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

Time: 3 Hours

Maximum Marks: 80

General Instructions:

- This Question Paper has 5 Sections A, B, C, D and E.
- Section A has 20 MCQs carrying 1 mark each
- Section B has 5 questions carrying 02 marks each.
- Section C has 6 questions carrying 03 marks each.
- Section D has 4 questions carrying 05 marks each.
- Section E has 3 case based integrated units of assessment (04 marks each) with sub parts of the values of 1, 1 and 2 marks each respectively.
- All Questions are compulsory. However, an internal choice in 2. Qs of 5 marks, 2.Qs of 3 marks and 2. Questions of 2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.
- Draw neat figures wherever required. Take n = 22/7 wherever required if not stated.

Section – A ( 20 Marks)

Section A Consists of Multiple Choice Type questions of 1 mark each

Question 1.

Let p and q be two natural numbers such that p > q. When p is divided by q, then remainder is r.

(i) r CANNOT be (p -q).

(ii) r CAN either be q or (p -q).

(iii) r is DEFINITELY less than q.

Which of the above statements is/are true?

(A) only (ii)

(B) only (iii)

(C) only (i) and (iii)

(D) cannot be determined without knowing the values of p, q and r

Solution:

(B) only (iii)

Explanation: q is divisor and r is remainder

∴ r is less than q.

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.

If the discriminant of the equation 6x^{2} – bx + 2 = 0 is 1, then value of b is:

(A) 7

(B) -7

(C) ±7

(D) None of these

Solution:

(C) ±7

Explanation: Given equation, 6x^{2} – bx + 2 = 0

On comparing with Ax^{2} + Bx + C = 0, we get A = 6, B= -b and C = 2

Now, Discriminant = B^{2} – 4AC

⇒ 1 = (0) – 4 × 6 × 2

⇒ 1 = b^{2} – 48

⇒ b^{2} = 49

⇒ b = ±7

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

Out of 100 marks, we did not get 101 marks, which is an impossible event. Hence, the probability of an impossible event is 0.

Question 5.

If 7 times the 7^{th} term of an A.P. is equal to 11 times its 11^{th} term, then its 18^{th} term will be:

(A) 7

(B) 11

(C) 18

(D) 0

Solution:

(D) 0

Explanation: According to the question,

7t_{7} = 11t_{11}

= 7(a + 6d) = 11(a + 10d)

⇒ 4a + 68d = 0

⇒ 4(a + 17d) = 0

⇒ (a + 17d) = 0

⇒ t_{18} = 0

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.

In what ratio, does the point (-4, 6) divide the line segment joining the points A(-6, 10) and B(3, -8)?

(A) 2:7

(B) 7:2

(C) 1:2

(D) 2:1

Solution:

(A) 2:7

Explanation:

∴ The required ratio is 2:7.

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.

In the given figure, if TP and TQ are the two tangents to a circle with centre O such that ∠POQ = 110°, then ∠PTQ is equal to:

(A) 60°

(B) 70°

(C) 80°

(D) 90°

Solution:

(B) 70°

Explanation: Given that, TP and TQ are tangents.

Since, the radius drawn to the tangents will be perpendicular on it.

∴ OP ⊥ TP and OQ ⊥ TQ

∠OPT = 90° and ∠OQT = 90°

In quadrilateral POQT,

Sum of all interior angles = 360°

∠OPT + ∠POQ + ∠OQT + ∠PTQ =360°

90° + 110° + 90° + ∠PTQ = 360°

∠PTQ = 70°

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 = 3 cm

Question 11.

sec θ when expressed in terms of cot θ, is equal to:

Solution:

(C) \(\frac{\sqrt{1+\cot ^2 \theta}}{\cot \theta}\)

Explanation:

[∵ cosec^{2} θ ∵ = 1 + cot^{2} θ]

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.

If sin A = \(\frac{12}{13}\), the value of cos A is :

(A) \(\frac{13}{4}\)

(B) \(\frac{13}{5}\)

(C) \(\frac{5}{13}\)

(D) \(\frac{4}{13}\)

Solution:

(C) \(\frac{5}{13}\)

Explanation:

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.

If the sum of the first n terms of an A.P. be 3n^{2} + n and its common difference is 6, then its first term is:

(A) 2

(B) 3

(C) 1

(D) 4

Solution:

(D) 4

Explanation: S_{n} = 3n^{2} + n

According to the formula,

S_{n} = \(\frac{n}{2}\)[2a + (n – 1)d]

3n^{2} + n = \(\frac{n}{2}\)[2a + 6n – 6]

6n^{2} + 2n = 2an + 6n^{2} – 6n

\(\frac{8n}{2n}\) = a

∴ a = 4

Thus, first term = 4

Question 16.

