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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. If two positive integers a and b are written as a = x ^{3}y^{2} and b = xy^{3}; x, y are prime numbers,, then HCF (a,b) is : (1 Mark)** Correct option : (b)

(a) xy

(b) xy^{2}

(c) x^{3}y^{3}

(d) x^{2}y^{2 }

Ans.

Explanation : Since a = x

Thus, HCF of a and b =

(a) 1

(b) 3

(c) 5

(d) 4

Ans.

[For any n ∈ N, 6

∴ 6

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Ans.

⇒ sinA = sin(90° - B)

⇒ A = 90° - B

⇒ A+ B = 90°

(a) 11.2 cm

(b) 5.71cm

(c) 8.14 cm

(d) 11.4 cm

Ans.

Given, ΔABC - ΔDEF, EF - 15.4 cm, ar(ΔABC) = 64 sq. cm and ar(ΔDEF) = 121 sq. cm.

The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides

(a) 2 : 3

(b) 4 : 9

(c) 81 :1 6

(d) 16 : 81

OR

In the given figure, AT is a tangent to the circle with centre 'O' such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to :

(a) 4 cm

(b) 2 cm

(c) 2√3 cm

(d) 4√3 cm

Ans.

Thus, required ratio of the areas of the two

∴ ∠OAT = 90°,

Given that, OT = 4 cm

Ans.

Now, area of a quadrant

Ans.

Distance between two parallel tangents = Diameter = 2 x Radius

= 2 x 6 cm

= 12 cm

(b) 16 : 9

(c) 4 : 3

(d) 16

Ans.

Explanation :

Now, volumes of two spheres,

(a) 0

(b) 1

(c) 2

(d) 3

Ans.

[Here, P(2, a) is the mid-point of the line segment joining L(6, -5) and M(-2, 1)

Ans.

Ans.

Let S

...(1)

⇒ a

Putting n = 2 m (1), we get

S_{2} = 2 x 2^{2}+ 6 x 2 = 8 +12 = 20

...(3)**Q.13. In figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20 √3 m long. Find the Sun's altitude. (1 Mark)**

Ans.

⇒ θ = 30°

Thus, the Sun's altitude is 30°.

Ans.

Ans.

Ans.

2 sin 2θ = 1

Ans.

Ans.

⇒ x + y = 1

⇒ T

Hence, the 9th term of the AP is 22.

Ans.

Since OTPQ is a cyclic quadrilateral

∴ ∠QOT + ∠QPT = 180°

⇒ ∠QOT + 70° = 180°

⇒ ∠QOT= 180° -70°

⇒ ∠QOT = 110° ...(1)

**Section** **B**

**Q.21. Find the smallest natural number by which 1,200 should be multiplied so that the square root of the product is a rational number. (2 Mark)Ans. **Since, 1,200 = 4 x 3 x (2 x 5)

= 2

Hence, the required smallest natural number is 3.

Ans.

Or

A parallelogram has vertices P(1,4), Q(7,11), R(a,4) and S(1,-3). Find the value of a.

Ans.

Hence, the given triangle is a scalene triangle.

Let P(1,4), Q(7,11), R(a,4) and S(1,-3) be the vertices of a parallelogram POPS, respectively. Join PR and QS. Let PR and QS intersect at the point T.

We know that, the diagonals of a parallelogram bisect each other.

So, T is the mid-point of PR as well as that of QS. Mid-point of PR = Mid-point of QS

On comparing x-coordinate from both sides, we get

Hence, the value of a is 7.

Ans.

∴ Total number of possible outcomes in the sample space = 8.

Let A be the event of getting exactly one head.

Then, cases favourable to the event A are HTT, THT and TTH.

Number of cases in favour of event A = 3.

Probability of getting exactly one head - P(A)

Or

If the total surface area of a solid hemisphere is 462 cm

Ans.

OR

Ans.

Shopkeeper is telling truth.

Difference in the quantity of the two tankers = 1.18 - 0.52 = 0.66 m

**Section C**

**Q.27. A bag contains cards numbered 1 to 49. Find the probability that the number on the drawn card is : (3 Mark)(i) an odd number(ii) a multiple of 5(iii) Even primeAns.** Total cards = 49

Odd numbers from 1 to 49 are 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47 and 49.

No. of favourable outcomes = 25

(ii) Multiple of 5 are 5,10, 15, 20, 25, 30, 35, 40 and 45.

No. of favourable outcomes = 9

(ii) P(multiple of 5) = 9/49

Only even prime number is 2 i.e., P(even prime) 1. D

(iii) P(even prime)= 1/49

Ans.

Ans.

Then, the speed for the journey of 72 km will be (x + 6) km/h.

[by factorisation method]

Since, x is the average speed of the train, so x cannot be negative.

Hence, the original average speed of the train is 42 km/h.

Let the point P(- 4, y) divides the line segment joining the points A(- 6,10) and 5(3, - 8) in the ratio k: 1.

Using section formula, the coordinates of point P are

It is given that point P is (-4,y).

∴ The point P divides AB in the ratio k : 1 i.e., 2/7: 1 i.e., 2 : 7 and y = 6.

Or

Let the given points be (-5,1),B(1,p) and C(4,-2). Since the points A(- 5, 1),B(1,p) and C(4,-2) are collinear, therefore, the area of the triangle formed by them is zero.

∴ Area of ΔABC = 0

Hence, the points are collinear when p = - l .

Ans.

Using Pythagoras theorem in the right AABC, we have

Q.32. Prove that:

Ans.

Ans.

∴ Height of multistoried building,

DC = 10.92 + 8 = 18.92 m

Hence, the distance between the two buildings is 18.92 m.

Ans.

Lower limit of modal class, I = 20; Class size, h = 5.

Frequency of modal class, f

Frequency of class succeeding the modal class, f

Using the formula :

**Section D**

**Q.35. Solve graphically the pair of linear equations :3x - 4y + 3 = 0 and 3x + 4y - 21 = 0.Find the co-ordinates of the vertices of the triangular region formed by these lines and x-axis. Also, calculate the area of this triangle.**

(i) These lines intersect each other at point (3, 3).

Hence, x = 3 and y = 3.

(ii) The vertices of triangular region are (3, 3), (-1, 0) and (7, 0).

Or

Ans.

Produce PM to N, such that PM = MN.

Join BE and QN.

Ans.

In ΔABC, it is given that

∴ PTCB is a parallelogram

⇒ ∠1= ∠2

On applying angle sum property in ΔAPQ, we get

...(1)

In ΔADB, we have

Time taken by the boat in moving C to D = 2 minutes

Or

...(1)

In ΔACB, we have

...(2)

Putting the value of x from (2) in (1), we get

Ans.

⇒

⇒ h = 20 m.

Ans.

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