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## CBSE Class 10 Maths Question Paper 2016 (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 of 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 Fig., the graph of a polynomial p(x) is shown.

Find the number of zeroes of p(x).

Answer:

The number of zeroes of p(x) is 1 as the graph intersect x-axis at one point.

Question 2.

In Fig., AD = 4 cm, BD = 3 cm and CB = 12 cm, the find the value of cot θ.

Answer:

In rt. ΔADB,

AB^{2} = AD^{2} + BD^{2} ….[Pythagoras’ theorem

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

AB^{2} = 16 + 9 ⇒ AB = 5 cm

∴ cot θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{12}{5}\)

Question 3.

In Fig., PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA.

Answer:

∠ACB = 90° …[Angle in the semi-circle

In ΔABC, ∠CAB + ∠ACB + ∠CBA = 180°

30° + 90° + ∠CBA = 180° ….[Angle-sum property of Δ

∠CBA = 180° – 30° – 90°

= 60°

∠PCA = ∠CBA …[Angle in the alte-rnate segment

∴ ∠PCA = 60°

Question 4.

For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P.?

Answer:

As we know, a_{2} – a_{1} = a_{3} – a_{2}

2k – 1 – (k + 9) = 2k + 7 – (2k – 1)

2k – 1 – k – 9 = 2k + 7 – 2k + 1

k – 10 = 8

∴ k = 8 + 10 = 18

Question 5.

A ladder, leaning against a wall, makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.

Answer:

Let AC be the ladder.

cos 60° = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)

\(\frac{1}{2}\) = \(\frac{2.5}{\mathrm{AC}}\)

∴ Length of ladder, AC = 5 m

Question 6.

A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.

Answer:

Total Sample space, S = 52

P (neither a red card nor a queen) = 1 – P(red card or a queen)

= 1 – (\(\frac{26+4-2}{52}\)) …..[Red cards = 26 Queen = 4 Red queen = 2

= 1 – \(\frac{28}{52}\) = \(\frac{24}{52}\) = \(\frac{6}{13}\)

Section B

Questions number 7 to 12 carry 2 marks each.

Question 7.

Is 7 × 5 × 3 × 2 + 3a composite number? Justify your answer.

Answer:

7 × 5 × 3 × 2 + 3

= 3 (7 × 5 × 2 + 1) = 3 × 71 ………(i)

By Fundamental theorem of Arithmetic, every composite number can be expressed as product of primes in a unique way, apart from the order of factions.

∴ (i) is a composite number.

Question 8.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 9.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 10.

If -5 is a root of the quadratic equation 2x^{2} + px – 15 = 0 and the quadratic equation p(x^{2} + x) + k = 0 has equal roots, find the value of k.

Answer:

2x^{2} + px – 15 = 0

Since (-5) is a root of the given equation

∴ 2(-5)^{2} + p(-5) – 15 = 0

⇒ 2(25) – 5p – 15 = 0

⇒ 50 – 15 = 5p = 0

⇒ 35 = 5p ⇒ p = 7 ……(i)

p(x^{2} + x) + k ⇒ px^{2} + px + k = 0

∴ 7x^{2} + 7x + k = 0 …[From (i)

Here, a – 7, b = 7, c = k

D = 0 …[Roots are equal

b^{2} – 4ac = 0

⇒ (7)^{2} – 4(7)k = 0 …[From (i)

⇒ 49 – 28k = 0 ⇒ 49 = 28k

∴ k = \(\frac{49}{28}\) = \(\frac{7}{4}\)

Question 11.

Let P and Q be the points of trisection of the line segment joining the points A (2, -2) and B (-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

Answer:

AP : PB = 1 : 2

By using section formula, coordinates of P are:

(\(\frac{1(-7)+2(2)}{1+2}\), \(\frac{1(4)+2(-2)}{1+2}\))

⇒ (\(\frac{-7+4}{3}\), \(\frac{4-4}{3}\)) = (-1, 0)

We know, Q is the mid-point of PB

Coordinates of Q are (\(\frac{-1+(-7)}{2}\), \(\frac{0+4}{2}\))

⇒ (\(\frac{-8}{2}\), \(\frac{4}{2}\)) = (-4, 2)

Question 12.

In Fig., a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively.

Prove that: AB + CD = BC + DA.

Or

Question 12.

