# CBSE Sample Papers for Class 10 Maths Standard Set 5 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 5 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions

Time: 3 hrs

Max. Marks: 80

Instructions:

1. This question paper has 5 Sections A-E.

2. Section A has 20 MCQs carrying 1 mark each.

3. Section B has 5 questions carrying 2 marks each.

4. Section C has 6 questions carrying 3 marks each.

5. Section D has 4 questions carrying 5 marks each.

6. Section E has 3 Case Based integrated units of assessment (4 marks each).

7. 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.

8. 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.

If (-1)^{n} +(-1)^{4n} = 0, then n is [1]

(a) any positive

(b) any negative integer

(c) any odd natural number

(d) any even natural number

Answer:

(c) any odd natural number

(-1)^{n} +(-1)^{4n} = 0 will be possible, when n is any odd natural number.

Question 2.

If tan²45° – cos²30° = x sin 45°cos 45°, then the value of x is [1]

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

(b) 1

(c) 2

(d) \(\frac{1}{3}\)

Answer:

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

We have, tan²45° – cos²30° = x sin 45° cos45°

⇒ (1)² – (\(\frac{\sqrt{3}}{2}\))^{2} = x × \(\frac{1}{\sqrt{2}}\) × \(\frac{1}{\sqrt{2}}\)

⇒ 1 – \(\frac{3}{4}\) = \(\frac{x}{2}\)

⇒ \(\frac{1}{4}\) = \(\frac{x}{2}\)

⇒ x = \(\frac{1}{2}\)

Question 3.

The ratio of the total surface area to the lateral surface area of a cylinder with base radius 80 cm and height 20 cm is [1]

(a) 1 : 2

(b) 2 : 1

(c) 3 : 1

(d) 5 : 1

Answer:

(d) 5 : 1

Given, radius of the base, r = 80 cm

and height of the cylinder h = 20 cm

∴ \(\frac{\text { Total surface area }}{\text { Lateral surface area }}\) = \(\frac{2 \pi r(h+r)}{2 \pi r h}\)

= \(\frac{h+r}{h}\) = \(\frac{20+80}{20}\)

= \(\frac{100}{20}\) = \(\frac{5}{1}\) = 5 : 1

Question 4.

The probability of getting 101 marks in out of 100 marks is

(a) 1

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

(c) 0

(d) 2

Answer:

(c) 0

In out of 100 marks, we do not get 101 marks, which is an impossible event. Hence, the probability of impossible event is 0.

Question 5.

The ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7) is [1]

(a) 3 : 4

(b) 4 : 3

(c) 2 : 3

(d) 3 : 2

Answer:

(a) 3 : 4

Suppose the line 3x + y – 9 = 0 divides the line segment joining A(1, 3) and 6(2,7) in the ratio k : 1 at point C.

Then, coordinates of C are (\(\frac{2k+1}{k+1}\), \(\frac{7k+3}{k+1}\))

But C, lies on 3x + y – 9 = 0.

Therefore, 3(\(\frac{2k+1}{k+1}\)) + \(\frac{7k+3}{k+1}\) – 9 = 0

⇒ (6k + 3) + (7k + 3) – 9k – 9 = 0

⇒ 4k – 3 = 0

∴ k = \(\frac{3}{4}\)

So, the required ratio is 3 : 4 internally.

Question 6.

If median = 137 and mean = 137.05, then the value of mode is [1]

(a) 156.90

(b) 136.90

(c) 186.90

(d) 206.90

Answer:

(b) 136.90

Given, median = 137 and mean = 137.05 We know that

Mode = 3 Median – 2 Mean = 3(137) – 2 (137.05)

= 411 – 274.10 = 136.90

Question 7.

