# CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions

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## CBSE Class 10 Maths Question Paper 2017 (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 – 1 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.

If the sum of the zeroes of the polynomial p(x) = (k^{2} – 14)x^{2} – 2x – 12 is 1, then find the value of k.

Answer:

p(x) = (k^{2} – 14)x^{2} – 2x – 12

Here a = k^{2} – 14, b = -2, c = -12

Sum of the zeroes, (α + β) = 1 …[Given

⇒ \(\frac{-b}{a}\) = 1 ⇒ \(\frac{-(-2)}{k^2-14}\) = 1

⇒ k^{2} – 14 = 2 ⇒ k^{2} = 16 ∴ k = ±4

Question 2.

ΔABC ~ ΔDEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm. Find the perimeter of ΔDEF.

Answer:

∴ Peri(ΔDEF) = 30 cm

Question 3.

If tan α = √3 and tan β = \(\frac{1}{\sqrt{3}}\), 0 < α, β < 90° then find the value of cot (α + β).

Answer:

tan α = √3 = tan 60°; tan β = \(\frac{1}{\sqrt{3}}\) = tan 30°

∴ α = 60° and p = 30°

Hence, cot (a + β) = cot (60° + 30°) = cot 90° = 0

Question 4.

If sin θ – cos θ = 0, then find the value of sin^{4} θ + cos^{4} θ.

Answer:

sin θ – cos θ = 0 ⇒ sin θ = cos θ

\(\frac{\sin \theta}{\cos \theta}\) = 1 ⇒ tan θ = 1 ⇒ θ = 45°

Now, sin^{4} θ + cos^{4} θ

= sin^{4} 45° + cos^{4} 45°

= (\(\frac{1}{\sqrt{2}}\))^{4} + (\(\frac{1}{\sqrt{2}}\))^{4} = \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

Question 5.

Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?

Answer:

Volume of hemisphere = Surface area of hemisphere …[Given

\(\frac{2}{3}\)πr^{3} = 3πr^{2} ⇒ \(\frac{2}{3}\) r = 3

r = \(\frac{9}{2}\)

∴ Diameter of hemisphere = 2r = 2(\(\frac{9}{2}\)) = 9 cm

Question 6.

A number is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1?

Answer:

(-3)^{2} = 9; (-2)^{2} = 4; (-1)^{2} = 1; (0)^{2} = 0

(1)^{2} = 1; (2)^{2} = 4; (3)^{2} = 9

∴ P(Sq. of nos. of ≤ 1) = \(\frac{3}{7}\)

Section B

Questions number 7 to 12 carry 2 marks each.

Question 7.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 8.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 9.

The mean of the following data is 18.75. Find the value of P:

Or

Question 9.

The distribution given below gives the daily income of 50 workers in a factory:

Convert the above distribution to a less than type cumulative frequency distribution.

Answer:

Gass marks (x_{i}) |
frequency (f_{i}) |
f_{i}x_{i} |

10 | 5 | 50 |

5 | 10 | 50 |

p | 7 | 7P |

25 | 8 | 200 |

30 | 2 | 60 |

Σf_{i} = 32 |
Σf_{i}x_{i} = 360 + 7P |

Now, Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = 18.75

\(\frac{18.75}{1}\) = \(\frac{360+7 P}{32}\) ⇒ 360 + 7p = 600

7P = 600 – 360 = 240

∴ P = \(\frac{240}{7}\) = 34.29 (approx.)

Or

Daily Wages (C.I.) | Less than type | No. of workers (f) | No. of Workers (c.f.) |

200 – 220 | Less than 220 | 14 | 14 |

220 – 240 | Less than 240 | 12 | 26 |

240 – 260 | Less than 260 | 8 | 34 |

260 – 280 | Less them 280 | 6 | 40 |

280 – 300 | Less than 300 | 10 | 50 |

Question 10.

Find the roots of the quadratic equation √2x^{2} + 7x + 5√2 = 0.

Answer:

√2x^{2} + 7x + 5√2 = 0

⇒ √2x^{2} + 2x + 5x + 5√2 = 0

⇒ √2x(x + √2) + 5(x + √2) = 0

⇒ (x + √2)(√2x + 5) = 0

⇒ x + √2 =0 or √2x + 5= 0

⇒ x = – √2 or x = \(\frac{-5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{-5 \sqrt{2}}{2}\)

Question 11.

Find the value of k for which the equation x^{2} + k(2x + k – 1) + 2 = 0 has real and equal roots.

Answer:

x^{2} + k(2x + k – 1) + 2 = 0

x^{2} + 2kx + k^{2} – k + 2 = 0

Here a = 1, b = 2k, c = k^{2} – k + 2

D = 0 …[real & equal roots

∴ b^{2} – 4ac = 0

⇒ (2k)^{2} – 4 x 1(k^{2} – k + 2) = 0

⇒ 4k^{2} – 4(k^{2} – k + 2) = 0

⇒ 4(k^{2} – k^{2} + k – 2) = 0

⇒ 4(k – 2) = 0

⇒ k – 2 = 0 ⇒ k = 2

Question 12.

