Students can use CBSE Previous Year Question Papers Class 10 Maths with Solutions and CBSE Class 10 Maths Question Paper 2019 (Series: JMS/1) to familiarize themselves with the exam format and marking scheme.

## CBSE Class 10 Maths Question Paper 2019 (Series: JMS/1) with Solutions

Time allowed: 3 hours

Maximum marks: 80

General Instructions:

Read the following instructions carefully and follow them:

- All questions are compulsory.
- This question paper consists of 30 questions divided into four sections – A, B, C and D.
- Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in 2 questions of 1 mark, 2 questions of 2 marks each, 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
- Use of calculator is not permitted.

SET I Code NO. 30/1/1

Section – A

Questions number 1 to 6 carry 1 mark each.

Question 1.

Find the coordinates of a point A, where AB is a diameter of a circle whose centre is (2, – 3) and B is the point (1, 4).

Answer:

Let coordinates of Point A be (a, b).

As O is the mid point of AB, using mid point formula

∴ (2, -3) = [\(\frac{a+1}{2}\), \(\frac{b+4}{2}\)]

⇒ 2 = \(\frac{a+1}{2}\)

⇒ a + 1 = 4,

⇒ a = 4 – 1 = 3,

– 3 = \(\frac{b+4}{2}\)

b + 4 = -6

b = – 6 – 4 = -10

∴ Coordinates of Point A = (3, -10)

Question 2.

For what values of k, the roots of the equation x^{2} + 4x + k = 0 are real?

Or

Question 2.

Find the value of k for which the roots of the equation 3x^{2} – 10x + k = 0 are reciprocal of each other.

Answer:

Given: x^{2} + 4x + k = 0

Here a = 1, b = 4, c = k

D ≥ 0 …[When roots are real)

D = b^{2} – 4ac ≥ 0

⇒ (4)^{2} – 4(1) (k) ≥ 0

⇒ 16 – 4k ≥ 0

⇒ 4k ≤ 16

⇒ k ≤ \(\frac{16}{4}\) ∴ k ≤ 4

Or

Given: 3x^{2} – 10x + k = 0

Let roots be α and \(\frac{1}{\alpha}\).

Here a = 3, b = – 10, c = k

As we know, Product of Roots (α × \(\frac{1}{\alpha}\)) = \(\frac{c}{a}\)

⇒ 1 = \(\frac{k}{3}\) ∴ k = 3

Question 3.

Not in Current Syllabus

Answer:

Not in Current Syllabus

Question 4.

How many two digit numbers are divisible by 3?

Answer:

Two digit numbers divisible by 3 are 12, 15, 18,…. 99.

Here a = 12, d = 15 – 12 = 3, a_{n} = 99

Now, a_{n} = a + (n – 1) d = 99

⇒ 12 + (n – 1) (3) = 99

⇒ (n – 1) (3) = 99 – 12

⇒ (n – 1) (3) = 87

⇒ (n – 1) = \(\frac{87}{3}\) = 29

⇒ n = 29 + 1 = 30

∴ There are 30 such numbers.

Question 5.

Not in Current Syllabus

Answer:

Not in Current Syllabus

Question 6.

Find a rational number between √2 and √3.

Answer:

√2 = 1.414 and √3 = 1.732

∴ √2 < 1.5 < √3 ⇒ √2 < \(\frac{3}{2}\) < √3

∴ \(\frac{3}{2}\) is a rational number between √2 and √3.

Section – B

Questions number 7 to 12 carry 2 marks each.

Question 7.

Not in Current Syllabus

Answer:

Not in Current Syllabus

Question 8.

Which term of the AP 3, 15, 27, 39, will be 120 more than its 21^{st} term?

Or

Question 8.

If S_{n}, the sum of first n terms of an AP is given by S_{n} = 3n^{2} – 4n, find the n^{th} term.

