MCQ - Maths STD 6 Questions - Vidyadip

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22 questions · auto-graded multiple-choice test.

MCQ 11 Mark

Mark the correct alternative in the following question:
The area of a rectangle is $650cm^2$ and its breadth is $13cm.$ The perimeter of the rectangle is:

A
$63cm$
B
$130cm$
C
$100cm$
✓
$126cm$

Answer

Correct option: D.

$126cm$

$126cm$
Area of the rectangle $=650 cm^2$
Breadth of the rectangle $=13 cm$
As, length of the rectangle $=\frac{\text { Area }}{\text { Breadth }}$
$=\frac{650}{13}$
$=50 cm$
So, the perimeter of the rectangle $=2$ (length + breadth)
$=2(13+50)$
$=2 \times 63$
$=126 cm$

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MCQ 21 Mark

If the length of the diagonal of a square is $20\ cm,$ then its perimeter is:

A
$10\sqrt{2}\ \text{cm}$
B
$40\ \text{cm}$
✓
$40 \sqrt{2}\ \text{cm}$
D
$200\ \text{cm}$

Answer

Correct option: C.

$40 \sqrt{2}\ \text{cm}$

Length of diagonal $= 20\ cm$
Length of side of a square $=\frac{\text{Length of diagonal}}{\sqrt{2}}$
$=\frac{20}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$
$=10\sqrt{2}$
Therefore, perimeter of the square is $4 \times$ side $=4\times10\sqrt{2}\ \text{cm}$
$=40\sqrt{2}\ \text{cm}$

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MCQ 31 Mark

Mark the correct alternative in the following question:
The length of the diagonal of a square is $20\ cm$. Its area is:

A
$400\ cm^2$
✓
$200\ cm^2$
C
$300\ cm^2$
D
$100\sqrt{2}\text{cm}^{2}$

Answer

Correct option: B.

$200\ cm^2$

The area of the square $=\frac{1}{2}\times\text{Diagonal}\times\text{Diagonal}$

$=\frac{1}{2}\times20\times20$
$=\frac{400}{2}$
$=200\text{cm}^{2}$

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MCQ 41 Mark

The cost of putting a fence around a square field at As $2.50$ per metre is As $200.$ The length of each side of the field is:

A
$80m$
B
$40m$
✓
$20m$
D
None of these

Answer

Correct option: C.

$20m$

$20m$

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MCQ 51 Mark

The length of a rectangle is three tmies of its width. If the length of the diagonal is $8\sqrt{10}\text{m}$, then the perimeter of the rectangle is:

A
$15\sqrt{10}\text{m}$
B
$16\sqrt{10}\text{m}$
C
$24\sqrt{10}\text{m}$
✓
$64\text{m}$

Answer

Correct option: D.

$64\text{m}$

Let us consider a rectangle $\ce{ABCD}.$

Also, let us assume that the width of the rectangle, i.e., $BC$ be $\times m.$

It is given that the length is three times width of the rectangle.
Therefore, length of the rectangle, i.e., $AB = 3 \times m$
Now, $AC$ is the diagonal of rectangle.
In right angled triangle $\ce{ABC}.$
$\text{AC}^{2}=\text{AB}^{2}+\text{BC}^{2}$
$\big(8\sqrt{10}\big)^{2}=\big(3\text{x}\big)^{2}+\text{x}^{2}$
$640=9\text{x}^{2}+\text{x}^{2}$
$640=10\text{x}^{2}$
$\text{x}^{2}=\frac{64}{10}=64$
$\text{x}=\sqrt{64}=8\text{m}$
Thus, breadth of the rectangle $= x = 8m$
Similarly, length of the rectangle $= 3x = 3 \times 8 = 24m$
Perimeter of the rectangle $= 2($Length $+$ Breadth$)$
$= 2(24 + 8)$
$= 2 \times 32 $
$= 64m$

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MCQ 61 Mark

If the cost of fencing a rectangular field at $Rs. 7.50$ per metre is $Rs. 600,$ and the length of the field is $24m,$ then the breadth of the field is:

A
$8m$
B
$18m$
C
$24m$
✓
$16m$

Answer

Correct option: D.

