# Oswal 36 Sample Question Papers CBSE Class 10 Maths Solutions

## Section - A

1. (b) (2, 5)

Explanation: Given A(0, 3) and B(6, 9) and let (x, y) be the coordinates of P. Since the line is trisected then AP : PB = 1 : 2
Using section formula,
$$(x,y)=\lbrack\frac{m_1x_2+m_2x_1}{m_1+m_2}\frac{m_1y_2+m_2y_1}{m_+m_2}\rbrack\\x_1 = 0, y_1 = 3, x_2=6\\y_2=9,m_1=1,m_2=2\\(x,y)=(\frac{1×6+2×0}{1+2},\frac{1×9+2×3}{1+2})\\=(\frac{6+0}{3}\frac{9+6}{3})\\=(2,5)\\\text{So, coordinates of P are (2, 5).}$$
So, coordinates of P are (2, 5).

2. (c) A = 52.5°, = 7.5°

Explanation: Given, tan (A + B) = √3=tan60°

⇒ A + B = 60° ...(i)

and tan (A – B) = 1 = tan 45°

or A – B = 45° ...(ii)

Adding (i) and (ii), we get 2A = 105°

A = 52.5°

Put A = 52.5° in equation (i), we get

52.5º + B = 60°

B = 60° – 52.5°

⇒ B = 7.5°

Hence A = 52.5° and B = 7.5°.

3. (b) – 4

Explanation: Given : nth term of an A.P.,

an = 7 – 4n

⇒ an + 1 = 7 – 4(n + 1)

= 7 – 4n – 4 = 3 – 4n

So, common difference,

d = an + 1 – an

= (3 – 4n) – (7 – 4n) = – 4

4. (c) 0.7

Explanation: Given, P (winning) = 0·3

So, P(losing) = 1 – 0·3 = 0·7

5. (a) 5.5

Explanation:

From the table,  Σfi = 75

Σfi xi = 412·5

We know,
$$\overline{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}\\⇒\overline{x}=\frac{412.5}{75}\\⇒\overline{x}=5.5\\\text{Mean of given distribution is 5.5.}$$

6. (d) 30°

Explanation: In ΔABC, tan 60° =(h/x)

⇒ √3 = (h/x)

h = x√3      …(i)

⇒ tan θ = (x√3x/3x)

⇒ tan θ = (1/√3)

⇒ tan θ = tan 30°

⇒ θ = 30°

7. (b) 4

Explanation: A composite number is a natural number which has more than two factors. Here, factors of 4 = 1, 2, 4.

∴ 4 is a smallest composite number.

∴ 4 is a smallest composite number.

8. (c) 10 m

Explanation: Let, DC be the tower of height 15m. and AB be the height of pole

Now, in ΔAED, tan 30° =(DE/AE)

⇒(1/√3)=(DE/AE)

⇒ AE = √3 DE  …(i)

In ΔBCD, tan 60° =(DC/BC)

⇒ √3 =(15/BC)

⇒ BC =(15/√3)

⇒ BC = 5 √3  …(ii)

As, AE = BC

∵ √3 DE = 5 √3 [using (i) and (ii)]

⇒ DE = 5 m

Height of pole, AB = EC = DC – DE

= 15 – 5 = 10 m.

Thus, height of the pole is 10m.

9. (a) 32

Explanation: Given : OA = OB = 10.5 cm

and ∠AOB = 60°

Perimeter of the sector = OA + OB + AB
$$= 10.5+10.5+\frac{60°}{360°}×2×\frac{22}{7}×10.5\space\space\space\space [l=\frac{\theta}{360°}×2\pi r]\\= 21 + 11 = 32 cm.$$

