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## CBSE Class 10 Maths Question Paper 2017 (Outside Delhi) 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 contains 30 questions.
- Question No. 1 – 6 in Section A are very short answer type questions carrying 1 mark each.
- Question No. 7 – 12 in Section B are short answer type questions carrying 2 marks each.
- Question No. 13 – 22 in Section C are long answer – I type questions carrying 3 marks each.
- Question No. 23 – 30 in Section D are long answer – II type questions carrying 4 marks each.

Section A

Questions number 1 to 6 carry 1 mark each.

Question 1.

In the given figure, XY || QR, \(\frac{\mathrm{PQ}}{\mathrm{XQ}}\) = \(\frac{7}{3}\) and PR = 6.3 cm, find YR.

Answer:

Let YR = x

\(\frac{\mathrm{PQ}}{\mathrm{XQ}}\) = \(\frac{\mathrm{PR}}{\mathrm{YR}}\) …[Thales’ theorem

\(\frac{7}{3}\) = \(\frac{6.3}{x}\)

⇒ x = \(\frac{6.3 \times 3}{7}\) = 2.7

∴ YR = 2.7 cm

Question 2.

If sec θ + tan θ = 7, then evaluate sec θ – tan θ.

Answer:

We know that,

sec^{2} θ – tan^{2} θ = 1

(sec θ + tan θ) (sec θ – tan θ) = 1

(7) (sec θ – tan θ) = 1 …[sec θ + tan θ = 7; (Given)

sec θ – tan θ = \(\frac{1}{7}\)

Question 3.

For what value of k, the pair of equations 4x – 3y = 9, 2x + ky = 11 has no solution.

Answer:

4x – 3y = 9; 2x + ky = 11

\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\) (No solution)

\(\frac{4}{2}\) = \(\frac{-3}{k}\) ≠ \(\frac{9}{11}\) ⇒ 2 = \(\frac{-3}{k}\)

2k = -3 ⇒ k = \(\frac{-3}{2}\)

Question 4.

What is the common difference of an A.P. in which a_{21} – a_{7} = 84?

Answer:

a_{21} – a_{7} = 84 … [Given

∴ (a + 20d) – (a + 6d) = 84 …[a_{n} = a + (n – 1)d

20d – 6d = 84

14d = 84 ⇒ d = \(\frac{84}{14}\) = 6

Question 5.

If a tower 30 m high, casts a shadow 10√3 m long on the ground, then what is the angle of elevation of the sun?

Answer:

Let required angle be θ.

tan θ = \(\frac{30}{10 \sqrt{3}}\)

tan θ = √3

⇒ tan θ = tan60° ∴ θ = 60°

Question 6.

The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?

Answer:

P(rotten apples) = \(\frac{\text { No. of rotten apples }}{ Total apples}\)

0.18 = \(\frac{\text { No. of rotten apples }}{900}\)

∴ No. of rotten apples = 900 × 0.18 = 162

Section B

Questions number 7 to 12 carry 2 marks each.

Question 7.

Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons.

Answer:

No,

LCM = Product of highest power of each factor involved in the numbers.

HCF = Product of smallest power of each common factor.

∴ We can conclude that LCM is always a multiple of HCF i.e., LCM = k × HCF

We are given that, LCM = 175 and HCF = 15

∴ 175 = k × 15 ⇒ 11.67 = k

But in this case, LCM ≠ k × HCF

Therefore, 2 numbers cannot have LCM as 175 and HCF as 15.

Question 8.

Find the zeroes of p(x) = 2x^{2} – x – 6 and verify the relationship of zeroes with these co-efficients.

Or

Question 8.

Find the condition that zeroes of polynomial p(x) = ax^{2} + bx + c are reciprocal of each other.

Answer:

p(x) = 2x^{2} – x – 6 …[Given

= 2x^{2} – 4x + 3x – 6

= 2x (x – 2) + 3 (x – 2) = (x – 2) (2x + 3)

Zeroes are: x – 2 = 0 or 2x + 3 = 0

x = 2 or x = \(\frac{-3}{2}\)

Here a = 2, b = -1, c = -6

Sum of zeroes = 2 + (\(\frac{-3}{2}\)) = \(\frac{1}{2}\) = -(\(\frac{-1}{2}\))

= -(\(\frac{b}{a}\)) = \(\frac{\text {-Coefficient of } x}{\text { Coefficient of } x^2}\)

Product of zeroes = 2 × (\(\frac{-3}{2}\)) = \(\frac{-6}{2}\) = \(\frac{c}{a}\)

= \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

⇒ Relationship holds.

