Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 10 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 10 with Solutions

Time: 3 hrs

Max. Marks: 80

Instructions

- This question paper has 5 Sections A-E.
- Section A has 20 MCQs carrying 1 mark each.
- Section B has 5 questions carrying 2 marks each.
- Section C has 6 questions carrying 3 marks each.
- Section D has 4 questions carrying 5 marks each.
- Section E has 3 Case Based integrated units of assessment (4 marks each).
- All questions are compulsory. However, an internal choice in 2 questions of 2 marks, 2 questions of 3 marks and 2 questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.
- Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

Section A

(Section A consists of 20 questions of 1 mark each.)

Question 1.

Which of the following is not an irrational number? [1]

(a) 7\(\sqrt{5}\)

(b) \(\sqrt{2}\) + 2\(\sqrt{2}\)

(c)(\(\sqrt{7}\)– 3)- \(\sqrt{7}\)

(d) \(\sqrt{3}\) + 2

Answer:

(c)(\(\sqrt{7}\)– 3)- \(\sqrt{7}\)

We have, (\(\sqrt{7}\) – 3) – \(\sqrt{7}\) = \(\sqrt{7}\) – 3 – \(\sqrt{7}\) = -3

Since, – 3 is a rational number.

So, (\(\sqrt{7}\) – 3) – \(\sqrt{7}\) is a rational number.

Question 2.

The value of \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}\) is [1]

(a) -1

(b) 1

(c) 0

(d) 2

Answer:

(c) 0

We have, \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}\) = \(\frac{1-(1)^2}{1+(1)^2}\) = \(\frac{1-1}{1+1}\) = \(\frac{1-1}{2}\) = 0

Question 3.

If the HCF of 408 and 1032 is expressed in the form 1032 × 2 + 408 × p, then the value of p is [1]

(a) 5

(b) -5

(c) 4

(d) -4

Answer:

(b) -5

Prime factors of given numbers are

408 = 2^{3} × 3 × 17

and 1032 = 2^{3} × 3 × 43

HCF (408, 1032) = 2^{3} × 3^{1} = 24

But HCF expressed in the form

1032 × 2 + 408 × p = 24

⇒ 2064 + 408 × p = 24

⇒ 408p = – 2040

⇒ p = –\(\frac{2040}{408}\) = -5

Question 4.

The value of x and y in the given figure are [1]

(a) 7, 13

(b) 13, 7

(c) 9, 12

(d) 12, 9

Answer:

(a) 7, 13

Given number is 1001. Then, the factor tree of 1001 is given as below

On comparing with given factor tree, we get

x = 7 and y = 13

Question 5.

The discriminant of the quadratic equation 6x^{2} – 7x + 2 = 0 is [1]

(a) 0

(b)-1

(c) 1

(d) 2

Answer:

(c) 1

Given, 6x^{2} – 7x + 2 = 0 …….. (i)

On comparing Eq. (i) with ax^{2} + bx + c = 0, we get

a = 6, b = -7 and c = 2

We know that, D = b^{2} – 4ac

= (-7)^{2} – 4(6)(2)

= 49 – 48 = 1

Hence, the value of discriminant is 1.

Question 6.

Two dice are thrown together. Then, the probability that sum of the two numbers will be multiple of 4, is [1]

(a)-\(\frac{1}{4}\)

(b)\(\frac{1}{2}\)

(c)-\(\frac{1}{2}\)

(d)\(\frac{1}{4}\)

Answer:

(d)\(\frac{1}{4}\)

Let E be the event of getting the sum of two numbers as a multiple of 4

i.e., E = {(1, 3), (2, 2), (3, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}

∴ n(E) = 9

Here, total number of events, n(S) = 36

∴ Required probability = \(\frac{9}{36}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

Question 7.

In the given figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, then the value of CD is [1]

(a) 16 cm

(b) 8 cm

(c) 4 cm

(d) 10 cm

Answer:

(a) 16 cm

Given, BD = 8 cm and AD = 4 cm

In ∆ADB and ∆BDC,

∠BDA = ∠CDB [each 90°]

∠DBA = ∠DCB [each (90°-∠A)]

∆ADB~∆BDC [by AA similarity criterion]

⇒ \(\frac{B D}{C D}\) = \(\frac{A D}{B D}\)

⇒ CD = \(\frac{(B D)^2}{A D}\) = \(\frac{8^2}{4}\) = \(\frac{64}{4}\) = 16 cm

Question 8.

The mean of n observations is \(\bar{x}\). If the first term is increased by 1, second by 2 and so

on, then the new mean is [1]

(a) \(\bar{x}\) + \(\frac{n+1}{2}\)

(b) \(\bar{x}\) – \(\frac{n+1}{2}\)

(c) \(\frac{n+1}{2}\)

(d) None of these

Answer:

(a) \(\bar{x}\) + \(\frac{n+1}{2}\)

∴ Required mean

Question 9.

