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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
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$$79\ cm$$
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$$1224.5\ cm$$
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$$141\ cm$$
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$$70.5\ cm$$
Explanation
Length of photograph = 39.5 cm
Breadth of photograph = 31 cm
$$\therefore$$ 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:
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28$$m^{2}$$
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14$$m^{2}$$
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11$$m^{2}$$
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21$$m^{2}$$
Explanation
Let $$l$$ and $$b$$ be the length and breadth of rectangle.
According to the question-
$$l = 3b - 5 ..... \left( 1 \right)$$
Perimeter of rectangle $$= 22 \; m \; \left( \text{Given} \right)$$
As we know that the perimeter of rectangle is given as-
$$P = 2 \left( l + b \right)$$
Therefore,
$$2 \left( l + b \right) = 22$$
$$l + b = 11$$
$$\left( 3b - 5 \right) + b = 11 \; \left( \text{from } \left( 1 \right) \right)$$
$$4b = 11 + 5$$
$$\Rightarrow b = \cfrac{16}{4} = 4 \; m$$
Substituting the value of $$b$$ in $${eq}^{n} \left( 1 \right)$$, we have
$$l = 3 \times 4 - 5 = 7 \; m$$
Therefore,
Area of rectangle $$= l \times b = 7 \times 4 = 28 \; {m}^{2}$$
Hence the area of rectangle is $$28 \; {m}^{2}$$.
What is the perimeter of a square with side $$6 cm$$?
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$$36\ cm^{2}$$
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$$216\ cm^{3}$$
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$$24\ cm$$
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$$12\ cm$$
Explanation
Perimeter of a square is $$4*side$$
Given $$side=6cm$$
Perimeter of given square $$=4\times 6=24cm$$
A tabletop measures 315 cm by 90 cm. The perimeter of the top of the table is:
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4 m 5 cm
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8 m 10 cm
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24 m 30 cm
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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 $$\displaystyle 80\frac{244}{729}$$sq. m. The length of each side of the filed, is ___________.
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$$8.96$$m
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$$10.26$$m
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$$13.54$$m
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$$12.26$$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
$$x^2=\dfrac{729*80+244}{729}$$
$$x^2=\dfrac{58564}{729}$$
$$x=\dfrac{242}{27}$$
$$x=8.96$$
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.
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Rs. $$48,000$$; Rs. $$45,000$$
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Rs. $$52,000$$; Rs. $$75,000$$
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Rs. $$66,000$$; Rs. $$35,000$$
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Rs. $$17,000$$; Rs. $$85,000$$
Explanation
Given, length of hall is $$45$$ m and width is $$32$$ m
Then area of hall $$=$$ $$45\times 32=1440 m^{2}$$
Length of carpet $$=$$ $$\dfrac{1440}{1.2}$$ .....(given width of carpet is $$1.2$$ m)
Then cost of carpet $$=$$
$$\dfrac{1440}{1.2}\times40=$$ Rs. $$48000$$
When cost of $$80$$ cm width at rate Rs. $$25$$ per meter
Then cost carpet $$=$$
$$\dfrac{1440}{0.80}\times 25=$$ Rs. $$45000$$
The cost of enclosing a rectangular garden with a fence all around, at the rate of $$75$$ paise per metre, is $$Rs. 300$$. If the length of the garden is $$120$$ metres, find the area of the field in square metres.
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$$9600sq.m$$
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$$8400sq.m$$
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$$7500sq.m$$
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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 $$=\cfrac{300}{0.75}$$$$=400$$ m
Perimeter of a rectangle $$=2(length + breadth)$$
$$\therefore 2(length + breadth)=400$$
$$\Rightarrow 2(120+breadth)=400$$
$$\Rightarrow 120 + breadth = 200$$
$$\Rightarrow breadth = 80 $$ m
Thus, the area of the rectangle $$=\ell \times b =80 \times 120$$
$$=9600$$ 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 $$Rs$$. $$8$$ per meter, the total amount spent is ?
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$$Rs. 768$$
0%
$$Rs. 7680$$
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$$Rs. 76800$$
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$$Rs. 768000$$
Explanation
$$l = 900, b = 700$$
Perimeter = $$2 (900 + 700)$$
$$=(1600)$$
$$=3200$$
$$3$$ rounds of fence :
$$=3 (3200)$$
$$=9600 m%$$
$$9600 \times 8$$
$$=Rs. 76800$$
The perimeter of a rectangle is $$170$$ m and its length is $$50$$ m. Then the breadth is:
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$$80$$ m
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$$65$$ m
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$$55$$ m
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$$35$$ m
Explanation
Let the breadth of the rectangle is $$b$$.