10. (b) 13.5 m

Explanation: In ∆ABC,

$$\frac{AP}{AB}=\frac{3.5}{3.5+7}=\frac{3.5}{10.5}=\frac{1}{3}\\\text{and}\space\frac{AQ}{AC}=\frac{3}{3+6}=\frac{3}{9}=\frac{1}{3}\\⇒\frac{AP}{AB}=\frac{AQ}{AC}=\frac{1} {3}\\\text{Also}\space∠A=∠A[Common]\\⇒ ∆APQ ~ ∆ABC [SAS criterion]\\\text{Now,}\space \frac{AP}{AB}=\frac{PQ}{BC}\text{[Corresponding sides of similar triangles]}\\⇒\frac{3.5}{10.5}=\frac{4.5}{BC}\\⇒ BC = 4·5 × 3 = 13.5 cm.$$

11. (a) 3 median–2 mean

Explanation: The relationship between three measures of central tendencies is given by

Mode = 3 median – 2 mean.

12. (b) 2

Explanation: A circle can have maximum two parallel tangents.

13. (b) 30°

Explanation: Given,
$$\text{Explanation: Given,}\space sin\alpha=\frac{ \sqrt{3}}{2}\space\space\space[sin60°\frac{ \sqrt{3}}{2}]\\⇒sin\alpha=sin60°$$

⇒ α = 60°

and cos β = 0

cos β = cos 90°

⇒ β = 90°

Now, β – α = 90° – 60° = 30°

14. (b) 2r cm

Explanation: Here, maximum diameter of sphere is equal to the diameter of cylinder.

·.· Radius of cylinder = r

∴ Diameter of cylinder = Diameter of sphere = 2r

15. (a) 18

Explanation: Prime factors of 36 = 3 × 3 × 2 × 2

Prime factors of 54 = 3 × 3 × 2 × 3

Hence HCF (35, 54) = 3 × 3 × 2 = 18

16. (c) k ≠ 10

Explanation: Given system of equations are,

3x – y – 5 = 0

and 6x – 2y – k = 0

For the system to have no solution,
$$\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}\\⇒\frac{3}{6}=\frac{-1}{-2}≠\frac{-5}{-k}\\or\frac{5}{k}≠\frac{1}{2}⇒ k ≠ 10.$$

17. (a) (7/2)

Explanation: Here,
$$\text{Explanation: Here,}\space\alpha+\beta=\frac{-(-7)}{5}\\\alpha\beta=\frac{2}{5}\\\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{\frac{7}{5}}{\frac{2}{5}}=\frac{7}{2}$$

18. (d) no solution

Explanation: Given, equations are x + 2y + 5 = 0 and – 3x – 6y + 1 = 0

Here, a1 = 1, b1 = 2, c1 = 5

and a2 = – 3, b2 = – 6, c2 = 1
$$\frac{a_1}{a_2}=-\frac{1}{3},\frac{b_1}{b_2}=-\frac{2}{6}=-\frac{1}{3},\frac{c_1}{c_2}=\frac{5}{1}\\\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}\\\text{Hence, the pair of equations has no solution.}$$

19. (c) Assertion is true but the Reason is false.

Explanation: Suppose the given system of equations has infinitely many solutions if
$$\frac{2}{2a}=\frac{3}{a+b}=-\frac{-7}{-28}\\\frac{1}{a}=\frac{3}{a+b}=-\frac{1}{4}\\⇒3a = a + b\\⇒ 2a – b = 0\\\text{Also clearly}\space a=4\space\text{and}\space a+b=12\\⇒\\b = 8\\2a – b = 8 – 8 = 0\\2a – b = 8 – 8 = 0\\\text{So, the assertion is true.}\\\text{On the other hand, a pair of equations have a unique solution, if}\\\frac{a_1}{a_2}≠\frac{b_1}{b_2}\\\text{Here,}\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}\\\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}\\\frac{a_1}{a_2}≠\frac{b_1}{b_2}$$
So, reason is false. Hence correct option is (c).

20. (d) Assertion is false but the Reason is true.

