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The scalar product of $5 \hat{i}+\hat{j}-3 \hat{k}$ and $3 \hat{i}-4 \hat{j}+7 \hat{k}$ is (A) 10 (B) $-10$ 6 (C) 15 (D) $-15$

The scalar product of $5 \hat{i}+\hat{j}-3 \hat{k}$ and $3 \hat{i}-4 \hat{j}+7 \hat{k}$ is (A) 10 (B) $-10$ 6 (C) 15 (D) $-15$ Sol. Correct answer option is (B) $-10$ 6

Construct a triangle with sides 3 cm, 4 cm and 5 cm. Draw its circumcircle and measure its radius.

CBSE, Class 10, Constructions, Exercise 1, maths, Maths, ML Aggarwal

Solution: Steps to construct: Step 1: Draw a line segment BC = 4cm. Step 2: With Center as B and radius 3cm, with center as C and radius 5cm draw two arcs which intersect each other at point A. Step...

Using a ruler and a pair of compasses only, construct: (i) A triangle ABC given AB = 4 cm, BC = 6 cm and ∠ABC = 90°. (ii) A circle which passes through the points A, B and C and mark its centre as O. (2008)

CBSE, Class 10, Constructions, Exercise 1, maths, Maths, ML Aggarwal

Solution: Steps to construct: Step 1: Draw a line segment AB = 4cm. Step 2: At point B, draw a ray BX making an angle of 90o and cut off BC = 6cm. Step 3: Join AC. Step 4: Draw the perpendicular...

Draw an equilateral triangle of side 4 cm. Draw its circumcircle.

CBSE, Class 10, Constructions, Exercise 1, maths, Maths, ML Aggarwal

Solution: Steps to construct: Step 1: Draw a line segment BC = 4cm. Step 2: With centers B and C, draw two arcs of radius 4cm which intersects each other at point A. Step 3: Join AB and AC. Step 4:...

Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

CBSE, Class 10, Constructions, Exercise 1, maths, Maths, ML Aggarwal

Solution: Steps to construct: Step 1: Draw a line segment AB = 8cm. Step 2: With center as A and radius 4cm, with center as B and radius 3cm, draw circles. Step 3: Draw the third circle AB as...

(a) In the figure given below, O is the center of the circle. If ∠BAD = 30°, find the values of p, q and r.

CBSE, Circles, Class 10, Exercise 1, Maths, maths, Maths, ML Aggarwal, ML Aggarwal

(a) In the figure given below, two circles intersect at points P and Q. If ∠A = 80° and ∠D = 84°, calculate (i) ∠QBC (ii) ∠BCP Solution: (i) ABCD is a cyclic quadrilateral ∠A + ∠C = 180o 30o + p =...

(a) In the figure given below, ABCD is a cyclic quadrilateral. If ∠ADC = 80° and ∠ACD = 52°, find the values of ∠ABC and ∠CBD.

CBSE, Circles, Class 10, Exercise 1, maths, Maths, ML Aggarwal

(b) In the figure given below, O is the center of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC. Solution: (a) In the given figure, ABCD is a cyclic quadrilateral ∠ADC =...

(a) In the figure, (i) given below, if ∠DBC = 58° and BD is a diameter of the circle, calculate: (i) ∠BDC (ii) ∠BEC (iii) ∠BAC

CBSE, Circles, Class 10, Exercise 1, maths, Maths, ML Aggarwal

(b) In the figure (if) given below, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find: (i) ∠CAD (ii) ∠CBD (iii) ∠ADC (2008) Solution: (a) ∠DBC = 58° BD is diameter ∠DCB = 90° (Angle in...

(a) In the figure (i) given below, O is the center of the circle. If ∠AOC = 150°, find (i) ∠ABC (ii) ∠ADC (b) In the figure (i) given below, AC is a diameter of the given circle and ∠BCD = 75°. Calculate the size of (i) ∠ABC (ii) ∠EAF.

CBSE, Circles, Class 10, Exercise 1, maths, Maths, ML Aggarwal

Solution: (a) Given, ∠AOC = 150° and AD = CD We know that an angle subtends by an arc of a circle at the center is twice the angle subtended by the same arc at any point on the remaining part of the...

If O is the center of the circle, find the value of x in each of the following figures (using the given information)

CBSE, Circles, Exercise 1, maths, Maths, ML Aggarwal

Solution: From the figure (i) ABCD is a cyclic quadrilateral Ext. ∠DCE = ∠BAD ∠BAD = xo Now arc BD subtends ∠BOD at the center And ∠BAD at the remaining part of the circle. ∠BOD = 2 ∠BAD = 2 x 2 x =...

If O is the center of the circle, find the value of x in each of the following figures (using the given information):

CBSE, Circles, Class 10, Exercise 1, maths, Maths, ML Aggarwal

Solution: (i) ∠ACB = ∠ADB (Angles in the same segment of a circle) But ∠ADB = x° ∠ABC = xo Now in ∆ABC ∠CAB + ∠ABC + ∠ACB = 180o 40o + 900 + xo = 180o (AC is the diameter) 130o + xo = 180o xo =...

If $\int_{0}^{\frac{\pi}{2}} \log \cos x d x=\frac{\pi}{2} \log \left(\frac{1}{2}\right)$ then $\int_{0}^{\frac{\pi}{2}} \log \sec x d x=$
(A) $\frac{\pi}{2} \log \left(\frac{1}{2}\right)$
(B) $1-\frac{\pi}{2} \log \left(\frac{1}{2}\right)$
(C) $1+\frac{\pi}{2} \log \left(\frac{1}{2}\right)$
(D) $\frac{\pi}{2} \log 2$

maths, Maths

Correct Option is (A) $\frac{\pi}{2} \log \left(\frac{1}{2}\right)$

The value of $\cos ^{1}\left(\cot \left(\frac{\pi}{2}\right)\right)+\cos ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ is
(A) $\frac{2 \pi}{3}$
(B) $\frac{\pi}{3}$
(C) $\frac{\pi}{2}$
(D) $\pi$

maths, Maths

Correct option is (A) $\frac{2 \pi}{3}$ $ \cos ^{-1}\left(\cot \frac{\pi}{2}\right)+\cos ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\cos ^{-1}(0)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) $ $...

If the distance of points $2 \hat{i}+3 \hat{j}+\lambda \hat{k}$ from the plane $\bar{r}(3 \hat{i}+2 \hat{j}+6 \hat{k})=13$ is 5 units then $\lambda=$
(A) $6,-\frac{17}{3}$
(B) $6, \frac{17}{3}$
(C) $-6,-\frac{17}{3}$
(D) $-6, \frac{17}{3}$

maths, Maths

Correct option is (A) $6,-\frac{17}{3}$ Equation of plane $\bar{r} \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})=13$ i.e. $3 x+2 y+6 z-13=0$ Given point $(2,3, \lambda)$ distance of plane from the points...

Probability that a person will develop immunity after vaccination is $0.8$. If 8 people are given the vaccine then probability that all develop immunity is $=$
(A) $(0.2)^{8}$
(B) $(0.8)^{8}$
(C) 1
(D) ${ }^{8} C_{6}(0.2)^{6}(0.8)^{2}$

maths, Maths

Correct option is (B) $(0.8)^{8}$ Probability that a person develops immunity after vaccination is $0.8$ 8 people are given vaccine Probability of developing immunity after vaccination of two people...

The equation of the plane through $(-1,1,2)$, whose normal makes equal acute angles with coordinate axes is
(A) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$
(B) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$
(C) $\bar{r} \cdot(3 \hat{i}-3 \hat{j}+3 \hat{k})=2$
(D) $\bar{r} \cdot(\hat{i}-\hat{j}+\hat{k})=3$

maths, Maths

Correct option is (A) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$ Equation plane passing through $\mathrm{A}(\overrightarrow{\mathrm{a}})$ and $\perp$ to $\overrightarrow{\mathrm{n}}$ is...

If $f(x)=x$ for $x \leq 0$ $=0$ for $x>0$ then $f(x)$ at $x=0$ is
(A) Continuous but not differentiable
(B) Not continuous but differentiable
(C) Continuous and differentiable
(D) Not continuous and not differentiable

maths, Maths

Correct option is (A) Continuous but not differentiable Continuity at $\mathrm{x}=0$ $ \begin{array}{l} \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} x=0 \\ \lim _{x \rightarrow 0^{+}}...

If the angle between the planes $\bar{r} \cdot(m \hat{i}-\hat{j}+2 \hat{k})+3=0$ and $\bar{r} \cdot(2 \hat{-} m \hat{j}-\hat{k})-5=0$ is $\frac{\pi}{3}$ then $m=$
(A) 2
(B) $\pm 3$
(C) 3
(D) $-2$

maths, Maths

Correct option is (C) 3 Direction ratios $\overline{\mathrm{n}_{1}}$ are $\mathrm{m},-1,2$ Direction ratios $\overline{\mathrm{n}_{2}}$ are $2,-\mathrm{m},-1$ $ \begin{array}{l} \theta=\frac{\pi}{3}...

Which of the following statement pattern is a tautology ?
(A) $p \vee(q \rightarrow p)$
(B) $\sim q \rightarrow \sim p$
(C) $(q \rightarrow p) \vee(\sim p \leftrightarrow q)$
(D) $p \wedge \sim p$

maths, Maths

Correct option is (C) $(q \rightarrow p) \vee(\sim p \leftrightarrow q)$ (A) $p \vee(q \rightarrow p) \equiv p \vee(\sim q \vee p) \equiv p \vee p \vee \sim q$ $\equiv \mathrm{p} \vee \sim...

