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CBSE Questions for Class 6 Maths Mensuration Quiz 2 - MCQExams.com
CBSE
Class 6 Maths
Mensuration
Quiz 2
The length of the wooden strip required to frame a photograph of length and breadth 39.5 cm and 31 cm respectively, is
Report Question
0%
79
c
m
0%
1224.5
c
m
0%
141
c
m
0%
70.5
c
m
Explanation
Length of photograph = 39.5 cm
Breadth of photograph = 31 cm
∴
Required length of the wooden strip
= Perimeter of photograph
= 2(39.5 + 31) = 2(70.5) = 141 cm
If length of a rectangle is 5 less than 3 times it's breath and perimeter of the rectangle is 22m, then area of the rectangle is:
Report Question
0%
28
m
2
0%
14
m
2
0%
11
m
2
0%
21
m
2
Explanation
Let
l
and
b
be the length and breadth of rectangle.
According to the question-
l
=
3
b
−
5
.
.
.
.
.
(
1
)
Perimeter of rectangle
=
22
m
(
Given
)
As we know that the perimeter of rectangle is given as-
P
=
2
(
l
+
b
)
Therefore,
2
(
l
+
b
)
=
22
l
+
b
=
11
(
3
b
−
5
)
+
b
=
11
(
from
(
1
)
)
4
b
=
11
+
5
⇒
b
=
16
4
=
4
m
Substituting the value of
b
in
e
q
n
(
1
)
, we have
l
=
3
×
4
−
5
=
7
m
Therefore,
Area of rectangle
=
l
×
b
=
7
×
4
=
28
m
2
Hence the area of rectangle is
28
m
2
.
What is the perimeter of a square with side
6
c
m
?
Report Question
0%
36
c
m
2
0%
216
c
m
3
0%
24
c
m
0%
12
c
m
Explanation
Perimeter of a square is
4
∗
s
i
d
e
Given
s
i
d
e
=
6
c
m
Perimeter of given square
=
4
×
6
=
24
c
m
A tabletop measures 315 cm by 90 cm. The perimeter of the top of the table is:
Report Question
0%
4 m 5 cm
0%
8 m 10 cm
0%
24 m 30 cm
0%
None of these
Explanation
Length of top of the table =
315 cm
Breadth of top of the table = 90 cm
Perimeter = 2(315 + 90) = 2(405) = 810 cm
= 8 m 10 cm
The area of a square field is
80
244
729
sq. m. The length of each side of the filed, is ___________.
Report Question
0%
8.96
m
0%
10.26
m
0%
13.54
m
0%
12.26
m
Explanation
area of any square with side
a
m is
=
a
2
sq.m.
let the length of the side of the square field is
x
m
x
2
=
729
∗
80
+
244
729
x
2
=
58564
729
x
=
242
27
x
=
8.96
length of the side is
8.96
m
The rate for a
1.2
m wide carpet is Rs.
40
per metre; find the cost of covering a hall
45
m long and
32
m wide with this carpet. Also, find the cost of carpeting the same hall if the carpet,
80
cm wide, is at Rs.
25
per metre.
Report Question
0%
Rs.
48
,
000
; Rs.
45
,
000
0%
Rs.
52
,
000
; Rs.
75
,
000
0%
Rs.
66
,
000
; Rs.
35
,
000
0%
Rs.
17
,
000
; Rs.
85
,
000
Explanation
Given, length of hall is
45
m and width is
32
m
Then area of hall
=
45
×
32
=
1440
m
2
Length of carpet
=
1440
1.2
.....(given width of carpet is
1.2
m)
Then cost of carpet
=
1440
1.2
×
40
=
Rs.
48000
When cost of
80
cm width at rate Rs.
25
per meter
Then cost carpet
=
1440
0.80
×
25
=
Rs.
45000
The cost of enclosing a rectangular garden with a fence all around, at the rate of
75
paise per metre, is
R
s
.
300
. If the length of the garden is
120
metres, find the area of the field in square metres.
Report Question
0%
9600
s
q
.
m
0%
8400
s
q
.
m
0%
7500
s
q
.
m
0%
None of these
Explanation
The cost of enclosing a rectangular garden with a fence all around, at the rate of
75
paise (=
0.75
rupee) per meter, is Rs.