Explanation: The highest power of x in the polynomial 4x3 – x2 + 5x4 + 3x – 2 is 4.

Therefore, the degree of the polynomial p(x) is 4.

Hence, assertion is false but reason is true.

(Any two)

## Section - B

21. Let p(x) = 6x2 – 7x – 3

Put p(x) = 0

6x2 – 7x – 3 = 0

⇒ 6x2 – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (3x + 1) (2x – 3) = 0

⇒ (3x + 1) = 0 or (2x – 3) = 0

$$⇒x=\frac{-1}{3}\space or\space x=\frac{3}{2}\\\text{So, zeroes of the polynomial are α}=\frac{-1}{3}\space and β=\frac{3}{2}\\\text{Now,}α+β=-\frac{1}{3}+\frac{3}{2}=-\frac{-2+9}{6}=\frac{7}{6}\\⇒ αβ=−\frac{1}{3}×\frac{3}{2}=-\frac{1}{2}\\\text{Here, in this polynomial, a = 6, b = – 7, c = – 3}\\So,-\frac{b}{a}=\frac{(-7)}{6}=\frac{7}{6}\\\text{and}\space\frac{c}{a}=\frac{-3}{6}=\frac{-1}{2}\\⇒α +β=\frac{-b}{a},αβ=\frac{c}{a}\\\text{Relationship holds.}$$

22. Given A = (1, 4), B = (– 1, 2) and P = (x, y)

Also, AP = BP
$$\sqrt{(x − 1)^2 + (y − 4)^2}=\sqrt{(x + 1)^2 + (y − 2)^2}$$

on squaring both sides,

⇒ (1 – x)2 + (4 – y)2 = (x + 1)2 + (y – 2)2

⇒ 1 + x2 – 2x + 16 + y2 – 8y = x2 + 1 + 2x + y2 + 4 – 4y

– 2x – 8y + 16 = 2x – 4y + 4

⇒ – 2x – 2x – 8y + 4y = 4 – 16

⇒ – 4x – 4y = – 12

⇒ – 4(x + y) = – 12

⇒ x + y = 3.     Ans.

23. Given : Sample space = {1, 2, 3, 4, 5, 6}.

(i) Prime numbers are 2, 3, 5.

∴ P(Prime number) =(3/6)=(1/2)  Ans.

(ii) Numbers divisible by 2 are 2, 4, 6.

∴ P(number divisible by 2) =(3/6)=(1/2)  Ans.

OR

Total number of possible outcomes = 1, 2, 3, 4, 5, 6.

Let E be the event of getting an even number.

Then, the favourable outcomes are 2, 4, 6.

Hence, the total number of favourable outcomes = 3

∴ P(E) =(3/6)=(1/2)    Ans.

24. Consider part of curve between x = – 2 and x = 2

The curve cuts X-axis at x = – 1 and x = 1

Hence, required zeroes are – 1 and 1.    Ans.

OR

Given equations are, x – y = – 1          ...(i)

and 3x + 2y = 12       ...(ii)

Multiplying equation (i) with 2, we get

2x – 2y = – 2        ...(iii)

Adding equations (ii) and (iii), we get

5x = 10

⇒ x = 2

Putting x = 2 in equation (i), we get

2 – y = – 1

⇒ – y = – 3

⇒ y = 3.

Hence, the values of x and y are 2 and 3 respectively.         Ans.

25. Since, jacks, queens and kings of red colour are removed. Then,

Total number of possible outcomes = 52 – 6 = 46

(i) Let E1 be the event of getting a black king.

∴ Favourable outcomes = king of spade and king of club.

No. of favourable outcomes = 2

P(E1) =(2/46)=(1/23). Ans.

(ii) Let E2 be the event of getting a card of red colour.

∴ Favourable outcomes = 10 cards of heart and 10 cards of diamond.

No. of favouable outcomes = 20

P(E2) =(20/46)=(10/23). Ans.