If cosec A = 2, then the value of \(\frac{1}{\tan A}\) + \(\frac{\sin A}{1+\cos A}\) is [1]

(a) 0

(b) 1

(c) 2

(d) 3

Answer:

(c) 2

Given, cosec A = 2 ⇒ A = 30°

Question 10.

The minute hand of a clock is 10 cm long. The area swept by the minute hand between 8:00 am to 8:25 am, is [1]

(a) 130.95 cm^{2}

(b) 130 cm^{2}

(c) 131.95 cm^{2}

(d) 131 cm^{2}

Answer:

(a) 130.95 cm^{2}

Length of minute hand = Radius of circle

= 10 cm

Angle swept by minute hand in 60 min = 360°.

Angle swept by minute hand in 25 min

= \(\frac{360^{\circ}}{60}\) × 25 = 150°

∴ Area of sector = \(\frac{\theta}{360^{\circ}}\) × πr^{2}

= \(\frac{150^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 10 × 10

= 130.95 cm^{2}

Question 11.

The number of solutions of the pair of equations x = 0 and x = 3 is [1]

(a) infinite many solutions

(b) no solution

(c) one solution

(d) None of the above

Answer:

(b) no solution

x = 0 is the Y-axis and x = 3 is the line parallel to Y-axis at a distance of 3 units from it. These lines do not meet anywhere. So, no solution exists.

Question 12.

The distance of the point (3, 5) from the X-axis is [1]

(a) 3 units

(b) 5 units

(c) 8 units

(d) 4 units

Answer:

(b) 5 units

The point (3, 5) is shown in the figure

From the figure, the required distance is 5 units.

Question 13.

If a cylinder is filled with water and spherical ball of radius r is dropped into the cylinder, then the quantity of water spread out, is [1]

(a) 2 × volume of sphere

(b) volume of sphere

(c) 3 × volume of sphere

(d) None of the above

Answer:

(b) volume of sphere

When, we draped the ball into a cylinder full of water, the quantity of spreading water is equal to the volume of sphere i.e. \(\frac{4 \pi}{3}\)r^{3}

Question 14.

The diameter of a wheel is 1.26 m. How long will it travel in 500 revolutions? [1]

(a) 1492 m

(b) 2530 m

(c) 1980 m

(d) 2880 m

Answer:

(c) 1980 m

Given, Diameter of the wheel = 1.26 m

Radius of the wheel, r = \(\frac{1.26}{2}\) = 0.63 m

Distance travelled in one revolution = Perimeter of the wheel

= 2πr = 2 × \(\frac{22}{7}\) × 0.63 = 3.96 m

Thus, distance travelled in 500 revolutions = 500 × 3.96 = 1980 m

Question 15.

The simplified value of(l -cos^{2} A) cosec^{2} A is [1]

(a) -1

(b) 1

(c) 0

(d) 2

Answer:

(b) 1

We have, (1 – cos^{2} A) cosec^{2}A

= cosec^{2}A – cos^{2}A cosec^{2}A

= cosec^{2}A – cos^{2}A∙\(\frac{1}{\sin ^2 A}\)

= cosec^{2}A – cot^{2}A = 1

Question 16.

In the given figure, O is the centre of the circle with PA and PB as tangents. [1]

If measure of ∠ADB = 60°, then ∆PAB is an

(a) isosceles triangle

(b) equilateral triangle

(c) scalene triangle

(d) None of these

Answer:

(b) equilateral triangle

Given, ∠ADB = 60° ⇒ ∠AOB = 2∠ADB = 120°

[since, angle made by an arc at the centre is double the angle made at the circumference]

In ∆AOS, OA = OB [radii]

∴ ∠OAB = ∠OBA

[angle opposite to equal sides]

So, we have ∠OAB = ∠OBA = 30°

Question 17.

If the points A(4,3) and B(x, 5) are on the circle with centre O (2, 3), then the value of x is [1]

(a) 5

(b) 6

(c) 2

(d) 4

Answer:

(c) 2

Since, A and B lie on the circle having centre O. Then, the distance between points A and B from the centre are same as they are radius of the circle.

i.e. OA = OB

From the distance formula,

\(\sqrt{(4-2)^2+(3-3)^2}\) = \(\sqrt{(x-2)^2+(5-3)^2}\)

⇒ \(\sqrt{4+0}\) = \(\sqrt{(x-2)^2+4}\)

On squaring both sides, we get

4 = (x – 2)^{2} + 4

⇒ (x – 2 )^{2} = 0 ⇒ x – 2 = 0 ⇒ x = 2

Question 18.

If the difference of mode and median of a data is 28, then the difference of median and mean is [1]

(a) 10

(b) 12

(c) 14

(d) 16

Answer:

(c) 14

We have, mode – median = 28 We know that

∴ Mode = 3 (Median) – 2 Mean

Mode – Median = 2 (Median) – 2 (Mean)

⇒ 28 = 2 (Median – Mean)

⇒ Median – Mean = 14

Directions In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Question 19.