Answer:

Given: A.P.: 3, 15, 27, 39…

Here I^{st} term, a = 3

Common difference, d = 15 – 3 = 12

Now, a_{n} = a_{21} + 120

a + (n – 1) d = a + 20 d + 120

(n – 1) (12) = 20 (12) + 120

(n – 1) (12) = 240 + 120

(n – 1) 12 = 360

n – 1 = \(\frac{360}{12}\) = 30 ∴ n = 30 + 1 = 31

∴ 31^{st} term is 120 more than its 21^{st} term.

Or

We have, S_{n} = 3n^{2} – 4n

When n = 1, S_{1} = 3(1)^{2} – 4 (1) = 3 – 4 = -1

When n = 2, S_{2} = 3(2)^{2} – 4 (2) = 12 – 8 = 4

Here a_{1} = S_{1} = -1

a_{2} = S_{2} – S_{-1} = 4 – (- 1) = 5

∴ d = a_{2} – a_{1} = 5 – (- 1) = 6

Now, a_{n} = a + (n – 1) d = – 1 + (n – 1) (6)

= -1 + 6n – 6 = 6n – 7

Question 9.

Find the ratio in which the segment joining the points (1, -3) and (4, 5) is divided by r-axis? Also find the coordinates of this point on r-axis.

Answer:

Let C(x, 0) be on r-axis. Let AC : CB = k: 1 Coordinates of C = Coordinates of C

(x, 0) = (\(\frac{4 k+1}{k+1}\), \(\frac{5 k-3}{k+1}\))

x = \(\frac{4 k+1}{k+1}\)

x = \(\frac{4\left(\frac{3}{5}\right)+1}{\frac{3}{5}+1}\) …[using (i)]

x = \(\frac{\frac{12+5}{5}}{\frac{3+5}{5}}\) = \(\frac{17}{8}\)

0 = \(\frac{5 k-3}{k+1}\)

5k – 3 = 0

5k = 3

k = \(\frac{3}{5}\) ……….. (i)

Therefore, Required Ratio = 3 : 5 and the

coordinates of required point C are (\(\frac{17}{8}\), 0).

Question 10.

A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.

Answer:

Number of outcomes (S) = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT] = 8

Same result in all the tosses (success) can be obtained as (HHH), (TTT), i.e., 2 ways.

P(wins) = \(\frac{2}{8}\) = \(\frac{1}{4}\)

∴ P(losing the game) = 1 – P(wins)

1 – \(\frac{1}{4}\) = \(\frac{4-1}{4}\) = \(\frac{3}{4}\)

Question 11.

A die is thrown once. Find the probability of getting a number which (i) is a prime number (ii) lies between 2 and 6.

Answer:

S = {1, 2, 3, 4, 5, 6}

(i) Prime numbers are 2, 3, 5 i.e., 3

∴ P(prime number) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

(ii) Numbers lying between 2 and 6 are 3,4, 5 i.e., 3

∴ P(Number lies between 2 and 6)

= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 12.

Find c if the system of equations cx + 3y + (3-, c) = 0; 12x + cy – c = 0 has infinitely many solutions?

Answer:

We have, cx + 3y + (3 – c) = 0

12x + cy – c = 0

For infinitely many solutions,

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

\(\frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c}\)

Taking 1st two

\(\frac{c}{12}=\frac{3}{c}\)

c^{2} = 36

⇒ c = ±6

Taking last two

\(\frac{3}{c}=\frac{3-c}{-c}\)

3c – c^{2} = – 3c

– c^{2} + 3c + 3c = 0

– c^{2} + 6c = 0

c (- c + 6) = 0

c = 0or-c + 6 = 0

c = 0 or c = 6

∴ c = 6 is the only common solution.

Section – C

Questions number 13 to 22 carry 3 marks each.

Question 13.

Prove that √2 is an irrational number.

Answer:

Let us assume, to the contrary, that √2 is rational.

That is, we can find integers a and b (≠ 0) such that √2 = \(\frac{a}{b}\).

Suppose a and b have a common factor other than 1, then we can divide by the common factor and assume that a and b are coprime.