(iii) Let E3 be the event of getting a card of black colour

∴ Favourable outcomes = 13 cards of spade and 13 cards of club.

No. of favourable outcomes = 26

P(E3) =(26/46)=(13/23)      Ans.

## Section - C

26. Let O be the centre of the given circle. Suppose the tangent at P meets BC at Q. Join BP.

To Prove : BQ = QC

Proof : Since, tangent at any point of circle is perpendicular to radius through the point of contact.

∴ ∠ABC = 90°

In ∆ABC, ∠1 + ∠5 = 90° [y angle sum property, ∠ABC = 90°]

and, ∠3 = ∠1

[Angle between tangent and the chord equals angle made by the chord in alternate segment]

Now, ∠3 + ∠5 = 90°        ...(i)

Also, ∠APB = 90°    [Angle in semi-circle]

∴ ∠3 + ∠4 = 90° [∠APB + ∠BPC = 180°, linear pair] ...(ii)

From (i) and (ii), we get

∠3 + ∠5 = ∠3 + ∠4

or ∠5 = ∠4

∴ PQ = QC [Sides opposite to equal angles are equal]

Also, QP = QB

[Tangents drawn from an internal point to a circle are equal]

∴ QB = QC    Hence Proved.

27.

$$L.H.S. =\frac{cot A-cos A}{cot A+cos A}\\=\frac{\frac{cos A}{sin A}-cos A}{\frac{cos A}{sin A}+cos A}\\=\frac{cos A(\frac{1}{sin A}-1)}{cos A(\frac{1}{sin A}+1)}\\=\frac{cosec A-1}{cosec A+1}\\= R.H.S.$$
Hence Proved.

OR

L.H.S. = (1 + cot A – cosec A) (1 + tan A + sec A)
$$L.H.S. =(1+\frac{cos A}{sin A}-\frac{1}{sin A})(1+\frac{sin A}{cos A}+\frac{1}{cos A})\\=[\frac{(sin A+cos A)-1}{sin A}]×[\frac{(cos A+sin A)+1}{cos A}]\\=\frac{(sin A+cos A)^2-1}{sin A\space cos A}\\=\frac{sin^2 A+cos^2A+2sin A cos A-1}{sin A\space cos A}\\=2\\=R.H.S$$

28. (i) House = (2, 4), Bank = (5, 8), School = (13, 14) and Office = (13, 21)

$$∴\text{Distance between house and office}=\sqrt{(13-2)^2+(21-4)^2}\\=\sqrt{(11)^2+(17)^2}\\=\sqrt{(121)+(289)}\\=\sqrt{410}\text{unit}$$

$$\text{(ii) Distance covered from house to bank}=\sqrt{(5-2)^2+(8-4)^2}\\=\sqrt{9+16}=\sqrt{25}=5\text{unit}\\\text{Distance covered from bank to school}=\sqrt{(13-5)^2+(14-8)^2}\\=\sqrt{64+36}=\sqrt{100}=10 \text{unit}\\\text{Distance covered from school to office}=\sqrt{(13-13)^2+(21-14)^2}\\=\sqrt{0+49}=\sqrt{49}=7\text{units}\\\text{Extra distance covered =}(5 + 10 + 7)-\sqrt{410}\\=(22-\sqrt{410})\text{units \space\space Ans}$$

29. Given : Quadrilateral ABCD, AC and BD meet at O.

and (AO/ OC) = (BO/ OD)

To prove : ABCD is a trapezium.

Proof : (AO /OC) = (BO/ OD)  [Given]

∠1 = ∠2       [Vertically opposite angles]

⇒ ∆AOB ~ ∆COD [SAS criterion]

⇒ ∠3 = ∠4     [Corresponding angles of similar triangles]

⇒ AB || DC        [∵ A pair of alternate angles are equal]

⇒ ABCD is a trapezium.    Hence Proved.