$16m$

Cost of fencing the rectangular field $= Rs. 600$
Rate of fencing the field $= Rs. 7.50m$
Therefore, perimeter of the field $=\frac{\text{Cost of fencing}}{\text{Rate of fencing}}$
$=\frac{600}{7.50}=80\text{m}$
Now, length of the field $= 24m$
Therefore, breadth of the field $=\frac{\text{Perimeter}}{2}-\text{Length}$
$=\frac{80}{2}-24=16\text{m}$

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MCQ 71 Mark

If a diagonal of a rectangle is thrice its smaller side, then its length and breadth are in the ratio.

A
$3:1$
B
$\sqrt{3}:1$
C
$\sqrt{2}:1$
✓
$2\sqrt{2}:1$

Answer

Correct option: D.

$2\sqrt{2}:1$

Let us assume that the length of the smaller side of the rectangle,
i.e., $BC$ be $x$ and length of the larger side ,
i.e., $AB$ be $y.$
It is given that the length of the diagonal is three times that of the smaller side.
Therefore, diagonal $= 3x = AC$

Now, applying Pythagoras theorem, we get:
$(\text { Diagonal })^2=(\text { Smaller side })^2+(\text { Larger side })^2$
$(AC)^2=(AB)^2+(BC)^2$
$(3 x)^2=(x)^2+(y)^2$
$9 x^2=x^2+y^2$
$8 x^2=y^2$
Now, taking square roots of both sides, we get:
$22 x=y$
or, $\frac{ y }{ x }=\frac{22}{1}$
Thus, the ratio of the larger side to the smaller side $=22: 1$

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MCQ 81 Mark

Mark the correct alternative in the following question:
The area of the shaded path in the following figure is:

A
$16\ m^2$
✓
$18\ m^2$
C
$14\ m^2$
D
$20 \ m^2$

Answer

Correct option: B.

$18\ m^2$

Area of the region $=$ Area of the rectangle $+$ Area of the isosceles right angled triangle
$=\text{length}\times\text{breadth}+\frac{1}{2}\times\text{base}\times\text{height}$
$=8\times2+\frac{1}{2}\times2\times2$
$=16+2$
$=18\text{m}^{2}$

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MCQ 91 Mark

The sides of a rectangle are in the ratio $5 : 4.$ If its perimeter is $72\ cm,$ then its length is:

A
$40\ cm$
✓
$20\ cm$
C
$30\ cm$
D
$60\ cm$

Answer

Correct option: B.

$20\ cm$

Let the sides of the rectangle be $5x$ and $4x. ($Since, they are in the ratio $5 : 4)$
Now, perimeter of rectangle $= 2($Length $+$ Breadth$)$
$72 = 2(5x + 4x)$
$72 = 2 \times 9x$
$72 = 18x$
$x = 4$
Thus, the length of the rectangle $= 5x$
$= 5 \times 4$
$= 20\ cm$

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MCQ 101 Mark

A rectangular carpet has area $120\ m^2$ and perimeter $46$ metres. The length of its diagonal is:

A
$15 \ m$
B
$16 \ m$
✓
$17 \ m$
D
$20 \ m$

Answer

Correct option: C.

$17 \ m$

Area of the rectangle $=120 m^2$

Perimeter $=46 m$
Let the sides of the rectangle be $I$ and $b$.
Therefore,
Area $= lb$
$=120 m^2 \ldots(1)$
Perimeter $=2(1+b)=46$
Or, $(l+b)=\frac{46}{2}$
$=23 m \ldots(2)$
Now, length of the diagonal of the rectangle $=I^2+b^2$
So, we first find the value of $\left(l^2+b^2\right)$
Using identity:
$\left(l^2+b^2\right)=(1+b)^2-2(lb)[\text { From (1) and (2)] }$
Therefore,
$\left(1^2+b^2\right)=(23)^2-2(120)$
$=529-240$
$=289$
Thus, length of the diagonal of the rectangle $=R^2+b^2=289$
$=17 m$

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MCQ 111 Mark

If the ratio between the length and the perimeter of a rectangular plot is $1 : 3,$ then the ratio between the length and breadth of the plot is:

A
$1 : 2$
✓
$2 : 1$
C
$3 : 2$
D
$2 : 3$

Answer

Correct option: B.