For an event E, P(E) + P(Ē) = x, then the value of x^{3} – 3 is:

(A) -2

(B) 2

(C) 1

(D) -1

Solution:

(A) -2

Explanation: Given

P(E) + P(Ē) = x …(i)

Also, according to the law of probability,

P(E) + P(Ē) = 1 … (ii)

From (i) and (ii), we get x = 1

Put value of x in x^{3} – 3, we get

x^{3} – 3 = (1)^{3} – 3 = 1 – 3 = -2

Question 17.

Look at the numbers shown below:

(i) -0.5

(ii) 0.00001

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

(iv) 1

(v) 1.00001

(vi) 99%

Which of the above numbers represent probabilities of events?

(A) only (i) and (iii)

(B) only (i), (ii), (iii) and (iv)

(C) only (ii), (ii), (iv) and (v)

(D) only (ii), (ii), (iv) and (vi)

Solution:

(D) only (ii), (ii), (iv) and (vi)

Explanation: Since, we know that the probability can neither be negative nor be greater than one.

So, -0.5 arid 1.00001 can not represent probabilities of events.

Hence, only (ii), (iii), (iv) and (vi) represent probabilities of events.

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 \(\frac{h_1}{h_2}=\frac{4}{1}\)

DIRECTIONS: Two statements are given below – one labelled Assertion (A) and the other labelled Reason (R). Read the statements carefully and choose the option that correctly describes statements (A) and (R).

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

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

(C) (A) is true but (R) is false.

(D) (A) is false but (R) is true.

Question 19.

Assertion (A): The area of a quadrant of a circle whose circumference is 22 cm is \(\frac{77}{8}\) cm^{2}

Reason (R): Area of a quadrant of a circle with radius, r is = \(\frac{\pi r^2}{4}\).

Solution:

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

Explanation: Given circumference of a circle = 22 cm

2πr = 22

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 (10 Marks)

Section B consists of 5 questions of 2 marks each

Question 21.

Given that √2 is irrational, prove that (5+3√2) is an irrational number.

Solution:

Given that √2 is irrational number.

Let, 5 + 3√2 be rational.

So 5 + 3√2 = \(\frac{p}{q}\),

where′a′and′b′are integers and q ≠ 0

3√2 = \(\frac{p}{q}\) – 5

√2 = \(\frac{p-5q}{3q}\)

Since ‘a’ and ‘b’ are integers p – 5q is also an integer \(\frac{p-5q}{3q}\) is rational. So RHS is rational.

√2 is irrational number.

So, our supposition is wrong.

So 5 + 3√2 is an irrational number.

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.

Find the number of natural numbers between 102 and 998 which are divisible 45 by 2 and 5 both.

Solution:

The number which ends with 0 is divisible by 2 and 5 both.

∴ Such numbers between 102 and 998 are:

110, 120,130, ……., 990.

Last term, a_{n} = 990

a + (n – 1)d =990

110 + (n – 1) × 10 = 990

110 + 10n -10 = 990

10n + 100 = 990

10n = 990 – 100

10n = 890

n = \(\frac{890}{10}\) = 89

n = 89

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) Show below are two overlapping sectors of a circle. The radii of the sectors are 6 cm and 8 cm. The figure is divided into three regions -I, II and III

(Note: The figure is not to scale)

Find the difference in the areas of regions I and III. Show your work.

(Note: Take π = \(\frac{22}{7}\))

Solution:

Let radius of sector (I + II) = 8 cm

And radius of setor (II + III) = 6 cm

OR

(B) The length of the minute hand of a clock is 6 cm. Find the area swept by it when it moves from 7:05 p.m. to 7:40 p.m.

Solution:

We know that, in 60 minutes, the tip of minute hand moves 360°

In 1 minute, it will move = \(\frac{360°}{60}\) = 60°

∴ From 7: 05 p.m. to 7: 40 p.m. i.e., 35 min, it will move through = 35 × 6° = 210°

∴ Area swept by the minute hand in 35 min = Area of sector with sectorial

angle 210° and radius of 6 cm

= \(\frac{210°}{360°}\) × π × 6^{2}

= \(\frac{7}{12}\) × \(\frac{22}{7}\) × 36

= 66 cm^{2}

Section – C (18 Marks)

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.

f(x) = x^{3} – ax^{2} + (a – 3)x + 6, where a is a non-zero real number. When f(x) is divided by (x + 1), there is no remainder.

If f(x) is completely factorable, find the zeroes of f(x). Show your steps.