Assertion (A) If α and β are the zeroes of the polynomial x^{2} + 2x – 15, then \(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) is \(\frac{2}{15}\).

Reason (R) If α and β are the zeroes of a quadratic polynomial ax^{2} + bx + c, then

α + β = –\(\frac{b}{a}\) and αβ = \(\frac{c}{a}\). [1]

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Let f(x) = x^{2} + 2x – 15

On comparing f(x) with ax^{2} + bx + c = 0, we get a = 1, b = 2 and c = -15

Then, α + β = -2 and α∙β = -15

⇒ \(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) = \(\frac{\alpha+\beta}{\alpha \cdot \beta}\) = \(\frac{-2}{-15}\) = \(\frac{2}{15}\)

So, both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Question 20.

Assertion (A) AABC is an isosceles right triangle, right angled at C. Then, AB^{2} = 2 AC^{2}.

Reason (R) In a right angled triangle, the cube of the hypotenuse is equal to the sum of the squares of the other two sides. [1]

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true.

Answer:

(c) Assertion (A) is true but Reason (R) is false.

In right angled ∆ABC,

AB^{2} = AC^{2} + BC^{2} [by Pythagoras theorem]

= AC^{2} + AC^{2} [∵ BC = AC]

= 2AC^{2}

∴ AB^{2} = 2AC^{2}

Hence, Assertion (A) is true but Reason (R) is false.

Section B

(Section B consists of 5 questions of 2 marks each.)

Question 21.

Solve the following system of equations by elimination method [2]

x + y = a – b

ax – by = a^{2} + b^{2}

Answer:

The given equations are

x + y = a – b ………..(i)

ax – by = a^{2} + b^{2} ……….(ii)

On multiplying Eq. (i) by b, we get

bx + by = ab – b^{2} ……….(iii)

On adding Eqs. (ii) and (iii), we get

(ax – by) + (bx + by) = (a^{2} + b^{2}) + (ab – b^{2})

(a + b)x = a^{2} + ab = a(a + b)

⇒ x = \(\frac{a(a+b)}{a+b}\) = a

On putting x = a in Eq . (i), we get

a + y = a – b ⇒ y = a – b – a = -b

Or

In a two-digit number, the ten’s digit is three times the unit’s digit. When the number is decreased by 54, then the digits are reversed. Find the number. [2]

Answer:

Let the digit in unit’s place be x and the digit in ten’s place be y.

Then, number = 10y + x

According to the given condition,

y = 3x ……. (i)

Number obtained by reversing the digits = 10x + y

If the number is decreased by 54, the digits are reversed.

∴ Number – 54 = Number of obtained by reversing the digits

⇒ 10y + x – 54 = 10x + y

⇒ 9x – 9y = -54

⇒ x – y = -6

[dividing on both side by 9] ……… (ii)

On putting y = 3x from Eq. (i) to Eq. (ii), we get

x – 3x = -6

⇒ x = 3

On putting x = 3 in Eq. (i), we get

y = 9

Hence, number = 10y + x

= 10 × 9 + 3

= 93

Question 22.

Find the length of a tangent drawn to a circle, with radius 5 cm, from a point 13 cm away from the centre of the circle. [2]

Answer:

According to the given information, we have the following figure.

Clearly, OT ⊥ PT

[∵ radius is perpendicular to the tangent at the point of contact]

∴ ∠OTP = 90°

Now, in right angled ∆OTP, we have

OP^{2} = OT^{2} + PT^{2} [by Pythagoras theorem]

⇒ 13^{2} = 5^{2} + PT^{2}

⇒ PT^{2} = 169 – 25 = 144

∴ PT = 12 cm

[taking positive square roots on both sides]

Hence, length of the tangent is 12 cm.

Question 23.

If cos α = \(\frac{1}{2}\) and tan β = \(\frac{1}{\sqrt{3}}\), then find sin (α + β), where α and β are both acute angles. [2]

Answer:

Or

Find an acute angle θ, when

\(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\) = \(\frac{1-\sqrt{3}}{1+\sqrt{3}}\) [2]

Answer:

Alternate Method

We have, \(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\) = \(\frac{1-\sqrt{3}}{1+\sqrt{3}}\)

⇒ \(\frac{\frac{\cos \theta-\sin \theta}{\cos \theta}}{\frac{\cos \theta+\sin \theta}{\cos \theta}}\) = \(\frac{1-\sqrt{3}}{1+\sqrt{3}}\)

[dividing numerator and denominator by cos θ]

⇒ \(\frac{1-\tan \theta}{1+\tan \theta}\) = \(\frac{1-\sqrt{3}}{1+\sqrt{3}}\)

⇒ 1 + \(\sqrt{3}\) – tan θ – \(\sqrt{3}\)tan θ = 1 + tan θ – \(\sqrt{3}\) – \(\sqrt{3}\)tan θ

⇒ 2tan θ = 2\(\sqrt{3}\)

⇒ tan θ = \(\sqrt{3}\) ⇒ tan θ = tan 60°

⇒ θ = 60°

Question 24.