In Fig., from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°.

Answer:

AP = AS (i)

BP = BQ (ii) …[Tangents drawn from an external point are equal in length

CR = CQ (iii)

DR = DS (iv)

By adding (i) to (iv)

(AP + BP) + (CR + DR) = AS + BQ + CQ + DS

AB + CD = (BQ + CQ) + (AS + DS)

∴ AB + CD = BC + AD (Hence proved)

Or

Given: OP = 2r

Let ∠TOP = θ

∠OTP = 90° ….[∵ Tangent is ⊥ to the radius through the point of contact

Let OT = OS = r

In rt. ΔOTP, cos θ = \(\frac{\mathrm{OT}}{\mathrm{OP}}\)

⇒ cos θ = \(\frac{r}{2 r}\) ⇒ cos θ = \(\frac{4}{2}\)

⇒ cos θ = cos 60°

⇒ θ = 60°

∴ ∠TOS = 60° + 60° = 120°

In ΔTOS,

∠OTS = ∠OST …(i)

…[Angles opposite to equal sides

In ΔTOS, ∠TOS + ∠OTS + ∠OST = 180°

…[Angle-sum-property of a Δ

120° + ∠OTS + ∠OTS = 180° …[From (i)

2∠OTS = 180° – 120°

∠OTS = 60°/2 = 30°

From (i), ∠OTS = ∠OST = 30° (Hence proved)

Section C

Questions number 13 to 22 carry 3 marks each.

Question 13.

Not in Current Syllabus

Answer:

Not in Current Syllabus.

Question 14.

A person can row a boat at the rate of 5 km/hour in still water. He takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.

Answer:

Let the speed of stream be x km/ hour

∴ Speed of the boat rowing upstream = (5 – x) km/hour

Speed of boat rowing downstream = (5 + x) km/hour

A.T.Q.

⇒ 3 × \(\frac{40}{5+x}\) = \(\frac{40}{5-x}\) [∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)

⇒ \(\frac{120}{5+x}\) = \(\frac{40}{5-x}\)

⇒ 600 – 120x = 200 + 40x

⇒ -160x = -400 ⇒ x = \(\frac{-400}{-160}\) = 2.5

∴ Speed of the stream = 2.5 km/hour

Question 15.

(a) If α, β are zeroes of the polynomial x^{2} – 2x – 15, then form a quadratic polynomial whose zeroes are (2α) and (2β ).

(b) If α and β are the zeroes of x^{2} – 2x – 1, form a quadratic polynomial whose zeroes are 2α – 1, 2β – 1

(c) If α and β are the zeroes of x^{2} – x – 2, form a quadratic polynomial whose zeroes are 2α + 1, 2β + 1

Answer:

(a) p(x) = x^{2} – 2x – 15 ……. (i)

As α, β are zeroes of (i)

∴ α + β = 2 and aβ = -15

For the given zeroes (2α) and (2β),

Sum of zeroes = 2α + 2β = 2(α + β) = 4

Product of zeroes = 4(αβ) = 4(-15) = -60

∴ The required polynomial is x^{2} – 4x – 60.

(b) p(x) = x^{2} – 2x – 1 …… (i)

As, α, β are zeroes of (i)

∴ α + β = \(\frac{-(-b)}{a}\) = \(\frac{1}{2}\);

αβ = \(\frac{c}{a}\) = \(\frac{-1}{1}\)

For the given zeroes (2α – 1), (2β – 1)

Sum of zeroes = 2α + 2β – 2

= 2(α + β – 1)

= 2(2 – 1) = 2(1) = 2

Product of zeroes = 4αβ – 2α – 2β + 1

= 4αβ – 2(α + β) + 1

= 4(-1) – 2(2) + 1 = -7

∴ The required polynomial is x^{2} – 2x – 7.

(c) p(x) = x^{2} – x – 2 …….(i)

As, α, β are zeroes of (i)

∴ α + β = \(\frac{-(-b)}{a}\) = 1; αβ = \(\frac{c}{a}\) = -2

For the given zeroes (2α + 1), (2β + 1)

Sum of zeroes = 2α + 2β + 2

= 2(α + β + 1)

= 2(1 + 1) = 4

Product of zeroes = 4αβ + 2α + 2β + 1

= 4αβ + 2(α + β) + 1

= 4(-2) + 2(1) + 1 = -5

∴ Required polynomial is x^{2} – 4x – 5.