300
.
Thus, the perimeter of the rectangular garden
=
300
0.75
=
400
m
Perimeter of a rectangle
=
2
(
l
e
n
g
t
h
+
b
r
e
a
d
t
h
)
∴
2
(
l
e
n
g
t
h
+
b
r
e
a
d
t
h
)
=
400
⇒
2
(
120
+
b
r
e
a
d
t
h
)
=
400
⇒
120
+
b
r
e
a
d
t
h
=
200
⇒
b
r
e
a
d
t
h
=
80
m
Thus, the area of the rectangle
=
ℓ
×
b
=
80
×
120
=
9600
sq. m
The length and breadth of a rectangular plot are
900
m
and
700
m
respectively.
If three rounds of fence is fixed around the field at the cost of
R
s
.
8
per meter, the total amount spent is ?
Report Question
0%
R
s
.
768
0%
R
s
.
7680
0%
R
s
.
76800
0%
R
s
.
768000
Explanation
l
=
900
,
b
=
700
Perimeter =
2
(
900
+
700
)
=
(
1600
)
=
3200
3
rounds of fence :
=
3
(
3200
)
=
9600
m
9600
×
8
=
R
s
.
76800
The perimeter of a rectangle is
170
m and its length is
50
m. Then the breadth is:
Report Question
0%
80
m
0%
65
m
0%
55
m
0%
35
m
Explanation
Let the breadth of the rectangle is
b
.
Perimeter of the rectangle
=
2
(
l
+
b
)
=
170
m
{Given}
⇒
2
(
50
+
b
)
=
170
⇒
50
+
b
=
85
⇒
b
=
35
m
A rectangular field is to be fenced on three sides leaving a side of
20
m
uncovered. If the area of the field is
680
m
2
, how many metres of fencing will be required.
Report Question
0%
34
0%
40
0%
68
0%
88
Explanation
If length of one side is
20
m
and area is
680
m
2
⇒
Length of other side
=
680
20
=
34
m
Now, length of fencing required for the three sides
=
34
+
34
+
20
=
88
m
The length of a rectangle which is
25
c
m
is equal to the side of a square, and the area of the rectangle is
125
c
m
2
less than the area of the square. What is the breadth of the rectangle?
Report Question
0%
20
c
m
0%
25
c
m
0%
40
c
m
0%
15
c
m
Explanation
Area of the square
=
(
25
×
25
)
c
m
2
=
625
c
m
2
∴
Area of the rectangle
=
(
625
−
125
)
c
m
2
=
500
c
m
2
∴
Breadth of the rectangle
=
A
r
e
a
L
e
n
g
t
h
=
500
25
c
m
=
20
c
m
The floor of a rectangular room is
15
m
long and
12
m
wide. The room is surrounded by a verandah of width
2
m
on all its sides. The area of verandah is:
Report Question
0%
124
m
2
0%
120
m
2
0%
108
m
2
0%
58
m
2
Explanation
Area of outer rectangle
=
19
×
16
=
304
m
2
Area of inner rectangle
=
15
×
12
=
180
m
2
∴
Reqd. area
=
(
304
−
180
)
m
2
=
124
m
2
.
The perimeter of a rectangular plot whose length is
75
m
and breadth is
50
m
is ...........
Report Question
0%
125
m
0%
250
m
2
0%
25
m
0%
250
m
Explanation
The perimeter of the rectangular plot =
2
×
(
l
e
n
g
t
h
+
b
r
e
a
d
t
h
)
=
2
×
(
75
+
70
)
=
2
×
145
=
250
m
The cost of levelling a rectangular ground at Rs
1.25
per sq meter is Rs
900
. If the length of the ground is
30
meters, then the width is
Report Question
0%
330
meters
0%
34
meters
0%
24
meters
0%
18
meters
Explanation
Surface area of floor=
t
o
t
a
l
c
o
s
t
c
o
s
t
p
e
r
m
2
=
900
1.25
=
720
m
2
30
×
(width)=
720
m
2
w
i
d
t
h
=
24
m
The length and breadth of a rectangular plot are 900 m and 700 m respectively If three rands of fence is fixed around the field at the cost of Rs.8 per meter the total amount spend is
Report Question
0%
Rs 768
0%
Rs 7680
0%
Rs 76,800
0%
Rs 768,000
Explanation
l
=
900 b
=
700
Perimeter
=
2(900 + 700)
=
2(1600)
=
3200
3 rands fence:
=
3(3200)
=
9600m
=
9600
×
8
=
Rs 76,800
The perimeter of a square is ____ times the length of the side.