In an isosceles ∆ABC, AB ⊥ AC and BD ⊥ AC. Prove that BD^{2} – CD^{2} = 2 CD ∙ AD. [2]

Answer:

Given, ∆ABC in which AB = AC and BD ⊥ AC.

To prove (BD^{2} – CD^{2}) = 2CD ∙ AD

Proof From right ∆ADB, we have

AB^{2} = AD^{2} + BD^{2} [by Pythagoras Theorem]

⇒ AC^{2} = AD^{2} + BD^{2} [∵ AB = AC]

⇒ (CD + AD)^{2} = AD^{2} + BD^{2}

[∵ AC = CD + AD]

⇒ CD^{2} + AD^{2} + 2CD.AD = AD^{2} + BD^{2}

⇒ (BD – CD)^{2} = 2CD.AD

Hence, (BD^{2} – CD^{2}) = 2CD.AD

Hence proved.

Question 25.

The short and long hands of a clock are 6 cm and 8 cm long, respectively. Then, find the sum of the distance travelled by their tips in 1 day. [take π = 22/7] [2]

Answer:

In 1 day, i.e. 24 h, short (hour) hand of the clock make 2 revolutions and long (minute) hand make 24 revolutions.

In 1 revolution, distance travelled by tip of hour hand = Circumference of circle of radius 6 cm = 2 × \(\frac{22}{7}\) × 6

In 1 revolution distance travelled by tip of minute hand = Circumference of circle of radius 8 cm = 2 × \(\frac{22}{7}\) × 8

∴ Sum of distances travelled by tips of both hand in 1 day = 2 × 2 × \(\frac{22}{7}\) × 6 + 24 × 2 × \(\frac{22}{7}\) × 8

= 2 × \(\frac{22}{7}\)(12 + 192)

= 2 × \(\frac{22}{7}\) × 204

= 1282.29 cm (approx).

Section C

(Section C consists of 6 questions of 3 marks each.)

Question 26.

Prove that 7\(\sqrt{5}\) is irrational. [3]

Answer:

Let us suppose that 7\(\sqrt{5}\) is rational.

Let there be two co-prime integers a and b.

Such that, 7\(\sqrt{5}\) = \(\frac{a}{b}\), where b ≠ 0

Now, 7\(\sqrt{5}\) = \(\frac{a}{b}\)

⇒ \(\sqrt{5}\) = \(\frac{a}{7 b}\)

= \(\frac{\text { integer }}{7 \text { (integer) }}\) = rational

⇒ \(\sqrt{5}\) is a rational.

This contradicts the fact that \(\sqrt{5}\) is irrational.

∴ We conclude that 7\(\sqrt{5}\) is irrational.

Question 27.

If sin θ + sin^{2} θ = 1, then find the value of

cos^{12}θ + 3cos^{1o}θ + 3cos^{8}θ + cos^{6}θ + 2 cos^{4}θ + 2cos^{2}θ – 2. [3]

Answer:

Given sin θ + sin^{2}θ = 1

⇒ sinθ = 1 – sin2θ

⇒ sinθ = cos^{2}θ

Now, cos^{12}θ + 3cos^{10}θ + 3cos^{8}θ + cos^{6}θ + 2cos^{4}θ + 2cos^{2}θ – 2

= (cos^{12}θ + 3cos^{10}θ + 3cos^{8}θ + cos^{6}θ) + 2(cos^{4}θ + cos^{2}θ – 1)

= (cos^{4}θ + cos^{2}θ)^{3} + 2 (cos^{4}θ + cos^{2}θ – 1)

[∵ a^{3} + 3a^{2}b+ 3ab^{2} + b^{3} = (a+ b)^{3}]

= (sin^{2}θ + cos^{2}θ) + 2(sin^{2}θ + cos^{2}θ – 1)

[∵ cos^{2}A = sinA ⇒ cos^{4}A = sin^{2}A]

= 1 + 2(1 – 1) = 1

Question 28.

