Mensuration Questions For SSC MTS
Download Top-20 SSC MTS Mensuration Questions PDF. Mensuration questions based on asked questions in previous year exam papers very important for the SSC MTS exam.
Download Mensuration Questions For SSC MTS
Take a free mock test for SSC MTS
SSC MTS Previous Papers (Download PDF)
Question 1: A solid cuboid of dimensions 8 cm x 4 cm x 2 cm is melted and cast into identical cubes of edge 2 cm. Number of such identical cubes is
a) 8
b) 4
c) 10
d) 16
Question 2: A platform of width 1 m is to be constructed around a rectangular park of dimensions 100 m x 80 m. If the cost of constructing 1 square metre of the platform is Rs.75, then the total cost incurred in the process will be
a) Rs. 23,700
b) Rs. 36,400
c) Rs. 27,300
d) Rs. 34,600
Question 3: The four walls of an empty room of dimensions 24ft x 20ft x 16ft are to be painted. There are two windows of dimensions 4ft x 6ft on the opposite walls. If a painter can paint 20 sq. ft area per hour and his charges are Rs. 18 per hour, how much money will be spent to paint the four walls of the room?
a) Rs. 1386
b) Rs. 1472
c) Rs. 1224
d) Rs. 1286
Question 4: Analyze the figure shown below in which DE || BC and the other dimensions are as follows AD=3 cm, BD=4 cm, AE=4.4 cm and DE=6 cm. Calculate the length (in cm) of BC.
a) 6
b) 8
c) 12
d) 14
Question 5: Arun has an ice cream vending machine of dimensions 1.1m x 1m x 0.6 m. The machine is completely filled with ice cream. The machine fills ice cream into cones of height 7 cm and radius 3 cm to the brim. The number of cones that can be filled is
(Take 1 m = 100 cm)
a) 1000
b) 10,000
c) 1,00,000
d) 10,00,000
Take a free mock test for SSC MTS
Question 6: There is a cylindrical vessel. Arjun, is trying to fill it with water using a conical vessel of the same dimensions (radius and height). How many times does he have to pour water into the cylindrical vessel to fill it to the brim. (He always pours water by filling the conical vessel to the brim.)
a) 2
b) 3
c) 5
d) Cannot be determined
Question 7: A footpath of width 2m runs outside a square lawn of side 6m. The foot path is to be paved with small paver blocks of dimensions 40 cm x 40 cm. How many paver blocks are required to complete the process ? ( 1 m = 100 cm).
a) 400
b) 200
c) 100
d) 175
Question 8: The residents of a city fetch water in buckets from a tank. The tank is cubical in shape. One side of the tank measures 10 m. The buckets are cuboidal in shape. The dimensions of a bucket is 0.5m x 1 m x 0.5m. How many buckets of water can be fetched from a half full tank?
a) 1000
b) 2000
c) 3000
d) 4000
Question 9: Ravi, a painter, charges Rs.1 to paint every 6 square metres of wall. Ravi gets a contract to paint the walls of an auditorium of dimensions 30m x 40 m x 50 m. The floor and the ceiling are not to be painted. If it is known that the floor has the largest area, then the amount earned by Ravi is
a) Rs. 32400
b) Rs. 5400
c) Rs. 6000
d) Rs. 900
Question 10: Ram has a house of area 1200 sqft. He wants to pave the floor with tiles of dimension 4 feet x 3 feet. If the cost of 1 tile is Rs. 16 and the cost of labour is Rs. 6 per tile, the total cost of paving the floor is
a) Rs. 2200
b) Rs. 1600
c) Rs. 2400
d) Rs, 600
SSC Study Material (18000 Solved Questions)
General Science Notes for SSC Exams
Question 11: Praveen wants to paint the walls of his rooms (except the floor). The dimensions of his room are 10 m x 12 m x 5 m. The floor (and the ceiling) has the maximum area as compared to the surface area of any of the wall. If the cost of painting 1 square metre is Rs. 8, the total cost of painting is
a) Rs. 3680
b) Rs. 1760
c) Rs. 2340
d) Rs. 2720
Question 12: Preeta has a rectangle of dimensions 12 cm by 16 cm. First, she rotates the rectangle by its smaller side and then by its larger side. The ratio of the volumes swept in the 2 cases is
a) 4:3
b) 3:4
c) 2:3
d) 3:2
Question 13: A factory plans to construct a cuboidal vessel of dimensions 18m x 12m x 8m. Due to space constraints, the factory remodels the vessel in the form of cube with no change in the volume. The side of the cube is
a) 12 m
b) 14 m
c) 16 m
d) 10 m
Question 14: The perimeter of a rectangular plot is 48 m and area is 108 m2. The dimensions of the plot are
a) 36 m and 3 m
b) 12 m and 9 m
c) 27 m and 4 m
d) 18 m and 6 m
Question 15: The perimeter of a rectangular plot is 48 m and area is 108 m2. The dimensions of the plot are
a) 36 m and 3 m
b) 12 m and 9 m
c) 27 m and 4 m
d) 18 m and 6 m
Question 16: A cylinder is made from a rectangular sheet of dimension 11 cm X 22 cm. What can be the maximum surface area of the base of such cylinder?
