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**Class X****Mathematics – Standard (041)****Sample Question Paper 2019-20**

**Max. Marks: 80****Duration : 3 hrs****General Instructions:**

(i) All the questions are compulsory.

(ii) The question paper consists of 40 questions divided into 4 sections A, B, C, and D.

(iii) Section A comprises of 20 questions of 1 mark each. Section B comprises of 6 questions of 2 marks each. Section C comprises of 8 questions of 3 marks each. Section D comprises of 6 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark each, two questions of 2 marks each, three questions of 3 marks each, and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.

(v) Use of calculators is not permitted.

**Section A**

**Q 1. A number is selected at random from the numbers 1 to 30. The probability that it is a prime number is : (1 Mark)(a) 2/3(b) 1/6(c) 1/3(d) 11/30Ans:** (c)

(a) 2

(b) 3

(c) 5

(d) 11

Ans:

[Here, least prime factor of 'a’ is 3.

The other factors are more than or equal to 3, but not divisible by 2.

∴ a must be an odd number

Similarly, b is also an odd number.

∴ (a + b) is an even number.

⇒ Least prime factor of (a + b) is 2]

(a) -10/3

(b) -20/3

(c) -25/3

(d) -15/4

Ans:

On comparing with a

For unique solution,

Thus, given lines have a unique solution for all real values of k, except -20/3

(a) (4, 0)

(b) (2, 0)

(c) (-8, 0)

(d) (-2, 0)

Ans:

The given points are A(- 2, 0) and 5(6, 0).

Let the required point lying on x-axis be P(x, 0).

Since point P is equidistant from A and B.

Hence, the required point is P(2, 0).

(a) 7 cm

(b) 12 cm

(c) 15 cm

(d) 24.5 cm

OR

If in triangles ABC and DEF, then they will be similar , when:

(a) ∠B = ∠E

(b) ∠A = ∠D

(c) ∠B =∠D

(d) ∠A = ∠F

Ans:

Let O be the centre of the circle.

Given that, OQ — 25 cm and PQ = 24 cm

We know that the radius is perpendicular to the tangent at the point of contact.

∴ OP ⊥ PQ

In ΔOPQ, we have By Pythagoras theorem,

Therefore, the radius of the circle is 7 cm.

To be similar of ΔABC and ΔDEF, we must have

(a) 2 cos 0

(b) 0

(c) 2 sin θ

(d) 1

Ans:

[sin (45° + θ) - cos (45° - θ)

= sin (45° + θ) - sin (90° - (45° - θ))

= sin (45° + θ) - sin (90° - 45° + θ)

= sin (45° + θ) - sin (45° + θ) = θ]

(a) 25 : 49

(b) 125 : 49

(c) 49 : 25

(d) 343 : 125

Ans:

Since, the ratio of the areas of two similar triangles is equal to the squares of the ratio of any two corresponding sides.

The ratio of the areas of two similar triangles, whose sides are in the ratio

5 :7 is 5

We have, cos 80° cos 10° - sin 80° sin 10°

= cos (90° - 10°) cos 10° - sin (90° - 10°) sin 10°

= sin 10° cos 10° - cos 10° sin 10° [∴ sin (90°-θ )= cos θ; cos (90°-θ ) - sin θ] = θ

6x - 3y + 10 = 0

2x - y + 9 = 0

represents two lines which are

(a) intersecting at exactly one point

(b) intersecting at exactly two points

(c) coincident

(d) parallel

Ans:

Here,

So, the system of linear equations is inconsistent (no solution) and graph will be a pair of parallel lines.

(a) 3

(b) 1

(c) 2

(d) 4

Ans:

Ans: =

Given, radius of the cirle, r = 21/2 cm

and angle of sector, θ = 150

∴ Area of the sector

Ans:

Let r

∴

Hence, Volume of cylinder : Volume of cone = 9:8.

Ans:

The equation of one line is 4x + 3y — 14. We know that if two lines a

or

Hence, one of the possible, second parallel line is 12x + 9y = 5.

Ans:

Ans:

Since, ΔABC and ΔBDE are two equilateral triangles

Reason (R) : cos

(a) Both A and R are true and R is the correct explanation for A.