In figure, if PQ || BC and PR || CD. Prove that \(\frac{A R}{A D}\) = \(\frac{A Q}{A B}\) [3]

Answer:

If ∆ABC, we have

PQ || SC [given]

Therefore, by basic proportionality theorem, we have,

\(\frac{A Q}{A B}\) = \(\frac{A P}{A C}\) ….. (i)

In ∆ACD, we have

PR || CD ………. (2)

Therefore, by basic proportionally theorem, we have

\(\frac{A P}{A C}\) = \(\frac{A R}{A D}\)

From Eqs. (i) and (ii), we get

\(\frac{A Q}{A B}\) = \(\frac{A R}{A D}\) or \(\frac{A R}{A D}\) = \(\frac{A Q}{A B}\)

Or

Two triangles ∆ABC and ∆DBC are on the same base BC and on the same side of BC in which ∠A = ∠D = 90°. If CA and BD meet each other at E, then show that AE ∙ EC = BE ∙ ED. [3]

Answer:

Given, ∆ABC and ∆DBC are on the same base BC and on the same side of BC in which ∠A = ∠D = 90° such that CA and BD meet each other at E.

To prove AE ∙ EC = BE ∙ ED

Proof In ∆AEB and ∆DEC, ∠BAC = ∠BDC = 90°

and ∠AEB = ∠DEF []

[vertically opposite angles]

∴ ∆AEB ~ ∆DEC

[by AA similarity criterion]

⇒ \(\frac{A E}{E D}\) = \(\frac{B E}{E C}\)

∴ AE ∙ EC = BE ∙ ED

Hence proved.

Question 29.

If α and β are zeroes of a quadratic polynomial x^{2} – 5, then form a quadratic polynomial whose zeroes are 1 + α and 1 + β. [3]

Answer:

Let p(x) = x^{2} – 5

For finding the zeroes of p(x), put p(x) = 0.

∴ x^{2} – 5 = 0 ⇒ x = ±5

Let α = 5 and β = -5

Now, 1 + α = 1 + 5 = 6 and 1 + β = 1 – 5 = -4

Thus, 6 and -4 are the zeroes of new quadratic polynomial.

Therefore, the new quadratic polynomial will be (x – 6) (x + 4) or x^{2} – 2x – 24.

Or

Can the quadratic polynomial x^{2} + kx + k have equal zeroes for some odd integer k > 1? [3]

Answer:

Let the given polynomial be p(x) = x^{2} + kx + k.

On comparing it with ax^{2} + bx + c, we get a = 1, b = k and c = k

Let equal zeroes be α and α.

Then, sum of zeroes, α + α = \(\frac{-b}{a}\) = \(\frac{-k}{1}\)

[∵ sum of zeroes = \(-\left(\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\right)\)]

⇒ 2α = \(\frac{-k}{1}\) ⇒ α = \(\frac{-k}{2}\)

and product of zeroes, α ∙ α = \(\frac{c}{a}\) = \(\frac{k}{1}\) ⇒ α^{2} = k

[∵ product of zeroes = \(=\frac{\text { constant term }}{\text { coefficient of } x^2}\)]

∴ \(\left(\frac{-k}{2}\right)^2\) = k

⇒ k^{2} – 4k = 0

⇒ k(k – 4) = 0 ⇒ k = 0 or 4

When k = 0, then p(x) = x^{2}

When k = 4, then p(x) = x^{2} + 4x + 4

∴ p(x) = (x – 0)^{2} or (x + 2)^{2}

So, zeroes of p(x) are 0, 0 or -2, -2.

[∵ (x – 0)^{2} = 0 ⇒ x = 0, 0

and (x + 2)^{2} = 0 ⇒ x = -2, -2]

Thus, the equal roots are possible only, when k is even.

Hence, the given quadratic polynomial have no equal zeroes for some odd integer k > 1.

Question 30.

Solve the following system of linear equations graphically

3x + y – 11 = 0 and x – y – 1 = 0

Also, find the area of the region bounded by these lines and the Y-axis. [3]

Answer:

Given, system of linear equations is 3x + y – 11 = 0 and x – y – 1 = 0.

Now, table for equation 3x + y – 11 = 0 i.e. y = 11 – 3x is

x | 0 | 2 | 3 |

y | 11 | 5 | 2 |

Points | A(0, 11) | B(2, 5) | C(3, 2) |

Now, plot all the points on a graph paper and join them to get a line ABC.

Table for equation of line x – y – 1 = 0, i.e. y = x – 1 is

x |
-3 | 0 | 3 |

y | -4 | -1 |
2 |

Points | D(-3, -4) | E(0, -1) |
C(3, 2) |

Now, plot all the points on a graph paper and join them to get a line DEC.

It is clear from the above figure that the two lines intersect at point C(3, 2).

Hence, the solution of the given system of equations is x = 3, y = 2.

∴ Area of ∆ACE = \(\frac{1}{2}\) × Base × Altitude

= \(\frac{1}{2}\) × AE × CF

= \(\frac{1}{2}\) × (11 + 1) × 3

= \(\frac{1}{2}\) × 12 × 3

= 18 sq units

Question 31.

From a well-shuffled pack of 52 cards, few cards of same colour are missing. If P (drawing a red card) = 2/3 and P (drawing a black card) = 1/3, then which colour cards are missing and how many? [3]

Answer:

Since, the probability of drawing a black card is less than that of a red card, therefore black cards are missing.

