CBSE Class 10 Maths Question Paper 2015 (Outside Delhi) with Solutions

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CBSE Class 10 Maths Question Paper 2015 (Outside 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 of 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 – I 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.
In the Fig., ΔABC, DE || BC, find the value of x.

Answer:
In ΔABC, DE || BC …[Given
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\) …[Thales’ theorem
\(\frac{x}{x+1}\) = \(\frac{x+3}{x+5}\)
x(x + 5) = (x + 3)(x + 1)
x² + 5x = x² + 3x + x + 3
x² + 5x – x² – 3x – x = 3
∴ x = 3 cm

Question 2.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Question 3.
In a continuous frequency distribution, the median of the data is 21. If each observation is increased by 5, then find the new median.
Answer:
New median = 21 + 5 = 26

Question 4.
If the quadratic equation px² – 2√5 px + 15 = 0 has two equal roots, then find the value of p.
Answer:
The given quadratic equation can be written as
px² – 2√ 5 px + 15 = 0
a = p, b = -2 √5p, c = 15
For equal roots, D = 0
D = b² – 4ac
0 = (-2√5p)² – 4 × p × 15
0 = 4 × 5p² – 60p
0 = 20p² – 60p ⇒ 20p² = 60p
P = \(\frac{60 p}{20 p}\) = 3 ∴ P = 3

Question 5.
Two different dices are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6.
Answer:
Total outcomes = 6ⁿ = 6² = 36
Possible outcomes having the product of the two numbers on the top of the dice as 6 are (3 × 2, 2 × 3, 6 × 1, 1 × 6), i.e, 4.
∴ P(Product of two numbers is 6) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

Question 6.
In Fig., PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ.

Answer:

PQ is the chord of the circle & PT is tangent.
∴ ∠OPT = 90° [∵ Tangent to a circle is ⊥ to the radius through the point of contact
Now, ∠QPT = 60°…[Given
∠OPQ = ∠OPT – ∠QPT
∠OPQ = 90° – 60° = 30°
In ∆OPQ,
OP = OQ …[radii
∠OQP = ∠OPQ = 30° …[In a ∆, equal sides have equal ∠s opp. them
In ∆OPQ, ∠OQP + ∠OPQ + ∠POQ = 180° …[Angle Sum property of ∆
∴ ∠POQ = 120° …[∠POQ = 180° – (30° + 30°)
⇒ Reflex ∠POQ = 360° – 120° = 240°
∴ Reflex ∠POQ = 2∠PRQ …..[We know that the angle substended by an arc at the centre of a circle is twice the angle subtended by it at any point on the remaining part of the circle
⇒ 240° = 2∠PRQ
∠PRQ = \(\frac{240^{\circ}}{2}\) = 120°

Section B

Questions number 7 to 12 carry 2 marks each.

Question 7.
Check whether 4ⁿ can end with the digit 0 for any natural number n.
Or
Question 7.
Find the largest number which divides 70 and 125 leaving remainder 5 and 8 respectively.
Answer:
4ⁿ = (2²)ⁿ = 2²ⁿ
The only prime in the factorization of 4ⁿ is 2.
There is no other primes in the factorization of 4ⁿ = 2²ⁿ (By uniqueness of the fundamental Theorem of Arithmetic).
5 does not occur in the prime factorization of 4ⁿ for any n.
∴ 4ⁿ does not end with the digit zero for any natural number n.
Or
It is given that on dividing 70 by the required number, there is a remainder 5. This means that 70 – 5 = 65 is exactly divisible by the required number.
Similarly, 125 – 8 = 117 is also exactly divisible by required number.

65 = 5 × 13
117 = 32 × 13
∴ HCF = 13
∴ Required number = 13

Question 8.
A 6.5 m long ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall. Find the height of the wall where the top of the ladder touches it.
Answer:

Let AC be the ladder and AB be the wall.
AC = 6.5 m = \(\frac{13}{2}\)m
BC = 2.5m = \(\frac{5}{2}\)m
In rt. ΔABC, AB² + BC² = AC² …[Pythagoras theorem
AB² + (\(\frac{5}{2}\))² = (\(\frac{13}{2}\))²
AB² = \(\frac{169}{4}\) – \(\frac{25}{4}\)
= \(\frac{169-25}{4}\) = \(\frac{144}{4}\) = 36
∴ Required height, AB = 6 m

Question 9.
If x = a cos θ – b sin θ and y = a sin θ + b cos θ, then prove that a² + b² = x² + y².
Answer:
RHS = x² + y²
= (a cos θ – b sin θ)² + (a sin θ + b cos θ)²
= a² cos² θ + b² sin² θ – 2ab cos θ sin θ + a²2 sin² θ + b² cos² θ + 2ab sin θ cos θ
= a² (cos² θ + sin² θ) + b² (sin² θ + cos² θ)
= a² + b² = LHS …[∵ cos² θ + sin² θ = 1

Question 10.
In Fig., two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ 120°, then prove that OR = PR + RQ.

