2 Marks Questions

Question 12 Marks

A carton consists of 60 shirts of which 48 are good, 8 have major defects and 4 have minor defects. Nigam, a trader, will accept the shirts which are good but Anmol, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. Find the probability that it is acceptable to Anmol.

Answer

Number of Shirts without major defects $=52$
$P($ Anmol will accept the shirt $)=\frac{52}{60}$ or $\frac{13}{15}$

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Question 22 Marks

Prove that the points $(3,0),(6,4)$ and $(-1,3)$ are the vertices of an isosceles triangle.

Answer

$\text { Let } A (3,0), B (6,4), C (-1,3)$
$AB =\sqrt{(3-6)^2+(0-4)^2}=5$
$BC =\sqrt{(6+1)^2+(4-3)^2}=\sqrt{50}$
$CA =\sqrt{(3+1)^2+(0-3)^2}=5$
$ As , AB = AC$
$\Rightarrow ABC \text { is an isosceles triangle }$

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Question 32 Marks

Find the ratio in which the point $P (-4,6)$ divides the line segment joining the points $A (-6,10)$ and $B (3,-8)$.

Answer

Let the ratio be $k: 1$
$-4=\frac{3 k-6}{k+1}$
$\Rightarrow k=\frac{2}{7}$
$\therefore$ required ratio is $2: 7$

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Question 42 Marks

If $\alpha, \beta$ are zeroes of the polynomial $p(x)=5 x^2-6 x+1$, then find the value of $\alpha+\beta+\alpha \beta$.

Answer

$\alpha+\beta=\frac{6}{5}$
$\alpha \beta=\frac{1}{5}$
$\alpha+\beta+\alpha \beta=\frac{6}{5}+\frac{1}{5}=\frac{7}{5}$

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Question 52 Marks

Evaluate :
$\frac{2 \tan 30^{\circ} \cdot \sec 60^{\circ} \cdot \tan 45^{\circ}}{1-\sin ^2 60^{\circ}}$

Answer

$\frac{2 \times \frac{1}{\sqrt{3}} \times 2 \times 1}{1-\frac{3}{4}}$
$=\frac{16}{\sqrt{3}} \text { or } \frac{16 \sqrt{3}}{3}$

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Question 62 Marks

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer

$
\angle OAY=\angle OBP=90^{\circ}
$
But they are forming alternate interior angles
Therefore, $\quad PQ \| XY$

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Question 72 Marks

If two tangents inclined at an angle of $60^{\circ}$ are drawn to a circle of radius $3 \ cm ,$ then find the length of each tangent.

Answer

$\angle APO =30^{\circ}$
$\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{3}{A P}$
$AP =3 \sqrt{3} \ cm$

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Maths 2 Marks Questions

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