A boy tosses fair coin 3 times. If he gets Rs. $2 X$ for $X$ heads then his expected gain equals to Rs. ….
(A) 1
(B) $\frac{3}{2}$
(C) 3
(D) 4

maths, Maths

Correct option is (C) 3 For $x$ heads, he gets $$y=2 \mathrm{x} Rs. \begin{tabular}{|c|c|c|c|c|} \hline $\mathrm{x}$ & 0 & 1 & 2 & 3 \\ \hline $\mathrm{y}$ & 0 & 2 & 4 & 6 \\ \hline...

Without using trigonometric tables, evaluate the following
(i) (sin 35o cos 55o + cos 35o sin 55 o)/ (cosec2 10o – tan2 80 o)
(ii) sin2 34o + sin2 56o + 2 tan18o tan 72o – cot2 30o

maths, Maths, ML Aggarwal, trigonometric identites

Given, (i) (sin 35o cos 55o + cos 35o sin 55 o)/ (cosec2 10o – tan2 80 o) (ii) sin2 34o + sin2 56o + 2 tan18o tan 72o – cot2 30o = sin2 34o + sin2 (90o – 34o) + 2 tan18o tan (90o – 18o) – cot2 30o =...

Without using trigonometric tables, evaluate the following
(i) (sin 65o/ cos 25o) + (cos 32o/sin 58o) – sin 28o sec 62o + cosec2 30o
(ii) (sin 29o/ cosec 61o) + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).

maths, Maths, ML Aggarwal, trigonometric identites

Given, (i) (sin 65o/ cos 25o) + (cos 32o/sin 58o) – sin 28o sec 62o + cosec2 30o = (sin 65o/ cos (90o – 65o)) + (cos 32o/sin (90o – 32o)) – sin 28o sec (90o – 28o) + 22 = (sin 65o/sin 65o) + (cos...

The equation of line equally inclined to co-ordinate axes and passing through $(-3,2,-5)$ is
(A) $\frac{x+3}{1}=\frac{y-2}{1}=\frac{z+5}{1}$
(B) $\frac{x+3}{1}=\frac{y-2}{1}=\frac{5+z}{1}$
(C) $\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{1}$
(D) $\frac{x+3}{-1}=\frac{2-y}{1}=\frac{z+5}{-1}$

maths, Maths

Correct option is (B) $\frac{x+3}{-1}=\frac{y-2}{1}=\frac{5+z}{-1}$ Equation of line passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}\right)$ and having d.c.s.l $\mathrm{m},...

Without using trigonometric tables, evaluate the following
(i) cos2 26o + cos 64o sin 26o + (tan 36o/ cot 54o)
(ii) (sec 17o/ cosec 73o) + (tan 68o/ cot 22o) + cos2 44o + cos2 46o

maths, Maths, ML Aggarwal, trigonometric identites

Given, (i) cos2 26o + cos 64o sin 26o + (tan 36o/ cot 54o) = cos2 26o + cos (90o – 16o) sin 26o + [tan 36o/ cot (90o – 54o)] = [cos2 26o + sin2 26o] + (tan 36o/ tan 36o) = 1 + 1 = 2 (ii) (sec 17o/...

If 12 cosec θ = 13, find the value of (2 sin θ – 3 cos θ)/ (4 sin θ – 9 cos θ)

maths, Maths, ML Aggarwal, trigonometric identites

Given, 12 cosec θ = 13 ⇒ cosec θ = 13/12 In right ∆ ABC, ∠A = θ So, cosec θ = AC/BC = 13/12 AC = 13 and BC = 12 By Pythagoras theorem, AB = √(AC2 – BC2) = √[(13)2 – (12)2] = √(169 – 144) = √25 = 5...

$\int \frac{\sec ^{8} x}{\operatorname{cosec} x} d x=$
(A) $\frac{\sec ^{8} x}{8}+c$
(B) $\frac{\sec ^{7} x}{7}+c$
(C) $\frac{\sec ^{6} x}{6}+c$
(D) $\frac{\sec ^{9} x}{9}+c$

maths, Maths

Option (B) $ \begin{array}{l} \text { Let } l=\int \frac{\sec ^{8} x}{\operatorname{cosec} x} d x=\int \frac{\sin x}{\cos ^{8} x} d x \\ {\left[\because \frac{1}{\operatorname{cosec} x}=\sin x \text...

If tan A = 1/√3, find all other trigonometric ratios of angle A.

maths, Maths, ML Aggarwal, trigonometric identites

In right ∆ ABC, tan A = BC/AB = 1/√3 So, BC = 1 and AB = √3 By Pythagoras theorem, AC = √(AB2 + BC2) = √[(√3)2 + (1)2] = √(3 + 1) = √4 = 2 Hence, sin A = BC/AC = ½ cos A = AB/AC = √3/2 cot A = 1/tan...

Express the ratios cos A, tan A and sec A in terms of sin A.

maths, Maths, ML Aggarwal, trigonometric identites

We know that, sin2 A + cos2 A = 1 So, cos A = √(1 – sin2 A) tan A = sin A/cos A = sin A/ √(1 – sin2 A) sec A = 1/cos A = 1/ (√1 – sin2 A)

If A is an acute angle and sec A = 17/8, find all other trigonometric ratios of angle A (using trigonometric identities).

maths, Maths, ML Aggarwal, trigonometric identites

sec A = 17/8 and A is an acute angle So, in ∆ ABC we have ∠B = 90o And, AC = 17 and AB = 8 By Pythagoras theorem, BC = √(AC2 – AB2) = √(172 – 82) = √(289 – 64) = √225 = 15 Now, sin A = BC/AC = 15/17...

The lines $\frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{1}=\frac{z}{1}$ intersect each other at point
(A) $(-2,-4,5)$
(B) $(-2,-4,-5)$
(C) $(2,4,-5)$
(D) $(2,-4,-5)$

maths, Maths

Option (B) Given lines are $\frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}=\lambda$ (say) and $\frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}$ Any point on the line (i) is $P(2 \lambda+1,2 \lambda-1,4...

For the following distribution function $\mathrm{F}(\mathrm{x})$ of a.r.v. $\mathrm{X}$
$$\begin{tabular}{|c|c|c|c|c|c|c|} \hline $\mathrm{x}$ & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline $\mathrm{F}(\mathrm{x})$ & $0.2$ & $0.37$ & $0.48$ & $0.62$ & $0.85$ & 1 \\ \end{tabular}$$ $P(3(A) $0.48$
(B) $0.37$
(C) $0.27$
(D) $1.47$

maths, Maths

Correct option is (B) $0.37$ $$\begin{tabular}{|l|c|c|c|c|c|c|} \hline $\mathrm{x}$ & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline $\mathrm{f}(\mathrm{x})$ & $0.2$ & $0.37$ & $0.48$ & $0.62$ & $0.85$ & 1 \\...

$\Delta A B C$ has vertices at $A=(2,3,5), B=(-1,3,2)$ and $C=(\lambda, 5, \mu)$. If the median through A is equally inclined to the axes, then the values of $\lambda$ and $\mu$ respectively are
(A) 10,7
(B) 9,10
(C) 7,9
(D) 7,10

maths, Maths

Correct option is (D) 7,10 $ \begin{array}{l} \mathrm{D} \equiv\left(\frac{\lambda-1}{2}, \frac{5+3}{2}, \frac{\mu+2}{2}\right) \\ \equiv\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}\right)...

The particular solution of the differential equation $x d y+2 y d x=0$, when $x=2, y=1$ is
(A) $x y=4$
(B) $x^{2} y=4$
(C) $x y^{2}=4$
(D) $x^{2} y^{2}=4$

maths, Maths

Correct option is (B) $x^{2} y=4$ $ \mathrm{xdy}+2 \mathrm{ydx}=0 $ $ \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}+\frac{2 \mathrm{dx}}{\mathrm{x}}=0 $ On Integrating both sides, we get $ \int...

If vector $\bar{r}$ with d.c.s. $1, \mathrm{~m}, \mathrm{n}$ is equally inclined to the co-ordinate axes, then the total number of such vectors is
(A) 4
(B) 6
(C) 8
(D) 2

maths, Maths

Correct option is (C) 8 $ \overline{\mathrm{r}}=|\overline{\mathrm{r}}|\left(\pm \frac{1}{\sqrt{3}} \hat{\mathrm{i}} \pm \frac{1}{\sqrt{3}} \hat{\mathrm{j}} \pm \frac{1}{\sqrt{3}}...

If slopes of lines represented by $K x^{2}+5 x y+y^{2}=0$ differ by 1 then $K=$
(A) 2
(B) 3
(C) 6
(D) 8

maths, Maths

Correct option is (C) 6 $\mathrm{Kx}^{2}+5 \mathrm{xy}+\mathrm{y}^{2}=0$ $\mathrm{m}_{1}+\mathrm{m}_{2}=-5, \mathrm{~m}_{1} \mathrm{~m}_{2}=\mathrm{K}$ Also, $\mathrm{m}_{1}-\mathrm{m}_{2}=1$.........

Let $\square P Q R S$ be a quadrilateral. If $\mathrm{M}$ and $\mathrm{N}$ are the midpoints of the sides $\mathrm{PQ}$ and RS respectively then $\overline{P S}+\overline{Q R}=$
(A) $3 \overline{M N}$
(B) $4 \overline{M N}$
(C) $2 \overline{M N}$
(D) $2 \overline{N M}$

maths, Maths

Correct option is (C) $2 \overline{\mathrm{MN}}$ Let $\overline{\mathrm{m}}, \overline{\mathrm{n}}, \overline{\mathrm{p}}, \overline{\mathrm{q}}, \overline{\mathrm{r}}$ and $\bar{s}$ be position...