$2 : 1$

It is given that, $\frac{\text{Length of the rectangle}}{\text{Perimeter of the rectangle}}=\frac{1}{3}$
$\Rightarrow\frac{\text{l}}{(2\text{l}+2\text{b})}=\frac{1}{3}$
After cross multiplying, we get:
$3\text{l}=2\text{l}+2\text{b}$
$\Rightarrow\text{l}=2\text{b}$
$\Rightarrow\frac{\text{l}}{\text{b}}=\frac{2}{1}$
Thus, the ratio of the length and the breadth is $2 : 1.$

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MCQ 121 Mark

The ratio of the areas of two squares, one having its diagonal double than the other, is:

A
$1 : 2$
B
$2 : 3$
C
$3 : 1$
✓
$4 : 1$

Answer

Correct option: D.

$4 : 1$

Let the two squares be $\ce{ABCD}$ and $\ce{PQRS}. $ Further, the diagonal of square $\ce{PQRS}$ is twice the diagonal of square $\ce{ABCD}.$

$PR = 2AC$
Now, area of the square $=\frac{(\text{diagonal})^{2}}{2}$
Area of $\ce{PQRS}=\frac{(\text{PR})^{2}}{2}$
Similarly, area of $\ce{ABCD}=\frac{(\text{AC})^{2}}{2}$
According to the question:
If $AC = x$ units, then, $PR = 2x$ units
Therefore, $\frac{\text{Area of PQRS}}{\text{Area of ABCD}}=\frac{(\text{PR})^{2}\times2}{2\times(\text{AC})^{2}}$
$=\frac{(\text{PR})^{2}}{(\text{AC})^{2}}=\frac{(2\text{x})^{2}}{(1\text{x})^{2}}=\frac{4}{1}$
$=4:1$
Thus, the ratio of the areas of squares $\ce{PQRS}$ and $\ce{ABCD} = 4 : 1$

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MCQ 131 Mark

Mark the correct alternative in the following question:
The length and breadth of a rectangle of area $A$ are doubled. The area of the new rectangle is:

A
$2A$
B
$A2$
✓
$4A$
D
None of these.

Answer

Correct option: C.

$4A$

Let the length and breadth of the given rectangle be $I$ and $b$, respectively.

We have,
$A=lb \ldots \text { (i) }$
Also, the length of the new rectangle, $|=2|$
the breadth of the new rectangle, $b^{\prime}=2 b$
Now, the area of the new rectangle $=1 \times b^*$
$=(21) \times(2 b)$
$=4 lb$
$=4 A[\text { Using }(i)]$

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MCQ 141 Mark

Mark the correct alternative in the following question:
How many envelopes can be made out of a sheet of paper $72 \ cm$ by $48 \ cm$, if each envelope requires a paper of size $18 \ cm$ by $12 \ cm$?

A
$4$
B
$8$
C
$12$
✓
$16$

Answer

Correct option: D.

$16$

We have,length of the sheet of the paper $=72 \ cm$
breadth of the sheet of the paper $=48 \ cm$
length of the envelope $=18 \ cm$
breadth of the enveolope $=12 \ cm$
The area of the sheet of the paper = length $\times$ breadth
$=(18 \times 12) \ cm ^2$
Now, the number of envelope that can be made out $=\frac{\text{Area of the sheet of the paper}}{\text{Area of the envelope}}$
$=\frac{(72\times48)}{(18\times12)}$
$=4\times4$
$=16$

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MCQ 151 Mark

If the ratio of areas of two squares is $225 : 256,$ then the ratio of their perimeters is:

A
$225 : 256$
B
$256 : 225$
✓
$15 : 16$
D
$16 : 15$

Answer

Correct option: C.

