms

3. An equilateral triangle and a regular hexagon have equal perimeters. The ratio of the area of the triangle and that of the hexagon is
A.1 : 1
B.2 : 3
C.3 : 2
D.3 : 4

4. ABCD is a parallelogram with area 120 cm2, and BX : XC =3:2, CY : YD = 2 : 1 and AZ : ZD = 3 : 1. Area (in cm2) of pentagon AXCYZ is
A.47
B.63
C.73
D.79

5. The perimeters of a square and a regular hexagon are equal. The ratio of the area of the hexagon to the area of the square is
A.2√3 : 3
B. √3 : 1
C.3√3 : 2
D.√2 : 3

6. The circumference of a park is 750 m. A and B start walking from the same direction at 6.75 kmph and 4.75 kmph. In what time will they meet each other again?
A.3 hours
B.2.5 hours
C.3.5 hours
D.4 hours

7. The length and breadth of the floor of a room are 20 feet and 10 feet respectively. Square tiles of 2 feet length of three different colours are to be laid on the floor. Black tiles are laid in the first row on all sides. If white tiles are laid in the one-third of the remaining and blue tiles in the rest, how many blue tiles will be there?
A.16
B.32
C.48
D.24

8. If the length of a rectangular field is increased by 20% and the breadth is reduced by 20%, the area of the rectangle will be 192m2. What is the area of the original rectangle?
A.184 m2
B.196 m2
C.204 m2
D.None of these

9. One of the angles of a right angled triangle is 15°, and he hypotenuse is 1 m. The area ofthe triangle(in square cm) is
A.1220
B.1200
C.1250
D.1215

Set 1 of Mensuration questions

Solutions for Mensuration Questions SSC CGL Set 3

Ques:1 Ans-C
Circumference of circle = 2Πr
=> 572 = 2 x 22/7 x r
=> r =(572 x7)/2 x 22 = 91
=>Length of the floor = 91 + 6 = 97 m
=>2(l+b) =356 ; 2 (97 + b) = 356
=> 97 + b = 178
=> B = 178 – 97 = 81 m
=> Area of the floor = 97 x 81 = 7857 sq. m
=> Required cost = 12 x 7857 = ₹944284

Ques:2 Ans – B
Side of a square = √Area
= √324 = 18 cm
Length of the rectangle = 20 cm and breadth = 18 – 5 = 13 cm
Area of rectangle = Length x breadth = 20 x 13 = 260 sq. cm

Ques:3 Ans –B
3x = 6y ; x = side of triangle and y = side of regular hexagon
=> x = 2y ; Area of triangle = √3/4 x2
=> Area of hexagon = 6 X √3/4y2
=> 6 X √3/4 X x2/4 = 3 √3/82
=> Required ratio = √3/4x2 : 3 √3/8 2 = 2 : 3

Ques:4 Ans – D
BC= BX + XC = 3x+ 2x = 5x cm
CD = CY + YD = 2y + y = 3y cm
5x X 3y =120; BC = 20 cm and CD = 6 cm ; BX = 3/5 x 20 = 12cm
YD = 1/3 x 6 = 2 cm
ZD = ¼ x 20 = 5 cm
Area of the shaded region = 120 – tri ABC – tri ZYD = 120 – ½ x 12 x 6 – ½ x 2 x 5 = 79 sq. cm

Ques:5 Ans – A
Side of the square = x units and the side of regular hexagon = y units
4x = 6y => x = 3/2 y;Area of the square = 9/4y2
Area of the hexagon = 3√3/2 x y2
= 3√3/2 x y2 : 9/4y2 = 2√3:3

Ques:6 Ans-A
Time take by A to complete one revolution = 750/ (6.75 x 5)/18 sec = (750×18)/(6.75×5)sec.
Time taken by B = (750 x 18)/(4.75 x 5) sec
Time required = LCM of numerators / HCF of denominators = 750 x 1800 / 125 = 1800 x 6 seconds
= (1800 x 6)/3600 = 3 hours

Ques:7 Ans-A
Area of the floor = 20 x 10 = 200 sq. ft.
Area covered by black tiles = 2(20 + 20 + 6 + 6) = 104 sq. ft
Remaining in Space = 200 – 104 = 96 sq. ft.
Area covered by white tiles = 1/3 x 96 = 32 sq. ft.
Area covered by blue tiles = 96 – 32 = 64 sq. ft.
Area of 1 tile = (2)2 = 4sq. ft.
No. of blue tile = 64/4 = 16

Ques:8 Ans – D
Let the length and breadth be x and y
Area =xy sq m
x X 120/100 X y X 80/100 = 192
=> x X 120/100 X y X 80/100 = 192
=> x X 1.2 X y X 0.8 = 192
=> xy = 192/1.2 X 0.8 = 200 square metre

Ques:9 Ans – C
Sin 15° = sin (45° – 30°) = sin 45° x cos 30° – cos 45° x sin 30°
=1/√2 x √3/2 – 1/√2 x 1/2 = √3/2√2 – 1/2√2 = (√3-1)/2√2
cos 15° = cos (45° – 30°) = cos 45°. cos 30° + sin 45°. sin 30° = 1/√2 x √3/2 + 1/√2 x ½
=√3/2√2 + 1/2√2 = (√3+1)/2√2; AB = AC sin 15°
= (√3 – 1)/2√2 m
BC =AC cos 15° = (√3+1)/2√2 m
Area of ABC = 1/2 x AB x BC = (1/2 x (√3-1)/2√2 x (√3+1)/2√2) sq m
(3-1)/16 sq m
1/8 sq m = 10000/8 = 1250 sq m

Set 2 Mensuration questions to practice