Let x be the number of missing black cards.

Then, P (drawing a black card) = \(\frac{26-x}{52-x}\) = \(\frac{1}{3}\)

⇒ 78 – 3x = 52 – x

⇒ -3x + x = 25 – 78

⇒ -2x = -26

⇒ x = \(\frac{26}{2}\) = 13

Hence, 13 black cards are missing.

Section D

(Section D consists of 4 questions of 5 marks each.)

Question 32.

If a hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that \(\frac{1}{8}\) space of the cube remains unfilled. Then, find the 1 number of marbles that the cube can accommodate. [5]

Answer:

Given, edge of cube 22 cm

Volume of cube = (22)^{3} = 10648 cm^{3}

Also, given diameter of marble = 0.5 = \(\frac{1}{2}\) cm

∴ Radius of a marble, (r) = \(\frac{\text { radius }}{2}\) = \(\frac{1}{4}\)cm

Volume of one marble = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{1}{4}\) × \(\frac{1}{4}\) × \(\frac{1}{4}\)

= \(\frac{22}{336}\) cm^{3}

Since, \(\frac{1}{8}\)th of the cube is empty.

Let n be the number of marbles required to fill \(\frac{1}{8}\)th part of the cube.

According to the question,

\(\frac{7}{8}\) volume of cube = n × volume of marble

⇒ \(\frac{7}{8}\) × 10648 = n × \(\frac{22}{336}\)

⇒ n = \(\frac{484 \times 336 \times 7}{8}\)

⇒ n = 142296

Hence, the number of marbles that the cube can accommodate ¡s 142296.

Or

A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. Find the capacity of the capsule. [5]

Answer:

Given, diameter of cylinder = Diameter of hemisphere = 0.5 cm

[since, both hemispheres are attach with cylinder]

∴ Radius of cylinder (r) = radius of hemisphere

⇒ r = \(\frac{0.5}{2}\) = 0.25 cm

[∵ diameter = 2 × radius]

and total length of capsule = 2 cm

∴ Length of cylindrical part of capsule,

h = Length of capsule – Radius of both hemispheres

= 2 – (0.25 + 0.25) = 1.5 cm

Now, capacity of capsule = Volume of cylindrical part + 2 × Volume of hemisphere

= πr^{2}h + 2 × \(\frac{2}{3}\)πr^{3}

[∵ volume of cylinder = π × (radius)^{2} × height and volume of hemisphere = \(\frac{2}{3}\)π(radius)^{3}]

= \(\frac{22}{7}\)[(0.25)^{2} + 1.5 + \(\frac{4}{3}\) × (0.25)^{3}]

= \(\frac{22}{7}\)[0.09375 + 0.0208]

= \(\frac{22}{7}\) × 0.11455 = 0.36 cm^{3}

Hence, the capacity of capsule is 0.36 cm^{3}.

Question 33.

Compute the Arithmetic Mean for the following data [5]

Marks obtained | Number of students |

Less than 10 | 14 |

Less than 20 | 22 |

Less than 20 | 37 |

Less than 40 | 58 |

Less than 50 | 67 |

Less than 60 | 75 |

Answer:

The above data may be written as

Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

Frequency | 14 | 8 | 15 | 21 | 9 | 8 |

Here, h = 10

Let assumed mean = midpoint of (30 – 40) = 35

Now, we form the table as under

Question 34.

Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O’P are two perpendicular tangents to the two circles. Find the length of the common chord PQ. [5]

Answer:

Here, two circles of radii

OP = 3 cm and PO’ = 4 cm

These two circles intersect at P and Q.

Here, OP and PO’are two tangents drawn at point P

But these two tangents make an angle 90°.

Join OO’ and PN.

In right ∆OPO’,

OO’^{2} = OP^{2} + PO’^{2} [by Pythagoras theorem]

= 3^{2} + 4^{2} = 25

⇒ OO’ = 5 cm

Also, PN ⊥OO’

Let ON = x, NO’ = 5 – x

In right ∆OPN,

OP^{2} = ON^{2} + NP^{2}

⇒ NP^{2} = 3^{2} × x^{2} = 9 – x^{2} ……… (i)

and in right ∆PNO’,

PO’^{2} = PN^{2} + NO’^{2}

⇒ 4^{2} = PN^{2} + (5 – x)^{2}

⇒ PN^{2} = 16 – (5 – x)^{2} ……… (ii)

From Eqs. (i) and (ii), we get

9 – x^{2} = 16 – (5 – x^{2})

⇒ 7 + x^{2} – (25 + x^{2} – 10x) = 0

⇒ 10x = 18

⇒ x = 1.8

Again, in right ∆OPN,

OP^{2} = ON^{2} + NP^{2}

⇒ 3^{2} = (1.8)^{2} + NP^{2}

⇒ NP^{2} = 9 – 3.24 = 5.76

⇒ NP = 2.4 cm

∴ Length of common chord,

PQ = 2PN = 2 × 2.4 = 4.8 cm

Question 35.