How many terms are there in the arithmetic series 1 + 3 + 5 + … + 73 + 75? [1]

(a) 28

(b) 30

(c) 36

(d) 38

Answer:

(d) 38

Given, arithmetic series is

1 + 3 + 5 + …. + 73 + 75

Here, first term a = 1, common differenced = 2,

Last term = t_{n} = 75

Let number of terms in given series = n

Then, t_{n} = a + (n – 1)d

⇒ 75 = 1 + (n – 1) × 2

⇒ 75 = 1 + 2n – 2

⇒ 2n = 76

⇒ n = 38

Hence, there are 38 times in given series.

Question 8.

If 8 tan θ = 15, then the value of sin θ – cos θ is [1]

(a) \(\frac{7}{17}\)

(b) \(\frac{17}{7}\)

(c) \(\frac{15}{17}\)

(d) \(\frac{8}{17}\)

Answer:

(a) \(\frac{7}{17}\)

We have, tan θ = \(\frac{15}{8}\) = \(\frac{B C}{A B}\)

Question 9.

The pair of linear equations x + 2 y = 5 and 3x + 12y = 10 has [1]

(a) unique solution

(b) no solution

(c) more than two solutions

(d) infinitely many solutions

Answer:

(a) unique solution

The pair of linear equations are

x + 2y = 5 ⇒ x + 2y – 5 = 0

and 3x + 12y = 10 ⇒ 3x + 12y – 10 = 0

Here, a_{1} = 1, b_{1} = 2, c_{1} = -5

and a_{2} = 2, b_{2}= 12, c_{2} = -10

Now,

\(\frac{a_1}{a_2}=\frac{1}{3}\), \(\frac{b_1}{b_2}=\frac{2}{12}=\frac{1}{6}\)

As, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

So, pair of equations has unique solution.

Question 10.

In the given figure, if ∆ACB ~ ∆APQ, BA = 6 cm, BC = 8 cm and PQ = 4 cm, then the length of AQ is [1]

(a) 3 cm

(b) 5 cm

(c) 4 cm

(d) 6 cm

Sol.

(a) 3 cm

As, ∆ACB ~ ∆APQ

So, \(\frac{A B}{B C}=\frac{A Q}{Q P}\)

⇒ \(\frac{6}{8}=\frac{A Q}{4}\)

[∵ AB = 6cm, SC = 8cm and PQ = 4 cm, (given)]

⇒ \(\frac{6×4}{8}\) = AQ

∴ AQ = 3cm

Question 11.

In the given figure, DE || BC and \(\frac{A D}{D B}=\frac{3}{5}\). If AC = 4.8 cm, then the length of AE is [1]

(a) 1.5 cm

(b) 1.8 cm

(c) 2 cm

(d) 4.2 cm

Answer:

(b) 1.8 cm

Let AE = x cm

Then, EC = (AC-AE)

= (4.8 – x)

Now, in ∆ABC, DE || BC

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

\(\frac{3}{5}=\frac{x}{4.8-x}\)

3(4.8 – x) = 5x

8x = 14.4

x = 1.8

Hence, AE = 1.8 cm

Question 12.

If radius of circle is 3 cm and tangent drawn from an external point to the circle is 4 cm, then the distance from centre of circle to the external point is [1]

(a) 3 cm

(b) 2 cm

(c) 5 cm

(d) 4 cm

Answer:

Sol. (a)

Given, OQ = 3 cm and PQ = 4 cm.

In right angled ∆OPQ, using Pythagoras theorem

OP = \(\sqrt{OQ^2 + QP^2}\)

= \(\sqrt{(3)^2 + (4)^2}\) = \(\sqrt{ 9+16}\) = \(\sqrt{25}\) = 5 cm

Question 13.

The difference between the circumference and radius of a circle is 37 cm. The area of the circle is [1]

(a) 111 cm²

(a) 111 cm²

(c) 154 cm²

(d) 259 cm²

Answer:

(c) 154 cm²

Let R be the radius of circle.