If lines represented by equation $p x^{2}-q y^{2}=0$ are distinct then
(A) $p q>0$
(B) $p q<0$
(C) $p q=0$
(D) $p+q=0$

maths, Maths

Correct option is (A) $\mathrm{pq}>0$ Given line is $\mathrm{px}^{2}-\mathrm{qy}^{2}=0$ General equation is $a x^{2}+b y^{2}+h=0$ Comparing above equation with (i), we get $\mathrm{a}=\mathrm{p},...

If the origin and the point $\mathrm{P}(2,3,4), \mathrm{Q}(1,2,3)$ and $\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ are co-planar then
(A) $x-2 y-z=0$
(B) $x+2 y+z=0$
(C) $x-2 y+z=0$
(D) $2 x-2 y+z=0$

maths, Maths

Correct option is (C) $x-2 y+z=0$ $\mathrm{O}, \mathrm{P}, \mathrm{Q}, \mathrm{R}$ are co-planar $\therefore[\overrightarrow{O R} \overrightarrow{O P} \overrightarrow{O Q}]=0$...

If $f(x)=\log \left(\sec ^{2} x\right)^{\cot 2} x$ for $x \neq 0$ $=K$ for $x=$ is continuous at $x=0$ then $K 1 \mathrm{~S}$
$(\mathrm{A}) e^{-1}$
(B) 1
(C) $e$
(D) 0

maths, Maths

Correct option is (B) 1 Since, $\mathrm{f}$ is continuous at $\mathrm{x}=0$ $ \therefore \mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0} \log \left(\sec ^{2} \mathrm{x}\right)^{\cot ^{2} \mathrm{x}}...

If the volume of spherical ball is increasing at the rate of $4 \pi \mathrm{cc} / \mathrm{sec}$ then the rate of change of its surface area when the volume is $288 \pi \mathrm{cc}$ is
(A) $\frac{4}{3} \pi \mathrm{cm}^{2} / \mathrm{sec}$
(B) $\frac{2}{3} \pi \mathrm{cm}^{2} / \mathrm{sec}$
(C) $4 \pi \mathrm{cm}^{2} / \mathrm{sec}$
(D) $2 \pi \mathrm{cm}^{2} / \mathrm{sec}$

maths, Maths

Correct option is (A) Given, volume $\mathrm{V}=288 \pi \mathrm{cc}, \frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=4 \pi \mathrm{cc} / \mathrm{sec}$ volume of sphere $ \begin{array}{l}...

A box contains 6 pens, 2 of which are defective. Two pens are taken randomly from the box. If r.v. X: Number of defective pens obtained, then standard deviation of $X=$
(A) $\pm \frac{4}{3 \sqrt{5}}$
(B) $\frac{8}{3}$
(C) $\frac{16}{45}$
(D) $\frac{5}{3 \sqrt{5}}$

maths, Maths

Correct option is (D) $\frac{4}{3 \sqrt{5}}$ $\mathrm{X}:$ no. of defective pens Two pens are taken from box $\therefore \mathrm{X}$ can take values $0,1,2$ $ \begin{array}{l}...

The area of the region bounded by the lines $y=2 x+1, y=3 x+1$ and $x=4$ is
(A) 16 sq. unit
(B) $\frac{121}{3}$ sq. unit
(C) $\frac{121}{6}$ sq. unit
(D) 8 sq. unit

maths, Maths

Option (D) $ \begin{array}{l} y=3 x+1 \Rightarrow \frac{y}{1}+\frac{x}{\frac{-1}{3}}=1 \\ y=2 x+1 \Rightarrow \frac{y}{1}+\frac{x}{\frac{-1}{2}}=1 \\ \operatorname{Area}(A B C)=\operatorname{Area}(A...

A $r . v . X \sim B(n, p)$. If values of mean and variance of $X$ are 18 and 12 respectively then total number of possible values of $X$ are
(A) 54
(B) 55
(C) 12
(D) 18

maths, Maths

Correct option is (B) 55 $ \begin{array}{l} \text { Mean }=n p=18 \\ \text { V ariance }=\mathrm{npq}=12 \\ \frac{\mathrm{npq}}{\mathrm{np}}=\frac{12}{18} \\ \therefore \mathrm{q}=\frac{2}{3}...

If $x-f(t)$ and $y=g(t)$ are differentiable functions of $t$ then $\frac{d^{2} y}{d x^{2}}$ is
(A) $\frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^{3}}$
(B) $\frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^{2}}$
(C) $\frac{g^{\prime}(t) \cdot f^{\prime \prime}(t)-f^{\prime}(t) g^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^{3}}$
(D) $\frac{g^{\prime}(t) \cdot f^{\prime \prime}(t)+f^{\prime}(t) g^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^{3}}$

maths, Maths

Correct option is (A) $\frac{\mathrm{f}^{\prime}(\mathrm{t}) \cdot \mathrm{g}^{\prime \prime}(\mathrm{t})-\mathrm{g}^{\prime}(\mathrm{t}) \cdot \mathrm{f}^{\prime...

In $\Delta A B C$ if $\sin ^{2} A+\sin ^{2} B=\sin ^{2} C$ and $l(A B)=10$ then the maximum value of the area of $\triangle A B C$ is
(A) 50
(B) $10 \sqrt{2}$
(C) 25
(D) $25 \sqrt{2}$

maths, Maths

Correct option is (C) 25 $ \begin{array}{l} \sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}=\sin ^{2} \mathrm{C} \\ \Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2} \text { (Sine Rule) } \\...

The solution of the differential equation $\frac{d y}{d x}=\tan \left(\frac{y}{x}\right)+\frac{y}{x}$ is
(A) $\cos \left(\frac{y}{x}\right)=c x$
(B) $\sin \left(\frac{y}{x}\right)=c x$
(C) $\cos \left(\frac{y}{x}\right)=c y$
(D) $\sin \left(\frac{y}{x}\right)=c y$

maths, Maths

Correct option is (B) $\sin \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\mathrm{cx}$ $ \frac{\mathrm{dy}}{\mathrm{dx}}=\tan...

For a invertible matrix $\mathrm{A}$ if $A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$ then $|A|=$
(A) 100 (B) $-100$ (C) 10 (D) $-10$

maths, Maths

Correct option is (C) 10 Given : $\mathrm{A}(\operatorname{adj} \mathrm{A})=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]=10\left[\begin{array}{ll}1 & 0 \\ 0 &...

If the function $f(x)=\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{1 / x}$ for $x \neq 0 \\{=K \quad \text { for } x=0}$ is continuous at $x=0$ then $K=$ ?
(A) $e$
(B) $e^{-1}$
(C) $e^{2}$
(D) $e^{-2}$

maths, Maths

Option (C) The function would be continuous if $\lim _{x \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)$, $ \begin{array}{l} \lim _{\mathrm{x} \rightarrow 0}...

If the inverse of the matrix $\left[\begin{array}{ccc}\alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3\end{array}\right]$ does not exist then the value of $\alpha$ is
(A) 1
(B) $-1$
(C) 0
(D) $-2$

maths, Maths

Correct option is (D) -2 $ \mathrm{A}=\left[\begin{array}{ccc} \alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{array}\right] $ $ |\mathrm{A}|=\alpha[9-2]-14[6-6]-1[4-18] $ $...

The objective function of LPP defined over the convex set attains its optimum value at
(A) At least two of the corner points
(B) All the corner points
(C) At least one of the corner points
(D) None of the corner points

maths, Maths

Correct option is (C) atleast one of the corner points Let $\mathrm{Z}=\mathrm{ax}+$ by be the objective function When $\mathrm{Z}$ has optimum value(maximum or minimum), where the variables...

$\int_{0}^{3}[x] d x=\ldots$ where $[x]$ is greatest integer function (A)3 (B)0 (C)2 (D) 1

maths, Maths

Correct option is (A) 3 $ \begin{array}{l} \int_{0}^{3}[\mathrm{x}] \mathrm{dx}=\int_{0}^{1} 0 \mathrm{dx}+\int_{1}^{2} 1 \mathrm{dx}+\int_{2}^{3} 2 \mathrm{dx} \\...

The differential equation of all parabolas whose axis is $y$-axis is
(A) $x \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0$
(B) $x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0$
(C) $\frac{d^{2} y}{d x^{2}}-y=0$
(D) $\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0$

maths, Maths

Correct option is (A) $x \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0$ axis $=$ y axis vertex is $(0, \mathrm{k})$ Equation of parabola is $ \begin{array}{l} (\mathrm{x}-0)^{2}=4...

If c denotes the contradiction then dual of the compound statement $\sim p \wedge(q \vee c)$ is
$(\mathrm{A}) \sim p \vee(q \wedge t)$
(B) $\sim p \wedge(q \vee t)$
(C) $p \vee(\sim q \vee t)$
(D) $\sim p \vee(q \wedge c)$

maths, Maths

Correct option is $A \sim p \vee(q \wedge t)$ In dual of statement the following changes $V$ to $\wedge$ $\wedge$ to $\mathrm{V}$ $\mathrm{T}$ to $\mathrm{F}$ $\mathrm{F}$ to $\mathrm{T}$...

$\int_{0}^{1} x \tan ^{-1} x d x=$
(A) $\frac{\pi}{4}+\frac{1}{2}$
(B) $\frac{\pi}{4}-\frac{1}{2}$
(C) $\frac{1}{2}-\frac{\pi}{4}$
(D) $-\frac{\pi}{4}-\frac{1}{2}$

maths, Maths

Option (B) Explanation: $ \begin{array}{l} \int \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{dx}=\mathrm{f}(\mathrm{x}) \int \mathrm{g}(\mathrm{x})...