(b) Both A and R are true and R is not the correct explanation for A.

(c) A is true but R is false.

(d) A is false but R is true.

Ans:

We have,

x

= a

= a

= a

(a) 2x

(b) - x

(c)

(d) 3x

Ans:

-x

On comparing with ax

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

∴ sum of roots

If two tangents inclined at an angle 60° are drawn to a c irc le of radius 3 cm , then find the length of each tangent.

Ans.

Now, in rt. ∠edΔPTO, we have

Hence, the length of each tangent is 3√3 cm.

Ans.

Then, perimeter of protractor = (πr + 2r) = 72

⇒ (π + 2)r = 72

⇒

⇒ 36/7 r = 72

⇒

Ans.

⇒

Squaring both sides, we get

2x = (6x-1)

which is a quadratic equation.

**SECTION B**

**Q.21. Find the value of k, if -1 is a zero of the polynomial p(x) = kx ^{2} - 4x + k.**

and p(x) = kx

then p (-1) = 0

∴ k (-1)

⇒ k + 4 + k = 0

⇒ 2k + 4 = 0

⇒ 2k = - 4

Hence, k = -2

Let income of A = 8x and income of B = 7x.

Also their expenditures be 19y and 16y.

⇒ 8x - 19y = 2550 ...(i)

and 7x - 16y = 2550 ...(ii)

Solving the equations

x = 1530 and y = 510

∴ Salary of A = 12240

Salary of B = 10710

Ans.

∴ OM = ON = radius of smaller circle. Thus, AB and CD are two chords of the larger circle such that they are equidistant from the centre.

Hence, AB = CD.

The students are to sow seeds of flowerin plants on the remaining area of the plot.

Then, taking A as origin, find the area of the triangle in this case.

Ans.

In this case, AD and AB taken as coordinate axes.

Then, area of ΔPQR

Ans.

The given system of linear equations is

This system of equations is of the form

where,

The given system of linear equations will have a unique solution if

Hence, the given system has a unique solution if k ≠ 6.

Ans.

Ans.

∴ Area of 64 squares = (x-4)

64 x 6.25 = x

400 =x

**SECTION C**

**Q.27. In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice? (ii) a total of 9 or 11? (3 Mark)**

P (a prime number on each die) = 9/36 or 1/4

(ii) Favourable outcomes are (3, 6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5) i.e. 6 outcomes

P (a total of 9 or 11) = 6/36 or 1/6

Ans.

∴ S

Subtracting (ii) from (i), we have

Ans.

So, ΔPQR is isosceles.

Thus, ΔPQR is a right angled triangle, right angled at R. [by convers of Pythogoras theorem]

Ans.

The equation of the given line is x - 3y = 0 ...(1)

Let given line divides the join of A and B in the ratio k : 1 at P(x, y).

Using section formula, the coordinates of point P are

But the point P lies on the line (1), so its coordinates satisfy equation (1).

∴

and

and ratio is k : 1 i.e., 13/3 : 1 i.e., 13 : 3.

Hence, the given line divides the join of pooints A(-2, -5) and B(6, 3) in the ratio 13 : 3 and point of intersection is

OR

The perpendicular AD on the base BC of a ΔABC intersects BC at D so that DB = 3CD. Prove that 2(AB)

Ans.

Let ∠OPQ be θ.

(Opposite angles of equal sides)

Given, in ΔADB,AB

In ΔADC, AC

Subtracting eqn. (ii) from eqn. (i),

AB

Or,

or,

Or

If tan

Ans.

y = sec A - sin A ...(ii)

Adding (i) and (ii), we obtain

x + y = 2 sec A

Subtracting (ii) from (i), we obtain

Squaring and adding (iii) and (iv), we obtain

Here, we have tan

Or

A car has wheels which are 80 cm in diameter. How many complete revolutions does each wheel make in 10 min when the car is travelling at a speed of 66 km/h?

Ans.