Given, 2πR – R = 37

R(2 × \(\frac{22}{7}\) – 1) = 37

R = 37 × \(\frac{7}{37}\) = 7 cm

Area of circle = πR² = (\(\frac{22}{7}\) × 7 × 7) = 154 cm²

Question 14.

If x_{i}‘ s are the mid-points of the class intervals of grouped data, f_{i}‘s are the corresponding frequencies and \(\bar{x}\) is the mean, then Σ(f_{i}x_{i} – \(\bar{x}\)) is equal to [1]

(a) 0

(b) -1

(c) 1

(d) 2

Answer:

(a) 0

We know that

Question 15.

The sum of two numbers is 137 and their difference is 43. This situation can be algebraically represented as [1]

(a) x – y = 137, x + y = 180

(b) 2(x + y) = 137, 2(x – y) = 43

(c) x + y = 137, x – y = 43

(d) x + y = 43, x – y = 137

Answer:

(c) x + y = 137, x – y = 43

Let the two numbers be x and y, where x > y

Given, sum of numbers = 137

∴ x + y = 137

and difference of numbers = 43

∴ x – y = 43

Question 16.

The smallest odd composite number is

(a) 3

(b) 9

(c) 5

(d) 7

Answer:

(b) 9

We know that composite numbers are those numbers which have atleast one factor other than 1 and the number itself. Number 3, 5 and 7 has no other factor than 1 and themselves, so they are not composite numbers. Number 9 is a composite number, because it has 3 factors other than 1, 3 and 9.

Question 17.

If a and B are the zeroes of the polynomial f(x) = x² – p(x + 1) – c such that (α + 1)(β + 1) = 0, then c is equal to [1]

(a) 1

(b) 0

(c) -1

(d) 2

Answer:

(a) 1

Given, a and p are the zeroes of the polynomial

f(x) = x² – p(x + 1) – c

= x² – px – p – c = x – px – (p + c)

= -(P + c)

Now, it is given that (α + 1)(β + 1) = 0

⇒ αβ + α + β + 1 = 0

⇒ -(p + c) + p + 1 = 0

⇒ -c +1 = 0

∴ c = 1

Question 18.

Two cones have their heights in the ratio 1 : 4 and radii in the ratio 4:1. The ratio of their volumes is [1]

(a) 1 : 4

(b) 4 : 1

(c) 2 : 1

(d) 1 : 2

Answer:

(b) 4 : 1

Let the radii of two cones are r_{1}, and r_{2} and their heights are h_{1} and h_{2}.

Then, \(\frac{r_1}{r_2}=\frac{4}{1}\) and \(\frac{h_1}{h_2}=\frac{4}{1}\)

Direction 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 the circumference of a circle is 176 cm, then its radius is 28 cm.

Reason (R) Circumference =2π × radius [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)

C = 2 × \(\frac{22}{7}\) × r = 176

⇒ r = \(\frac{176 \times 7}{2 \times 22}\) = 28 cm

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

Question 20.

Assertion (A) √2 is an irrational number.

Reason (R) If p be a prime, then √p is an irrational number. [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)

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

Section B

(Section B consists of 5 questions of 2 marks)

Question 21.

In the following figures, CM and RN are respectively the medians of ∆ABC and ∆PQR.

If ∆ABC ~ ∆PQR, then prove that ∆AMC ~ ∆PNR. [2]

Solution:

Given, ∆ABC ~ ∆PQR

\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{P R}\)

[Y in similar triangles, corresponding sides are proportional]

and ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R …(ii)

[v in similar triangles, corresponding angles are equal]

We know that the median bisects the opposite side.

Question 22.

For what value of p, will the following system of linear equations represent parallel lines?