So, b√2 = a

Squaring on both sides, and rearranging, we get 2b^{2} = a^{2}.

Therefore, a^{2} is divisible by 2 and a is also divisible by 2.

So, we can write a = 2c for some integer c.

Substituting for a, we get 2b^{2} = 4c^{2}, that is b^{2} = 2c^{2}.

This means that b^{2} is divisible by 2 and so b is also divisible by 2.

Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that √2 is rational. So we conclude that √2 is irrational.

Question 14.

Find the value of k such that the polynomial x^{2} – (k + 6)x + 2 (2k – 1) has sum of its zeroes equal to half of their product.

Answer:

Given: x^{2} – (k + 6)x + 2(2k – 1)

Here a = 1, b = – (k + 6), c = 2 (2k – 1)

According to Question,

Sum of zeroes = \(\frac{1}{2}\) (Product of zeroes) …[Given

∴ \(\frac{-b}{a}=\frac{1}{2}\left(\frac{c}{a}\right)\)

⇒ \(\frac{-[-(k+6)]}{1}\) = \(\frac{1}{2}\) × \(\frac{2(2 k-1)}{1}\)

⇒ k + 6 = 2k – 1

⇒ 6 + 1 = 2k – k

∴ k = 7

Question 15.

A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.

Or

Question 15.

A fraction becomes \(\frac{1}{3}\) 2 when 2 is subtracted from the numerator and it becomes \(\frac{1}{2}\) when 1 is subtracted from the denominator. Find the fraction.

Answer:

Let the present ages of his children be x years and y years. Then the present age of the father = 3 (x + y)

After 5 years, his children’s ages will be (x + 5) and (y + 5) years respectively.

After 5 years, the father’s age will be = [3 (x + y) + 5] years According to the Question,

3 (x + y) + 5 = 2 [(x + 5) + (y + 5)]

⇒ 3x + 3y + 5 = 2x + 10 + 2y + 10

⇒ 3x + 3y – 2x – 2y = 20 – 5

⇒ x + y = 15

∴ Present age of the father = 3 (x + y) = 3(15) = 45 years

Or

Let the numerator and denominator of required fraction be x and y.

∴ Fraction = \(\frac{x}{y}\)

\(\frac{x-2}{y}\) = \(\frac{1}{3}\)

3x – 6 = y ……… (i)

\(\frac{x}{y-1}\) = \(\frac{1}{2}\)

2x = y – 1

2x = 3x – 6 – 1 …[from (i)

3x – 2x = 7

x = 7

From (i), y = 3 (7) – 6 = 21 – 6 = 15

∴ The required Fraction = \(\frac{x}{y}\) = \(\frac{7}{15}\)

Question 16.

Find the point on y-axis which is equidistant from the points (5, -2) and (-3, 2).

Or

Question 16.

The line segment joining the points A (2, 1) and B (5, -8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by 2x – y + k = 0, find the value of k.

Answer:

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

Let A (5, -2) and B (-3, 2).

According to Question,

PA = PB

PA^{2} = PB^{2} …..[Squaring both sides

(5 – 0)^{2} +(- 2 – y)^{2} = (- 3 – 0)^{2} + (2 – y)^{2}

⇒ 25 + y^{2} + 4y + 4 = 9 + y^{2} – 4y + 4

⇒ 4y + 4y = 9 – 25

⇒ 8y = -16 ⇒ y = \(\frac{-16}{8}\) = -2

∴ Required Point = P (0, – 2).

Or

Here AP : PB = m : n = 1 : 2

Using Section formula,

= (3,-2)

Now, putting P (3, – 2) in 2x – y + k = 0

⇒ 2 (3) – (-2) + k = 0 ⇒ 6 + 2 + k = 0

⇒ 8 + k = 0

∴ k = – 8

Question 17.

Prove that: (sin θ + cosec θ)^{2} + (cos θ + sec θ)^{2} = 7 + tan^{2}θ + cot^{2}θ.