Perimeter of the rectangle $$=2(l+b)=170 m$$ {Given}
$$\Rightarrow2(50+b)=170$$
$$\Rightarrow 50+b=85$$
$$ \Rightarrow b=35m$$
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.
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$$34$$
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$$40$$
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$$68$$
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$$88$$
Explanation
If length of one side is $$ 20\ m $$ and area is $$680\ m^2$$
$$\Rightarrow$$ Length of other side $$ = \dfrac {680}{20} = 34\ m $$
Now, length of fencing required for the three sides $$ = 34 + 34 + 20 = 88\ m $$
The length of a rectangle which is $$25\ cm$$ is equal to the side of a square, and the area of the rectangle is $$125\ cm^2$$ less than the area of the square. What is the breadth of the rectangle?
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$$20\ cm$$
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$$25\ cm$$
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$$40\ cm$$
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$$15\ cm$$
Explanation
Area of the square $$= (25 \times 25)\ cm^2= 625\ cm^2$$
$$\therefore$$ Area of the rectangle $$= (625 - 125)\ cm^2$$
$$=500 \ cm^2$$
$$\therefore$$ Breadth of the rectangle $$=\dfrac{Area}{Length}=\displaystyle \frac{500}{25} cm= 20\, cm$$
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:
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$$124\, m^2$$
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$$120\, m^2$$
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$$108\, m^2$$
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$$58\, m^2$$
Explanation
Area of outer rectangle $$= 19\, \times\, 16\, =\, 304\, m^2$$
Area of inner rectangle $$= 15\, \times\, 12\, =\, 180\, m^2$$
$$\therefore$$ 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 ...........
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$$125\ m$$
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$$250\ m^2$$
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$$25\ m$$
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$$250\ m$$
Explanation
The perimeter of the rectangular plot = $$2\times(length + breadth)$$
$$=2\times(75+70)$$
$$=2\times 145$$
$$=250\ m$$
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
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$$330$$ meters
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$$34$$ meters
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$$24$$ meters
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$$18$$ meters
Explanation
Surface area of floor=$$\dfrac{total cost}{cost\quad per\quad m^2}$$
=$$\dfrac{900}{1.25}=720 m^2$$
$$30 \times$$ (width)=$$720 m^2$$
$$width=24 m$$
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
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Rs 768
0%
Rs 7680
0%
Rs 76,800
0%
Rs 768,000
Explanation
l $$\displaystyle = $$ 900 b $$\displaystyle = $$ 700
Perimeter $$\displaystyle = $$ 2(900 + 700)
$$\displaystyle = $$ 2(1600)
$$\displaystyle = $$ 3200
3 rands fence:
$$\displaystyle = $$ 3(3200)
$$\displaystyle = $$ 9600m
$$\displaystyle = $$ 9600 $$\displaystyle \times $$ 8 $$\displaystyle = $$ Rs 76,800
The perimeter of a square is ____ times the length of the side.
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$$2$$
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$$3$$
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$$4$$
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$$5$$
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?
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$$198$$ m
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$$298$$ m
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$$228$$ m
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$$328$$ m
Explanation
In $$1$$ round Raghu walks $$(66+75+102+85)m=328m$$
Find the perimeter of the following figure?
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$$800\ m$$
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$$600\ m$$
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$$1200\ m$$
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None of these
Explanation
Perimeter of figure = sum of all sides $$=100+100+100+100+100+100$$
$$=600m$$
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).
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34900
0%
35100
0%
35900
0%
36100
Explanation
area of square plot =$$123201\ m^2$$
area=$$ (Side)^2$$
A=$$(Side)^2=123201m^2$$
$$S^2=123201$$
$$S=\sqrt { 123201 } =\sqrt { 9\times 9\times 3\times 3\times 13\times 13 } $$
S= $$9\times 3\times 13m=351m$$
S= $$35100$$cm
Answer (B) $$35100$$ cm
Find the perimeter of a square.
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13.2 in
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23.2 in
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33.2 in
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43.2 in
Explanation
The perimeter of a square $$= 4a$$
Where, $$a = side$$
Perimeter $$= 4 \times 5.8$$
$$= 23.2$$ in
Find the sum of the perimeters of the figures given above.
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$$350$$ cm
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$$360$$ cm
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$$370$$ cm
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$$380$$ cm
Explanation
Perimeter is simply the sum of sides of figure.
Perimeter for first figure(triangle) $$=(60+50+60)=170\,cm$$
Perimeter for first figure(rectangle) $$=(60+40+60+40)=200\,cm$$
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?