Report Question
0%
2
0%
3
0%
4
0%
5
Explanation
As we know that all the sides of a squqre are equal.
So its perimeter will be
4
times the length of the side.
Hence, the answer is
4
.
Raghu walks around a park everyday. How far does he walk in one
round?
Report Question
0%
198
m
0%
298
m
0%
228
m
0%
328
m
Explanation
In
1
round Raghu walks
(
66
+
75
+
102
+
85
)
m
=
328
m
Find the perimeter of the following figure?
Report Question
0%
800
m
0%
600
m
0%
1200
m
0%
None of these
Explanation
Perimeter of figure = sum of all sides
=
100
+
100
+
100
+
100
+
100
+
100
=
600
m
A man purchased a plot which is in the shape of a square The area of the plot is
12
hectares and
3201
m
2
Find the length of each side of the plot (in cm).
Report Question
0%
34900
0%
35100
0%
35900
0%
36100
Explanation
area of square plot =
123201
m
2
area=
(
S
i
d
e
)
2
A=
(
S
i
d
e
)
2
=
123201
m
2
S
2
=
123201
S
=
√
123201
=
√
9
×
9
×
3
×
3
×
13
×
13
S=
9
×
3
×
13
m
=
351
m
S=
35100
cm
Answer (B)
35100
cm
Find the perimeter of a square.
Report Question
0%
13.2 in
0%
23.2 in
0%
33.2 in
0%
43.2 in
Explanation
The perimeter of a square
=
4
a
Where,
a
=
s
i
d
e
Perimeter
=
4
×
5.8
=
23.2
in
Find the sum of the perimeters of the figures given above.
Report Question
0%
350
cm
0%
360
cm
0%
370
cm
0%
380
cm
Explanation
Perimeter is simply the sum of sides of figure.
Perimeter for first figure(triangle)
=
(
60
+
50
+
60
)
=
170
c
m
Perimeter for first figure(rectangle)
=
(
60
+
40
+
60
+
40
)
=
200
c
m
So, the sum of perimeter
=
170
+
200
=
370
cm
A playground which is
250
m long and
20
m broad is to be fenced with wire. How much wire is needed?
Report Question
0%
270
m
0%
230
m
0%
540
m
0%
None
Explanation
Given that, a playground which is
250
m
long and
20
m
wide is fenced with wire.
To find out: The amount of wire.
Amount of wire needed
=
Perimeter of a rectangle =
2
×
(
l
e
n
g
t
h
+
b
r
e
a
d
t
h
)
Hence, wire needed
=
2
×
(
250
+
20
)
=
540
m
Hence, the amount of wire required to fence the playground is
540
m
.
Length and breadth of a rectangle are 3.2 m and 150 cm. Then the area is
Report Question
0%
64
s
q
.
m
0%
5.6
s
q
.
m
0%
4.8
s
q
.
m
0%
48
s
q
.
c
m
Explanation
We know that, a
rea of rectangle
=
l
×
b
, where
l
= length,
b
= breadth
l
=
3.2
m
b
=
1.5
m
∴
l
×
b
=
3.2
×
1.5
=
4.80
sq m.
A hall-room
39
m
10
cm long and
35
m
70
cm broad is to be paved with equal square tiles. Find the number of tiles required, so that the tiles exactly fit.