a) 30 sq. sm
b) 42 sq. cm
c) 37.5 sq. cm
d) 38.5 sq. cm
Question 17: A rectangular sheet of dimensions 80 cm by 30 cm is cut off at corners by square of side 5 cm. The remaining sheet is folded to form a rectangular box. Find the volume of the box
a) 10500 $\text{cm}^3$
b) 8000 $\text{cm}^3$
c) 12000 $\text{cm}^3$
d) 7000 $\text{cm}^3$
200+ SSC Important Practice Sets
General Knowledge Q&A for Competitive Exams (Download PDF)
Answers & Solutions:
1) Answer (A)
Length, breadth and height of cuboid respectively are = 8 cm, 4 cm and 2 cm
Side of cube = 2 cm
=> Number of cubes formed = Volume of cuboid/Volume of cube
= $\frac{l\times b\times h}{a^3}$
= $\frac{8\times4\times2}{2^3}$
= $4\times2=8$
=> Ans – (A)
2) Answer (C)
The area of the platform will be equal to the difference between the areas of the park with and without the platform.
Area of the park with the platform = 102*82 = 8364 square metre
Area of the park without the platform = 100*80 = 8000 square metre
Area of the platform = 364 square metre
Cost of constructing the platform = 364*75 = Rs. 27,300.
Therefore, option C is the right answer.
3) Answer (C)
Area of the four walls = 2 (lh + bh)
= 2 (24 * 16 + 20 * 16) sq. ft = 1408 sq. ft
Area of the two windows = 2 * 4 * 6 sq. ft = 48 sq. ft
Total area to be painted = (1408 – 48) sq. ft = 1360 sq.ft
Time required = 1360/20 hr = 68 hr
Money required = Rs. 68 * 18 = Rs. 1224
Hence, option C is the correct answer.
4) Answer (D)
5) Answer (B)
Volume of the vending machine $= 1.1$ x $0.6$ x $1$ = $0.66 m^3$
$0.66 m^3 = 0.66$ x $10^6 cm^3$ = $66$ x $10^4 cm^3$.
Volume of 1 cone = $\frac{\pi}{3} r^2 h$ = $\frac{22}{7} * \frac{1}{3} * 3^2 *7$
= $66 cm^3$
Number of cones that can be filled = $\frac{66 * 10^4}{66}$ = $10,000$ cones.
Hence, option B is the right answer.
6) Answer (B)
The volume of a cylinder = $\pi r^2 h$
The volume of a cone = $\frac{1}{3} \pi r^2 h$
We see that the volume of the cone is one-third the volume of a cylinder. Hence, he has to pour the water three times.
7) Answer (A)
Area of the lawn = $6^2 = 36 m^2$
Area of the lawn + footpath = $(6+2+2)^2 = 10^2 = 100 m^2$
Area of the foot path = $ 100 – 36 = 64 m^2$
Area of 1 paver block =$0.4*0.4 = 0.16m^2$
Number of paver blocks required = $64/0.16 = 400$
Therefore, option A is the right answer.