-x + py = 1 and px – y = 1 [2]

Solution:

Given, pair of equations is

-x + py -1 = 0 … (i)

and px – y -1 = 0 … (ii)

On comparing the given equations with standard form i.e.

a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0, we get

a_{1} = -1, b_{1} = p, c_{1} = -1

and a_{2} = p, b_{2}= -1, c_{2} = -1

For parallel lines,

\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

⇒ \(\frac{-1}{p}=\frac{p}{-1} \neq \frac{-1}{-1}\) ………(iii)

On taking I and II terms, we get

\(\frac{-1}{p}=\frac{p}{-1}\) ⇒ p² = 1 ⇒ p = ±1

Since, p = -1 does not satisfy the last two terms of Eq. (iii).

∴ p = 1 is the required value.

Hence, for p = 1, the given system of equations will represent parallel lines.

Question 23.

The length of minute hand of a clock is 14 cm. Then, find the area swept by the minute hand in one minute. [2]

Solution:

The minute hand of a clock describes a circle of radius equal to its length, i.e. 14 cm in 1 h. So, the angle described by minute hand in 60 min = 360°.

∴ Angle described by minute hand in

1 min = \(\frac{360°}{60°}\)= 6°

So, the area swept by the minute hand in

1 min is the area of a sector of angle 6? in a circle of radius 14 cm.

∴ Required area = \(\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{6^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(14)^2\)

= \(\frac{1}{60}\) × \(\frac{22}{7}\) × 14 × 14 = 10.27 cm²

Question 24.

Prove that

(tan^{2}A – tan^{2}B) = \(\frac{\left(\sin ^2 A-\sin ^2 B\right)}{\cos ^2 A \cos ^2 B}\) = \(\frac{\cos ^2 B-\cos ^2 A}{\cos ^2 B \cos ^2 A}\). [2]

Or

If √3 tan 2θ – 3 = 0, then find the value of cos θ.

Solution:

Or

Question 25.

A line through the centre O of a circle of radius 5 cm cuts the tangent at a point P on the circle at Q such that OQ =13 cm. Find the length of PQ.

Or

In the given figure, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O. If ∠QPR = 90°, then find the length of PQ.

Solution:

We know that tangent at a point on a circle is perpendicular to the radius through = 144

Therefore, OP is perpendicular to PQ.

In right angled ∆OPQ, we have

OQ^{2} = OP^{2} + PQ^{2} [by Pythagoras theorem]

⇒ PQ^{2} = OQ^{2} – OP^{2}

⇒ PQ^{2} = 13^{2} – 5^{2}

⇒ PQ^{2} = 169 – 25

⇒ PQ = 12 cm

Or

We know that if pair of tangents are drawn from an external point P, then line joining from centre O to the point P, bisects the angle P.

Also, radius of circle OQ is perpendicular to the tangent line QP.

Now, in right angled ∆OQP,

tan 45° = \(\frac{OQ}{QP}\)

⇒ 1 = \(\frac{4}{QP}\) [v OQ = 4 cm]

⇒ QP = 4cm

Hence, length of PQ is 4 cm.

Section C

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

Question 26.

Prove that if a, b,c and d are positive rationals such that, a + √b = c + √d, then either a-c and b – dor b and d are squares of rationals. [3]

Solution:

Given, a + √b = c + √d

If a = c, then a + 4b = a + 4d

⇒ √b = √d

⇒ b = d [squaring both sides]

If a *c, then there exists a positive rational number x such that a = c + x

Now, a + √b = c + √d

⇒ c + x + √b = c + √d [∵ a = c + x]

⇒ x + √b = √d …(i)

⇒ (x + √b)² = (√d)² [squaring both sides]

⇒ x² + b + 2x√b = d

⇒ x² + 2x√b + b – d = 0

[∵ (A + B)^{2} = A^{2} + B^{2} + 2AB]

⇒ 2x√b = d – b – x²

∴ √b = \(\frac{d – b – x²}{2x}\)

Since, d, x and b are rational numbers and x > 0.

So, \(\frac{d – b – x²}{2x}\) is a rational.

Then, √b is a rational number.

Hence, b is the square of a rational number.

