3 Marks Question

Question 13 Marks

The sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If ten’s digit is 3 more than the unit’s digit, then find the number.

Answer

Let the two-digit number be $10 x+y$
Therefore $(10 x+y)+(10 y+x)$ = 99
$\Rightarrow$ $x+y=9$ ……….(i)
Also, $x=3+y$ ……..(ii)
Solving (i) & (ii) to get 𝑦 = 3 , 𝑥 = 6
Therefore, required number is 63

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Question 23 Marks

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Answer

Correct Given, to prove, figure and construction
Correct proof

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Question 33 Marks

In the given figure, PA and PB are tangents to a circle centred at O. Prove that (i) OP bisects $\angle A P B$ (ii) OP is the right bisector of AB.

Answer

(i) $\triangle O A P \cong \triangle O B P$
$\angle A P O=\angle B P O$
Or OP bisects $\angle P$
(ii) $\triangle A Q P \cong \triangle B Q P$
⇒AQ=QB and $\angle A Q P=\angle B Q P$
AB is a straight line
therefore $\angle A Q P=\angle B Q P=90^{\circ}$
Hence OP is right bisector of AB

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Question 43 Marks

Find the mean using the step deviation method.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	6	10	15	9	10

Answer

Class	x	frequency(f)	$u=\frac{x-25}{10}$	𝑓u
0-10	5	6	-2	-12
10-20	15	10	-1	-10
20-30	25	15	0	0
30-40	35	9	1	9
40-50	45	10	2	20
		$\sum f=50$		$\sum f u=7$

Mean = 25 + 10 × $\left(\frac{7}{50}\right)$
= 26.4

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Question 53 Marks

Prove that: $(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$.

Answer

$LHS =\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)$
$=\frac{1-\sin ^2 A}{\sin A} \times \frac{1-\cos ^2 A}{\cos A}$
$=\frac{\cos ^2 A}{\sin A} \times \frac{\sin ^2 A}{\cos A}$
$=\cos A \sin A$
$RHS =\frac{\cos A \sin A}{\sin ^2 A+\cos ^2 A}$
$=\cos A \sin A= LHS$

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Question 63 Marks

Line 4x + y = 4 divides the line segment joining the points (−2, −1) and (3,5)in a certain ratio. Find the ratio.

Answer

Let the line 4x + y = 4 intersects AB at P(x1, y1) such that AP: PB=𝑘:1

$x_1=\frac{3 k-2}{k+1}$ and $y_1=\frac{5 k-1}{k+1}$
$\left(x_1, y_1\right)$ lies on $4 x+y=4$
Therefore, $4\left(\frac{3 k-2}{k+1}\right)+\left(\frac{5 k-1}{k+1}\right)=4$
⇒ 𝑘=1
Required ratio is 1:1

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Question 73 Marks

Find the ratio in which the y-axis divides the line segment joining the points (4, −5) and (−1,2). Also find the point of intersection.

Answer

Let the required point on the y axis be P(0,y).

Let AP : PB be k : 1
Therefore, $\frac{-k+4}{k+1}=0$
⇒k=4
Therefore, required ratio is 4:1
$\& y=\frac{8-5}{5}=\frac{3}{5}$
Hence point of intersection is $\left(0, \frac{3}{5}\right)$.

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Question 83 Marks

Prove that $\sqrt{5}$ is an irrational number.

Answer

Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5}=\frac{p}{q}$, where $q \neq 0$ and $p \& q$ are coprime.
$5 q^2=p^2 \Rightarrow p^2$ is divisible by 5
⟹ p is divisible by 5----- (i)
⟹ p = 3a, where ‘a’ is a postive integer
$25 a^2=5 q^2 \Rightarrow q^2=5 a^2 \Rightarrow q^2$ is divisible by 5
⟹ q is divisible by 5 ----- (ii)
(i) and (ii) leads to contradiction as ‘p’ and ‘q’ are coprime.
$\therefore \sqrt{5}$ is an irrational number.

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Maths 3 Marks Question

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