Or
Question 10.
In Fig., a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are
respectively of lengths 6 cm and 9 cm. If the area of ΔABC is 54 cm², then find the lengths of sides AB and AC.

Answer:

Const.: Join OP and OQ.
∠OPR = ∠OQR = 90° ….[Tangent ⊥ to the radius through the point of contact
PR = RQ …[Tangents drawn from an external point are equal
Given: ∠PRQ = 120°
∠PRO = \(\frac{1}{2}\)∠PRQ = \(\frac{1}{2}\) × 120° = 60°
Now, In ΔOPR,
⇒ ∠OPR + ∠POR + ∠PRO = 180° …[Δ Rule
⇒ 90° + POR + 60° = 180°
⇒ ∠POR + 150° = 180°
⇒ ∠POR = 30°
⇒ \(\frac{\mathrm{PR}}{\mathrm{OR}}\) = sin 30° = \(\frac{1}{2}\)
⇒ OR = 2PR
⇒ OR = PR + QR …[∵ PR = RQ …. Hence proved
Or
Given: OD = 3 cm
Construction: Join OA, OF, OE, OC, OD & OB.
Let AF = AE = x cm
BD = BF = 6 cm ….[Tangents drawn from an external point are equal
CD = CE = 9 cm
Let OF = OE = OD = 3 cm ….[Radii

∴ AB = AF + BF = x + 6 ………… (i)
AC = AE + CE = x + 9 ………… (ii)
BC = DB + CD = 6 + 9 = 15 cm …………. (iii)
In ΔABC, Area of ΔABC = 54 cm² …[Given
Ar(ΔAOB + ΔAOC + ΔBOC) = 54 …[Area of Δ = \(\frac{1}{2}\) × Base × height
⇒ \(\frac{1}{2}\) × AB × OF + \(\frac{1}{2}\) × AC × OE + \(\frac{1}{2}\) × BC × OD = 54
⇒ \(\frac{1}{2}\) × (6 + x) × 3 + \(\frac{1}{2}\) × (x + 9) × 3 + \(\frac{1}{2}\) × 15 × 3
⇒ \(\frac{1}{2}\) [x + 6 + x + 9 + 15] × 3 = 54 …[From (i), (ii) & (iii)
⇒ \(\frac{1}{2}\) [2x + 30] × 3 = 54
⇒ 6x + 90 = 108
⇒ 6x = 18 ⇒ x = 3
⇒ AB =x + 6 = 3 + 6 = 9 cm
⇒ AC = x + 9 = 3 + 9 = 12 cm
∴ AB = 9 cm, AC = 12 cm and BC = 15 cm

Question 11.
Solve the following quadratic equation for x:
4x² + 4bx – (a² – b²) = 0
Answer:
Given. 4x² + 4bx + b² – a² = 0
⇒ (2x + b)² – (a)² = 0 …[x² – y² = (x + y)(x – y)
⇒ (2x + b + a) (2x + b – a) = 0
⇒ (2x + b + a) = 0 or (2x + b – a) = 0
⇒ 2x = -(a + b) or 2x = (a – b)
⇒ x = –\(\frac{(a+b)}{2}\) or x = \(\frac{(a-b)}{2}\)
Hence solution for x = – \(\frac{(a+b)}{2}\), \(\frac{(a-b)}{2}\)

Question 12.
In an AP, if S₅ + S₇ = 167 and S₁₀ = 235, then find the AP, where Sn denotes the sum of its first n terms.
Answer:
Given: S₅ + S₇ = 167
⇒ \(\frac{5}{2}\)[2a + (5 – 1 )d] + \(\frac{7}{2}\) [2a + (7 – 1)d] = 167 ….[S_n = \(\frac{n}{2}\)(2a + (n – 1 )d
⇒ \(\frac{5}{2}\)[2a + 4d] + \(\frac{7}{2}\) [2a + 6d] = 167
⇒ 5(a + 2d) + 7(a + 3d) = 167
⇒ 5a + 10d + 7a + 21d = 167
⇒ 12a + 31d = 167 ……(i)
⇒ S₁₀ = \(\frac{10}{2}\) (2a + (10 – 1 )d) = 235
⇒ 5[2a + 9d] = 235
⇒ 10a + 45d = 235 …..(ii)
Solving (i) and (ii), a = 1, d = 5
∴ a₁
a₂ = a + d ⇒ a₂ = 1 + 5 = 6
a₃ = a + 2d ⇒ a₃ = 1 + 10 = 11
Hence A.P. is 1, 6,11…

Section C

Questions number 13 to 22 carry 3 marks each.