Question 16.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 17.

Find the median of the following data:

Answer:

Here, = \(\frac{N}{2}\) = \(\frac{53}{2}\) 26.5; Median class is 60 – 70.

∴ l = 60, c.f = 22, f = 7, h = 10

Median = l + (\(\frac{\frac{n}{2}-c . f .}{f}\)) × h

= 60 + (\(\frac{26.5-22}{7}\)) × 10

= 60 + \(\frac{4.5}{7}\) × 10 = 60 + \(\frac{45}{7}\) = 66.43

Question 18.

If the point P(x, y) is equidistant from the points A (a + b, b – a) and B(a – b, a + b), prove that bx = ay.

Answer:

PA = PB …[Given

PA^{2} = PB2 …[Squaring both sides

⇒ [(a + b) – x]^{2} + [(b – a) – y]^{2} = [(a – b) – x]^{2} + [(a + b) – y]^{2}

⇒ (a + b)^{2} + x^{2} – 2(a + b)x + (b – a)^{2} + y^{2}

– 2(b – a)y = (a – b)^{2} + x^{2} – 2(a – b)x + (a + b)^{2} + y^{2} – 2(a + b)y …[∵ (a – b)^{2} = (b – a)^{2}

⇒ -2(a + b)x + 2 (a – b)x = -2 (a + b)y + 2 (b – a)y

⇒ 2x(- a – b + a – b) = 2y(- a – b + b – a)

⇒ -2bx = – 2ay ⇒ bx = ay (Hence proved)

Question 19.

If the ratio of the sum of first n terms of two A.P’s is (7n + 1) : (4n + 27), find the ratio of their m^{th} terms.

Answer:

Let A be first term and D be the common difference of 1^{st} A.P.

Let a be the first term and d be the common difference of 2^{nd} A.P.

Question 20.

Solve for x: \(\frac{1}{(x-1)(x-2)}\) + \(\frac{1}{(x-2)(x-3)}\) = \(\frac{2}{3}\) x ≠ 1, 2, 3

Answer:

⇒ (x – 1)(x – 3) = 3 ⇒ x^{2} – 3x – x + 3 – 3 = 0

⇒ x^{2} – 4x = 0 ⇒ x(x – 4) = 0

⇒ x = 0 or x – 4 = 0

∴ x = 0 or x = 4

Question 21.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 22.

A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.

Answer:

x can be any one of 1, 2, 3, 4 i.e., 4 ways

y can be any one of 1, 4, 9,16 i.e., 4 ways

Total no. of cases of xy = 4 × 4 = 16 ways

No. of cases, where product is less than 16

= (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3,1), (3, 4), (4, 1) i.e., 8 ways

i.e.,

1 × 1 = 1; 1 × 4 = 4; 1 × 9 = 9

2 × 1 = 2; 2 × 4 = 8; 3 × 1 = 3

3 × 4 = 12; 4 × 1 = 4

{1, 4, 9, 2, 8, 3, 12, 4}

∴ Required probability = \(\frac{8}{16}\) = \(\frac{1}{2}\)

Section D

Questions number 23 to 30 carry 4 marks each.

Question 23.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 24.

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

Or

Question 24.

If sec θ + tan θ = p, prove that sin θ = \(\frac{p^2-1}{p^2+1}\)

Answer:

Or

Question 25.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 26.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 27.

In Fig, two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\).

Answer:

Given: Two equal circles, with centres O and O’, touch each other at point X. OO’ is produced to meet the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC.

To find = \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\)

Proof: ∠ACO = 90° …[Tangent is ⊥ to the radius through the point of contact

In ΔAO’D and ΔAOC

∠O’AD = ∠OAC …[Common

∴ ∠ADO = ∠ACO ….[Each 90°

∴ Δ∠AO’D ~ Δ∠AOC …..[AA similarity

\(\frac{\mathrm{AO}^{\prime}}{\mathrm{AO}}\) = \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\) …[In similar Δs, corresponding sides are proportional

\(\frac{r}{3 r}\) = \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\) …[Let AO’ = O’X = OX = r ⇒ AO = r + r + r = 3r

∴ \(\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}\) = \(\frac{1}{3}\)

Question 28.