∴ Diameter of each circle = 7 x 2 = 14 cm

Clearly, length of side of the square

= 3 x ( Diameter of a circle) = 3 x 14 = 42 cm

∴ Area of the square = (Side)

Now, area of one circle

So, area of 9 circles = 9 x 154 = 1386 cm

∴ Area of remaining portion

= Area of the square - Area of 9 circles

= 1764 - 1386 = 378 cm

Hence, the area of the remaining portion is 378 cm

Given, diameter of a wheel = 80 cm

Clearly, radius of car wheel = 40 cm

Now, distance covered by the car when its wheel take one complete revolution = circumference of wheel

∵ Speed of car = 66 km/h

∴ Distance covered by the car in 1 h i.e. 60 min

= 66 km

⇒ Distance covered by the car in 10 min

= 11 x 1000 m [∵ 1 km = 1000 m]

= 11 x 1000 x 100 cm [∵ 1 m = 100 cm]

Clearly, number of revolutions made by the wheels in 10 min

= Distance covered by the car in 10 min/Distance covered1 by the car when its wheel take one complete revolution

Hence, each wheel makes 4375 revolutions in 10 min.

f(x) = 3x

completely divisible by 3x

OR

Find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and the coefficients.

Ans.

f(x) = 3x

and g(x) = 3x

Dividing f(x) = 3x

Since, f(x) is divisible by g(x), remainder must be zero.

∴ k + 10 = 0

⇒ k = -10

The given polynomial is

Therefore, the value of f (y) is zero when 3y - 2 = 0 or 7y + 1 = 0 i.e., when y = 2/3 or y = -1/7.

∴ The zeroes of the given polynomial are 2/3 and -1/7.

Sum of zeroes

and Product of zeroes =

**SECTION D**

**Q.35. Draw the graphs of the pair of linear equations : x + 2y = 5 and 2x - 3y = -4.****Also, find the points where the lines meet the x-axis.****OR****Sum of the areas of two squares is 400 cm ^{2}. If the difference of their perimeters is 16 cm, find the sides of the two squares. (4 Mark)**

⇒

and 2x - 3y = - 4

⇒

Thus, the lines meet x-axis at (5, 0) and (-2, 0) respectively.

OR

Let the sides of two squares be a and b,

then a

and 4 (a - b) =16

⇒ a - b = 4

⇒ a = 4 + b

From equations (i) and (ii), we get

(4 + b)

b = - 16 or, b = 12

(Rejecting the negative value)

So, b = 12 cm

and a = 16cm.

Ans.

To Prove: ∠B = 90°

Const. : Construct a right triangle PQR, right-angled at Q, such that

PQ = AB

and QR = BC.

Proof : From right ΔPQR, we have

PR

[using Pythagoras Theorem]

⇒ PR

[∵ PQ = AB and QR = BC]

Also, we have

AC

From (i) and (ii), we have

**(i) ΔAMC - ΔPNRAns.** ΔABC and ΔPQR

CM is the median of ΔABC and RN is the median of ΔPQR.

Proof

(i) Given ΔABC ~ ΔPQR

∴ ...(i)

[∴ in similar triangle, corresponding sides are proportional]

[∴ in similar triangles, corresponding angles are equal]

We know that the median bisects the opposite side.

⇒ ...(iii)

In ΔAMC and ΔPNR, we have

∠A = ∠P [from Eq. (ii)]

and [from Eq. (iii)]

So, ΔAMC ~ ΔPNR

[by SAS similarity criterion]

(ii) We have, ΔAMC ~ ΔPNR

⇒

[∵ triangles are similar, so corresponding sides will be proportional]

∴

⇒ [from Eq. (i)]

Or

Prove that:

Ans.

We want to prove that

Now,

∴

Or

The internal and external diameters of a hollow hemispherical vessel are 16 cm and 12 cm respectively. If the cost of painting 1 cm

Ans.

R = 8 cm, r = 6 cm

Total Surface area = 2πR

∴ Total cost= 715.92 x ₹5 = ₹ 3579*60.

Ans.

Multiplying equation (i) by 5 and subtracting from (ii), we have

⇒ f

From (i), we have

f

⇒ f

⇒ f

Hence, missing frequencies are 10 and 8.

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