The maximum value of $f(x)=\frac{\log x}{x}(x \neq 0, x \neq 1)$ is (A) $e$ (B) $\frac{1}{e}$ (C) $e^{2}$ (D) $\frac{1}{e^{2}}$

maths, Maths

Correct option is (B) $\frac{1}{e}$ $f(x)=\frac{\log x}{x}$ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x} \cdot \frac{1}{\mathrm{x}}-\log \mathrm{x}}{\mathrm{x}^{2}}=\frac{1-\log...

If $z_{1}$ and $z_{2}$ are $z$ co-ordinates of the points of trisection of the segment joining the points $A(2,1,4), B(-1,3,6)$ then $z_{1}+z_{2}=$ (A) 1 (B) 4 (C) 5 (D) 10

maths, Maths

Correct option is D 10 For $\mathrm{z}_{1}$, the ratio of line segment is $1: 2$ For $\mathrm{z}_{2}$, the ratio of line segment is $2: 1$ By internal division formula,...

The objective function $z=4 x_{1}+5 x_{2}, \quad$ subject to $2 x_{1}+x_{2} \geq 7,2 x_{1}+3 x_{2} \leq 15$, $x_{2} \leq 3, x_{1}, x_{2} \geq 0$ has minimum value at the point. (A) On $x$-axis (B) On y-axis (C) At the origin (D) On the line parallel to $x$-axis

maths, Maths

Correct option is A On $\mathrm{x}$-axis Value of $z=4 \mathrm{x}_{1}+5 \mathrm{x}_{2}$ Convert the given inequalities into equalities to get the corner points $2 \mathrm{x}_{1}+\mathrm{x}_{2}=7...

The number of principal solutions of $\tan 2 \theta=1$ is (A) One (B) Two (C) Three (D) Four

maths, Maths

Correct option is B Two Let $\mathrm{y}=\tan 2 \theta$ $\tan 2 \theta=1$ $\Longrightarrow 2 \theta=\tan ^{-1}(1)$ $\Longrightarrow 2 \theta=\frac{\pi}{4}+\mathrm{k} \pi$, where $\mathrm{k}$ is a...

The point on the curve $y=\sqrt{x-1}$ where the tangent is perpendicular to the line $2 x+y-5=0$ is (A) $(2,-1)$ (B) $(10,3)$ (C) $(2,1)$ (D) $(5,-2)$

maths, Maths

Correct option is C $(2,1)$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \sqrt{\mathrm{x}-1}}=\mathrm{m}_{1}$ is the slope of tangent to $\mathrm{y}=\sqrt{\mathrm{x}-1}$ Slope of the line $2...

If $x=a\left(t-\frac{1}{t}\right), y=a\left(t+\frac{1}{t}\right)$ where $t$ be the parameter then $\frac{d y}{d x}=$ ? (A) $\frac{y}{x}$ (B) $\frac{-x}{y}$ (C) $\frac{x}{y}$ (D) $\frac{-y}{x}$

maths, Maths

Correct option is (C) $\frac{x}{y}$ $ x=a\left(t-\frac{1}{t}\right), y=a\left(t+\frac{1}{t}\right) $ $ y^{2}-x^{2}=a^{2}\left[\left(t+\frac{1}{t}\right)^{2}-\left(t-\frac{1}{t}\right)^{2}\right] $ $...

If $\alpha$ and $\beta$ are roots of the equation $x^{2}+5|x|-6=0$ then the value of $\left|\tan ^{-1} \alpha-\tan ^{-1} \beta\right|$ is (A) $\frac{\pi}{2}$ (B) 0 (C) $\pi$ (D) $\frac{\pi}{4}$

maths, Maths

Correct option is (A) $\frac{\pi}{2}$ $\mathrm{x}^{2}+5|\mathrm{x}|-6=0$ $|\mathrm{x}|^{2}+5|\mathrm{x}|-6=0$ $|\mathrm{x}|^{2}+6|\mathrm{x}|-|\mathrm{x}|-6=0$...

If $\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=A \tan ^{-1} \frac{x}{2}+B \tan ^{-1}\left(\frac{x}{3}\right)+C$ then $A-B=$
(A) $\frac{1}{6}$
(B) $\frac{1}{30}$
(C) $-\frac{1}{30}$
(D) $-\frac{1}{6}$

maths, Maths

Correct Option is (A) 1/6 Given $: \therefore \int \frac{1}{\left(x^{3}+4\right)\left(x^{2}+9\right)}^{4 x}=A \tan ^{-1} \frac{x}{2}+$ $\mathrm{B} \tan ^{-1}...

$O(0,0), A(1,2), B(3,4)$ are the vertices of $\Delta O A B$. The joint equation of the altitude and median drawn from $\mathrm{O}$ is
(A) $x^{2}+7 x y-y^{2}=0$
(B) $x^{2}+7 x y+y^{2}=0$
(C) $3 x^{2}-x y-2 y^{2}=0$
(D) $3 x^{2}+x y-2 y^{2}=0$

maths, Maths

Correct option is b $3 x^{2}+x y-2 y^{1}=0$ Equation of median $\mathrm{OD}=y=\mathrm{mx} \Rightarrow 3=2 \mathrm{~m}$ $\Rightarrow m=\frac{3}{2}$ $\therefore y=\frac{3}{2} x \Rightarrow 3 x-2 y=0$...

If $\int \frac{1}{\sqrt{9-16 x^{2}}} d x=\alpha \sin ^{-1}(\beta x)+c$ then $\alpha+\frac{1}{\beta}=$
(A) 1
(B) $\frac{7}{12}$
(C) $\frac{19}{12}$
(D) $\frac{9}{12}$

maths, Maths

Correct option is A 1 ∫19-16x2dx=asin-1(βx)+c…… (i) \int \frac{1}{\sqrt{9-16 \mathrm{x}^{2}}} \mathrm{dx}=\mathrm{a} \sin ^{-1}(\beta \mathrm{x})+\mathrm{c}...

The inverse of the matrix $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$ is (A) $-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 2 & -3\end{array}\right]$ (B) $-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$ (C) $-\frac{1}{3}\left[\begin{array}{ccc}3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$ (D) $-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ -3 & 1 & 0 \\ -9 & -2 & 3\end{array}\right]$

maths, Maths

Correct option is B $-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]$ $\mathrm{A}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3...

If $g(x)$ is the inverse function of $f(x)$ and $f^{\prime}(x)-\frac{1}{1+x^{4}}$ then $g^{\prime}(x)$ is (A) $1+[g(x)]^{4}$ (B) $1-[g(x)]^{4}$ (C) $1+[f(x)]^{4}$ (D) $\frac{1}{[g(x)]^{4}}$

maths, Maths

Correct option is A $1+[g(x)]^{4}$ g=f-1f(g(x))=x \begin{array}{l} \mathrm{g}=\mathrm{f}^{-1} \\ \mathrm{f}(\mathrm{g}(\mathrm{x}))=\mathrm{x} \end{array} Differentiate w.r.t.x...

The statement pattern $(\sim p \wedge q)$ is logically equivalent to $(\mathrm{A})(\sim p \vee q) \vee \sim p$ (B) $(p \vee q) \wedge \sim p$ (C) $(p \wedge q) \rightarrow p$ (D) $(p \vee q) \rightarrow p$

maths, Maths

Correct option is (B) $(p \vee q) \wedge \sim p$ $\mathrm{B} \rightarrow(\mathrm{p} \vee \mathrm{q}) \wedge \sim \mathrm{p}=(\mathrm{p} \wedge \sim \mathrm{p}) \vee(\mathrm{q} \wedge \sim...

A model of a ship is made to a scale of 1: 250 calculate:
(i) The length of the ship, if the length of model is 1.6 m.
(ii) The area of the deck of the ship, if the area of the deck of model is 2.4 m2.
(iii) The volume of the model, if the volume of the ship is 1 km3.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, a model of a ship is made to a scale of 1 : 250 (i) Given, the length of the model is 1.6 m Then, length of the ship = (1.6 × 250)/1 = 400 m (ii)...

In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O. (i) Now consider the ∆AOD and ∆EDC, ∠AOD = ∠EOC … [because Vertically...

In the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of ∆AOB and ∆COD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ABCD is a trapezium in which AB || DC. If 2AB = 3DC. So, 2AB = 3DC AB/DC = 3/2 Now, consider ∆AOB and ∆COD ∠AOB = ∠COD … [because vertically opposite...

In the adjoining figure, the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of ∆AOE : area of parallelogram ABCD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given figure, The diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E. In the figure four triangles have equal area. So, area of ∆OAB = ¼ area...

In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find
(i) BC
(ii) DC
(iii) area of ∆ACD : area of ∆BCA.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ∠ABD = ∠CAD AB = 5 cm, AC = 3 cm and AD = 4 cm Now, consider the ∆ABC and ∆ACD ∠C = ∠C … [common angle for both triangles] ∠ABC = ∠CAD … [from the...

If the areas of two similar triangles are 360 cm² and 250 cm² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, the areas of two similar triangles are 360 cm² and 250 cm². one side of the first triangle is 8 cm So, PQR and XYZ are two similar triangles, So, let...

In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, In a ∆ABC, D and E are points on the sides AB and AC respectively. DE || BC AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm Consider the ∆ABC, Given,...

In the adjoining figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:
(i) DF || BH
(ii) AH = 3 AF.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, 2 AD = BD, EC || BH (i) Given, E is mid-point of BD 2DE = BD … [equation (i)] 2AD = BD … [equation (ii)] From equation (i) and equation (ii) we get,...

In the figure given below, CD = ½ AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :
(i) CE || AG
(ii) 3 ED = GD

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Solution:- From the question it is given that, CD = ½ AC BF || AG (i) We have to prove that, CE || AG Consider, CD = ½ AC AC = 2BC … [because from the figure B is mid-point of AC] So, CD = ½ (2BC)...

