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**Class X****Mathematics â€“ 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 the HCF of 65 and 117 is expressible in the form 65m - 117, find the value of m : (1 Mark)Ans.** By the Euclid's division algorithm, HCF of (65, 117) = 13

Since 65m - 117 = 13 â‡’ m = 2

(a) exactly two solutions

(b) infinitely many solutions

(c) no solution

(d) a unique solution

Hence, the given system has no solution.

(b) 50.5

(c) 27.5

(d) 20.5

On comparing, we get

a = 20 and h = 10

âˆ´

(a) I quadrant

(c) III quadrant

(d) IV quadrant

Let (x, y) be the point

Here, x

So, the required point lies in IV

Q.6. If sin Î¸ + cos Î¸ = cos Î¸, (Î¸ â‰ 90Â°), then the value of tan Î¸ is :

(a)

(b)

(c)

(d)

Ans.

â‡’ 65m - 117 = 13 â‡’ 65m = 130

âˆ´ m = 2

[Subtracting â€˜1â€™ from both sides]

In Î”PQR, STâ•‘QR.

âˆ´ By Basic Proportionality Theorem

â‡’ [using (1), TR = 15 cm (given)]

â‡’

â‡’ PT = 10 cm

Take a point C on the circumference of the semi-circle and join it by the end points of diameter AB.

âˆ C = 90Â° [Angle in a semi-circle is right angle]

Or

Correct option: (a)

The sharpened part of the pencil is cone and unsharpened part is cylinder.

As from the ratios, values of c are not common. So, there is no common value of c for which lines have infinitely many solutions.

Or

Now, a + d = x

and a

2 + d = x

x + d = 26, 26 + d = y

2 + d + d = 26, 26 + 12 = y

â‡’ d = 12

Hence, 2 + 12 = 14 = x and 26 + 12 = 38 = y

Or

Proportional

âˆ´ 1.42 (terminating decimal) lies between 1.41 and 1.73 i.e.,

âˆ OAC = 90Â° (as radius âŠ¥ tangent)

âˆ BOC = âˆ OAC + âˆ ACO

(Exterior angle property)

or, 130Â° = 90Â° + âˆ ACO

or, âˆ ACO = 130Â° - 90Â° = 40Â°

Ans.

On comparing with ay

we get a = 5, b = - 4 and c = 3

Now, D = b

= 16 - 60

= - 44 < 0

Since, D < 0, so the given quadratic equation has no real roots.

Lower limit of median class = 40.

**Section B**

**Q.21. What should be added to the polynomial x ^{3} - 3x^{2} + 6x - 15 so that it is completely divisible by x - 3. **

OR

Solve the following pair of linear equations by cross multiplication method : (2 Mark)

x + 2y = 2

x - 3y = 7

Remainder = 3

Hence, - 3 must be added.

OR

x + 2y - 2 = 0

x - 3y - 7 = 0

Using the formula

âˆ´ AB = DC and AD = BC ...(i)

We know that length of the tangents drawn from an external point to the circle are of equal lengths.

âˆ´ AP = AS

BP = BQ

CR = CQ

DR = DS

Adding vertically, we have

AB + BP + CR + DR

= AS + BQ + CQ + DS

(AP + BP) + (CR + DR)

= (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

AB + AB = AD + AD [using (i)]

2AB = 2AD

â‡’ AB = AD

Since adjacent sides of a rectangle ABCD are equal.

Hence, ABCD is a square.

PA = PB â‡’ PA

â‡’ (0 + 5)

â‡’ 52 + (y - 2)

â‡’ 25 + (y - 2)

â‡’ (y + 2)

â‡’ 8y = - 56 â‡’ y = - 7.

Hence, the required point is P(0, - 7).

Or

Given the point P(x, y) is equidistant from the points A(5,1)and B(-1,5).

Hence, PA = PB â‡’ PA

â‡’ (x - 5)

â‡’ x

= x

â‡’ 12x - 8y = 0 â‡’ 12x = 8y

â‡’ 3x = 2y

A box contains 80 discs which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect square number.

