MCQ Questions for 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

0%
$79\ cm$
0%
$1224.5\ cm$
0%
$141\ cm$
0%
$70.5\ cm$

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:

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-

$l = 3b - 5 ..... \left( 1 \right)$

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

$6 cm$ ?

0%
$36\ cm^{2}$
0%
$216\ cm^{3}$
0%
$24\ cm$
0%
$12\ cm$

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:

0%
4 m 5 cm
0%
8 m 10 cm
0%
24 m 30 cm
0%
None of these

The area of a square field is

$\displaystyle 80\frac{244}{729}$ sq. m. The length of each side of the filed, is ___________.

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

$x^2=\dfrac{729*80+244}{729}$

$x^2=\dfrac{58564}{729}$

$x=\dfrac{242}{27}$

$x=8.96$

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.

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

$=$

$45\times 32=1440 m^{2}$

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.

0%
$9600sq.m$
0%
$8400sq.m$
0%
$7500sq.m$
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

$=\cfrac{300}{0.75}$

$=400$ 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 ?

0%
$Rs. 768$
0%
$Rs. 7680$
0%
$Rs. 76800$
0%
$Rs. 768000$

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:

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}

$\Rightarrow2(50+b)=170$

$\Rightarrow 50+b=85$

$\Rightarrow b=35m$

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.

0%
$34$
0%
$40$
0%
$68$
0%
$88$

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?

0%
$20\ cm$
0%
$25\ cm$
0%
$40\ cm$
0%
$15\ cm$

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:

0%
$124\, m^2$
0%
$120\, m^2$
0%
$108\, m^2$
0%
$58\, m^2$

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 ...........

0%
$125\ m$
0%
$250\ m^2$
0%
$25\ m$
0%
$250\ m$

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

0%
$330$ meters
0%
$34$ meters
0%
$24$ meters
0%
$18$ 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

0%
Rs 768
0%
Rs 7680
0%
Rs 76,800
0%
Rs 768,000

Explanation

$\displaystyle =$ 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.

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?

0%
$198$ m
0%
$298$ m
0%
$228$ m
0%
$328$ m

Explanation

$1$ round Raghu walks

$(66+75+102+85)m=328m$

Find the perimeter of the following figure?

0%
$800\ m$
0%
$600\ m$
0%
$1200\ m$
0%
None of these

Explanation

Perimeter of figure = sum of all sides

$=100+100+100+100+100+100$

$=600m$

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).

0%
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.

0%
13.2 in
0%
23.2 in
0%
33.2 in
0%
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.

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)

$=(60+50+60)=170\,cm$

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?

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 =

$2 \times (length+breadth)$

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

0%
$64 sq. \ m$
0%
$5.6 sq. \ m$
0%
$4.8 sq. \ m$
0%
$48 sq. \ cm$

Explanation

We know that, area 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.

0%
$480$
0%
$483$
0%
$467$
0%
$487$

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?

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

$=20\times 12=240\ m^2 -(1)$

$75\ cm=0.75\ m$

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?

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

$P = 2W + 2L$ . 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

0%
$12$ cm
0%
$14$ cm
0%
$15$ cm
0%
$20$ 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.

2063885_428653_ans_7232821d916242d892c1de93a17a6139.png

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.

0%
$40$
0%
$50$
0%
$60$
0%
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?

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

$=112m,$ 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.

0%
$80$ cm
0%
$90$ cm
0%
$70$ cm
0%
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

CBSE Questions for Class 6 Maths Mensuration Quiz 2 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

CBSE Questions for Class 6 Maths Mensuration Quiz 2 - MCQExams.com

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Practice Class 6 Maths Quiz Questions and Answers

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