Question 13.
Not in Current Syllabus
Answer:
Not in Current Syllabus.

Question 14.
7x – 5y – 4 = 0 is given. Write another linear equation, so that the lines represented by the pair are:
(i) intersecting
(ii) coincident
(iii) parallel
Answer:
7x – 5y – 4 = 0
(i) 7x + 3y + 2 = 0 …[∵ Here, \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\)
(ii) 14x – 10y – 8 = 0… [∵ Here, \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
(iii) 7x – 5y + 3 = 0 … [∵ Here, \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)

Question 15.
State whether the given pairs of triangles are similar or not. In case of similarity mention the criterion.

Answer:
(a) \(\frac{\mathrm{AO}}{\mathrm{DO}}\) = \(\frac{16}{9}\) and \(\frac{\mathrm{BO}}{\mathrm{CO}}\) = \(\frac{9}{5}\)
\(\frac{\mathrm{AO}}{\mathrm{DO}}\) ≠ \(\frac{\mathrm{BO}}{\mathrm{CO}}\)
∴ Given Δs are not similar.

(b) In ΔPQR, ∠P + ∠Q + ∠R = 180° …[Angle-Sum Property of a Δ
45° + 78° + ∠R = 180°
∠R = 180° – 45° – 78° = 57°
In ΔLMN, ∠L + ∠M + ∠N = 180° …[Angle-Sum Property of a Δ
57° + 45° + ∠N = 180°
∠N = 180° – 57° – 45° = 78°
Here, ∠P = ∠M …[each = 45°
∠Q = ∠N …[each = 78°
∠R = ∠L …[each = 57°
∴ ΔPQR – ΔMNL …[By AAA similarity theorem

Question 16.
Prove that: \(\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}\) = tan θ
Answer:

Question 17.
Find the mean and median for the following data: :

Answer:

Mean (\(\overline{\mathbf{X}}\)) = \(\frac{\Sigma f_i X_i}{\Sigma f_i}\) = \(\frac{250}{25}\) = 10
Here, = \(\frac{n}{2}\) = \(\frac{25}{2}\) = 12.5
∴ Median class is 8 – 12.
Median = l + \(\frac{\frac{n}{2}-c . f}{f}\) × h …[Here l = 8, h = 4, cf = 8, f = 9
= 8 + \(\frac{12.5-8}{9}\) × 4 = 8 + \(\frac{4.5}{9}\) × 4
= 8 + 2 = 10

Question 18.
The 14^th term of an AP is twice its 8^th term. If its 6^th term is -8, then find the sum of its first 20 terms.
Answer:
Let a = First term, d = Common difference
a₁₄ = 2.a₈ …[Given
⇒ a + 13d = 2 (a + 7d) …[∵ a_n = a + (n – 1)d
⇒ a + 13 d = 2a + 14 d
⇒ 1a – 2a = 14d – 13d
⇒ -1a = d ⇒ a = -d ……(i)
a₆ = -8 …[Given
-8 = a + 5d ⇒ -d + 5d = -8 …[From (i)
4d = -8 ⇒ d = -2
Value of d put in equation (i), we get
a = -d ⇒ a = -(-2)
Now, a = 2, d = -2
Now, Sum of first 20 terms,
S₂₀ = \(\frac{20}{2}\)[2 × 2 + (20 – 1)(-2)]
S_n = \(\frac{n}{2}\)(2a + (n – 1)d
S₂₀ = 10[4 + 19(-2)]
S₂₀ = 10[4 – 381 = -340

Question 19.
The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p.
Or
Question 19.
If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = \(\frac{3}{7}\) AB, where P lies on the line segment AB.
Answer:

In right ΔABC, AB² + BC² = AC² …[Pythagoras’ theorem
⇒ (p – 4)² + (3 – 7)² + (7 – p)² + (3 – 3)² = (7 – 4)² + (3 – 7)² …[by using distance formula
⇒ (p – 4)² + (-4)² + (7 – p)² = (3)² + (-4)²
⇒ 16 + p² – 8p + 16 + 49 + p² – 14p = 9 + 16
⇒ 81 + 2p² – 22p = 25
⇒ 2p² – 22p + 56 = 0 or p² – 11p + 28 = 0
⇒ p² – 7p – 4p + 28 = 0
⇒ p(p – 7) – 4(p – 7) = 0
⇒ (p – 7) (p – 4) = 0
⇒ p – 7 = 0 or p – 4 = 0
⇒ P = 7 or p = 4
Since p ≠ 7
∴ p = 4
Or
AP = \(\frac{3}{7}\) AB ⇒ AP : PB = 3 : 4 (2,-4)

Question 20.
The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is \(\frac{1}{4}\). The probability of selecting a blue ball at random from the same jar is \(\frac{1}{3}\). If the jar contains 10 orange balls, find the total number of balls in the jar.
Answer:
P(Red) = \(\frac{1}{4}\), P (blue) = \(\frac{1}{3}\)
As we know, Total Probability = 1
⇒ P(orange) = 1 – \(\frac{1}{4}\) – \(\frac{1}{3}\) = \(\frac{5}{12}\)

⇒ \(\frac{5}{12}\) = \(\frac{10}{\text { Total no. of balls }}\)
∴ Total no. of balls = \(\frac{10 \times 12}{5}\) = 24

Question 21.
Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°.
Also find the area of the corresponding major segment. [Use π = 22/7]
Answer:

r = 14 cm, θ = 60°
∴ Area of minor segment = πr² – \(\frac{1}{2}\) r² sin θ
= \(\frac{22}{7}\) × 14 × 14 × \(\frac{60}{360}\) – \(\frac{1}{2}\) × 14 × 14 × \(\frac{\sqrt{3}}{2}\)
= (\(\frac{308}{3}\) – 49√3) cm² = 17.9 cm² (Approx.)
∴ Area of major segment
= Area of circle – Area of minor segment
= πr² – (\(\frac{308}{3}\) – 49√3)
= \(\frac{22}{7}\) × 14 × 14 – (\(\frac{308}{3}\) – 49√3)
= 616 – \(\frac{308}{3}\) + 49√3
= (\(\frac{1540}{3}\) + 49√3) cm²
or 598.10 cm² (Approx.)

Question 22.
Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs ₹100 per sq. m, find the amount, the associations will have to pay. [Use π = 22/7]
Answer:

Let slant height of cone be l
Let height of cone be h = 2.8 m
Let radius of cone be r = \(\frac{4.2}{2}\)m = 2.1 m
Let height of cylinder be H = 4 m
Using pythagoras’ theorem,
Slant height (l)
= \(\sqrt{h^2+r^2}\) = \(\sqrt{(2.8)^2+(2.1)^2}\)
= \(\sqrt{7.84+4.41}\) = \(\sqrt{12.25}\) = 3.5 m
Area of canvas (1 tent) = Area of cone + Area of cylinder
= πrl + 2πrH ⇒ πr(l + 2H)
= \(\frac{22}{7}\) × 2.1(3.5 + 2 × 4) = \(\frac{46.2}{7}\) × (11.5)
= 6.6 × 11.5 = 75.9 m²
Area of 100 tents = 75.9 m² × 100 = 7590 m²
Cost of 100 tents = 7590 × ₹100 = ₹759000
∴ Associations have to pay 50% of the cost
= ₹7,59,000 × \(\frac{50}{100}\) = ₹3,79,500

Section D

Questions number 23 to 30 carry 4 marks each.

Question 23.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Question 24.
In the fig., ∠BED = ∠BDE and E divides BC in the ratio 2 : 1. Prove that AF × BE = 2 AD × CF.

Answer:
Given: ∠BED = ∠BDE and E divides BC in 2 : 1.
\(\frac{\mathrm{BE}}{\mathrm{EC}}\) = \(\frac{2}{1}\) ……… (i)
Construction:
Draw CG || DF