In the figure given below. ∠AED = ∠ABC. Find the values of x and y.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure it is given that, ∠AED = ∠ABC Consider the ∆ABC and ∆ADE ∠AED = ∠ABC … [from the figure] ∠A = ∠A … [common angle for both triangles] Therefore, ∆ABC ~ ∆ADE … [by AA axiom]...

In the adjoining figure, ∠1 = ∠2 and ∠3 = ∠4. Show that PT x QR = PR x ST.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ∠1 = ∠2 and ∠3 = ∠4 We have to prove that, PT x QR = PR x ST Given, ∠1 = ∠2 Adding ∠6 to both LHS and RHS we get, ∠1 + ∠6 = ∠2 + ∠6 ∠SPT = ∠QPR...

A model of a ship is made to a scale of 1 : 200.
(i) If the length of the model is 4 m, find the length of the ship.
(ii) If the area of the deck of the ship is 160000 m², find the area of the deck of the model.
(iii) If the volume of the model is 200 liters, find the volume of the ship in m³. (100 liters = 1 m³)

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, a model of a ship is made to a scale of 1 : 200 (i) Given, the length of the model is 4 m Then, length of the ship = (4 × 200)/1 = 800 m (ii) Given,...

The model of a building is constructed with the scale factor 1 : 30. (i) If the height of the model is 80 cm, find the actual height of the building in metres. (ii) If the actual volume of a tank at the top of the building is 27 m³, find the volume of the tank on the top of the model.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, The model of a building is constructed with the scale factor 1 : 30 So, Height of the model/Height of actual building = 1/30 (i) Given, the height of...

On a map drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD has the following measurements AB = 12 cm and BG = 16 cm. Calculate:
(i) the distance of a diagonal of the plot in km.
(ii) the area of the plot in sq. km.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, Map drawn to a scale of 1: 25000 AB = 12 cm, BG = 16 cm Consider the ∆ABC, From the Pythagoras theorem, AC2 = AB2 + BC2 AC = √(AB2 + BC2) = √((12)2 +...

On a map drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and ∠ABC = 90°. Calculate
(i) the actual length of AB in km.
(ii) the area of the plot in sq. km:

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, Map drawn to a scale of 1: 250000 AB = 3 cm, BC = 4 cm and ∠ABC = 90o (i) We have to find the actual length of AB in km. Let us assume scale factor K =...

Two isosceles triangles have equal vertical angles and their areas are in the ratio 7: 16. Find the ratio of their corresponding height.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Consider the two isosceles triangle PQR and XYZ, ∠P = ∠X … [from the question] So, ∠Q + ∠R = ∠Y + ∠Z ∠Q = ∠R and ∠Y = ∠Z [because opposite angles of equal sides] Therefore, ∠Q = ∠Y and ∠R...

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:
(i) ∆ADE ~ ∆ACB.
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of ∆ADE : area of quadrilateral BCED.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ∠ABC = 90° AB and DE is perpendicular to AC (i) Consider the ∆ADE and ∆ACB, ∠A = ∠A … [common angle for both triangle] ∠B = ∠E … [both angles are equal...

In the figure given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA
(ii) Find BC and CD
(iii) Find the area of ∆ACD : area of ∆ABC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ∠ABC = ∠DAC AB = 8 cm, AC = 4 cm, AD = 5 cm (i) Now, consider ∆ACD and ∆BCA ∠C = ∠C … [common angle for both triangles] ∠ABC = ∠CAD … [from the...

In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BB = 12 cm., find
(i) BP
(ii) the ratio of areas of ∆APB and ∆DPC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, DC is parallel to AB AB = 9 cm, DC = 6 cm and BB = 12 cm (i) Consider the ∆APB and ∆CPD ∠APB = ∠CPD … [because vertically opposite angles are equal]...

In the figure (ii) given below, AB || DC and AB = 2 DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find
(i) ED
(ii) BE
(iii) area of ∆EDC : area of trapezium ABCD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, AB || DC AB = 2 DC, AD = 3 cm, BC = 4 cm Now consider ∆EAB, EA/DA = EB/CB = AB/DC = 2DC/DC = 2/1 (i) EA = 2, DA = 2 × 3 = 6 cm Then, ED = EA – DA = 6 –...

The number of solutions of $\sin x+\sin 3 x+\sin 5 x=0$ in the interval $\left[\frac{\pi}{2}, 3 \frac{\pi}{2}\right]$ is
(A) 2
(B) 3
(C) 4
(D) 5

maths, Maths

CORRECT OPTION: B $ \begin{array}{l} (\sin x+\sin 5 x)+\sin 3 x=0 \\ 2 \sin 3 x \cos 2 x+\sin 3 x=0 \\ \sin 3 x(2 \cos 2 x+1)=0 \\ \sin 3 x=0 \text { and } 2 \cos 2 x+1=0 \\ \cos 2 x=-\frac{1}{2}...

In the figure (i) given below, DE || BC and the ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, DE || BC The ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5 Now, consider the ∆ABC and ∆ADE ∠A = ∠A … [common angle for both triangles] ∠D = ∠B...

If $X=e^{\theta}(\sin \theta-\cos \theta), y=e^{\theta}(\sin \theta+\cos \theta)$ then $\frac{d y}{d x}$ at $\theta=\frac{\pi}{4}$ is
(A) 1
(B) 0
(C) $\frac{1}{\sqrt{2}}$
(D) $\sqrt{2}$

maths, Maths

CORRECT OPTION: A $ \begin{array}{l} \frac{d x}{d \theta}=e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) e^{\theta}=2 e^{\theta} \sin \theta \\ \frac{d y}{d...

$\quad$ In $\Delta A B C$, with usual notations, if $a, b, c$ are in A.P. then $a \cos ^{2}\left(\frac{C}{2}\right)+c \cos ^{2}\left(\frac{A}{2}\right)=$
(A) $3 \frac{a}{2}$
(B) $3 \frac{c}{2}$
(C) $3 \frac{b}{2}$
(D) $\frac{3 a b c}{2}$

maths, Maths

CORRECT OPTION: C $ \begin{array}{l} 2 b=a+c \\ \therefore \frac{a}{2}\left(2 \cos ^{2}\left(\frac{C}{2}\right)\right)+\frac{C}{2}\left(2 \cos ^{2}\left(\frac{A}{2}\right)\right) \\...

In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q. If area of ∆CPQ = 20 cm², find
(i) area of ∆BPQ.
(ii) area ∆CDP.
(iii) area of parallelogram ABCD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ABCD is a parallelogram. BP: PC = 1: 2 area of ∆CPQ = 20 cm² Construction: draw QN perpendicular CB and Join BN. Then, area of ∆BPQ/area of ∆CPQ =...

In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find
(i) EF : AD
(ii) area of ∆BEF : area of ∆ABD In the figure
(iii) given below, ABCD is a parallelogram
(iv) area of ∆FEO : area of ∆OBC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O. AE : EB = 2 : 3 (i) We have to find EF : AD So, AB/BE = AD/EF EF/AD =...

In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find
(i) AB
(ii) BC
(iii) area of ∆ADM : area of ∆ANB.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ABCD is a parallelogram, AM ⊥ DC and AN ⊥ CB AM = 6 cm AN = 10 cm The area of parallelogram ABCD is 45 cm² Then, area of parallelogram ABCD = DC × AM =...

In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of ∆AOB and ∆COD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ABCD is a trapezium in which AB || DC and AB = 2 CD, Then, ∠OAB = ∠OCD … [because alternate angles are equal] ∠OBA = ∠ODC Then, ∆AOB ~ ∆COD So, area of...

In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:
(i) area ∆APO : area ∆ABC.
(ii) area ∆APO : area ∆CQO.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, PB = 2: 3 PO is parallel to BC and is extended to Q so that CQ is parallel to BA. (i) we have to find the area ∆APO: area ∆ABC, Then, ∠A = ∠A … [common...

In the adjoining figure, ABC is a triangle. DE is parallel to BC and AD/DB = 3/2,
(i) Determine the ratios AD/AB, DE/BC0
(ii) Prove that ∆DEF is similar to ∆CBF. Hence, find EF/FB.
(iii) What is the ratio of the areas of ∆DEF and ∆CBF?

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- (i) We have to find the ratios AD/AB, DE/BC, From the question it is given that, AD/DB = 3/2 Then, DB/AD = 2/3 Now add 1 for both LHS and RHS we get, (DB/AD) + 1 = (2/3) + 1 (DB + AD)/AD...

In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm, calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- (i) Consider the ∆ABC and ∆DEC, ∠ABC = ∠DEC … [both angles are equal to 90o] ∠C = ∠C … [common angle for both triangles] Therefore, ∆ABC ~ ∆DEC … [by AA axiom] (ii) AC/CD = AB/DE...

In the given figure, DE || BC.
(i) Prove that ∆ADE and ∆ABC are similar.
(ii) Given that AD = ½ BD, calculate DE if BC = 4.5 cm.
(iii) If area of ∆ABC = 18cm2, find the area of trapezium DBCE

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- (i) From the question it is given that, DE || BC We have to prove that, ∆ADE and ∆ABC are similar ∠A = ∠A … [common angle for both triangles] ∠ADE = ∠ABC … [because corresponding angles...

In the figure (ii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of ∆ADE and trapezium DBCE.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, DE || BC and AD : DB = 1 : 2, ∠D = ∠B, ∠E = ∠C … [corresponding angles are equal] Consider the ∆ADE and ∆ABC, ∠A = ∠A … [common angles for both...

In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm, find the area of ∆ABC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, DE || BC, DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm From the fig, ∠D = ∠B and ∠E = ∠C … [corresponding angles are equal] Now consider the ∆ADE...

In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm. (i) Prove that ∆OAB ~ ∆OCD. (ii) Find CD and OB. (iii) Find the ratio of areas of ∆OAB and ∆OCD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm (i) We have to prove that, ∆OAB ~ ∆OCD So, consider the ∆OAB and ∆OCD ∠AOB = ∠COD …...