From Eq. (i), we get

√d = x +√b

Also, √d is a rational.

So, d is the square of a rational number.

Hence, either a = c and b = d or b and d are the squares of rationals.

Hence proved.

Question 27.

Following table gives marks scored by students in an examination

Marks | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |

Number of Student | 3 | 7 | 15 | 24 | 16 | 8 | 5 | 2 |

Calculate the mean mark correct to 2 decimal places.

Solution:

We shall use step-deviation method. Construct the table as under, taking assumed mean a = 17.5.

Here, c (width of each class) = 5

Question 28.

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Or

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD. [3]

Solution:

Let PQ and PR be two tangents drawn from an external point P to a circle with centre O.

To prove ∠QOR = 180°- ∠QPR or ∠QOR + ∠QPR = 180°

Proof In AOQP and AORP,

PQ = PR [Y tangents drawn from an external point are equal in length]

OQ = OR [radii of circle]

OP = OP [common sides]

/. ∆OQP = ∆ORP [by SSS congruence rule]

Then, ∠QPO = ∠RPO [by CPCT]

and ∠POQ = ∠POR [by CPCT]

⇒ ∠QPR = 2 ∠OPQ and ∠QOR = 2∠POQ

Now, in right angled ∆OQP,

∠QPO + ∠QOP = 90°

⇒ ∠QOP = 90° – ∠QPO

⇒ 2 ∠QOP = 180° – 2 ∠QPO

[multiplying both sides by 2]

⇒ ∠QOR = 180°- ∠QPR

[from Eq. (i)]

⇒ ∠QOR + ∠QPR = 180°

Hence proved.

Or

Let the line BD intersects the bigger circle at E.

Now, Join AE.

, Let O be the centre of the bigger circle, then O is ‘ the mid-point of AB.

[∵ AB is a diameter of the bigger circle]

BD is a tangent to the smaller circle and OD is a radius through the point of contact D. Then,

OD ⊥ BD ⇒ OD ⊥ BE

Since, OD is perpendicular to the chord BE of bigger circle.

∴ BD = DE

[∵ perpendicular drawn from the centre to a chord bisects the chord]

D is the mid-point of BE.

∴ In ∆BAE, O is the mid-point of AB and D is the mid-point of BE.

OD = \(\frac{1}{2}\) AE

[∵ line segment joining the mid-points of any two sides of a triangle is half of the third side]

⇒ AE = 2(OD) = 2 × 8 = 16 cm

In right angled AOBD, using Pythagoras theorem,

OD^{2} + BD^{2} = OB^{2}

⇒ BD = \(\sqrt{OB^2 – OD^2}\) = \(\sqrt{13^2 -8^2}\) [∵ OB = 13cm]

= \(\sqrt{169-64}\) = \(\sqrt{105}\)

∴ DE = BD = \(\sqrt{105}\)

In right angled AAED, use Pythagoras theorem, we have

AD = \(\sqrt{(AE)^2 + (DE)^2}\)

= \(\sqrt{16^2 + \sqrt{105}^2}\) = \(\sqrt{256+105}\)

= \(\sqrt{361}\) = 19 cm

Question 29.

Given that the sum of the zeroes of the polynomial (a + 1)x^{2} + (2a + 3)x + (3a + 4) is -1 Find the product of its zeroes.

Or

If α, β, γ are the zeroes of the polynomial f(x) = ax^{3} + bx^{2} + cx + d, then find the value of \(\frac{1}{α}\) + \(\frac{1}{β}\) + \(\frac{1}{γ}\). [3]

Solution:

Since, the sum of the zeroes of the quadratic polynomial (a + 1)x^{2} + (2a + 3)x + (3a + 4) is -1.

∴ \(\frac{-(2a+3)}{a+1}\) = -1

Or

Given, polynomial is ax^{3} + bx^{2} + cx + d are α, β and γ.