8) Answer (B)
Volume of the tank = $10^3 = 1000 m^3$
Volume of a bucket = $ 0.5$ x $1 $x $0.5$ =$ 0.25 m^3$
A half full tank will contain $500 m^3$ of water.
Number of buckets that can be filled = $500/0.25$ = $2000$ buckets.
Therefore, option B is the right answer.
9) Answer (D)
The side with the largest area is $40m$ x $50 m $= $2000 m^2.$ This must be the area of the floor and the ceiling. We must exclude these 2 areas as they are not painted.
The area of the 4 walls are $2*(30*40 + 30*50)$ = $5400 m^2.$
Cost of painting $= 5400/6 = Rs. 900.$
Therefore, option D is the right answer.
10) Answer (A)
Area of 1 tile = 4*3 = 12 square feet.
Total number of tiles required = 1200/12 = 100
Cost of tiles = 16*100 = Rs. 1600
Cost of labour = 6*100 = Rs. 600
Total cost = Rs. 1600 + Rs. 600 = Rs. 2200
Therefore, option A is the right answer.
11) Answer (D)
The floor is of the maximum area. This means that the floor must have the dimensions 10m x 12m.
Total area to be painted = 2*10*5 + 2*12*5 + 10*12(ceiling)
= 100 + 120 + 120
= 340 square metres.
Cost of painting 1 square metre = Rs. 8
Cost of painting 340 square metres = 340*8 = Rs. 2720
Therefore, option D is the right answer.
12) Answer (A)
When a rectangle is rotated, the volume swept will be a cylinder.
The height of the cylinder will be the base about which the rectangle is rotated. The other side will be equal to the radius of the cylinder.
When the smaller side is the base, radius = $16 cm$ and height $= 12 cm.$
Volume $= \pi*16^2* 12 = \pi*16*16*12$
When the larger side is the base, radius $= 12 cm$ and height $= 16 cm.$
Volume $= \pi*12^2* 16 = \pi*16*12*12$
Ratio $= (\Pi*16*16*12)/(\pi*16*12*12) = 16/12 = 4:3$
Therefore, option A is the right answer.
13) Answer (A)
Volume of the cuboid $= 18*12*8 = 1728 m^3$
Volume of cube = volume of cuboid.
$a^3 = 1728 m^3$
$=> a = 12 m$
Hence, option A is the right answer.
14) Answer (D)
Let the length of the plot be $l$ and breadth be $b$
=> Perimeter of rectangular plot = $2(l + b) = 48$
=> $l + b = 24$
Area of plot = $lb = 108$
=> $l (24 – l) = 108$
=> $l^2 – 24l + 108 = 0$
=> $l^2 – 6l – 18l + 108 = 0$
=> $l(l-6) – 18(l-6) = 0$
=> $l = 6, 18$
When $l = 6$ => $b = 18$
and when $l = 18$ => $b = 6$
=> Dimensions of the plot are 18 m and 6 m.
15) Answer (D)
Let the length of the plot be $l$ and breadth be $b$
=> Perimeter of rectangular plot = $2(l + b) = 48$
=> $l + b = 24$
Area of plot = $lb = 108$
=> $l (24 – l) = 108$
=> $l^2 – 24l + 108 = 0$
=> $l^2 – 6l – 18l + 108 = 0$
=> $l(l-6) – 18(l-6) = 0$
=> $l = 6, 18$
When $l = 6$ => $b = 18$
and when $l = 18$ => $b = 6$
=> Dimensions of the plot are 18 m and 6 m.
16) Answer (D)
The cylinder can be formed by either rolling the sheet along the 11cm edge or along the 22 cm edge.
But the base area will be maximum when the base radius is maximum which will be maximum when the cylinder is rolled along the 22 cm edge.
In this case $2 \pi r = 22$ = > r = 7/2
So, area = $\pi r^2 = 77/2$ sq cm
17) Answer (D)
Dimension of the rectangular box would be ( 80 – 2(5) ) $\text{cm}$ x ( 30 – 2(5) ) $\text{cm}$ x 5 $\text{cm}$
Therefore , the volume of the box would be , V = 70 x 20 x 5 = 7000 $\text{cm}^3$
Hence, the correct Option is D