Or

Question 17.

Prove that: (1 + cot A – cosec A) (1 + tan A + sec A) = 2.

Answer:

L.H.S. (sin θ + cosec θ)^{2} + (cos θ + sec θ)^{2}

= sin^{2} θ + cosec^{2} θ + 2 sin θ. cosec θ + cos^{2} θ + sec^{2} θ + 2 cos θ sec θ

= (sin^{2} θ + cos^{2} θ) + (1 + cot^{2} θ) + 2 + (1 + tan^{2} θ) + 2 …[ ∵ cosec^{2} θ = 1 + cot^{2} θ, sec2 θ = 1 + tan^{2} θ]

= 1 + 1 + cot^{2} θ + 2 + 1 + tan^{2} θ + 2 …[∵ sin^{2} θ + cos^{2} θ = 1

= 7 + tan^{2} θ + cot^{2} θ = R.H.S.

Or

L.H.S. = (1 + cot A – cosec A) (1 + tan A + sec A)

= 1 + tan A + sec A + cot A + cot A tan A + cot A sec A – cosec A – cosec A tan A – cosec A sec A

Question 18.

In Fig., AB is a chord of length 8 cm of a circle of radius 5 cm and centre O. The tangents at A and B intersect at point P. Find the length of AP.

Answer:

Constriction. Join OP.

Let OP intersect AB at a point M. Then ΔAPB is isosceles and OP is the angle bisector of ∠APB.

So, OP⊥AB

AM = MB = 4 cm …[∵⊥ from the centre bisects the chord

In rt. ΔOMA,

OM = \(\sqrt{\mathrm{OA}^2-\mathrm{AM}^2}\) …[Pythagoras’ theorem

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

Let AP = x cm and PM = y cm

In rt. ΔAMP,

PA^{2} = AM^{2} + PM^{2} …[Pythagoras’ theorem

X^{2} = 4^{2} + y^{2}

⇒ x^{2} – y^{2} = 16

In rt. ΔOAP,

OA^{2} + PA^{2} = OP^{2} …[Pythagoras’ theorem

(5)^{2} + x^{2} = (y + 3)^{2}

25 + x^{2} = y^{2} + 6y + 9

x^{2} – y^{2} = 6y + 9 – 25

⇒ x^{2} – y^{2} = 6y – 16 ………… (ii)

From solving (i) & (ii),

6y – 16 = 16

⇒ 6y = 32 ⇒ y = \(\frac{32}{6}\) = \(\frac{16}{3}\)

From (i), x^{2} – (\(\frac{16}{3}\))^{2} = 16

⇒ x^{2} = 16 + \(\frac{256}{9}\) = \(\frac{144+256}{9}\) = \(\frac{400}{9}\)

⇒ x = \(\frac{230}{3}\) cm or \(6 . \overline{6}\) cm

Therefore, length of AP = x = \(6 . \overline{6}\) cm.

Question 19.

Not in Current Syllabus

Answer:

Not in Current Syllabus

Question 20.

Not in Current Syllabus

Answer:

Not in Current Syllabus

Question 21.

Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed much area will it irrigate in 30 minutes; if 8 cm standing water is needed?

Answer:

Speed of water = 10 km/hr. in the Canal

l = \(\frac{10}{2}\) = 5 km = 5000 m (in 30 min.)

b = 6 m and h = 1.5 m

Volume of standing water in the field = Volume of water that flows in 30 min.

(Area of field) × height = lbh

(Area of field) × \(\frac{8}{100}\) = 5000 × 6 × 1.5 …[8 cm = \(\frac{8}{100}\)

= 562500 m^{2} or 56.25 hectare …[1 hec. = 10000 m^{2}

Question 22.

Find the mode of the following frequency distribution.