OR

Given : ∠A = ∠D = 90°

To prove : AE × EC = BE × ED

Proof : ∠A = ∠D [Each 90°]

∠1 = ∠2 [Vertically opposite angles]

⇒ ∆AEB ~ ∆DEC

$$⇒\frac{AE}{DE}=\frac{EB}{EC}=\frac{AB}{DC}\\⇒\frac{AE}{DE}=\frac{EB}{EC}\\⇒AE × EC = EB × DE.$$  Hence Proved.

30. Given : Hexagon ABCDEF in which a circle is circumscribed.

Since tangents to a circle from an exterior point are equal in length.

AP = AU

BP = BQ

CR = CQ

DR = DS

ET = ES

FT = FU

(AP + BP) + (CR + DR) + (ET + FT ) = (AU + FU) + (BQ + CQ) + (DS + ES)

⇒ AB + CD + EF = AF + BC + DE.    Hence Proved.

31. Given, the dimensions of cuboid are 10 cm, 5 cm and 4 cm.

Radius of conical depression = 0·5 cm

Depth of conical depression = 2·1 cm

The edge of cubical depression = 3 cm

Volume of cuboid = 10 × 5 × 4 cm3 = 200 cm3

Volume of conical depression =(1/3)πr2h

=(1/3)×(22/7)×(1/2)×(1/2)×(21/10)cm3

⇒ = (11/20)cm3

Volume of cubical depression = 3 × 3 × 3 = 27 cm3

$$⇒\text{Volume of wood left}=(200-4×\frac{11}{20}-27)cm^2\\= (200 – 29.2) cm^3 = 170.8 cm^3.\space\space\space Ans.$$

## Section - D

32.

$$\text{(i) If r be the radius of the circle.}\\\text{Then, the perimeter of a quadrant of the circle}\\=\frac{1}{4}×2\pi r+2r\\=(\frac{\pi}{2}+2)r\\= 12.5\space\space\space\text{(Given)}\\⇒(\frac{1}{2}×\frac{22}{7}+2)r=\frac{25}{2}\\⇒\frac{25r}{7}=\frac{25}{2}\\⇒r=\frac{7}{2}\\\text{Area of quadrant}=\frac{1}{4}\pi r^2\\=\frac{1}{4}×\frac{22}{7}×(\frac{7}{2})^2\text{cm}^2\\= 9.625\text{cm}^2$$
$$\text{(ii) Given : Radius of circle, r = 42 cm}\\\text{Length of arc = 44 cm}\\⇒\frac{\theta}{360°}×2\pi r=44\\⇒θ =\frac{44×360}{2\pi r}=\frac{44×360×7}{2×22×42}=60°\\\text{Hence, θ = 60°}\\\text{Now, area of corresponding minor segment}=\frac{\theta}{360°}\pi r^2-\frac{1}{2}r^2sin\theta\\=\frac{60°}{360°}×\frac{22}{7}×42×42-\frac{1}{2}×42×42×sin60°\\= 22×42–21×42×\frac{\sqrt{3}}{2}\\=22×42-21×\sqrt{3}\\= 21(44 − 21\sqrt{3})cm^2\space Ans.$$

33. Given :
$$\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\\⇒\frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}\\⇒\frac{x-(a+b+x)}{x(a+b+x)}=\frac{b+a}{ab}\\⇒\frac{x-a-b-x)}{x^2+x(a+b)}=\frac{a+b}{ab}\\⇒\frac{-(a+b)}{x^2+x(a+b)}=\frac{a+b}{ab}\\⇒\frac{-1}{x^2+x(a+b)}=\frac{1}{ab}$$

⇒ x2 + x(a + b) = – ab

⇒ x2 + x(a + b) + ab = 0

⇒ x2 + ax + bx + ab = 0

⇒ x(x + a) + b(x + a) = 0

⇒ (x + b) (x + a) = 0

⇒ x = – a, – b.   Ans.

