M.C.Q. [1 Marks Each] - Maths STD 6 Questions - Vidyadip

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M.C.Q. [1 Marks Each]

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22 questions · timed · auto-graded

MCQ 11 Mark

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

A
$63\ cm$
B
$30\ cm$
C
$100\ cm$
✓
$126\ cm$

Answer

Correct option: D.

$126\ cm$

Area of the rectangle $= 650\ cm^2$
Breadth of the rectangle $= 13\ cm$
As, length of the rectangle $=\frac{\text{Area}}{\text{Breadth}}$
$=\frac{650}{13}$
$=50\text{ cm}$
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
$400cm^2$
✓
$200cm^2$
C
$300cm^2$
D
$100\sqrt{2}\text{cm}^{2}$

Answer

Correct option: B.

$200cm^2$

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.

$20m$

Cost of fencing the square field $= Rs. 200$

Rate of fencing the field $= Rs. 2.50$
Now, perimeter of the square field $=\frac{\text{Cost of fencing}}{\text{Rate of fencing}}$
$=\frac{200}{2..50}=80\text{m}$
Perimeter of square $= 4 \times $ Side of the square
Therefore, side of the square $=\frac{\text{Perimeter}}{4}$
$=\frac{80}{4}=20\text{m}$

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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 $ABCD.$

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

It is given that the length is three times width of the rectangle.
Therefore, length of the rectangle, i.e., $AB = 3x$ m
Now, $AC$ is the diagonal of rectangle.
In right angled triangle 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 $= Rs. 7.50/m$
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
$\sqrt{2}:1$
✓
$22 : 1$

Answer

Correct option: D.

$22 : 1$

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:
$\mathrm{(Diagonal)^2= (Smaller side)^2 + (Larger side)^2}$
$(\text{AC})^{2}=(\text{AB})^{2}+(\text{BC})^{2}$
$(3\text{x})^{2}=(\text{x})^{2}+(\text{y})^{2}$
$9\text{x}^{2}=\text{x}^{2}+\text{y}^{2}$
$8\text{x}^{2}=\text{y}^{2}$
Now, taking square roots of both sides, we get:
$22\text{x}=\text{y}$
or, $\frac{\text{y}}{\text{x}}=\frac{22}{1}$
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
$16m^2$
✓
$18m^2$
C
$14m^2$
D
$20m^2$

Answer

Correct option: B.

$18m^2$

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 $72cm$, 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 $120m^2$ and perimeter $46$ metres. The length of its diagonal is:

A
$15m$
B
$16m$
✓
$17m$
D
$20m$

Answer

Correct option: C.

$17m$

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

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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 $ABCD $and $PQRS$. Further, the diagonal of square $PQRS$ is twice the diagonal of square $ABCD.$

$PR = 2AC$
Now, area of the square $=\frac{(\text{diagonal})^{2}}{2}$
Area of PQRS $=\frac{(\text{PR})^{2}}{2}$
Similarly, area of 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 $PQRS $and $ABCD = 4 : 1$

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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
$A^2$
✓
$4A$
D
None of these.

Answer

Correct option: C.

$4A$

Let the length and breadth of the given rectangle be $l$ and $b,$ respectively.
We have,
$A = lb ...(i)$
Also, the length of the new rectangle,$ l = 2l$
the breadth of the new rectangle, $b' = 2b$
Now, the area of the new rectangle $= l × b'$
$= (2l) × (2b)$
$= 4lb$
$= 4A$ [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.

$15 : 16$

Let the two squares be $ABCD$ and $PQRS.$

Further, let the lengths of each side of $ABCD$ and $PQRS$ be $x$ and $y,$ respectively.
Therefore, $\frac{\text{Area of square ABCD}}{\text{Area of square PQRS}}=\frac{\text{x}^{2}}{\text{y}^{2}}$
$=\frac{225}{256}$
Taking square roots on both sides, we get:
$\frac{\text{x}}{\text{y}}=\frac{15}{16}$
Now, the ratio of their perimeters:
$\frac{\text{Perimeter of square ABCD}}{\text{Perimeter of square PQRS}}$
$\frac{4\times\text{side of square ABCD}}{4\times\text{side of square PQRS}}=\frac{4\text{x}}{4\text{y}}$
$\frac{\text{Perimeter of square ABCD}}{\text{Perimeter of square PQRS}}=\text{x}:\text{y}$
$\frac{\text{Perimeter of square ABCD}}{\text{Perimeter of square PQRS}}=\frac{15}{16}$
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
$8m^2$
✓
$4m^2$
C
$16m^2$
D
$6m^2$

Answer

Correct option: B.

$4m^2$

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 $225m^2$ is:

A
$15m$
✓
$60m$
C
$225m$
D
$30m$

Answer

Correct option: B.

$60m$

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

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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.

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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