In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.

Answer:

To find: PC + PD

= (PA – AC) + (PB – BD)

= (12 – 3) + (12 – 3) …[From (i), (ii) & (iii)

= 9 + 9 = 18 cm

Section C

Questions number 13 to 22 carry 3 marks each.

Question 13.

Verify whether 2, 3 and \(\frac{1}{2}\) are the zeroes of the polynomial p(x) = 2x^{2} – 11x^{2} + 17x – 6.

Or

Question 13.

Find the zeroes of the quadratic polynomial f(x) = x^{2} – 3x – 28 and verify the relationship between the zeroes and the co-efficients of the polynomial.

Answer:

p(x) = 2x^{3} – 11x^{2} + 17x – 6

When x = 2,

p(2) = 2(2)^{3} – 11(2)^{2} + 17(2) – 6

= 16 – 44 + 34 – 6 = 0

When x = 3,

p(3) = 2(3)^{3} – 11(3)^{2} + 17(3) – 6

= 54 – 99 + 51 – 6 = 0

When x = \(\frac{1}{2}\),

p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\))^{3} – 11(\(\frac{1}{2}\))^{2} + 17(\(\frac{1}{2}\)) – 6

= \(\frac{2}{8}\) – \(\frac{11}{4}\) + \(\frac{17}{2}\) – 6

= \(\frac{1}{4}\) – \(\frac{11}{4}\) + \(\frac{17}{2}\) – 6

= \(\frac{1-11+34-24}{4}\) = \(\frac{0}{4}\) = 0

Yes, x = 2, 3 and \(\frac{1}{2}\) all are the zeroes of the given polynomial.

Or

f(x) = x^{2} – 3x – 28

= x^{2} – 7x + 4x – 28

= x(x – 7) + 4(x – 7) = (x – 7) (x + 4)

Zeroes are:

x – 7 = 0 or x + 4 = 0

x = 7 or x = -4

In the given quadratic polynomial,

a = 1, b = -3, c = -28

Sum of zeroes,

= 7 + (-4) = 3 = \(\frac{-(-3)}{1}=\frac{-b}{a}\)

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

Product of zeroes,

= 7(-4) = \(\frac{-28}{1}\) = \(\frac{c}{a}\) = \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

⇒ Relaticinship holds.

Question 14.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 15.

Solve for x and y: \(\frac{10}{x+y}+\frac{2}{x-y}\) = 4; \(\frac{15}{x+y}-\frac{5}{x-y}\) = -2; x + y ≠ 0 x – y ≠ 0

Or

Question 15.

Solve for x and y: \(\frac{x}{a}\) = \(\frac{y}{b}\); ax + by = a^{2} + b^{2}

Answer:

\(\frac{10}{x+y}+\frac{2}{x-y}\) = 4 ………. (i)

\(\frac{15}{x+y}-\frac{5}{x-y}\) = -2 ………. (ii)

Multiplying (i) by 5 and (ii) by 2,

Putting the value of x in (iii), we get

3 + y = 5 y = 5 – 3 = 2

∴ x = 3, y = 2

Or

\(\frac{x}{a}\) – \(\frac{y}{b}\) = 0 ⇒ \(\frac{b x-a y}{a b}\) = \(\frac{0}{1}\)

bx – ay = 0 ………… (i)

ax + by = a^{2} + b^{2} ………… (ii)

Multiplying (i) by b and (ii) by a,

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

b(a) – ay = 0 ⇒ ba = ay

\(\frac{b a}{a}\) = y ∴ b = y

∴ x = a, y = b

Question 16.

In the figure, ΔAABC is right-angled at C and DE ⊥ AB. Prove that ΔABC ~ ΔADE and hence find the lengths of AE and DE.

Or

Question 16.

ΔABC ~ ΔPQR. AD is th median to BC and PM is the median to QR. Prove that \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\).

Answer:

Given: ΔABC is rt. angled at C

and DE⊥AB.

AD = 3 cm

DC = 2 cm,

BC = 12 cm

To prove:

(i) ΔABC ~ ΔADE

(ii) AE = ? & DE = ?