Let P = \(\frac{1}{α}\) + \(\frac{1}{β}\) + \(\frac{1}{γ}\)

P = \(\frac{\beta \gamma+\gamma \alpha+\alpha \beta}{\alpha \beta \gamma}\)

∴ α β γ = \(\frac{-d}{a}\)

and α β + β γ + γ α = \(\frac{c}{a}\)

Now from Eq. (i) P = \(\frac{c / a}{-\frac{d}{a}}\) = \(-\frac{c}{a}\)

Question 30.

If the zeroes of the quadratic polynomial x^{2} + (a + l)x + b are 2 and – 3, then find the value of a and b. [3]

Solution:

Let p(x) = x^{2} + (a + 1)x + b

Given that 2 and -3 are the zeroes of the quadratic polynomial p(x).

∴ p(2) = 0 and p(-3) = 0

⇒ 2^{2} + (a + 1) (2) + b = 0

⇒ 4 + 2a + 2 + b = 0

⇒ 2a + b = – 6 …(i)

and (-3)^{2} + (a + 1)(-3) + b = 0

⇒ 9 – 3a – 3 + b = 0

⇒ 3a- b = 6 …(ii)

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

5a = 0

⇒ a = 0

On putting the value of a in Eq. (i), we get

2 × 0 + b = -6 ⇒ b = -6

So, the required values are a = 0 and b = – 6

Question 31.

Prove that

Solution:

Section D

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

Question 32.

From a solid cylinder whose height is 12 cm and diameter is 10 cm, a conical cavity of same height and same diameter is hollowed out. Find the volume and total surface area of the remaining solid.

Or

A right angled triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone, so formed, [choose the value of n as found appropriate] [5]

Solution:

Given, diameter of the cylinder, d = 10 cm ⇒ r = 5cm

and height of the cylinder, h = 12 cm

∴ Volume of the cylinder = πr²h

= \(\frac{22}{7}\) × 5 × 5 × 12 = \(\frac{6600}{7}\) cm³

and volume of the cone = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5 × 5 × 12

= \(\frac{2200}{7}\) cm³

Now, volume of remaining solid

= Volume of the cylinder – Volume of the cone

= \(\frac{6600}{7}\) – \(\frac{2200}{7}\)

= \(\frac{4400}{7}\)

= 628.57 cm³

Since, slant height of the cone,

l = \(\sqrt{r^2 + h^2}\) = \(\sqrt{(5)^2 + (12)^2}\) = \(\sqrt{169}\) = 13cm

∴ Cun/ed surface area of the cone = πrl

= \(\frac{22}{7}\) × 5 × 13 = \(\frac{1430}{7}\) cm²

Clearly, curved surface area of the cylinder = 2πrh

= 2 × \(\frac{22}{7}\) × 5 × 12 = \(\frac{2640}{7}\) cm²

and area of upper base of the cylinder = πr²

= \(\frac{22}{7}\) × 5 × 5 = \(\frac{550}{7}\) cm²

Now, total surface area of the remaining solid

= Curved surface area of the cylinder

+ Curved surface area of the cone + Area of upper base of the cylinder

= \(\frac{2640}{7}\) + \(\frac{1430}{7}\) + \(\frac{550}{7}\)

= \(\frac{4620}{7}\) = 660 cm²

Or

Let ABC be a right angled triangle, right angled at A and BC is the hypotenuse.

Also, let AB = 3 cm

and AC = 4 cm.

Then, BC = \(\sqrt{3^2 + 4^2}\)

[by Pythagoras theorem]

= \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 cm

As, ∆ABC revolves about the hypotenuse BC. It forms two cones ABD and ACD.