Solve for x: \(\frac{1}{x+1}\) + \(\frac{2}{x+2}\) = \(\frac{4}{x+4}\) x ≠ -1, -2, -4

OR

Question 28.

A motor boat whose speed is 24 km/h in still water takes 1 hr more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream.

Answer:

\(\frac{1}{x+1}\) + \(\frac{2}{x+2}\) = \(\frac{4}{x+4}\)

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

⇒ \(\frac{3 x+4}{x^2+3 x+2}\) = \(\frac{4}{x+4}\)

⇒ 4(x^{2} + 3x + 2) = (3x + 4)(x + 4)

⇒ 4(x^{2} + 3x + 2) = 3x^{2} + 12x + 4x + 16

⇒ 4x^{2} + 12x + 8 – 3x^{2} – 12x – 4x – 16 = 0

⇒ x^{2} – 4x – 8 = 0

Here, a = 1, b = -4, c = -8

D = b^{2} – 4ac = (-4)^{2} – 4 × 1 × (-8)

= 16 + 32 = 48 ⇒ D > 0 (+ve)

∴ The given equation has real roots.

x = \(\frac{-b \pm \sqrt{D}}{2 a}\) ∴ x = \(\frac{-(-4) \pm \sqrt{48}}{2 \times 1}\)

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

⇒ x = 2(1 ± √3)

∴ Roots are 2(1 + √3) and 2(1 – √3).

Or

Let the speed of the stream = x km/hr

Speed of boat in still water = 24 km/hr

Speed of boat upstream = (24 – x) km/hr

Speed of boat downstream = (24 + x) km/hr

Distance = 32 km

According to the question,

\(\frac{32}{24-x}\) – \(\frac{32}{24+x}\) = 1 ….[ Time = \(\frac{\text { Distance }}{\text { Speed }}\)

⇒ \(\frac{32[24+x-(24-x)]}{(24-x)(24+x)}\) = 1 ⇒ \(\frac{32(2 x)}{576-x^2}\) = 1

⇒ 64x = 576 – x^{2} ⇒ x^{2} + 64x – 576 = 0

⇒ x^{2} + 72x – 8x – 576 = 0

⇒ x(x + 72) – 8(x + 72) = 0

⇒ (x – 8)(x + 72) = 0

⇒ x – 8 = 0 or x + 72 = 0

⇒ x = 8 or x = -72

But speed of stream cannot be negative.

Therefore, speed of stream = 8 km/hr.

Question 29.

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. (Use √3 = 1.73)

Answer:

Let QR = x m

QP(Height) = (40 + x) m

In rt. ΔQPX,

\(\frac{\mathrm{QP}}{\mathrm{XP}}\) = tan 60°

⇒ \(\frac{40+x}{\mathrm{XP}}\) = √3 ….. (i)

In rt. ΔQRY, \(\frac{\mathrm{QR}}{\mathrm{YR}}\) = tan 45°

Hence, height of the tower PQ is 94.64 m and the distance PX is 54.64 m.

Question 30.

In Fig., is shown a sector OAP of a circle with centre O, containing ∠θ. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is r [tan θ + sec θ + \(\frac{\pi \theta}{180}\) – 1]

Answer:

Length of arc \(\widehat{\mathrm{AP}}\) (minor)

= \(\frac{\theta}{360^{\circ}}\) × 2πr = \(\frac{\theta}{180^{\circ}}\) × πr ………. (i)

∴ ∠OAB = 90°

In rt. ΔOAB, ….[Tangent is ⊥ to the radius through the point of contact

tan θ = \(\frac{\mathrm{AB}}{\mathrm{OA}}\) ⇒ AB = OA . tan θ

∴ AB = r . tan θ ………. (ii)

In rt. ΔOAB,

sec θ = \(\frac{\mathrm{OB}}{\mathrm{OA}}\) ⇒ OB = OA . sec θ

OB = r . sec θ

∴PB = OB – OP = r sec θ – r …….. (iii)

The perimeter of shaded region

= \(\widehat{\mathrm{AP}}\) + AB + PB

= \(\frac{\theta}{180^{\circ}}\)πr + r tan θ + r sec θ – r …[From (i), (ii) & (iii)

= r (tan θ + r sec θ + \(\frac{\pi \theta}{180^{\circ}}\) – 1) (Hence proved)