Or

Let α, \(\frac{1}{\alpha}\) be the zeroes of P(x).

P(x) = ax^{2} + bx + c …[Given

Product of zeroes = \(\frac{c}{a}\)

α × \(\frac{1}{\alpha}\) = \(\frac{c}{a}\) ⇒ 1 = \(\frac{c}{a}\)

∴ a = c

∴ Required condition

∴ Coefficient of x^{2} = Constant term

Question 9.

In the given figure, QA⊥AB and PB⊥AB. If AO = 20 cm, BO = 12 cm, PB = 18 cm, find AQ.

Answer:

In ΔOAQ and ΔOBP,

∠OAQ = ∠OBP …[Each 90°

∠AOQ = ∠BOP …[vertically opposite angles

∴ ΔOAQ ~ ΔOBP …[By AA corollary

\(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{AQ}}{\mathrm{PB}}\) …[sides are proportional

\(\frac{20}{12}\) = \(\frac{A Q}{18}\) ⇒ AQ = \(\frac{18 \times 20}{12}\)

∴ AQ = 30 cm

Question 10.

Find the value of p, for which one root of the quadratic equation px^{2} – 14x + 8 = 0 is 6 times the other.

Answer:

Given equation is px^{2} – 14x + 8 = 0

Here a = p b = -14 c = 8

Let roots be α and 6α.

Sum of roots

α + 6α = \(\frac{-b}{a}\)

7α = \(\frac{-(-14)}{p}\)

α = \(\frac{2}{p}\) ………. (i)

Product of roots

α(6α) = \(\frac{c}{a}\)

6α^{2} = \(\frac{8}{p}\)

6(\(\frac{2}{p}\))^{2} = \(\frac{8}{p}\) ….[From (i)

\(\frac{24}{p^2}\) = \(\frac{8}{p}\)

⇒ 8p^{2} = 24p

⇒ 8p^{2} – 24p = 0

⇒ 8p(p – 3) = 0

⇒ 8p = 0 or p – 3 = 0

⇒ p = 0 or p = 3

But p ≠ 0 …[∵ coefficient of x^{2} ≠ 0

∴ p = 3

Question 11.

Which term of the progression 20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\), … is the first negative term?

Answer:

Given: A.P.: 20, \(\frac{77}{4}\), \(\frac{37}{4}\), \(\frac{71}{4}\)

Here a = 20, d = \(\frac{77-80}{4}\) = –\(\frac{3}{4}\)

For first negative term, a_{n} < 0

⇒ a + (n – 1)d < 0 ⇒ 20 + (n – 1) (-\(\frac{3}{4}\)) < 0

⇒ –\(\frac{3}{4}\) (n – 1) < -20 ⇒ 3(n – 1) > 80

⇒ 3n – 3 > 80 ⇒ 3n > 83

⇒ n > \(\frac{83}{3}\) ⇒ n > 27.\(\overline{6}\)

∴ Its negative term is 28^{th} term.

Question 12.

A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid point of PQ, then find the coordinates of P and Q.

Answer:

We know that at y-axis coordinates of points are (0, y) and at x-axis (x, 0).

Let P(0, b) and Q(a, 0)

Mid point of PQ = (2, -5) …[Given

(\(\frac{0+a}{2}, \frac{b+0}{2}\)) = (2, -5)

\(\frac{a}{2}\) = 2, \(\frac{b}{2}\) = -5

a = 4, b = -10 ∴ P(0, -10), Q (4, 0)

Section C

Questions number 13 to 22 carry 3 marks each.

Question 13.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 14.

In a school there are two Sections A and B of class X. There are 48 students in Section A and 60 students in Section B.

Determine the least number of books required for the library of the school so that the books can be distributed equally among all students of each Section.

Answer:

Since the books are to be distributed equally among the students of Section A and Section By therefore, the number of books must be a multiple of 48 as well as 60. Hence, required number of books is the LCM of 48 and

60.

48 = 2^{4} × 3

60 = 2^{2} × 3 × 5

∴ LCM = 2^{4} × 3 × 5

= 16 × 15 = 240

Hence, required number of books is 240.

