2 Marks Questions

Question 12 Marks

The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be ₹ x and that of a pen to be ₹ y).

Answer

Let the cost of a notebook be ₹ x.
Let the cost of a pen be ₹ y.
We need to write a linear equation in two variables to represent the statement, “Cost of a notebook is twice the cost of a pen”.
Therefore, we can conclude that the required statement will be $x = 2y.$

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Question 22 Marks

A three-wheeler scooter charges ₹ 15 for first kilometer and ₹ 8 each for every subsequent kilometer. For a distance of x km, an amount of y is paid. Write the linear equation representing the above information.

Answer

Given, a three-wheeler scooter charges ₹ 15 for first kilometre and ₹ 8 each for every subsequent kilometre. For a distance of x km, an amount of ₹ y is paid.
= 15 × 1 + (x – 1) × 8 = y
= 8x – y + 7 = 0

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Question 32 Marks

In given figure, $\angle A B C=69^{\circ}, \angle A C B=31^{\circ}$, find $\angle B D C$.

Answer

From the given figure, in $\triangle A B C$, we can write
$\angle ABC +\angle ACB +\angle BAC =180^{\circ} ($ by angle sum property$)$
$69^{\circ}+31^{\circ}+\angle BAC =180^{\circ}$
$\Rightarrow \angle BAC =180^{\circ}-100^{\circ}=80^{\circ}$
$\angle BDC =\angle BAC ($Angles in the same segment$)$
$\therefore \angle BDC =80^{\circ}$

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Question 42 Marks

In a given figure, $O$ is the centre of a circle and $P Q$ is a diameter. If $\angle R O S=40^{\circ}$, find $\angle R T S$.

Answer

It is given that $O$ is the center and $\angle R O S=40^{\circ}$

We have $\angle RQS =\frac{1}{2} \angle ROS =20^{\circ}$
In right angled triangle $RQT$ we have
$\angle RQT +\angle QTR +\angle TRQ =180^{\circ}$
$\Rightarrow 20^{\circ}+\angle QTR +90^{\circ}=180^{\circ}$
$\Rightarrow \angle QTR =180^{\circ}-20^{\circ}-90^{\circ}$
$\Rightarrow \angle QTR =70^{\circ}$
$\angle QTR =\angle RTS =70^{\circ} [$Same angles$]$
Hence, $\angle RTS =70^{\circ}$

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Question 52 Marks

The surface areas of two spheres are in the ratio of $4 : 25$. Find the ratio of their volumes.

Answer

Surface areas of two spheres $=\frac{4}{25}$
$\Rightarrow \frac{4 R^2}{4 r^2}=\frac{4}{25}$
$\Rightarrow \frac{R^2}{r^2}=\frac{4}{25}$
$\Rightarrow \frac{R}{r}=\frac{2}{5}$
Ratio of their volumes $=\frac{\frac{4}{3} \pi R^3}{\frac{4}{3} \pi r^3}$
$=\left(\frac{R}{r}\right)^3$
$=\left(\frac{2}{5}\right)^3$
$=\frac{8}{125}$
Hence, the ratio of their volumes is $8:125$

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Question 62 Marks

In given figure, if $O A=10, A B=16$ and $O D \perp$ to $A B$. Find the value of the $C D$.

Answer

Given : In given figure $OA=10\ cm$ and $Ab=16 \ cm$
To find : Length of $CD$

Solution : As $OD \perp AB$
$\Rightarrow AC = CB$
$(\perp$ from the centre to the chord bisects the chord$)$
$\therefore AC =\frac{A B}{2}=8$
In right $\triangle O C A$,
$OA ^2= AC ^2+ OC ^2$
$(10)^2=8^2+O C^2$
$O C^2=100-64$
$ OC ^2=36$
$\therefore OC =\sqrt{36}$
$OC = 6 \ cm$
$CD = OD - OC $
$= 10 - 6 $
$= 4 \ cm.$

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Question 72 Marks

The perimeter of an isosceles triangle is $42 \ cm$ and its base is $1 \frac{1}{2}$ times each of the equal sides. Find the height of the triangle. $($Given, $\sqrt{7}=2.64)$.

Answer

Let the equal sides of the isosceles triangle be $a \ cm$ each.
$\therefore$ Base of the triangle, $b=\frac{3}{2} a \ cm$
Perimeter of triangle $=42 \ cm$
$\Rightarrow a+a+\frac{3}{2} a=42$
$\Rightarrow \frac{7}{2} a=42$
$\Rightarrow a=12 \ cm$
and $b=\frac{3}{2}(12) \ cm =18 \ cm$
Area of triangle $=71.42 \ cm^2$
$\Rightarrow \frac{1}{2} \times \text { Base } \times \text { Height }=71.42$
$\Rightarrow$ Height $=\frac{71.42 \times 2}{18}$
$=7.94 \ cm$

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Maths 2 Marks Questions

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