Prove that:
AF × BE = 2AD × CF
To Prove: ∠BED = ∠BDE …[Given
BD = BE ……(ii) …[Sides opposite to equal angles
In ΔCBG, DE || CG …[By construction
\(\frac{\mathrm{BD}}{\mathrm{DG}}\) = \(\frac{\mathrm{BE}}{\mathrm{EC}}\)
\(\frac{\mathrm{BE}}{\mathrm{EC}}\) …[Thales’ theorem
\(\frac{\mathrm{BD}}{\mathrm{DG}}\) = \(\frac{2}{1}\) …[From (i)
2DG = BD ⇒ 2DG = BE …[From (ii)
DG = \(\frac{1}{2}\)BE ………. (iii)
In ΔADF, CG || DF …[By construction
⇒ \(\frac{\mathrm{AG}}{\mathrm{GD}}\) = \(\frac{\mathrm{AC}}{\mathrm{CF}}\) …[Thales’ theorem
⇒ \(\frac{\mathrm{AG}}{\mathrm{GD}}\) + 1 = \(\frac{\mathrm{AC}}{\mathrm{CF}}\) + 1 …[Adding one on both sides
⇒ \(\frac{A G+G D}{G D}\) = \(\frac{\mathrm{AC}+\mathrm{CF}}{\mathrm{CF}}\)
⇒ \(\frac{\mathrm{AD}}{\mathrm{GD}}\) = \(\frac{\mathrm{AF}}{\mathrm{CF}}\) ⇒ ⇒ AF × GD = AD × CF
⇒ AF × BE/2 = AD × CF …[From (iii)
⇒ AF × BE = 2AD × CF …(Hence proved)

Question 25.
If sec θ – tan θ = x, show that sec θ + tan θ = \(\frac{1}{x}\) and hence find the values of cos θ and sin θ.
Or
Question 25.
Prove the following trigonometric identities:
sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A.
Answer:
Part I: As we know, sec² θ – tan² θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1
(sec θ + tan θ) (x) = 1 …[Given
(sec θ + tan θ) = 1/x …(Hence proved)

Part II:

Or
sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A.


= sec A + cosec A = RHS (Hence proved)

Question 26.
Mode of the following frequency distribution is 65 and sum of all the frequencies is 70. Find the missing frequencies x and y.

Answer:

Now, 54 + x + y = 70
⇒ x + y = 70 – 54 = 16
Mode = 65
∴ Modal Class is 60 – 80.

⇒ x = 12 – 2 = 10
x + y = 16 ….[From (i)
10 + y = 16 ⇒ y = 16 – 10 = 6
∴ x = 10, y = 6

Question 27.
Find the 60^th term of the AP 8,10,12,…, if it has a total of 60 terms and hence find the sum of its last 10 terms.
Answer:
a = 8, d = a₂ – a₁ = 10 – 8 = 2, n = 60
a₆₀ = a + 59d = 8 + 59(2) = 126
∴ Sum of its last 10 terms = S₆₀ – S₅₀
= \(\frac{n}{2}\)(a + a_n) – \(\frac{n}{2}\) (2a + (n – 1)d
= \(\frac{60}{2}\) (8 + a₆₀) – \(\frac{50}{2}\)(2 × 8 + (50 – 1)2)
= 30 (8 + 126) – 25 (16 + 98)
= 4020 – 25 × 114
= 4020 – 2850 = 1170

Question 28.
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Answer:

B is the mid point of arc (ABC).
In ΔOAF and ΔOCF,
OA = OC ….[Radius
OF = OF …[Common
∠1 = ∠2 …[Equal angles opposite equal sides
∴ ΔOAF ☐ ΔOCF (SAS)
∴ ∠AFO = ∠CFO = 90° …[CPCT
⇒ ∠AFO = ∠DBO = 90° …[CPCT …[Tangent is ⊥ to the radius through the point of contact But these are corresponding angles.
∴ AC || DE

Question 29.
The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a constant height of 1500 √3 m, find the speed of the plane in km/hr.
Answer:
Let AL = x,

BL = CM = 1500 √3
In ΔABL, \(\frac{\mathrm{BL}}{\mathrm{AL}}\) = tan 60°
⇒ \(\frac{1500 \sqrt{3}}{x}\) = √3
⇒ x = 1500 m ………. (i)
In ΔAMC,
\(\frac{\mathrm{CM}}{\mathrm{AL}+\mathrm{LM}}\) = tan 30°
⇒ \(\frac{1500 \sqrt{3}}{1500+\mathrm{LM}}\) = \(\frac{1}{\sqrt{3}}\) …[From (i)
⇒ (1500 + LM) = 1500(3)
⇒ LM = 4500 – 1500 = 3000 m
Speed = \(\frac{\text { Distance }}{\text { Speed }}\) = \(\frac{3000}{15}\) = 200 m/s.
or, \(\frac{200}{1000}\) × 60 × 60
= 720 km/hr.

Question 30.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.