Question 15.

Prove that: \(\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}\) = 2 cosec θ

Answer:

Question 16.

Solve for x and y:

27x + 31y = 85; 31x + 27y = 89

Or

Question 16.

Solve: \(\frac{x}{a}\) + \(\frac{y}{b}\) = a + b; \(\frac{x}{a^2}\) + \(\frac{y}{b^2}\) = 2, a, b ≠ 0

Answer:

Adding (i) and (ii), we have

x + y + x – y = 3 + 1

⇒ 2x = 4 ⇒ x = 2

Putting the value of ‘x’ in (i), we get

2 + y = 3 ⇒ y = 3 – 2 = 1

∴ x = 2, y = 1

Or

Multiplying (i) by b and (in) by 1, we get

b^{2}x + aby = a^{2}b^{2} + ab^{3} ……….. (iv)

Subtracting (iii) from (iv)

∴ x = a^{2}, y = b^{2}

Question 17.

In the given figure, DE || BC and AD : DB = 7 : 5, find DE : BC.

Answer:

Given: In ΔABC, DE || BC

and AD : DB = 7 : 5.

To find: DE : BC

Proof: Let AD = 7k

and BD = 5k then

AB = 7k + 5k = 12 k

In ΔADE and ΔABC,

∠1 = ∠1 …[Common

∠2 = ∠ABC …[Corresponding angles

∴ ΔADE ~ ΔABC …[AA similarity criterion

\(\frac{\mathrm{AD}}{\mathrm{AB}}\) = \(\frac{\mathrm{AE}}{\mathrm{AC}}\) = \(\frac{\mathrm{DE}}{\mathrm{BC}}\)

\(\frac{7 k}{12 k}\) = \(\frac{\mathrm{DE}}{\mathrm{BC}}\) ∴ \(\frac{\mathrm{DE}}{\mathrm{BC}}\) = \(\frac{7}{2}\)

Question 18.

The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P.

Answer:

First term, a = 5, Last term, a_{n} = 45

Let the number of terms = n

S_{n} = 400

\(\frac{n}{2}\)(a + a_{n}) = 400

\(\frac{n}{2}\)(5 + 45) = 400

\(\frac{n}{2}\)(50) = 400

n = \(\frac{400}{25}\) = 16 = Number of terms

Now, a_{n} = 45

a + (n – 1)d = a_{n}

5 + (16 – 1 )d = 45

15d = 45 – 5

∴ d = \(\frac{40}{15}\) = \(\frac{8}{3}\)

Question 19.

A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.

Answer:

Let the number of black balls = x

the number of white balls = 15

∴ Total number of balls = x + 15

∴ P(black ball) = 3P(White balls)

\(\frac{x}{x+15}\) = 3(\(\frac{15}{x+15}\))

\(\frac{x}{x+15}\) = \(\frac{45}{x+15}\) [∵ x + 15 ≠ 0

∵ x = 45

Hence, the number of black balls = 45.

Question 20.

In what ratio does the point (\(\frac{24}{11}\), y) divide the line segment joining the points P(2, -2) and Q(3, 7)? Also find the value of y.

Answer:

Let P(2, -2), Q(3, 7), R(\(\frac{24}{11}\), y)

Let PR : RQ = k : 1

Coordinates of R= Coordinates of R

(\(\frac{3 k+2}{k+1}\), \(\frac{7 k-2}{k+1}\)) = (\(\frac{24}{11}\), y)

∴ \(\frac{3 k+2}{k+1}\) = \(\frac{24}{11}\)

33k + 22 = 24k + 24

33k – 24k = 24 – 22

9k = 2 ⇒ k = \(\frac{2}{9}\) ………. (i)

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

= \(\frac{7\left(\frac{2}{9}\right)-2}{\frac{2}{9}+1}\) …. [From(i)

= \(\frac{14-18}{2+9}\)

y = \(\frac{-4}{11}\)

∴ Required ratio = k : 1 = \(\frac{2}{9}\) : 1 or 2 : 9

Question 21.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 22.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Section D

Questions number 23 to 30 carry 4 marks each.

Question 23.

If a α and β are the zeroes of the polynomial p(x) = 2x^{2} + 5x + k, satisfying the relation, α^{2} + β^{2} + αβ = then find the value of k.