Answer:

Maximum frequency = 16

∴ Modal class is 30 – 40

Mode = l + \(\frac{f_1-f_0}{2 f_1-f_0-f_2}\) × h …..[l = 30, f_{0} = 10, f_{1} = 16, f_{2} = 12, h = 10

= 30 + (\(\frac{16-10}{32-10-12}\)) × 10

= 30 + \(\frac{6 \times 10}{10}\) = 30 + 6 ∴ Mode = 36

Section – D

Questions number 23 to 30 carry 4 marks each.

Question 23.

Two water taps together can fill a tank in 1\(\frac{7}{8}\) g hours. The tap with longer diameter takes hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Or

Question 23.

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.

Answer:

Let V be the total volume of the tank.

Let smaller water tap alone can fill the tank in x hrs.

Then larger water tap alone can fill the tank in (x – 2) hrs.

According to the Question

\(\frac{\mathrm{V}}{x}+\frac{\mathrm{V}}{x-2}=\frac{\mathrm{V}}{\frac{15}{8}}\) [∵ Time = 1\(\frac{7}{8}\)hrs = \(\frac{15}{8}\)hrs

V(\(\frac{1}{x}+\frac{1}{x-2}\)) = V\(\frac{8}{15}\)

⇒ \(\frac{1}{x}+\frac{1}{x-2}\) = \(\frac{8}{15}\)

⇒ 8x (x – 2) = 15 (2x – 2)

⇒ 8x^{2} – 16x = 30x – 30

⇒ 8x^{2} – 16x – 30x + 30 = 0

⇒ 8x^{2} – 46x + 30 = 0

⇒ 4x^{2} – 23x + 15 = 0 …[divided both sides by 2

⇒ 4x^{2} – 20x – 3x + 15 = 0

⇒ 4x (x – 5) – 3 (x – 5) = 0

⇒ (4x – 3) (x – 5) = 0

⇒ x – 5 = 0 or 4x – 3 = 0

x = 5 or x = \(\frac{3}{4}\)

Rejecting x = \(\frac{3}{4}\) hr, because one tap alone will fill the tank in more than 1\(\frac{7}{8}\) hours.

∴ x = 5 hours

Therefore, Smaller tap can fill the tank in 5 hours and the Larger tap can fill the tank in (5 – 2) = 3 hours.

Or

Let the speed of the boat and stream in still water be x km/hr and y km/hr.

Then the speed of the boat downstream and upstream is (x + y) km/hr and (x – y) km/ hr respectively

According to the Question,

\(\frac{30}{x-y}+\frac{44}{x+y}\) = 10 ………. (i) [∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)

\(\frac{40}{x-y}+\frac{55}{x+y}\) = 13 ………. (ii)

Let \(\frac{1}{x-y}\) = p and \(\frac{1}{x+y}\) = q

30p+ 44q = 10 ……. (iii)

40p + 55q = 13 …… (iv)

Multiplying Eq. (iii) by 4 and (iv) by 3, we have

120p + 176q = 40 ………. (v)

120p + 165q = 39 …………. (vi)

Subtracting (vi) from (v), we get

Putting the value of q in (iii), p = \(\frac{1}{5}\)

∴ \(\frac{1}{x-y}\) = \(\frac{1}{5}\)

x – y = 5

x = 5 + y …….. (A)

\(\frac{1}{x+y}\) = \(\frac{1}{11}\)

x + y = 11

5 + y + y = 11 …[From (A)

2y = 11 – 5 = 6

y = \(\frac{6}{2}\) = 3

Putting the value of y in (A), x = 5 + 3 = 8

Hence, the speed of the boat in still water is 8 km/hr. and the speed of the stream is 3 knyhr.

Question 24.

If the sum of first four terms of an AP is 40 and that of its first 14 terms is 280. Find the sum of its first n terms.