Proof: (i) In ΔABC and ΔADE,

∠ACB = ∠AED …[Each 90°

∠BAC = ∠DAE …[Common

∴ ΔABC ~ ΔADE …[AA Similarity Criterion

Or

ΔABC ~ ΔPQR …..[Given

∠1 = ∠2

…[In ~ Δs corresponding angles are equal

\(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BC}}{\mathrm{QR}}\) = \(\frac{\mathrm{AC}}{\mathrm{PR}}\) ….[In ~ Δs corresponding sides are proportional

In ΔABD and ΔPQM,

\(\frac{A B}{P Q}=\frac{2 B D}{2 Q M}\) …[∵ AD and PM are medians

⇒ \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BD}}{\mathrm{QM}}\) =

∠1 = ∠2 …[∵ ΔABC ~ ΔPQR (Given)

∴ ΔABD ~ ΔPQM …[SAS

Hence \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\)

…[In ~ Δs corresponding sides are proportional

Question 17.

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

Answer:

Question 18.

Find the sum of n terms of the series (4 – \(\frac{1}{n}\)) + (4 – \(\frac{2}{n}\)) + (4 – \(\frac{3}{n}\)) + ………. .

Answer:

Question 19.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 20.

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

Answer:

We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact (see Figure).

We need to prove that:

∠PTQ = 2∠OPQ

Let ∠PTQ = θ

Now, TP = TQ. …[∵ Lengths of tangents drawn from an external pt. to a circle are equal

So, TPQ is an isosceles triangle.

∴ ∠TPQ = ∠TQP = \(\frac{1}{2}\) (180° – θ) = 90° – \(\frac{1}{2}\) θ …………. (i)

Also, ∠OPT = 90° ……..[∵ Tangent at any point of a circle is ⊥ to the radius through the pt. of contact

So,∠OPQ = ∠OPT – ∠TPQ = 90° – (90° – \(\frac{1}{2}\)θ)

∠OPQ = \(\frac{1}{2}\)θ ⇒ ∠OPQ =\(\frac{1}{2}\) ∠OPQ

∴∠PTQ = 2∠OPQ ….. (Proved)

Question 21.

Not in Current Syllabus

Answer:

Not in Current Syllabus

Question 22.

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

(i) have a sum less than 6

(ii) have a product less than 16

(iii) is a doublet of odd numbers.

Answer:

Two dice can be thrown in (6 × 6) = 36 ways

(i) Sum less than ‘6’ are

(1, 1) (1, 2) (1, 3) (1, 4), (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (4, 1) i.e., 10 ways

∴ p(sum < 6)= \(\frac{10}{36}\) = \(\frac{5}{18}\)

(ii) Product less than ’16’ are

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5)

(4, 1) (4, 2) (4, 3)

(5, 1) (5, 2) (5, 3)

(6.1) (6, 2), i.e., 25 ways

∴ P(product less than 16) = \(\frac{25}{36}\)

(iii) (1, 1) (3, 3) (5, 5) i.e., 3 ways

∴ P(doublet of odd nos.) = \(\frac{3}{36}\) =\(\frac{1}{12}\)

Section D

Questions number 23 to 30 carry 4 marks each.

Question 23.

Prove that 3 + 2√5 is irrational.

Answer:

Let us assume, to the contrary, that 3 + 2√5 is rational.

So that we can find integers a and b(b ≠ 0),

such that 3 + 2√5 = \(\frac{a}{b}\),

where a and b are coprime

Rearranging this equation, we get

√5 = \(\frac{a-3 b}{2 b}\)

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

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

But this contradicts the fact that √5 is irrational.

So we conclude that 3 + 2√5 is irrational.

Question 24.

A man travels 300 km partly by train and partly by car. He takes 4 hours if the travels 60 km by train and the rest by car. If he travels 100 km by train and the remaining by car, he takes 10 minutes longer. Find the speeds of the train and the car separately.

Answer:

Let the speed of the train = x km/hr

Let the speed of the car = y km/ hr

According to the question,

Putting the value of x in (i),

\(\frac{60}{60}\) + \(\frac{240}{y}\) = 4 ⇒ \(\frac{240}{y}\) = 4 – 1

3y = 240 ⇒ y = 80

∴ Speed of the train = 60 km/hr

and Speed of the car = 80 km/hr

Question 25.

The median of the following data is 525. Find x and y if the sum of all frequencies is 100:

Answer:

∴ 68 + x + y = 100

x + y = 100 – 68 = 32

y = 32- x …….(i) Here, \(\frac{n}{2}\) = \(\frac{100}{2}\) = 50

Median = 525

Median class is 500 – 600.

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

525 = 500 + \(\frac{50-(33+x)}{20}\) × 100

⇒ 525 – 500 = (50 – 33 – x)5

⇒ \(\frac{25}{5}\) = 17 – x

⇒ 5 = 17 – x ∴ x = 17 – 5 = 12

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

y = 32 – 12 = 20

∴ x = 12, y = 20

Question 26.

Find the value of cos 60° geometrically. Hence find cosec 60°.

Or

Question 26.

If \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 and \(\frac{x}{a}\) sin θ – \(\frac{y}{b}\) cos θ = 1, prove that \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 2

Answer:

Let ΔABC be an equilateral Δ.

Let each side be 2a.