Solution:

Since,f(x) is divisible by (x + 1)

∴ f(-1) = 0

Thus f(-1) = (-1)^{3} – a(-1)^{2} + (a – 3) – 1 + 6 = 0

– 1 – a – a + 3 + 6 = 0

8 – 2a = 0

8 = 2a

a = 4

After substituting value of ‘a’ we get

f(x) = x^{3} – 4x^{2} + x + 6

On factorising x^{2} – 5x + 6, we get

= x^{2} – 3x – 2x + 6

= x(x – 3) -2 (x – 3)

= (x – 3)(x – 2)

x = 3 and x = 2

Thus, zeroes of f(x) are (-1), 2 and 3.

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 intersect 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.

(A) In given figure, AB is the diameter of a circle with centre O and AT is a tangent ∠AOQ = 68°, find ∠ATQ.

Solution:

∠AOQ = 68° (Given)

[Angle on the circumference of the circle by the same arc]

= \(\frac{1}{2}\) × 68°

= 34°

∠BAT = 90° [∵ OA ⊥ AT]

∠ATQ = 90° – 34°

= 56°

OR

(B) Ifacircle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively.

prove that AQ = \(\frac{1}{2}\)(BC + CA + AB)

Solution:

BC + CA + AB

= (BP + PQ + (AR – CR) + (AQ – BQ)

= AQ + AR – BQ + BP + PC – CR

From the same external point, the tangent segments drawn to a circle are equal.

From the point B, BQ = BP

From the point A, AQ = AR

From the point C, CP = CR

Perimeter of AABC,

AB + BC + CA = 2AQ – BQ + BQ + CR – CR

⇒ = 2AQ

⇒ AQ = \(\frac{1}{2}\)(BC + CA + AB)

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.

The frequency distribution of daily rainfall in a town during a certain period is shown below.

Rainfall (in mm) |
Number of days |

0 – 20 | 10 |

20 – 40 | x |

40 – 60 | 12 |

60 – 80 | 8 |

Unfortunately, due to manual errors, the information on the 20-40 mm range got deleted from the data.

If the mean daily rainfall for the period was 32 mm, find the number of days when the rainfall ranged between 20-40 mm. Show your work.

Solution:

Rainfall (in mm) |
Number of days frequency (f_{i}) |
Class Marks(x_{i}) |
(f_{i}x_{i}) |

0 – 20 | 10 | 10 | 100 |

20 – 40 | x | 30 | 30x |

40 – 60 | 12 | 50 | 600 |

60 – 80 | 8 | 70 | 560 |

Total | 30 + x | 1260 + 30x |

Mean = 32 mm (Given)

And, Mean = \(\frac{1260+30 x}{30+x}\)

32 = \(\frac{1260+30 x}{30+x}\)

960 + 32x = 1260 + 30x

2x = 300

x = 150

Hence, number of days are 23.

Section – D (20 Marks)

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

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.

In the given figure, ∠ACB = 90° and CD ⊥ AB, prove that CD^{2} = BD × AD.

Solution:

In ΔACB and ΔADC

∠C = ∠D (both 90°)

∠A = ∠A (common)

∴ ΔACB ~ ΔADC (AA similarity)

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.

Find the unknown values in the following table:

Class |
Frequency (f) |
Cumulative frequency |

0 – 10 | 5 | 5 |

10 – 20 | 7 | x_{1} |

20 – 30 | x_{2} |
18 |

30 – 40 | 5 | x_{3} |

40 – 50 | x_{4} |
30 |

Solution:

x_{1} = 5 + 7 = 12

x_{2} = 18 – x_{1} = 18 – 12 = 6

x_{3} = 18 + 5 = 23

and x_{4} = 30 – x_{3} = 30 – 23 = 7

Section – E (12 Marks)

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.

Read the following text and answer the following questions:

Aayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office.

(Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5,8), school at (13, 14) and office at (13, 26) and co-ordinates are in km.

(i) What is the distance between house and bank?

OR

What is the distance between the bank and daughter’s school?

Solution:

Since,

Distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}).

(ii) What is the distance daughter’s school and office?

Solution:

Distance between daughter’s school and office,

(iii) What is the extra distance travelled by Aayush?

Solution:

Distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}).

= 24.59

= 24.6 km

Total distance = (House + Bank + School + Office travelled)

= (5 + 10 + 12) km

= 27 km

Extra distance travelled by Aayush in reaching his office = 27 – 24.6 = 2.4 km

Question 38.

Tree Plantation to Control Pollution

The class X students of a secondary school in Krishnagar have been allotted 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