In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm², find the area of ∆QOA.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm² From the figure, Consider the ∆AOQ and ∆BOP, ∠OAQ = ∠OBP … [both angles are equal to 90o] ∠AOQ =...

The area of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, The area of two similar triangles are 36 cm² and 25 cm². Let us assume ∆PQR ~ ∆XYZ, PM and XN are their altitudes. So, area of ∆PQR = 36 cm2 Area of...

∆ABC ~ ∆DEF. If BC = 3 cm, EF = 4 cm and area of ∆ABC = 54 sq. cm. Determine the area of ∆DEF.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ∆ABC ~ ∆DEF BC = 3 cm, EF = 4 cm Area of ∆ABC = 54 sq. cm. We know that, Area of ∆ABC/ area of ∆DEF = BC2/EF2 54/area of ∆DEF = 32/42 54/area of ∆DEF =...

∆ABC ~ DEF. If area of ∆ABC = 9 sq. cm., area of ∆DEF =16 sq. cm and BC = 2.1 cm., find the length of EF.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ∆ABC ~ DEF Area of ∆ABC = 9 sq. cm Area of ∆DEF =16 sq. cm We know that, area of ∆ABC/area of ∆DEF = BC2/EF2 area of ∆ABC/area of ∆DEF = BC2/EF2 9/16 =...

In the figure (2) given below AD is bisector of ∠BAC. If AB = 6 cm, AC = 4 cm and BD = 3cm, find BC

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, AD is bisector of ∠BAC AB = 6 cm, AC = 4 cm and BD = 3cm Construction, from C draw a straight line CE parallel to DA and join AE ∠1 = ∠2 … [equation...

In the figure (1) given below, AB || CR and LM || QR.
(i) Prove that BM/MC = AL/LQ
(ii) Calculate LM : QR, given that BM : MC = 1 : 2.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, AB || CR and LM || QR (i) We have to prove that, BM/MC = AL/LQ Consider the ∆ARQ LM || QR … [from the question] So, AM/MR = AL/LQ … [equation (i)] Now,...

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem, prove that AO/BO = CO/DO

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O Now consider the ∆OAB and ∆OCD, ∠AOB = ∠COD [because vertically...

In the adjoining given below, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. show that BC || QR.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Consider the ∆POQ AB || PQ … [given] So, OA/AP = OB/BQ … [equation (i)] Then, consider the ∆OPR AC || PR OA/AP = OC/CR … [equation (ii)] Now by comparing both equation (i) and equation...

In the give figure, ∠D = ∠E and AD/BD = AE/EC. Prove that BAC is an isosceles triangle.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given figure, ∠D = ∠E and AD/BD = AE/EC, We have to prove that, BAC is an isosceles triangle So, consider the ∆ADE ∠D = ∠E … [from the question] AD = AE … [sides opposite to...

(a) In the figure (i) given below, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given figure, CD || LA and DE || AC, Consider the ∆BCA, BE/BC = BD/BA By using the corollary of basic proportionality theorem, BE/(BE + EC) = BD/AB 4/(4 + 2) = BD/AB … [equation...

In figure (ii) given below, AB || DE and BD || EF. Prove that DC² = CF x AC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure it is given that, AB || DE and BD || EF. We have to prove that, DC² = CF x AC Consider the ∆ABC, DC/CA = CE/CB … [equation (i)] Now, consider ∆CDE CF/CD = CE/CB …...

In figure (i) given below, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, DE || BC and BD = CE So, we have to prove that ABC is an isosceles triangle. Consider the triangle ABC, AD/DB = AE/EC Given, DB = EC … [equation (i)]...

E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR: of a ∆PQR. For each of the following cases, state whether EF || QR:PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the dimensions given in the question, Consider the ∆PQR So, PQ/PE = 1.28/0.18 = 128/18 = 64/9 Then, PR/PF = 2.56/0.36 = 256/36 = 64/9 By comparing both the results, 64/9 = 64/9...

E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR: (i) PE = 3.9 cm, EQ = 3 cm, PF = 8 cm and RF = 9 cm.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given dimensions, Consider the ∆PQR So, PE/EQ = 3.9/3 = 39/30 = 13/10 Then, PF/FR = 8/9 By comparing both the results, 13/10 ≠ 8/9 Therefore, PE/EQ ≠ PF/FR So, EF is not parallel...

In the given figure, DE || BC. (i) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x. (ii) If DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3, find the value of x.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- (i) From the figure, it is given that, Consider the ∆ABC, AD/DB = AE/EC x/(x – 2) = (x + 2)/(x – 1) By cross multiplication we get, X(x – 1) = (x – 2) (x + 2) x2 – x = x2 – 4 -x = -4 x =...

In the figure (iii) given below, if XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, find PY and XY.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, So, PX/QX = PY/YR 1/3 = PY/4.5 By cross multiplication we get, (4.5 × 1)/3 = PY PY = 45/30 PY = 1.5 Then, ∠X =...

In the figure (i) given below if DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm. Find
(i) AE : EC
(ii) DE.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm (i) AE: EC So, AD/BD = AE/EC AE/EC = AD/BD AE/EC = ¾ AE: EC = 3: 4 (ii) consider ∆ADE and ∆ABC ∠D = ∠B ∠E = ∠C Therefore,...

A street light bulb is fixed on a pole 6 m above the level of street. If a woman of height casts a shadow of 3 m, find how far she is away from the base of the pole?

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, Height of pole (PQ) = 6m Height of a woman (MN) = 1.5m So, shadow NR = 3m Therefore, pole and woman are standing in the same line PM ||MR ∆PRQ ~ ∆MNR...

A 15 metres high tower casts a shadow of 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, Height of a tower PQ = 15m It’s shadow QR = 24 m Let us assume the height of a telephone pole MN = x It’s shadow NO = 16 m Given, at the same time,...

In the given figure, ∠A = 90° and AD ⊥ BC If BD = 2 cm and CD = 8 cm, find AD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, consider ∆ABC, So, ∠A = 90o And AD ⊥ BC ∠BAC = 90o Then, ∠BAD + ∠DAC = 90o … [equation (i)] Now, consider ∆ADC ∠ADC = 90o So, ∠DCA + ∠DAC = 90o … [equation (ii)] From...

In the figure given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm, BE = x and AE = y. Find the values of x and y.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, AF, BE and CD are parallel lines. Consider the ∆AEF and ∆CED ∠AEF and ∠CED [because vertically opposite angles are equal] ∠F = ∠C [alternate angles are equal] Therefore,...

(a) In the figure given below, AB, EF and CD are parallel lines. Given that AB =15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate
(i) EF
(ii) AC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure it is given that, AB, EF and CD are parallel lines. (i) Consider the ∆EFG and ∆CGD ∠EGF = ∠CGD [Because vertically opposite angles are equal] ∠FEG = ∠GCD [alternate angles...

In the adjoining figure, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. Prove that
(i) EF = FC
(ii) AG : GD = 2 : 1

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure it is given that, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. (i) We have to prove that, EF = FC From the figure, D is the midpoint of...

The altitude BN and CM of ∆ABC meet at H. Prove that (i) CN × HM = BM × HN (ii) HC/HB = √[(CN × HN)/(BM × HM)] (iii) ∆MHN ~ ∆BHC

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Consider the ∆ABC, Where, the altitude BN and CM of ∆ABC meet at H. and construction: join MN (i) We have to prove that, CN × HM = BM × HN In ∆BHM and ∆CHN ∠BHM = ∠CHN [because vertically...

In the figure (2) given below, PQRS is a parallelogram; PQ = 16 cm, QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. (i) Prove that triangle RLQ is similar to triangle PLN. Hence, find PN. Sol

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is give that, Consider the ∆RLQ and ∆PLN, ∠RLQ = ∠NLP [vertically opposite angles are equal] ∠RQL = ∠LNP [alternate angle are equal] Therefore, ∆RLQ ~ ∆PLN So, QR/PN...

In the figure (1) given below, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that ∆ABE ~ ∆CFB.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, ABCD is a parallelogram, Then, E is a point on AD and produced and BE intersects CD at F. We have to prove that ∆ABE ~ ∆CFB Consider ∆ABE and ∆CFB ∠A = ∠C [opposite...

In ∆ABC, ∠A is acute. BD and CE are perpendicular on AC and AB respectively. Prove that AB x AE = AC x AD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Consider the ∆ABC, So, we have to prove that, AB × AE = AC × AD Now, consider the ∆ADB and ∆AEC, ∠A = ∠A [common angle for both triangles] ∠ADB = ∠AEC [both angles are equal to 90o] ∆ADB...

In the adjoining figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that AO/OC = BO/ODUsing the above result, find the values of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given figure, ABCD is a trapezium in which AB || DC, The diagonals AC and BD intersect at O. So we have to prove that, AO/OC = BO/OD Consider the ∆AOB and ∆COD, ∠AOB = ∠COD …...

Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Consider the two triangles, ∆MNO and ∆XYZ From the question it is given that, two triangles are similar triangles So, ∆MNO ~ ∆XYZ If two triangles are similar, the corresponding angles...

In the figure (3) given below, ∠PQR = ∠PRS. Prove that triangles PQR and PRS are similar. If PR = 8 cm, PS = 4 cm, calculate PQ.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, ∠P = ∠P (common angle for both triangles) ∠PQR = ∠PRS [from the question] So, ∆PQR ~ ∆PRS Then, PQ/PR = PR/PS = QR/SR Consider PQ/PR = PR/PS PQ/8 = 8/4 PQ = (8 × 8)/4 PQ...