Answer:

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

As per question

S_{4} = 40

\(\frac{4}{2}\)[2a + (4 – 1)d] = 40

2(2a + 3d) = 40

2a + 3d = 20 …….. (i)

S_{14} = 280

\(\frac{14}{2}\)[2a + (4 – 1)d] = 280

7(2a + 13d) = 280

2a + 3d = 40 ……… (ii)

Putting the value of d =2 in (ii), we get a = 7

∴ S_{n} = \(\frac{n}{2}\) [2(7) + (n – 1) . 2]

= \(\frac{n}{2}\)2[7 + n – 1]

= n (n +6) or n^{2} + 6n [Hence Proved

Question 25.

Prove that \(\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{1}{\sec A-\tan A}\)

Answer:

L.H.S. = \(\frac{\sin A-\cos A+1}{\sin A+\cos A-1}\)

Dividing Numerator and Denominator by cos A, we have

Question 26.

A man in a boat rowing away from a light house 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60° to 30°. Find the speed of the boat in metres per minute. [Use √3 =1.732]

Or

Question 26.

Two poles of equal heights are standing opposite to each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.

Answer:

Let AB be the light house, C and D be the two positions of the boat

In rt. ΔABC,

BD = 100√3 ; Now, CD = BD – BC

CD = 100√3 – \(\frac{100 \sqrt{3}}{3}\)

= 100√3(1 – \(\frac{1}{3}\) = 100√3(\(\frac{2}{3}\))

= 100(1.732) × \(\frac{2}{3}\) = \(115.4 \overline{6}\) m

Distance covered by boat in 2 minutes = 115.46 m.

∴ Speed of the boat in 1 min. = \(\frac{\text { Distance }}{\text { Time }}\)

= \(\frac{115.4 \overline{6}}{2 \min .}\) = 57.73 m/min.

Or

Let AB = DE = y m be the two poles such that BD = 80 m, CD = xm and BC = 80 – x m

In rt. ΔCDE,

tan 30° = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)

\(\frac{1}{\sqrt{3}}=\frac{y}{x}\) ⇒ x = √3 y ……… (i)

In rt. ΔABC,

tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)

\(\frac{\sqrt{3}}{1}=\frac{y}{80-x}\) ⇒ y = √3(80 – x)

⇒ y = √3(80 – √3 y) ….[From (i)

⇒ y = 80 √3 – 3y ⇒ 4y = 80√3

⇒ y = \(\frac{80 \sqrt{3}}{4}\) = 20√3m = 20(1.73) = 34.6m

Putting the value of y in (i),

x= √3(20√3) ⇒ x = 60 m

∴ CD, x = 60 m

BC = 80 – x = 80 – 60 = 20 m

∴ Height of poles = 20 √3 m & Distance of poles from the point are 20 m and 60 m.

Question 27.

Not in Current Syllabus

Answer:

Not in Current Syllabus

Question 28.

Not in Current Syllabus

Answer:

Not in Current Syllabus

Question 29.

Not in Current Syllabus

Answer:

Not in Current Syllabus

Question 30.

If the median of the following frequency distribution is 32.5. Find the values of f_{1} and f_{2}.

Answer:

∴ f_{1} + f_{2} + 31 = 40 ⇒ f_{1} + f_{2} 40 – 31 = 9

f_{1} = 9 – f_{2} ………… (i)

Here, \(\frac{n}{2}\) = \(\frac{40}{2}\) = 20

Median = 32.5 …[Given

∴ Median class is 30 – 40.

Median = l + \(\frac{\frac{n}{2}-c . f .}{f}\) × h …[l = 30, c.f = f_{1} + 14, f = 12, n = 10

32.5 = 30 + \(\frac{20-\left(f_1+14\right)}{12}\) × 10

32.5 – 30= \(\frac{\left(20-f_1-14\right)}{6}\) × 5

2.5= \(\frac{\left(6-f_1\right)}{6}\) × 5

⇒ 15 = 30 – 5f_{1} ⇒ 5f_{1} = 30 – 15

⇒ 5f_{1} = 15 ⇒ f_{1} = \(\frac{15}{5}\)

Putting the value of fa in (i), 3 + f_{2} = 9

⇒ f_{2} = 9 – 3 = 6 ∴ f_{1} = 3 f_{2} = 6

SET II Code. 30/1/2

Note: Except for the following questions, all the remaining questions have been asked in Set – I.