$15 : 16$

Therefore, $\frac{\text{Area of square ABCD}}{\text{Area of square PQRS}}=\frac{\text{x}^{2}}{\text{y}^{2}}$
$=\frac{225}{256}$
Thus, the ratio of their perimeters $= 15 : 16$

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MCQ 161 Mark

Mark the correct alternative in the following question: The maximum length of the side of a square sheet that can be cut off from a rectangular sheet of size $8m \times 3m$ is:

✓
$3m$
B
$4m$
C
$6m$
D
$4m$

Answer

Correct option: A.

$3m$

The maximum length of the side of a square sheet that can be cut off from a rectangular sheet of size $8m \times 3m$ is $3m.$

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MCQ 171 Mark

Mark the correct alternative in the following question:
If the diagonal of a square is $\sqrt{18}$ metre, then its area is:

A
$8\ m^2$
✓
$4\ m^2$
C
$16\ m^2$
D
$6\ m^2$

Answer

Correct option: B.

$4\ m^2$

We have,
length of the diagonal of the square $=\sqrt{8}\text{cm}$

Now, the area of the square $=\frac{1}{2}\times\text{diagonal}\times\text{diagonal}$
$=\frac{1}{2}\times\sqrt{8}\times\sqrt{8}$
$=\frac{8}{2}$
$=4\text{m}^{2}$

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MCQ 181 Mark

The cost of fencing a rectangular field $34m$ long and $18m$ wide at As $2.25$ per meter is:

A
$Rs. 243$
✓
$Rs. 234$
C
$Rs. 240$
D
$Rs. 334$

Answer

Correct option: B.

$Rs. 234$

For fencing the rectangular field, we need to find the perimeter of the rectangle.
Length of the rectangle $= 34m$
Breadth of the rectangle $= 18m$
Perimeter of the rectangle $= 2($Length $+$ Breadth$) = 2(34 + 18)m$
$= 2 \times 52m$
$= 104m$
Cost of fencing the field at the rate of $Rs. 2.25$ per meter $= Rs. 104 \times 2.25$
$= Rs. 234$

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MCQ 191 Mark

Mark the correct alternative in the following question: If the perimeter of a square is $40\ cm,$ then the length of its each side is:

A
$20\ cm$
✓
$10\ cm$
C
$5\ cm$
D
$40\ cm$

Answer

Correct option: B.

$10\ cm$

The length of the each side of the square $=\frac{\text{Perimeter of the square}}{4}$
$=\frac{40}{4}$
$=10\ \text{cm}$

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MCQ 201 Mark

Mark the correct alternative in the following question: The perimeter of a square whose area is $225\ m^2$ is:

A
$15\ m$
✓
$60\ m$
C
$225\ m$
D
$30\ m$

Answer

Correct option: B.

$60\ m$

We have,

Area of the square = $225\ m^2$
As, the side of the square $=\sqrt{\text{Area}}$
$=\sqrt{225}$
$=15\text{m}$
So, the perimeter of the square $= 4 \times side$
$=4 \times 15$
$=60\ m$

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MCQ 211 Mark

If the sides of a square are halved, then its area.

A
Remains same.
B
Becomes half.
✓
Becomes one fourth.
D
Becomes double.

Answer

Correct option: C.

Becomes one fourth.

Becomes one fourth.

Let the side of the square be x.
Then, area $= (Side \times Side) = (x \times x) = x^2$
If the sides are halved, new side $=\frac{\text{x}}{2}$
Now, new area $=\big(\frac{\text{x}}{2}\big)^{2}$
$=\frac{(\text{x})^{2}}{4}$
It is clearly visible that the area has become one-fourth of its previous value.

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MCQ 221 Mark

Mark the correct alternative in the following question: The area of a square of side $14\ cm$ is:

A
$49\ cm^2$
B
$156\ cm^2$
C
$56\ cm^2$
✓
$196\ cm^2$

Answer

Correct option: D.

$196\ cm^2$

The area of the square $= ($Side $\times$ Side$)$

$= 14 \times 14$
$= 196\ cm^2$

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