In ∆AEB and ∆CAB,

∠AEB = ∠CAB [each 90°]

∠ABE = ∠ABC [common]

∴ ∆AES ~ ∆CAS [by AA similarity criterion]

∴ \(\frac{A E}{C A}=\frac{A B}{B C}\)

[∵ in similar triangles, corresponding sides are proportional]

\(\frac{A E}{4}\) = \(\frac{3}{5}\)

⇒ AE = \(\frac{12}{5}\) = 2.4

So, radius of the base of each cone, AE = 2.4cm

Now, in right angled ∆AEB,

BE = \(\sqrt{AB^2 + AE^2}\) [by Pythagoras theorem]

= \(\sqrt{(3)^2 + (2.4)^2}\)

= \(\sqrt{9 + 5.76}\) = \(\sqrt{3.24}\) = 1.8

So, height of the cone ABD = BE = 1.8 cm

∴ Height of the cone, ACD =CE = BC – BE

= 5 – 1.8 = 3.2 cm

Now, volume of the cone ABD

Question 33.

Ram and Shyam together have 55 marbles. Both of them lost 5 marbles each and the product of the number of marbles they have is 164. Find out how many marbles they had to start with? [5]

Solution:

Given, Ram and Shyam together have 55 marbles. Let Ram has x marbles.

Then, number of marbles Shyam had = 55 – x

∵ Both of them lost 5 marbles each.

∴ The number of marbles Ram had = x – 5

and the number of marbles Shyam had = 55 – x – 5

= 50 – x

Now, product of the number of marbles = 164

⇒ (x – 5)(50 – x) = 164

⇒ 50x – x² – 250 + 5x = 164

⇒ -x² + 55x – 250 – 164 = 0

⇒ -x² + 55x – 414 = 0

⇒ x² – 55x + 414 = 0 [multiplying by (-1)]

which is the required quadratic equation.

Now, by factorisation method, we get

x² – 46x – 9x + 414 = 0

⇒ x(x – 46) – 9(x – 46) = 0

⇒ (x – 46)(x – 9) = 0

⇒ x – 46 = 0 or x -9 = 0

⇒ x = 46 or x = 9

When Ram has 46 marbles, then Shyam has = 55 – 46 = 9 marbles

When Ram has 9 marbles, then Shyam has = 55 – 9 = 46 marbles

Question 34.

The mean of the following frequency table is 50 but the frequencies and f2 in class interval 20-40 and 60-80 are missing. Find the missing frequencies.

Or

Compute the median from the following data.

Mid-value | 115 | 125 | 135 | 145 | 155 | 165 | 175 | 185 | 195 |

Frequency | 6 | 25 | 48 | 72 | 116 | 60 | 38 | 22 | 3 |

Solution:

Let assumed mean be A = 50.

Table for the given data is

Here N = Σf_{i} = 120 [given]

⇒ 68 + f_{1} + f_{2} = 120

⇒ f_{1} + f_{2} = 52 …….(i)

But mean = 50

⇒ A + \(\frac{\sum f_i d_i}{N}\) = 50

⇒ 50 + \(\frac{80-20 f_1+20 f_2}{120}\) = 50

⇒ 80 – 20f_{1} + 20f_{2} = 0

⇒ 4 – f_{1} + f_{2} = 0

f_{1} – f_{2} = 4 …….(ii)

On solving Eqs. (i) and (ii), we get

f_{1} = 28 and f_{2} = 24

Or

We have the mid-values, then firstly we should find the upper and lower limits of the various classes. The difference between two consecutive values is h = 125 – 115 = 10.

Lower limit of a class = Mid-value – h/2

Upper limit = Mid-value + h/2

Table for cumulative frequency is given below

Here, N = 390

Now, \(\frac{N}{2}=\frac{390}{2}\) = 195

The cumulative frequency just greater than N/2 i.e. 195 is 267 and the corresponding class is 150-160.

So, 150-160 is the median class.

Here, l = 150,f = 116, h = 10 and cf = 151

∴ Mean = l + \(\left(\frac{\frac{N}{2}-c f}{f}\right) \times h\)

= 150 + \(\left(\frac{195-151}{116}\right) \times 10\)

= 150 + \(\frac{44×10}{116}\)

= 150 + \(\frac{440}{116}\)

= 150 + 3.79

= 153.79

Question 35.