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$$270\ m$$
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$$230\ m$$
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$$540\ m$$
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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 = $$2 \times (length+breadth)$$
Hence, wire needed $$=$$ $$2 \times(250+20)$$
$$=$$ $$540 \ m$$
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
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$$64 sq. \ m$$
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$$5.6 sq. \ m$$
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$$4.8 sq. \ m$$
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$$48 sq. \ cm$$
Explanation
We know that, a
rea of rectangle $$=l \times b$$, where $$l$$ = length, $$b $$= breadth
$$l \displaystyle = 3.2m$$
$$b \displaystyle = 1.5m$$
$$ \therefore l \displaystyle \times b\displaystyle = 3.2 \displaystyle \times 1.5$$ $$\displaystyle = 4.80$$ 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.
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$$480$$
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$$483$$
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$$467$$
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$$487$$
Explanation
We convert the measurements to cm
Length $$= 39 m 10cm = 3910 cm$$
Width $$= 35 m 70 cm = 3570 cm$$
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 $$= 3910 \times 3570 = 13958700 cm^2$$
Area of one square tile $$= 170 \times 170 = 28900 cm^2$$
The number of tiles $$= \dfrac{13958700}{28900}$$
$$= 483 tiles$$
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?
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$$220$$ m
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$$320$$ m
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$$360$$ m
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$$400$$ m
Explanation
Given ,
For room , length $$=20\ m$$ and breadth $$=12\ m$$
Area of room $$=20\times 12=240\ m^2 -(1)$$
$$75\ cm=0.75\ m$$
Let $$l$$ be the length of carpet
Area of carpet required=$$l\times 0.75\ m^2~~~~~~~~-(2)$$
From eqs.(1) and (2)
$$\therefore l\times 0.75=240\\\to l=\dfrac{240}{0.75}=320\ m$$
The perimeter of a rectangular garden is $$30$$ feet. If its length is $$6$$ feet, what is its width?
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$$9$$ feet
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$$10$$ feet
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$$18 $$feet
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$$21 $$feet
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$$24 $$feet
Explanation
The perimeter of a shape is the distance around it. In particular, the perimeter of a rectangle is given by the formula $$P = 2W + 2L$$. Substitute the correct values of the variables into this formula ($$P = 30$$ and $$L = 6$$) and then solve for the width W:
$$30= 2W + 2(6)$$
$$30= 2W+12$$
$$18= 2W$$
$$W=9$$
Therefore, the width of the garden is $$9$$ feet.
Distance between two dots is $$1\,cm$$
Find the perimeter of Green Rectangle
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$$12$$ cm
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$$14$$ cm
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$$15$$ cm
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$$20$$ cm
Explanation
$$\therefore$$ This is the given rectangle with distance between two dots $$=1cm$$
Length $$(l)=4cm$$
Breadth $$(b)=3cm$$
Perimeter of given rectangle $$=2(l+b)$$
$$=2(4+3)$$
$$=14cm$$
$$\therefore$$ Hence, option (B) is correct.
The area of the floor of a rectangular shape of length $$20 m$$ is $$1200\:cm^2$$. Carpets of 4 cm $$\times$$ 6 cm are available. Find how many carpets are required to cover the hall.
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$$40$$
0%
$$50$$
0%
$$60$$
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None
Explanation
The area of the rectangular floor is $$1200 { cm }^{ 2 }$$.
The dimension of the carpet is $$4 cm \times 6 cm$$.
Hence the area of the carpet is $$24$$
$${ cm }^{ 2 }$$.
Let the number of carpets be n.
For the carpets to completely cover the hall
$$\Rightarrow n \times 24 = 1200$$
$$\Rightarrow n = 50$$
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?
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$$923$$ sq. m, Rs. $$1846$$
0%
$$925$$ sq. m, Rs. $$1850$$
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$$940$$ sq. m, Rs. $$1880$$
0%
none
Explanation
Given,
Length of plot $$=112m,$$ Breadth $$=78m$$
Area of plot=$$112\times 78=8736m^2$$
Now After forming the path
The lenth of remaining plot $$l=112-2(2.5)=108m$$
Breadth $$b=78-2(2.5)=73m$$
Area of remaining plot $$=108\times 73=7811m^2$$
Area of path $$=8736-7811=925m^2$$
Cost of constructing
$$=Area\times rate\\=925\times 2\\=Rs1850$$
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.
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$$80$$ cm
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$$90$$ cm
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$$70$$ cm
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None
Explanation
$$\displaystyle { P }_{ S }={ P }_{ R }$$
$${ P }_{ S }=40+40+40+40=160cm$$
$$\displaystyle { P }_{ R }=2\left( l \right) +\left( b \right) 2=2\left( l+b \right) =2\left( 10+b \right) cm$$
$$\therefore \displaystyle 160=2\left( 10+b \right) $$
$$\Rightarrow \displaystyle 80=10+b$$
$$\Rightarrow \displaystyle b=70$$cm
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