OR

(i) Let the fixed charges = ₹ x

Let the charge per day = ₹ y

Latika kept a book for 6 days

She pays fixed charges for 2 days and additional charges for 4 days.

x + 4y = 22    ...(i)

Anand kept book for 4 days.

He pays fixed charges for 2 days and additional charges for 2 days.

x + 2y = 16    ...(ii)

Subtracting equation (ii) from (i), we get

⇒ (x + 4y) – (x + 2y) = 22 – 16

⇒ x + 4y – x – 2y = 6

⇒ 2y = 6

y = 3

Therefore, the charges per day is ₹ 3.      Ans.

(ii) Put the value of y in equation (i),

x + 4y = 22

⇒ x + 4(3) = 22

⇒ x + 12 = 22

⇒ x = 22 – 12

⇒ x = 10

Thus, fixed charge for first two days is ₹  10. Ans.

(iii) The graph of the situation is given below :

34. Let BC be the pedestal, DC be the statue and A be the point of observation.

$$Let \space BC = h, AB = x.\\In ∆ABC,\space\frac{BC}{AB}=tan45°\\⇒\frac{h}{x}=1\\⇒ h = x …(i)\\In\space∆ABD,\space\frac{BD}{AB}=tan\space60°\\⇒\frac{h+1.6}{x}=\sqrt{3}\\⇒\frac{h+1.6}{\sqrt{3}}=x\space …(ii)\\\text{From equations (i) and (ii),}\space\\h=\frac{h+1.6}{\sqrt{3}}\\⇒h\sqrt{3}-h=1.6\\⇒h(\sqrt{3}-1)=1.6\\⇒h=\frac{1.6}{\sqrt{3}-1}×\frac{\sqrt{3}+1}{\sqrt{3}+1}\\⇒h=\frac{1.6(1.732+1)}{2}= 0.8 × 2.732\\⇒ h = 2.1856 ≈ 2.19 m.\space Ans.$$

35. Let the usual speed be x km/hr.

$$\text{We know, Time }=\frac{\text{Distance}}{\text{Distance}}\\So,\text{Usual time taken}=\frac{1500}{x}hr.\\\text{Increased speed = (x + 100) km/hr}\\\text{Thus, New time taken}=\frac{1500}{x+100}\\\text{Now, according to the question}\\\frac{1500}{x+100}+\frac{1}{2}=\frac{1500}{x}\\⇒\frac{1500}{x}-\frac{1500}{x+100}=\frac{1}{2}\\⇒1500[\frac{1}{x}-\frac{1}{x+1000}]=\frac{1}{2}\\⇒15000(\frac{x+100-x}{x(x+100)})=\frac{1}{2}\\⇒1500(\frac{100}{x(x+100)})=\frac{1}{2}$$

⇒ x(x + 100) = 300000

⇒ x2 + 100x – 300000 = 0

⇒ x2 + 600x – 500x – 300000 = 0

⇒ x(x + 600) – 500(x + 600) = 0

⇒ (x + 600) (x – 500) = 0

x = 500 and – 600

As speed cannot be negative, so the original speed of the plane is 500 km/hr.    Ans.

OR

Let total time be n minutes.

Since, policeman runs after 1 minute so he will catch the thief in (n –1) minutes.

Total distance covered by thief = 100 m/minute × n minute

= (100 n) m

Now, total distance covered by the policeman

= (100)m + (100 + 10)m + (100 + 10 + 10)m + ......+ (n – 1) terms

i.e., 100 + 110 + 120 + ..... + (n – 1) terms

∴ Sn – 1 =(n-1/2)[2×100+(n-2)10]

Now, (n-1/2)[200+(n-2)×10]=100n.

⇒ (n – 1) (200 + 10n – 20) = 200 n.