In the figure (2) given below, ∠ADE = ∠ACB.
(i) Prove that ∆s ABC and AED are similar.
(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given figure, (i) ∠A = ∠A (common angle for both triangles) ∠ACB = ∠ADE [given] Therefore, ∆ABC ~ ∆AED (ii) from (i) proved that, ∆ABC ~ ∆AED So, BC/DE = AB/AE = AC/AD AD = AB –...

(a) In the figure (i) given below, ∠P = ∠RTS. Prove that ∆RPQ ~ ∆RTS.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given figure, ∠P = ∠RTS So we have to prove that ∆RPQ ~ ∆RTS In ∆RPQ and ∆RTS ∠R = ∠R (common angle for both triangle) ∠P = ∠RTS (from the question) ∆RPQ ~ ∆RTS (b) In the figure...

In the figure (2) given below, CA || BD, the lines AB and CD meet at G.
(i) Prove that ∆ACO ~ ∆BDO.
(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- (i) We have to prove that, ∆ACO ~ ∆BDO. So, from the figure Consider ∆ACO and ∆BDO Then, ∠AOC = ∠BOD [from vertically opposite angles] ∠A = ∠B Therefore, ∆ACO = ∆BDO Given, BD = 2.4 cm,...

In the figure given below, AB || DE, AC = 3 cm, CE = 7.5 cm and BD = 14 cm. Calculate CB and DC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, AB||DE AC = 3 cm CE = 7.5 cm BD = 14 cm From the figure, ∠ACB = ∠DCE [because vertically opposite angles] ∠BAC = ∠CED [alternate angles] Then, ∆ABC ~...

Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4 cm, 7 cm and 8 cm.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Let us assume that, ∆ABC ~ ∆DEF ∆ABC is BC = 6cm ∆ABC ~ ∆DEF So, AB/DE = BC/EF = AC/DF Consider AB/DE = BC/EF AB/8 = 6/4 AB = (6 × 8)/4 AB = 48/4 AB = 12 Now, consider BC/EF = AC/DF 6/4 =...

If ∆ABC ~ ∆PQR, Perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then find the length of AC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ∆ABC ~ ∆PQR Perimeter of ∆ABC = 32 cm Perimeter of ∆PQR = 48 cm So, AB/PQ = AC/PR = BC/QR Then, perimeter of ∆ABC/perimeter of ∆PQR = AC/PR 32/48 =...

If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, then find the perimeter of ∆ABC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution; Now, we have to find out the perimeter of ΔABC Let ΔABC ~ ΔDEF So, AB/DE = AC/DF = BC/EF Consider, AB/DE = AC/DE 4/6 = AC/12 By cross multiplication we get, AC = (4 × 12)/6 AC = 48/6 AC =...

It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

From the question it is given that, ΔDEF ~ ΔLMN So, AB/ED = AC/EF = BC/DF Consider AB/ED = AC/EF 5/12 = 7/EF By cross multiplication, EF = (7 × 12)/5 EF = 16.8 cm Now, consider AB/ED = BC/DF 5/12 =...

If in two right triangles, one of the acute angle of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles are similar? Why?

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, two line segments are intersecting each other at P. In ΔBCP and ΔDPE 5/10 = 6/12 Dividing LHS and RHS by 2 we get, ½ = ½ Therefore, ΔBCD ~ ΔDEP

It is given that ∆DEF ~ ∆RPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P ? Why?

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question is given that, ∆DEF ~ ∆RPQ ∠D = ∠R and ∠F = ∠Q not ∠P No, ∠F ≠ ∠P

State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm):

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Similarity

Solution:- (i) From the ΔABC and ΔPQR AB/PQ = 3.2/4 = 32/40 Divide both numerator and denominator by 8 we get, = 4/5 AC/PR = 3.6/4.5 = 36/45 Divide both numerator and denominator by 9 we get, = 4/5...

Two different dice are thrown at the same time. Find the probability of getting :
(iii) sum divisible by 5
(iv) sum of at least 11.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii)Let E be an event of getting a sum divisible by 5. Favourable outcomes = {(1,4),(2,3), (3,2), (4,1),(4,6), (5,5), (6,4)} Number of favourable outcomes = 7 P(E) = 7/36 Probability of getting a...

Two different dice are thrown at the same time. Find the probability of getting :
(i) a doublet
(ii) a sum of 8

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When two dice are thrown simultaneously, the sample space of the experiment is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1),(3,2), (3,3),...

Two different dice are thrown simultaneously. Find the probability of getting:
(i) a number greater than 3 on each dice
(ii) an odd number on both dice.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When two dice are thrown simultaneously, the sample space of the experiment is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1),(3,2), (3,3),...

Two different coins are tossed simultaneously. Find the probability of getting :
(iii) no tail
(iv) at most one tail.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii)Let E be an event of getting no tails. Favourable outcomes = HH Number of favourable outcomes = 1 P(E) = 1/4 Probability of getting no tails is 1/4 . (iv)Let E be an event of getting atmost one...

Two different coins are tossed simultaneously. Find the probability of getting :
(i) two tails
(ii) one tail

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When 2 coins are tossed, the possible outcomes are HH. HT, TH, TT. Number of possible outcomes = 4 (i)Let E be an event of getting 2 tails. Favourable outcomes = TT Number of favourable...

Two coins are tossed once. Find the probability of getting:
(i) 2 heads
(ii) at least one tail.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When 2 coins are tossed, the possible outcomes are HH. HT, TH, TT. Number of possible outcomes = 4 (i)Let E be an event of getting 2 heads. Favourable outcomes = HH Number of favourable...

From a pack of 52 cards, a blackjack, a red queen and two black kings fell down. A card was then drawn from the remaining pack at random. Find the probability that the card drawn is
(i) a black card
(ii) a king
(iii) a red queen.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of cards = 52-4 = 48 [∵4 cards fell down] So number of possible outcomes = 48 (i) Let E be the event of getting black card. There will be 23 black cards since a black jack and...

All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting
(v) a spade
(vi) ‘9’ of black colour.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(v) Let E be the event of getting a spade. There will be 10 spades. Number of favourable outcomes = 10 P(E) = 10/49 Hence the probability of getting a spade is 10/49. (vi) Let E be the event of...

All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting
(iii) a black card
(iv) a heart

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of getting a black card. There will be 23 black cards remaining since 3 spades are removed. Number of favourable outcomes = 23 P(E) = 23/49 Hence the probability of getting...

All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting
(i) a black face card
(ii) a queen

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of cards = 52-3 = 49. [since 3 face cards of spade are removed] So number of possible outcomes = 49. (i) Let E be the event of getting black face card. There will be 3 black...

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(xi) neither a spade nor a jack
(xii) neither a heart nor a red king

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(xi) Let E be the event of getting a neither a spade nor a jack. There are 13 spades and 3 other jacks. So remaining cards = 52-13-3 = 36 There will be 36 cards which are neither a spade nor a jack....

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(ix) a non-ace
(x) non-face card of black colour

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(ix) Let E be the event of getting a non ace card. There will be 48 non ace cards. Number of favourable outcomes = 48 P(E) = 48/52 = 24/26 = 12/13 Hence the probability of getting a non ace card is...

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(vii) a black face card
(viii) a black card

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(vii) Let E be the event of getting a black face card. There will be 6 black face cards. Number of favourable outcomes = 6 P(E) = 6/52 = 3/26 Hence the probability of getting a black face card is...

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(iii) a king of red colour
(iv) a card of diamond

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of getting a king of red colour. There will be 2 cards of king of red colour. Number of favourable outcomes = 2 P(E) = 2/52 = 1/26 Hence the probability of getting a king of...

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(i) ‘2’ of spades
(ii) a jack .

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of cards = 52. So number of possible outcomes = 52. (i) Let E be the event of getting ‘2’ of spades. There will be only one card of ‘2’ spades. Number of favourable outcomes =...

A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. Find the probability that it is
(i) white
(ii) not red.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of balls = 24 Number of red balls = x. Number of white balls = 2x. Number of blue balls = 3x. x+2x+3x = 24 6x = 24 x = 24/6 = 4 Number of red balls = x = 4 Number of white...

A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball is twice that of a red ball, find the number of balls in the bag.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of red balls = 6 Let number of blue balls be x. Total number of balls = 6+x Probability of drawing a red ball = 6/(6+x) Probability of drawing a blue ball = x/(6+x) Given the...

A bag contains 15 balls of which some are white and others are red. If the probability of drawing a red ball is twice that of a white ball, find the number of white balls in the bag.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of balls in the bag = 15. Let the number of white balls be x. Then number of red balls = 15-x. The probability of drawing a white ball = x/15. Probability of drawing a red...

Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is
(iii) a number which is a perfect square
(iv) a prime number less than 30.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of getting the number on the card is a perfect square. Outcomes favourable to E are {4,9,16,25,36,49,64,81,100} Number of favourable outcomes = 9 P(E) = 9/100 Hence the...

Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is
(i) an even number
(ii) a number less than 14

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {2,3,…101} Number of possible outcomes = 100 (i) Let E be the event of getting the number on the card is an even number. Outcomes favourable to E are...

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,…90} Number of possible outcomes = 90 (i) Let E be the event of getting the number on the disc is a two-digit number. Outcomes favourable to E are...

Tickets numbered 3, 5, 7, 9,…., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is
(i) a prime number
(ii) a number less than 16
(iii) a number divisible by 3.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {3,5,7,9..…29} Number of possible outcomes = 14 (i) Let E be the event of getting the number on the ticket is a prime number. Outcomes favourable to E are...

Cards marked with numbers 13, 14, 15, …, 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card drawn is
(i) divisible by 5
(ii) a perfect square number.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {13,14,15,…60} Number of possible outcomes = 48 (i) Let E be the event of getting the number on the card is divisible by 5. Outcomes favourable to E are...