Question 1.

Find the coordinates of a point A, where AB is a diameter of the circle with centre (- 2, 2) and B is the point with coordinates (3, 4).

Answer:

Let A(a, b).

As O is the mid point of AB, using mid¬point formula

\(\frac{a+3}{2}[latex] = -2,

a + 3 = -4

a = – 4 – 3 = – 7

[latex]\frac{b+4}{2}[latex] = 2

b + 4 = 4

b = 4 – 4 = 0

∴ A(-7, 0)

Question 7.

Find the value of k for which the following pair of linear equations have infinitely many solutions. 2x + 3y = 7, (k + 1)x + (2k – 1)y = 4k + 1

Answer:

We have, 2x + 3y = 7

(k + 1)x + (2k – 1)y = 4k + 1

For infinitely many solutions, [latex]\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

\(\frac{2}{k+1}=\frac{3}{2 k-1}=\frac{7}{4 k+1}\)

Taking 1^{st} two

\(\frac{2}{k+1}=\frac{3}{2 k-1}\)

4k – 2 = 3k + 3

4k – 3k = 3 + 2

k = 5

Taking last two

\(\frac{3}{2 k-1}=\frac{7}{4 k+1}\)

14k – 7 = 12k + 3

14k – 12k – 3 + 7

2k = 10

⇒ k = \(\frac{10}{2}\) = 5

∴ k = 5 is the solution as it is common in both the cases.

Question 13.

The arithmetic mean of the following frequency distribution is 53. Find the value of k.

Answer:

Class | Frequency (f_{i}) |
Mid-value (x_{i}) |
f_{i}x_{i} |

0 – 20 | 12 | 10 | 120 |

20 – 40 | 15 | 30 | 450 |

40 – 60 | 32 | 50 | 1600 |

60 – 80 | k | 70 | 70 k |

80 – 100 | 13 | 90 | 1170 |

Σf_{i} = 72 + k |
Σf_{i}x_{i} = 3340 + 70k |

As we know, Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)

⇒ \(\frac{3340+70 k}{72+k}\)

⇒ 3340 + 70k = 53 (72 + k)

⇒ 340 + 70k = 3816 + 53k

⇒ 70k – 53k = 3816 – 3340

⇒ 17k = 476

⇒ k = \(\frac{476}{17}\) = 28 ∴ k = 28

Question 14.

Find the area of the segment shown in given figure, if radius of the circle is 21 cm and ∠AOB = 120°. [Use π = \(\frac{22}{7}\)]

Answer:

Given: Radius, r = 21 cm

∴ Area of Shaded Region

= Area of sector (AMBO) – Area of (ΔAOB)

Question 16.

In Fig., a circle is inscribed in a ΔABC having sides BC = 8 cm, AB = 10 cm and AC = 12 cm. Find the lengths BL, CM and AN.

Answer:

AB = 10 cm,

BC = 8 cm and

AC = 12 cm

As we know,

AN = AM

BN = BL

CL = CM

Let BL = BN = x cm,

∴ CL = BC – BL

= (8 – x) cm = CM

Similarly, AN = AM

⇒ AN = AB – BN = (10 – x) cm

AC = 12 cm …(Given)

⇒ AM + CM = 12 ⇒ 10 – x + 8 – x = 12

⇒ 18 – 2x = 12 ⇒ 6 = 2x ∴ x = 3

∴ BL = 3 cm, CM = 5 cm, AN = 7 cm

Question 23.

Prove that \(\frac{\tan ^2 \mathrm{~A}}{\tan ^2 \mathrm{~A}-1}+\frac{{cosec}^2 \mathrm{~A}}{\sec ^2 \mathrm{~A}-{cosec}^2 \mathrm{~A}}=\frac{\cdot 1^{\prime}}{1-2 \cos ^2 \mathrm{~A}}\)

Answer:

Question 24.