⇒ 200 n – 200 + 10n2 – 10n + 20 – 20n = 200n

⇒ 10n2 – 30n – 180 = 0

⇒ n2 – 3n – 18 = 0

⇒ n2 – (6 – 3) n – 18 = 0

⇒ n2 – 6n + 3n – 18 = 0

⇒ n(n – 6) + 3(n – 6) = 0

⇒ (n + 3) (n – 6) = 0

∴ n = 6 or n = – 3 (neglect)

Hence, policeman will catch the thief in (6 – 1) i.e., 5 minutes.         Ans.

## Section - E

$$\text{36. (i) Each guest will get number of apples}=\frac{36}{12}=3\\\text{Each guest will get no. of bananas =}\frac{60}{12}=5\\\text{Out of 36 apples and 60 bananas each guest will get 3 apples, 5 bananas. Ans.}\\\text{(ii) H.C.F. (36, 42, 60) = 6 So, fruits will be equally distributed among 6 guests. So she could invite max. of 6 guests Ans.}\\\text{(iii) Now Vedika has 30 apples, 60 bananas and 45 mangoes. H.C.F. (30, 45, 60) = 15. Ans.}$$

OR

$$\text{Total apples =}\frac{36}{6}=6,\text{bananas =}\frac{60}{6}= 10\text{ and mangoes}\frac{42}{6}=7\text{So, total no. of fruits will each guest get 6 + 10 + 7 = 23 fruits.}$$

37. (i) The upper limit of a class and the lower class of its succeeding class differ by 1. Ans.

(ii) Here, class intervals are in inclusive from. So, we first convert them in exclusive form. The frequency distribution table in exclusive form is as follows:

 Class interval Frequency (fi) Frequency cumulative c.f. 199.5 – 209.5 4 4 209.5 – 219.5 14 18 219.5 – 229.5 26 44 229.5 – 239.5 10 54 239.5 – 249.5 6 60
Here, Σfi = N = 60

∴ (N /2) = (60 /2) = 30

Now, the class interval whose cumulative frequency is just greater than 30 is 219.5 – 229.5

∴ Median class is 219.5 – 229.5             Ans.

OR

$$\text{Median}=l+(\frac{\frac{N}{2}-cf}{f})×h\\=219.5+(\frac{\frac{60}{2}-18}{26})×10\\= 224.12\\\text{∴ Median of the distance travelled is 224.12 km.\space Ans.}\\\text{(iii) We know, Mode = 3 Median – 2 Mean}\\\text{Mean}=\frac{(3 Median-Mode)}{2}\\=\frac{(672.36-223.75)}{2}\\=224.29 km.\space Ans.$$

38. (i) ∠OCA = 90°

[Since, radius at the point of contact is perpendicular to tangent]

In ΔOAC, ∠OCA + ∠OAC + ∠AOC = 180° (Angle sum property)

⇒ 90° + ∠OAC + ∠AOC = 180°

⇒ ∠OAC + ∠AOC = 90°

⇒ x + y = 90°. Ans.

(ii) Since, OB ⊥ PB [Since, radius at the point of contact is perpendicular to tangent]

and ∠PBA = 50°    (Given)

∴ ∠OBA = 90° – 50°

= 40°

Also, OA = OB [Radii of circle]

∴ ∠OAB = ∠OBA

= 40° [Angle opposite to equal sides are equal]

OR

Here, ∠ABC = 90° (Angle in a semicircle)

So, in ∆ABC, ∠BAC = 180° – 90° – 55° = 35°

Also, ∠OAT = 90°

⇒ ∠BAT + ∠OAB = 90°

⇒ ∠BAT = 90° – 35°

= 55° Ans.

(iii) Here, ∠PAB = 72°

∴ ∠OAB = 90° – 72° = 18°

Also, ∠AOB = 132° [Given]

Now, in DOAB, ∠ABC = 180° – 132° – 18°

= 30°.  Ans.

#### CBSE 36 Sample Question Papers All Subject Combined

All Subjects Combined for Class 10 Exam 2024