Answer:

Given polynomial is p(x) = 2x^{2} + 5x + k

Here a = 2, b = 5, c = k

α + β = \(\frac{-b}{a}\) = \(\frac{-5}{2}\); αβ = \(\frac{c}{a}\) = \(\frac{k}{2}\)

α^{2} + β^{2} + αβ = \(\frac{21}{4}\) ….[Given

(α + β)^{2} – 2αβ + αβ = \(\frac{21}{4}\)

(α + β)^{2} – αβ = \(\frac{21}{4}\) ⇒ (\(\frac{-5}{2}\))^{2} – \(\frac{k}{2}\) = \(\frac{21}{4}\)

\(\frac{-k}{2}\) = \(\frac{21}{4}\) – \(\frac{25}{4}\) = \(\frac{-4}{4}\)

⇒ \(\frac{-k}{2}\) = -1

∴ k = 2

Question 24.

Prove that: \(\frac{\cos ^2 \theta}{1-\tan \theta}+\frac{\sin ^3 \theta}{\sin \theta-\cos \theta}\) = 1 + sin θ cos θ.

Or

Question 24.

If x = r sin A cos C, y = r sin A sin C and z = r cos A, then prove that x^{2} + y^{2} + z^{2} = r^{2}.

Answer:

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

= cos^{2} θ + sin^{2} θ + cos θ sin θ

= 1 + sin θ cos θ

…[∵ cos^{2} θ + sin^{2} θ = 1

= R.H.S.

Or

x = r sin A cos C; y = r sin A sin C; z = r cos A

Squaring and adding

L.H.S. x^{2} + y^{2} + z^{2} = r^{2} sin^{2} A cos^{2} C + r^{2} sin^{2} A sin^{2} C + r^{2} cos^{2} A

= r^{2} sin^{2} A(cos^{2} C + sin^{2} C) + r^{2} cos^{2} A

= r^{2} sin^{2} A + r^{2} cos^{2} A ….[cos^{2} θ + sin^{2} θ = 1

= r^{2} (sin^{2} A + cos^{2} A) = r^{2} = R.H.S.

Question 25.

If sides AB, BC and median AD of ΔABC are proportional to the corresponding sides PQ, QR and median PM of PQR, show that ΔABC ~ ΔPQR.

Answer:

Given: \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BC}}{\mathrm{QR}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\)

To prove: ΔABC ~ ΔPQR

Proof: In ΔABD and ΔPQM

Given: \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{1 / 2 B C}{1 / 2 Q R}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\) ⇒ \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BD}}{\mathrm{QM}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\)

….[D & M are the mid points of BC & QR respectively

∵ Sides of Δs are proportional.

∴ ΔABD ~ ΔPQM …[SSS similarity

∠ABD = ∠PQM …[CPST

⇒ ∠ABC = ∠PQR

In ΔABC and ΔPQR,

\(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BC}}{\mathrm{QR}}\) (i) …[Given

⇒ ∠ABC = ∠PQR ……(ii)

From (i) & (ii), ΔABC ~ ΔPQR …[SAS similarity

Question 26.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 27.

If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27), then find the ratio of their 9^{th} terms.

Answer:

Question 28.

An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. [Use √3 = 1.732]

Answer:

Let CD be height of an aeroplane flying above the ground and AB be the two banks of the river in opposite directions.

In rt. ΔADC, tan 45° = \(\frac{\mathrm{CD}}{\mathrm{AD}}\)

1 = \(\frac{300}{\mathrm{AD}}\) ⇒ AD = 300 m

In rt. ΔCDB, tan 60° = \(\frac{\mathrm{CD}}{\mathrm{DB}}\)

√3 = \(\frac{300}{\mathrm{DB}}\) ⇒ DB = \(\frac{300}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\) = 100√3

= 100 (1.732) = 173.2 m ,..[∵ √3 = 1.732

∴ The width of the river, AB = AD + DB

= 300 + 173.2

= 473.2 m

Question 29.

Two different dice are thrown together. Find the probability that the numbers obtained have

(i) even sum, and

(ii) even product.

Answer:

Two dice can be thrown as 6 × 6 = 36 ways.

P(even sum) = \(\frac{18}{36}\) = \(\frac{1}{2}\)

i.e., 27 wayes.

P(even sum) = \(\frac{27}{36}\) = \(\frac{3}{4}\)

Question 30.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.