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

Students can use CBSE Previous Year Question Papers Class 10 Maths with Solutions and CBSE Class 10 Maths Question Paper 2017 (Delhi) to familiarize themselves with the exam format and marking scheme.

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:

  1. All questions are compulsory.
  2. This question paper contains 30 questions.
  3. Question No. 1 – 6 in Section A are very short answer type questions carrying 1 mark each.
  4. Question No. 7 – 12 in Section B are short answer type questions carrying. 2 marks each.
  5. Question No. 13 – 22 in Section C are long answer – 1 type questions carrying 3 marks each.
  6. 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) = (k2 – 14)x2 – 2x – 12 is 1, then find the value of k.
Answer:
p(x) = (k2 – 14)x2 – 2x – 12
Here a = k2 – 14, b = -2, c = -12
Sum of the zeroes, (α + β) = 1 …[Given
⇒ \(\frac{-b}{a}\) = 1 ⇒ \(\frac{-(-2)}{k^2-14}\) = 1
⇒ k2 – 14 = 2 ⇒ k2 = 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:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 1
∴ 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 sin4 θ + cos4 θ.
Answer:
sin θ – cos θ = 0 ⇒ sin θ = cos θ
\(\frac{\sin \theta}{\cos \theta}\) = 1 ⇒ tan θ = 1 ⇒ θ = 45°
Now, sin4 θ + cos4 θ
= sin4 45° + cos4 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
\(\frac{2}{3}\)πr3 = 3πr2 ⇒ \(\frac{2}{3}\) 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:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 2
Or
Question 9.
The distribution given below gives the daily income of 50 workers in a factory:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 3
Convert the above distribution to a less than type cumulative frequency distribution.
Answer:

Gass marks (xi) frequency (fi) fixi
10 5 50
5 10 50
p 7 7P
25 8 200
30 2 60
Σfi = 32 Σfixi = 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 √2x2 + 7x + 5√2 = 0.
Answer:
√2x2 + 7x + 5√2 = 0
⇒ √2x2 + 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 x2 + k(2x + k – 1) + 2 = 0 has real and equal roots.
Answer:
x2 + k(2x + k – 1) + 2 = 0
x2 + 2kx + k2 – k + 2 = 0
Here a = 1, b = 2k, c = k2 – k + 2
D = 0 …[real & equal roots
∴ b2 – 4ac = 0
⇒ (2k)2 – 4 x 1(k2 – k + 2) = 0
⇒ 4k2 – 4(k2 – k + 2) = 0
⇒ 4(k2 – k2 + 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.
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 4
Answer:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 5
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) = 2x2 – 11x2 + 17x – 6.
Or
Question 13.
Find the zeroes of the quadratic polynomial f(x) = x2 – 3x – 28 and verify the relationship between the zeroes and the co-efficients of the polynomial.
Answer:
p(x) = 2x3 – 11x2 + 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) = x2 – 3x – 28
= x2 – 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 = a2 + b2
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,
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 6
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 = a2 + b2 ………… (ii)
Multiplying (i) by b and (ii) by a,
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 7
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.
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 8
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
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 9
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
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 10
Or
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 11
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:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 12

Question 18.
Find the sum of n terms of the series (4 – \(\frac{1}{n}\)) + (4 – \(\frac{2}{n}\)) + (4 – \(\frac{3}{n}\)) + ………. .
Answer:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 13

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:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 14
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
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 15
(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,
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 16
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:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 17
Answer:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 18
∴ 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:
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 19
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
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 20
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 = 12 + 12
\(\frac{x^2}{a^2}\) cos2 θ + \(\frac{y^2}{b^2}\) sin2 θ + 2(\(\frac{x}{a}\) cos θ)(\(\frac{y}{b}\) sin θ) + (\(\frac{x^2}{a^2}\) sin2 θ) + (\(\frac{y^2}{a^2}\) cos2 θ)
– 2 (\(\frac{x}{a}\) sin θ)(\(\frac{y}{b}\) cos θ) = 1 + 1
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 21
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)
= sin2 θ + cos2 θ + 2 sin θ cos θ + sin2 θ + cos2 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.
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 22
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 – x2) x2 ⇒ 200 = 225 – x2
⇒ x2 = 225 – 200 ⇒ x2 = 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
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 23
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.
CBSE Class 10 Maths Question Paper 2017 (Delhi) with Solutions 24
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 = πr2
= (28)2 + \(\frac{3}{2}\) × \(\frac{22}{7}\) × 14 × 14
= 784 + 924 = 1708 cm2


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