.A box contains 19 balls bearing numbers 1, 2, 3,…., 19. A ball is drawn at random from the box. Find the probability that the number on the ball is :
(iii) neither divisible by 5 nor by 10
(iv) an even number.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of getting the number on the ball is neither divisible by 5 nor by 10. Outcomes favourable to E are {1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19} Number of favourable outcomes =...

.A box contains 19 balls bearing numbers 1, 2, 3,…., 19. A ball is drawn at random from the box. Find the probability that the number on the ball is :
(i) a prime number
(ii) divisible by 3 or 5

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,4…19} Number of possible outcomes = 19 (i) Let E be the event of getting the number on the ball is a prime number. Outcomes favourable to E are...

A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is :
(v) divisible by 3 or 2
(vi) a perfect square number.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(v) Let E be the event of getting the number on the card is divisible by 3 or 2 Outcomes favourable to E are {2,3,4,6,8,9,10,12,14,15} Number of favourable outcomes = 10 P(E) = 10/15 = 2/3 Hence the...

A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is :
(iii) divisible by 3
(iv) divisible by 3 and 2 both

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of getting the number on the card is divisible by 3. Outcomes favourable to E are {3,6,9,12,15} Number of favourable outcomes = 5 P(E) = 5/15 = 1/3 Hence the probability of...

A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) Odd
(ii) prime

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,4…15} Number of possible outcomes = 15 (i) Let E be the event of getting the number on the card is odd. Outcomes favourable to E are {1,3,5,7,9,11,13,15}...

A box contains 25 cards numbered 1 to 25. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) even
(ii) prime
(iii) multiple of 6.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,4 ….25} Number of possible outcomes = 25 (i) Let E be the event of getting the number on the card is an even number. Outcomes favourable to E are...

Cards marked with numbers 1, 2, 3, 4,…20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is
(i) a prime number
(ii) divisible by 3
(iii) a perfect square ? (2010)

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,….20} Number of possible outcomes = 20 (i) Let E be the event of getting the number on the card is a prime number. Outcomes favourable to E are...

An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7?
(ii) not divisible by 7?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of integers between 0 and 100 = 99 Number of possible outcomes = 99 (i) Let E be the event of getting an integer divisible by 7. Outcomes favourable to E are...

Sixteen cards are labeled as a, b, c,…, m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is:
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p} Number of possible outcomes = 16 (i) Let E be the event of getting a vowel. Outcomes favourable to E are { a,e,i,o} Number of...

Find the probability that the month of February may have 5 Wednesdays in
(i) a leap year
(ii) a non-leap year.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: There are 7 ways in which the month of February can occur, each starting with a different day of the week. (i)Only 1 way is possible for 5 Wednesdays to occur in February with 29 days....

Find the probability that the month of January may have 5 Mondays in
(i) a leap year
(ii) a non-leap year.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: For a leap year there are 366 days. Number of days in January = 31 Total number of January month types = 7 Number of January months with 5 Mondays = 3 (i)Probability that the month of...

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the adjoining figure) and these are equally likely outcomes. What is the probability that it will point at
(iii) a number greater than 2?
(iv) a number less than 9?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of arrow pointing a number greater than 2. Outcomes favourable to E are {3,4,5,6,7,8} Number of favourable outcomes = 6 P(E) = 6/8 = 3/4 Hence the probability of arrow...

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the adjoining figure) and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes of the game are {1,2,3,4,5,6,7,8} Number of possible outcomes = 8 (i) Let E be the event of arrow pointing 8. Outcomes favourable to E is 8. Number of favourable...

A die has 6 faces marked by the given numbers as shown below: The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than – 3.
(iii) the smallest integer ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When a die is thrown, the possible outcomes are {1,2,3,-1,-2,-3} Number of possible outcomes = 6 (i) Let E be the event of getting a positive integer. Outcomes favourable to E are {1,2,3}...

In a single throw of a die, find the probability of getting:
(vii) a number between 3 and 6
(viii) a number divisible by 2 or 3.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(vii) Let E be the event of getting a number between 3 and 6. Outcomes favourable to E is 4,5. Number of favourable outcomes = 2 P(E) = 2/6 = 1/3 Hence the probability of getting a number between 3...

In a single throw of a die, find the probability of getting:
(v) a number less than 8
(vi) a number divisible by 3

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(v) Let E be the event of getting a number less than 8. Outcomes favourable to E is 1,2,3,4,5,6. Number of favourable outcomes = 6 P(E) = 6/6 = 1 Hence the probability of getting a number less than...

In a single throw of a die, find the probability of getting:
(iii) a number greater than 5
(iv) a prime number

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii)Let E be the event of getting a number greater than 5. Outcomes favourable to E is 6. Number of favourable outcomes = 1 P(E) = 1/6 Hence the probability of getting a number greater than 5 is...

In a single throw of a die, find the probability of getting:
(i) an odd number
(ii) a number less than 5

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When a die is thrown, the possible outcomes are 1,2,3,4,5,6. Number of possible outcomes = 6 (i) Let E be the event of getting an odd number. Outcomes favourable to E are 1,3,5. Number of...

A die is thrown once. What is the probability that the
(i) number is even
(ii) number is greater than 2 ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When a die is thrown, the possible outcomes are 1,2,3,4,5,6. So Sample space = { 1,2,3,4,5,6} Number of possible outcomes = 6 Even numbers are (2,4,6). Number of favourable outcomes = 3...

A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Peter, a trader, will only accept the shirts which are good, but Salim, another trader, will only reject the shirts which have major defects. One shirts is drawn at random from the carton. What is the probability that
(i) it is acceptable to Peter ?
(ii) it is acceptable to Salim ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of shirts = 100 Number of good shirts = 88 Number of shirts with minor defects = 8 Number of shirts with major defects = 4 Peter accepts only good shirts. So number of shirts...

A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. It is equally likely that one of the coins will fall down when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be Rs 5 coin?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of 50 paisa coins = 100 Number of 1 rupee coins = 50 Number of 2 rupee coins = 20 Number of 5 rupee coins = 10 Total number of coins = 100+50+20+10 = 180 (i) Probability of getting...

A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :
(iii) not green
(iv) neither white nor black.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii)Probability of not green = Probability of getting red, white and black = (6+8+3)/22 = 17/22 Hence the probability of not green is 17/22. (iv) Probability of neither white nor black =...

A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :
(i) white
(ii) red or black

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of red balls = 6 Number of white balls = 8 Number of green balls = 5 Number of black balls = 3 Total number of marbles = 6+8+5+3 = 22 (i)Probability of white balls, = 8/22 = 4/11...

A box contains 7 blue, 8 white and 5 black marbles. If a marble is drawn at random from the box, what is the probability that it will be
(iii) not black?
(iv) green?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Probability of not black = Probability of white and blue = (7+8)/20 = 15/20 = 3/4 Hence the probability of not black is 3/4. (iv) Since there are no green marbles in the box, the probability...

A box contains 7 blue, 8 white and 5 black marbles. If a marble is drawn at random from the box, what is the probability that it will be
(i) black?
(ii) blue or black?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of blue marbles = 7 Number of white marbles = 8 Number of black marbles = 5 Total number of marbles = 7+8+5 = 20 (i) Probability of getting black , = 5/20 = 1/4 Hence the...

A bag contains 5 black, 7 red and 3 white balls. A ball is drawn at random from the bag, find the probability that the ball drawn is: (i) red (ii) black or white (iii) not black.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of black balls = 5 Number of red balls = 7 Number of white balls = 3 Total number of balls = 5+7+3 = 15 (i)Probability that the ball drawn is red, = 7/15 (ii) Probability of black...

A letter of English alphabet is chosen at random. Determine the probability that the letter is a consonant.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of alphabets = 26 Number of vowels = 5 Total number of consonants = 26-5 = 21 Probability that the letter chosen is a consonant , = 21/26 Hence the required probability is...

A letter is chosen from the word ‘TRIANGLE’. What is the probability that it is a vowel ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of vowels in the word ‘TRIANGLE’ = 3 Total number of letters = 8 Probability that the letter chosen is a vowel , P(E) = 3/8 Hence the probability that the letter chosen is a vowel...

There are 40 students in Class X of a school of which 25 are girls and the others are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of
(i) a girl ?
(ii) a boy ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of students = 40 Number of girls = 25 Number of boys = 40-25 = 15 (i) Probability of getting a girl, P(E) = 25/40 = 5/8 Hence the probability of getting a girl is 5/8. (ii)...

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from a bag. What is the probability that the ball drawn is .
(i) red ?
(ii) not red ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: (i) Number of red balls = 3 Number of black balls = 5 Total number of balls = 3+5 = 8 Probability that the ball drawn is red , P(E) = 3/8 Hence the probability that the ball drawn is red...

Two players, Sania and Sonali play a tennis match. It is known that the probability of Sania winning the match is 0.69. What is the probability of Sonali winning ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Probability of Sania winning the match, P(E) = 0.69 Probability of Sonali winning = Probability of Sania losing, = 1-0.69 = 0.31 Hence the probability of Sonali winning is...

If the probability of winning a game is 5/11, what is the probability of losing ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Given probability of winning the game, P(E) = 5/11 We know that, Probability of losing game, = 1-5/11 = (11-5)/11 = 6/11 Hence the probability of losing game is 6/11.

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of defective pens = 12 Number of good pens = 132. Total number of pens = 132+12 = 144 Probability of getting a good pen, P(E) P(E) = 132/144 = 11/12 Hence the required probability...

In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of prized tickets = 5 Number of blank tickets = 995 Total number of tickets = 5+995 = 1000 The probability of winning a prize, P(E) P(E) = 5/1000 = 1/200 Hence the required...