Find the solution of the following equation by factorisation method.

\(\frac{x+1}{x-1}\) – \(\frac{x-1}{x+1}\) = \(\frac{5}{6}\), x ≠ 1, -1 [5]

Answer:

Given, equation is \(\frac{x+1}{x-1}\) – \(\frac{x-1}{x+1}\) = \(\frac{5}{6}\)

⇒ \(\frac{(x+1)^2-(x-1)^2}{(x+1)(x-1)}\) = \(\frac{5}{6}\)

⇒ \(\frac{x^2+1+2 x-\left(x^2+1-2 x\right)}{\left(x^2-1\right)}\) = \(\frac{5}{6}\)

⇒ 6[4x] = 5(x^{2} – 1)

⇒ 5x^{2} – 24x – 5 = 0

⇒ 5x^{2} – 25x + x – 5 = 0

[splitting middle term]

⇒ 5x(x – 5) + 1(x – 5) = 0

⇒ (x – 5)(5x + 1) = 0

⇒ x – 5 = 0 and 5x + 1 = 0

⇒ x = 5 and x = –\(\frac{1}{5}\) ….. (2)

Or

Solve the quadratic equation 9x^{2} – 9(a + b)x + (2a^{2} + 5ab + 2b^{2}) = 0 by factorisation method.

Answer:

Given, quadratic equation is 9x^{2} – 9(a + b)x + (2a^{2} + 5ab + 2b^{2}) = 0 … (i)

Now, 2a^{2} + 5ab + 2b^{2} = 2a^{2} + 4ab + ab + 2b^{2} = 2a(a + 2b) + b(a + 2b)

= (2a + b)(a + 2b)

∴ Eq. (i), can be rewritten as 9x^{2} – 9(a + b)x + (2a + b)(a + 2b) = 0

⇒ 9x^{2} – 3{(2a + b) + (a + 2b)}x + (2a + b)(a + 2b) = 0

⇒ 9x^{2} – 3(2a + b)x – 3(a + 2b)x + (2a + b)(a + 2b) = 0

⇒ 3x{3x – (2a + b)} – (a + 2b){3x – (2a + b)} = 0

⇒ {3x-(2a + b)}{3x-(a + 2b)} = 0

⇒ 3x – (2a + b) = 0 or 3x – (a + 2b) = 0

⇒ x = \(\frac{2 a+b}{3}\) or x = \(\frac{a+2 b}{3}\)

Section E

(Case Study Based Questions)

(Section E consists of 3 questions. All are compulsory.)

Question 36.

Balloon Elevation from Windows

Suppose, there are two windows in a house. A window of the house is at a height of 1.5 m above the ground and the other window is 3 m vertically above the lower window.

Amit and Manjeet are sitting in the two windows. At an instant, the angles of elevation of a balloon from these windows are observed as 45° and 30°, respectively.

On the basis of above information, answer the following questions.

(i) Find the height of the balloon from the ground. (2)

Answer:

(i) Let AG be the height of the balloon from the ground and C, D be the position of the windows, such that SC = 1.5 m and CD = 3 m

At points C and D, angles of elevation of balloon are 4S° and 30°.

Draw perpendicular lines EC and ED on AG. Then, ∠ECG = 45° and ∠FDG = 30°,

BC = AE = 1.5 m and CD = EE = 3 m

Let AB = CE = DF = x m and FG = h m

In right angled ∆ECG,

tan 45° = \(\frac{E G}{E C}\) = \(\frac{3+h}{x}\)

⇒ 1 = \(\frac{3+h}{x}\)

⇒ x = 3 + h [∵ tan45° = 1] ……(i)

In right angled ∆FDG,

tan 30° = \(\frac{G F}{D F}\) = \(\frac{h}{x}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x}\) = x ⇒ x = \(\sqrt{3}\)h

[∵ tan 30° = \(\frac{1}{\sqrt{3}}\)]

On putting x = \(\sqrt{3}\)h in Eq. (i), we get 3 + h = \(\sqrt{3}\)h ⇒ h(\(\sqrt{3}\) – 1) = 3

⇒ h = \(\frac{3}{(\sqrt{3}-1)}\)

⇒ h = \(\frac{3}{1.732-1}\)

= \(\frac{3}{0.732}\)

∴ h = 4.098 m

Hence, height of balloon from the ground

= 4.098 + 3 + 1.5 = 8.598 m ≈ 8.6 m

Or

Find the distance between Manjeet and balloon. (2)

(ii) If the height of any tower is double and the distance between the observer and foot of the tower is also doubled, then what is the angle of elevation. (1)

(iii) Suppose a tower and a pole is standing on the ground and the angle of elevation from bottom of pole is θ_{1} and elevation from top of pole to the top of tower is θ_{2}, then show that θ_{1} > θ_{2}. (1)

Answer:

Now, distance between Sanjeev and balloon is given by

GD = \(\sqrt{(G F)^2+(F D)^2}\)

= \(\sqrt{h^2+x^2}\) = \(\sqrt{h^2+3 h^2}\) [∵ x = \(\sqrt{3}\)h]

= \(\sqrt{4 h^2}\) = 2h = 2 × 8.6 = 17.2m

(ii) If the height of any tower is double and the distance between the observer and foot of the tower is also double, then the angle of elevation remain same. (1)

(iii) It is clear from the figure that θ_{1} > θ_{2}.

