Students can use CBSE Previous Year Question Papers Class 10 Maths with Solutions and CBSE Class 10 Maths Question Paper 2015 (Outside Delhi) to familiarize themselves with the exam format and marking scheme.

## 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:

- 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 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)

x^{2} + 5x = x^{2} + 3x + x + 3

x^{2} + 5x – x^{2} – 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 px^{2} – 2√5 px + 15 = 0 has two equal roots, then find the value of p.

Answer:

The given quadratic equation can be written as

px^{2} – 2√ 5 px + 15 = 0

a = p, b = -2 √5p, c = 15

For equal roots, D = 0

D = b^{2} – 4ac

0 = (-2√5p)^{2} – 4 × p × 15

0 = 4 × 5p^{2} – 60p

0 = 20p^{2} – 60p ⇒ 20p^{2} = 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 = 6^{n} = 6^{2} = 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 4^{n} 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:

4^{n} = (2^{2})^{n} = 2^{2n}

The only prime in the factorization of 4^{n} is 2.

There is no other primes in the factorization of 4^{n} = 2^{2n} (By uniqueness of the fundamental Theorem of Arithmetic).

5 does not occur in the prime factorization of 4^{n} for any n.

∴ 4^{n} 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, AB^{2} + BC^{2} = AC^{2} …[Pythagoras theorem

AB^{2} + (\(\frac{5}{2}\))^{2} = (\(\frac{13}{2}\))^{2}

AB^{2} = \(\frac{169}{4}\) – \(\frac{25}{4}\)

= \(\frac{169-25}{4}\) = \(\frac{144}{4}\) = 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 a^{2} + b^{2} = x^{2} + y^{2}.

Answer:

RHS = x^{2} + y^{2}

= (a cos θ – b sin θ)^{2} + (a sin θ + b cos θ)^{2}

= a^{2} cos^{2} θ + b^{2} sin^{2} θ – 2ab cos θ sin θ + a^{2}2 sin^{2} θ + b^{2} cos^{2} θ + 2ab sin θ cos θ

= a^{2} (cos^{2} θ + sin^{2} θ) + b^{2} (sin^{2} θ + cos^{2} θ)

= a^{2} + b^{2} = LHS …[∵ cos^{2} θ + sin^{2} 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 cm^{2}, 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 cm^{2} …[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:

4x^{2} + 4bx – (a^{2} – b^{2}) = 0

Answer:

Given. 4x^{2} + 4bx + b^{2} – a^{2} = 0

⇒ (2x + b)^{2} – (a)^{2} = 0 …[x^{2} – y^{2} = (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 S_{5} + S_{7} = 167 and S_{10} = 235, then find the AP, where Sn denotes the sum of its first n terms.

Answer:

Given: S_{5} + S_{7} = 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)

⇒ S_{10} = \(\frac{10}{2}\) (2a + (10 – 1 )d) = 235

⇒ 5[2a + 9d] = 235

⇒ 10a + 45d = 235 …..(ii)

Solving (i) and (ii), a = 1, d = 5

∴ a_{1}

a_{2} = a + d ⇒ a_{2} = 1 + 5 = 6

a_{3} = a + 2d ⇒ a_{3} = 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

a_{14} = 2.a_{8} …[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)

a_{6} = -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,

S_{20} = \(\frac{20}{2}\)[2 × 2 + (20 – 1)(-2)]

S_{n} = \(\frac{n}{2}\)(2a + (n – 1)d

S_{20} = 10[4 + 19(-2)]

S_{20} = 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, AB^{2} + BC^{2} = AC^{2} …[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 + p^{2} – 8p + 16 + 49 + p^{2} – 14p = 9 + 16

⇒ 81 + 2p^{2} – 22p = 25

⇒ 2p^{2} – 22p + 56 = 0 or p^{2} – 11p + 28 = 0

⇒ p^{2} – 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 = πr^{2} – \(\frac{1}{2}\) r^{2} sin θ

= \(\frac{22}{7}\) × 14 × 14 × \(\frac{60}{360}\) – \(\frac{1}{2}\) × 14 × 14 × \(\frac{\sqrt{3}}{2}\)

= (\(\frac{308}{3}\) – 49√3) cm^{2} = 17.9 cm^{2} (Approx.)

∴ Area of major segment

= Area of circle – Area of minor segment

= πr^{2} – (\(\frac{308}{3}\) – 49√3)

= \(\frac{22}{7}\) × 14 × 14 – (\(\frac{308}{3}\) – 49√3)

= 616 – \(\frac{308}{3}\) + 49√3

= (\(\frac{1540}{3}\) + 49√3) cm^{2}

or 598.10 cm^{2} (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 m^{2}

Area of 100 tents = 75.9 m^{2} × 100 = 7590 m^{2}

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, sec^{2} θ – tan^{2} 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 = a_{2} – a_{1} = 10 – 8 = 2, n = 60

a_{60} = a + 59d = 8 + 59(2) = 126

∴ Sum of its last 10 terms = S_{60} – S_{50}

= \(\frac{n}{2}\)(a + a_{n}) – \(\frac{n}{2}\) (2a + (n – 1)d

= \(\frac{60}{2}\) (8 + a_{60}) – \(\frac{50}{2}\)(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.