The first term of an AP is 3, the last term is 83 and the sum of all its terms is 903. Find the number of terms and the common difference of the AP.

Answer:

We know, a = 3, a_{n} = 83, S_{n} = 903

S_{n} = \(\frac{n}{2}\)(a + a_{n}) = 903

⇒ \(\frac{n}{2}\)(3 + 83) = 903

⇒ \(\frac{n}{2}\)(86) = 903 ⇒ 43n = 903

∴ n = \(\frac{903}{43}\) = 21

Now, a_{n} = 83

⇒ a + (n – 1) d = 83 ⇒ 3 + (21 – 1) d = 83

⇒ 20d = 83 – 3 ⇒ 20d = 80

∴ d = \(\frac{80}{20}\) = 4

Therefore, The number of terms, n = 21 and Common difference, d = 4.

SET III Code No. 30/1/3

Note: Except for the following questions, all the remaining questions have been asked in Set – I and Set II.

Question 1.

Two positive integers a and b can be written as a = x^{3}y^{2} and b = xy^{3}, where x and y are prime numbers. Find LCM (a, b).

Answer:

We have, a = x^{3}y^{2} and b = xy^{3}

∴ L.C.M. (a, b) = x^{3}y^{3}

Question 7.

Find, how many two digits natural numbers are divisible by 7.

Or

Question 7.

If the sum of first n terms of an AP is n^{2}, then find its 10^{th} term.

Answer:

Two digit natural numbers divisible by 7 are 14, 21, 28 98.

First term, a = 14

Common difference, d = 21 – 14 = 7

Last term, a_{n} = 98

a + (n – 1) d = 98

14 + (n – 1) 7 = 98

⇒ (n – 1)7 = 98 – 14

⇒ (n – 1) = \(\frac{84}{7}\) y ⇒ n -1 = 12

∴ n = 12 + 1 = 13

Therefore, there are 13 two digit numbers that are divisible by 7.

Or

S_{n} = n^{2} …[Given

When n = 10, S_{10} = 10^{2} = 100

When n = 9, S_{9} = 9^{2} = 81

∴ 10^{th} term, a_{10} = S_{10} – S_{9}

= 100 – 81 = 19

Thus the 10^{th} term of the AP is 19.

Question 15.

Prove that \(\frac{2+\sqrt{3}}{5}\) is an irrational number, given that √3 is an irrational number.

Answer:

Let us assume, to the contrary, that \(\frac{2+\sqrt{3}}{5}\) is rational.

That is, we can find co-primes a and b,

(b ≠ 0) such that \(\frac{2+\sqrt{3}}{5}\) = \(\frac{a}{b}\).

⇒ 2 + √3 = \(\frac{5 a}{b}\) ⇒ √3 = \(\frac{5 a}{b}\) – 2

⇒ √3 = \(\frac{5 a-2 b}{b}\)

Since a and b are integers, we get (\(\frac{5 a}{b}\)– 2) is rational, and so √3 is rational.

But this contradicts the fact that √3 is irrational.

This contradiction has arisen because of our incorrect assumption that \(\) is rational.

So, we conclude that \(\) is irrational.

Question 23.

If sec θ = x + \(\frac{1}{4 x}\) , x ≠ 0, find (sec θ + tan θ).

Answer:

We know that, tan^{2} θ = sec^{2} θ – 1

Question 25.

The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food.

Answer:

C.I. Daily expenditure (₹) | f_{i} |
Class mark mid-value (x) | f_{i}x_{i} |

100 – 150 | 4 | 125 | 500 |

150 – 200 | 5 | 175 | 875 |

200 – 250 | 12 | 225 | 2700 |

250 – 300 | 2 | 275 | 550 |

300 – 350 | 2 | 325 | 650 |

Σf_{i }= 25 |
Σf_{i}x_{i }= 5275 |

∴ Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{5275}{25}\) = ₹211

Mean daily expenditure = ₹211