Question 37.

Sports Day

In a sports day celebration, the school decided to make activity between students in such a way that Vicky, Garvit and Shalini are standing at positions A, B and C whose coordinates are (3, 3), (5, 9) and (9, 3), respectively.

The teacher asked Deepanshi to fix the country flag at the mid-point of the line joining point A and B.

On the basis of above information, answer the following questions.

(i) Find the coordinates of the mid-point of line joining A and B. (1)

(ii) Find the distance between Garvit and Vicky. (1)

(iii) What is the space covered by these persons? (2)

Answer:

(i) Given points are A(3, 3) and B (5, 9).

Let Deepanshi fix the country flag at point D (x, y).

∴ x = \(\frac{3+5}{2}\) = \(\frac{8}{2}\) = 4 and y = \(\frac{3+9}{2}\) = \(\frac{12}{2}\) = 6

So, country flag is situated at point D(4, 6).

(ii) The distance between Gravit and Vicky

= \(\sqrt{(5-3)^2+(9-3)^2}\)

[∵ distance between two points = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)]

= \(\sqrt{(2)^2+(6)^2}\)

= \(\sqrt{4+36}\) = \(\sqrt{40}\)

= 2\(\sqrt{10}\) units

(iii) The space covered by three persons

= Area of ∆ABC

= \(\frac{1}{2}\)(base × height)

= \(\frac{1}{2}\) × (6 × 6) = \(\frac{36}{2}\)

= 18 sq units

Or

Find the point on X-axis, which is equidistant from points A and C. (2)

Answer:

Let point on X-axis be P(x, 0)

Then, AP^{2} = CP^{2}

∴ (x – 3)^{2} + (0 – 3)^{2} = (x – 9)^{2} + (0 – 3)^{2}

⇒ x^{2} – 6x + 9 + 9 = x^{2} + 81 – 18x + 9

⇒ 12x = 72 ⇒ x = 6

Hence, point on X-axis is (6, 0).

Question 38.

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

On the basis of above information, answer the following questions.

(i) Find the production during first year. (2)

Answer:

Let the production of TV sets in first year be a units. Then, production in the next consecutive years are a + d, a + 2d,………

Thus, we get the sequence, a ,a + d, a + 2d, …

This is an AP sequence, whose first term = a and common difference = d.

Given, T_{6} = 16000 and T_{9} = 22600

∴ a + (6 – 1)d = 16000

and a + (9 – 1)d = 22600 [∵ T_{n} = a + (n – 1)d]

⇒ a + 5d = 16000 ……(i)

and a + 8d = 22600 ……(ii)

On subtracting Eq. (i) from Eq. (ii), we get 3d = 22600 – 16000

⇒ 3d = 6600 ⇒ d = 2200

On putting d = 2200 in Eq. (i), we get

a + 5 × 2200 = 16000

⇒ a = 16000 – 11000

= 5000

Hence, the production during first year is 5000 sets.

Or

Find the production during 8th yr and first 3 yr. (2)

(ii) In which year, the production is ₹ 29200. (1)

(iii) Find the difference of the production during 7th yr and 4th yr. (1)

Answer:

The production during 8th yr is

T_{8} = a + (8 – 1) d

= 5000 + 7 × 2200

= 5000 + 15400

= 20400

Hence, production during 8th yr is 20400 sets.

The production during first 3 yr,

S_{3} = \(\frac{3}{2}\)[2a + (3-1)d]

= \(\frac{3}{2}\)[2 × 5000 + 2 × 2200]

= 3 [5000 + 2200]

= 3 × 7200

= 21600

(ii) Let in nth year, the production is 29200.

∵ T_{n} = a + (n – 1)d

∴ 29200 = 5000 + (n – 1)2200

⇒ (n – 1)2200 = 24200

⇒ (n – 1) = \(\frac{24200}{2200}\)

⇒ n – 1 = 11

⇒ n = 12

Hence, production is ₹ 29200 in 12th yr.

(iii) The difference of the production during 7th year and 4th yr = T_{7} – T_{4}

= a + (7 – 1)d – [a + (4 – 1)d]

= 6d – 3d = 3d

= 3 × 2200

= 6600