Report Question
0%
480
0%
483
0%
467
0%
487
Explanation
We convert the measurements to cm
Length
=
39
m
10
c
m
=
3910
c
m
Width
=
35
m
70
c
m
=
3570
c
m
To get the length of the sides of the square tile, we find the Greatest Common Divisor of
3910
and
3570
The GCD of
3910
and
3570
=
170
Area of the hall
=
3910
×
3570
=
13958700
c
m
2
Area of one square tile
=
170
×
170
=
28900
c
m
2
The number of tiles
=
13958700
28900
=
483
t
i
l
e
s
How many metres of a carpet
75
cm wide will be required to cover the floor of a room which is
20
metres long and
12
metres broad?
Report Question
0%
220
m
0%
320
m
0%
360
m
0%
400
m
Explanation
Given ,
For room , length
=
20
m
and breadth
=
12
m
Area of room
=
20
×
12
=
240
m
2
−
(
1
)
75
c
m
=
0.75
m
Let
l
be the length of carpet
Area of carpet required=
l
×
0.75
m
2
−
(
2
)
From eqs.(1) and (2)
∴
l
×
0.75
=
240
→
l
=
240
0.75
=
320
m
The perimeter of a rectangular garden is
30
feet. If its length is
6
feet, what is its width?
Report Question
0%
9
feet
0%
10
feet
0%
18
feet
0%
21
feet
0%
24
feet
Explanation
The perimeter of a shape is the distance around it. In particular, the perimeter of a rectangle is given by the formula
P
=
2
W
+
2
L
. Substitute the correct values of the variables into this formula (
P
=
30
and
L
=
6
) and then solve for the width W:
30
=
2
W
+
2
(
6
)
30
=
2
W
+
12
18
=
2
W
W
=
9
Therefore, the width of the garden is
9
feet.
Distance between two dots is
1
c
m
Find the perimeter of Green Rectangle
Report Question
0%
12
cm
0%
14
cm
0%
15
cm
0%
20
cm
Explanation
∴
This is the given rectangle with distance between two dots
=
1
c
m
Length
(
l
)
=
4
c
m
Breadth
(
b
)
=
3
c
m
Perimeter of given rectangle
=
2
(
l
+
b
)
=
2
(
4
+
3
)
=
14
c
m
∴
Hence, option (B) is correct.
The area of the floor of a rectangular shape of length
20
m
is
1200
c
m
2
. Carpets of 4 cm
×
6 cm are available. Find how many carpets are required to cover the hall.
Report Question
0%
40
0%
50
0%
60
0%
None
Explanation
The area of the rectangular floor is
1200
c
m
2
.
The dimension of the carpet is
4
c
m
×
6
c
m
.
Hence the area of the carpet is
24
c
m
2
.
Let the number of carpets be n.
For the carpets to completely cover the hall
⇒
n
×
24
=
1200
⇒
n
=
50
A rectangular grassy plot is
112
m by
78
m. If has a gravel path
2.5
m wide all round on the inside. Find the area of the path and the cost of constructing it at Rs.
2
per square metre?
Report Question
0%
923
sq. m, Rs.
1846
0%
925
sq. m, Rs.
1850
0%
940
sq. m, Rs.
1880
0%
none
Explanation
Given,
Length of plot
=
112
m
,
Breadth
=
78
m
Area of plot=
112
×
78
=
8736
m
2
Now After forming the path
The lenth of remaining plot
l
=
112
−
2
(
2.5
)
=
108
m
Breadth
b
=
78
−
2
(
2.5
)
=
73
m
Area of remaining plot
=
108
×
73
=
7811
m
2
Area of path
=
8736
−
7811
=
925
m
2
Cost of constructing
=
A
r
e
a
×
r
a
t
e
=
925
×
2
=
R
s
1850
A square and a rectangle have same perimeter. The side of square is
40
cm and length of rectangle is
10
cm, find breadth of rectangle.
Report Question
0%
80
cm
0%
90
cm
0%
70
cm
0%
None
Explanation
P
S
=
P
R
P
S
=
40
+
40
+
40
+
40
=
160
c
m
P
R
=
2
(
l
)
+
(
b
)
2
=
2
(
l
+
b
)
=
2
(
10
+
b
)
c
m
∴
160
=
2
(
10
+
b
)
⇒
80
=
10
+
b
⇒
b
=
70
cm
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