BL and CM are medians of ∆ABC right angled at A. Prove that

4(BL^{2}+ CM^{2}) = 5BC^{2}.

Solution:

In figure, ∠A = 90°. Since, BL and CM are medians, so the lines BL and CM divide AC and AB respectively, into two equal parts.

Section E

(Case study based questions are compulsory)

Question 36.

Your friend Veer wants to participate in a 200 m race. He can currently run that ‘ distance in 51 sec and with each day of practice it takes him 2 sec less. He wants to do in 31 sec.

(i) If nth term of an AP is given by a_{n} = 2n + 3, then find the common difference of an AP. [1]

(ii) Find the terms of AP for the given situation and determine the 10th term from the end. [2]

Or

What is the minimum number of days he needs to practice till his goal is achieved? [2]

(iii) Find the value of x, for which 2x, x + 10,3x +2 are three consecutive terms of an AP. [1]

Solution:

(i) Given, a_{n} = 2n + 3

Common difference = a_{n+1} – a_{n}

= 2(n + 1) + 3 – (2 n + 3)

= 2n + 2 + 3 – 2n – 3

= 2

(ii) In first day, Veer takes 51 sec to complete the 200 m race. But in every next day he takes 2 sec iesser than the previous day.

Thus, AP series will formed

51, 49, 47, …. 31

Here, l = 31 and d = 49 – 51 = -2

∴ 10th term fromthe end = l – (10 – 1)d

= l – 9d

= 31 – 9(-2)

= 31 + 18 = 49

Or

Since, Veer wants to achieve the goal in 31 sec.

Let Veer takes n days to achieve the target.

∴ T_{n} = a + (n – 1)d

Here, a = 51, d = 49 – 51 = -2

∴ 31 = 51 + (n – 1)(-2)

⇒ (n – 1)2 = 20

⇒ (n – 1) = 10 ⇒ n = 11

Hence, he needs minimum 11 days to achieve the goal.

(iii) Given, terms 2x, x + 10, 3x +2 are in AP.

∴ x + 10 = \(\frac{2x+(3x+2)}{2}\)

⇒ 2x + 20 = 5x + 2

⇒ 3x = 18 ⇒ x = 6

Question 37.

A cyclist is climbing through a 20 m long rope which is highly stretched and tied from the top of a vertical pole to the ground as shown below

Based on the above information, answer the following questions

(i) Find the height of the pole, if angle made by rope with the ground level is 600. [1]

(ii) If the angle made by the rope with the ground level is 45°, then find the height of the pole. [2]

Or

If the angle made by the rope with the ground level is 450 and 3 m rope is broken, then what will be the height of the pole. [2]

(iii) If the angle made by the rope with the ground level is 600, then calculate the distance between artist and pole at ground level. [1]

Solution:

Or

Question 38.

Tree Platation to Control Pollution

The class X students of a secondary school in Krishnagar have been alloted a rectangular plot of land for this gardening activity

Sapling of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a rectangular gracy lawn in , the plot as shown in above figure.

The students sowing seeds of flowering plants on the remaining area of the plot.

(i) Find the coordinates of point Q and S. [1]

(ii) If the point m divides the line QS in the ratio 3:2, then find the coordinates of m. [2]

Or

If the point G divides the line QR in the ratio 1:2, then find the coordinates of G. [2]

(iii) Find the distance between the vertices of diagonal Q and S. [1]

Solution:

(i) The coordinates of points 0 and S are (2, 3) and (6,6).

(ii) By using internal division formula,

Or

By using internal division formula,

(iii) Distance between the vertices of diagonal Q and S = \(\sqrt{(6-2)^2 + (6-3)^2}\)

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

= \(\sqrt{16 + 9}\)

= \(\sqrt{25}\) = 5 units