Since each angle in an equilateral Δ is 60°.

∴ ∠A = ∠B = ∠C = 60°

Draw AD ⊥ BC

In ΔADB and ΔADC,

AB = AC …[Each = 2 a

AD = AD …[Common

∠1 = ∠2 …[Each 90°

∴ ΔADB = ΔADC …[RHS congruency rule

BD = DC = \(\frac{2 a}{2}\) = a

Or

\(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 ………. (i)

\(\frac{x}{a}\) sin θ – \(\frac{y}{b}\) cos θ = 1

Squaring and adding (i) and (ii)

(\(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ)^{2} + (\(\frac{x}{a}\) sin θ – \(\frac{y}{b}\) cos θ)^{2} = 1^{2} + 1^{2}

\(\frac{x^2}{a^2}\) cos^{2} θ + \(\frac{y^2}{b^2}\) sin^{2} θ + 2(\(\frac{x}{a}\) cos θ)(\(\frac{y}{b}\) sin θ) + (\(\frac{x^2}{a^2}\) sin^{2} θ) + (\(\frac{y^2}{a^2}\) cos^{2} θ)

– 2 (\(\frac{x}{a}\) sin θ)(\(\frac{y}{b}\) cos θ) = 1 + 1

L.H.S = (\(\frac{x^2}{a^2}\) + (\(\frac{y^2}{b^2}\)

= (sin θ + cos θ)^{2} + (sin θ – cos θ)^{2} …… [From (iii) and (iv)

= sin^{2} θ + cos^{2} θ + 2 sin θ cos θ + sin^{2} θ + cos^{2} θ – sin θ cos θ

= 1 + 1 = 2 = R.H.S.

Question 27.

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Answer:

Given: PT and PS are tangents from an external point P to the circle with centre O.

To prove: PT = PS

Const.: Join O to P, T and S.

Proof: In ΔOTP and ΔOSP

OT = OS …[radii of same circle

OP = OP …[common

∠OTP = ∠OSP…[each 90°

∴ ΔOTP ☐ ΔOSP …[R.H.S.

∴ PT = PS …[c.p.c.t.

Question 28.

Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.

Answer:

Let the speed of the stream = x km/ hr

Speed of the boat in still water = 15 km/hr

then, the speed of the boat upstream = (15 – x) km/hr

and the speed of the boat downstream = (15 + x) km/hr

According to question: \(\frac{30}{15-x}+\frac{30}{15+x}\) = 4\(\frac{1}{2}\) hrs.

\(\frac{30[15+x+15-x]}{(15-x)(15+x)}\) = \(\frac{9}{2}\) …[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)

⇒ \(\frac{30 \times 30}{225-x^2}\) = \(\frac{9}{2}\)

⇒ 1800 = 9(225 – x^{2}) x^{2} ⇒ 200 = 225 – x^{2}

⇒ x^{2} = 225 – 200 ⇒ x^{2} = 25

⇒ x = 5 km/hr …….[∵ Speed cannot be -ve.

∴ Speed of stream = 5 km/hr

Question 29.

The angle of elevation of a cloud from a point 60 m above the surface of the water of a lake is 30° and the angle of depression of its shadow in water of lake is 60°. Find the height of the cloud from the surface of water.

Answer:

Let the height of the cloud from the surface of lake EC = H m = EF (image in water)

and A be the point, the distance between A and perpendicular of cloud AD = x m

In ΔACD, tan 30° = \(\frac{\mathrm{DC}}{\mathrm{AD}}\)

\(\frac{1}{\sqrt{3}}=\frac{H-60}{x}\)

x = √3H – 60√3 ……… (i)

In rt. ΔADF,

tan 60° = \(\frac{\mathrm{DF}}{\mathrm{AD}}\)

√3 = \(\frac{\mathrm{H}+60}{x}\)

x = \(\frac{H+60}{\sqrt{3}}\) ……… (ii)

Comparing (i) and (ii), we have H + 60

√3H – 60√3 = \(\frac{H+60}{\sqrt{3}}\)

⇒ 3H – 180 = H + 60

⇒ 3H – H = 60 + 180 ⇒ 2H = 240

⇒ H = 120

∴ Height of the cloud = 120 m

Question 30.

In the figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where O and O’ are centres of the circles. Find the area of shaded region.

Answer:

Side = 28 cm, Radius = \(\frac{28}{2}\) cm 14 cm

The area of the shaded region = Area of square + \(\frac{3}{4}\)(Ar. of circle) + \(\frac{3}{4}\)(Ar. of circle)

= Area of square + \(\frac{3}{2}\) (Area of circle) …[Area of square = (Side)^{2}; Area of circle = πr^{2}

= (28)^{2} + \(\frac{3}{2}\) × \(\frac{22}{7}\) × 14 × 14

= 784 + 924 = 1708 cm^{2}