Total number of possible outcomes in the sample space = 8, i.e., n(S) = 8

Let A be the event of getting either all three heads or all the three tails.

Then, the favourable outcomes are HHH, TTT.

Number of favourable outcomes for the event A = 2, i,eâ€ž n(A) = 2

Probability of winning the game

Or

A box contains 80 discs which are numbered from 1 to 80.

Total number of possible outcomes in the sample space = n(S) = 80

Let â€˜Eâ€™ denote the event that disc drawn from the box is a perfect square number.

E = {1,4, 9, 16, 25,36,49, 64}

âˆ´ n(E) = 8

Probability of getting perfect square number

or, cosec Î¸ = 7/5

or,

sin Î¸ + cos

= sin Î¸ - sin

(âˆµ sin

Ans.

Second Type :

Capacity of glass

Thus, capacity of second type of glass is more.

Hence, Suresh got more quantity of juice.

**Section C**

**Q.27. Two unbiased coins are tossed simultaneously. Find the probability of getting : (3 Mark)**

S = {HH, HT, TH, TT} (optional)

(ii) P (at most one head) = 3/4

(iii) P (no head) = 1/4

âˆ´

Ans.

There are 6 possible outcomes (1, 2, 3, 4, 5 and 6) in a single throw of a die.

(i) We know that, even prime number is only 2.

So, number of favourable outcomes = 1

âˆ´ P(getting a even prime number) = 1/6.

(ii) The numbers divisible by 2 are 2, 4 and 6.

So,number of favourable outcomes = 3

âˆ´ P (getting a even number divisible by 2)

= 3/6 = 1/2

Number of red cards = 26

Number of queens = 4

But, out of these 4 queens, 2 are red.

âˆ´ Number of queens which are not red = 2

Now, number of cards which are red or queen = 26+2 = 28

âˆ´ P (getting either red card or queen)

= 28/52 = 7/13

Now, P (not getting either red card or queen)

= 1 - P (getting either red card or queen)

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

Since point P is equidistant from A and B.

âˆ´ PA = PB

â‡’ PA

â‡’ (0 - 5)

â‡’ (- 5)

â‡’ 25 + y

â‡’ 8y = - 16

â‡’ y = - 2

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

Or

The given points are A(2, 1) and 5(5, - 8). P and Q are points of trisection such that P is nearer to A. Therefore, point P divides AS internally in the ratio 1 : 2.

Using section formula, the coordinates of point P are

It is given that point 5(3, - 2) lies on the line 2x - y + k = 0

âˆ´ 2 x 3 - (- 2) + k = 0

â‡’ 6 + 2 + k = 0

â‡’ k = - 8**Q.31. In the figure, PQ is a tangent to a circle with centre O. If âˆ OAB = 30Â°, find âˆ ABP and âˆ AOB. (3 Mark)**

Î”

Then,

âˆ OAB = âˆ OBA (âˆµ OA = OB)

âˆ OBA = 30Â°

âˆ´ âˆ AOB = 180Â° - (30Â° + 30Â°)

âˆ´ âˆ AOB = 120Â°

âˆ ABP = âˆ OBP - âˆ OBA

= 90Â° - 30Â° = 60Â°

Or

In Î”ACB and Î”CDB

âˆ ACB = âˆ CDB = 90Â°

âˆ B = âˆ B (common)

âˆ´ Î”ACB âˆ¼ Î”CDB (by AA Si.milarity)

or, b/p = c/a

or, 1/p = c/ab

Squaring on both sides,

or,

âˆ´

Or

Ans.

P(6,1), Q(8,2), R(9, 4) and S(P, 3).

We know that diagonals of a parallelogram bisect each other.

Coordinates of mid-point of diagonal PR = Coordinates of mid-point of diagonal QS.

On equating x-coordinates from both sides, we get

â‡’ P = 15 - 8 â‡’ P = 7

Hence the value of P is 7.

Or

Let the line segment joining P(-3, - 2) and Q(1, 4) be divided by the X- axis in the ratio k : 1 at some point A.

PA: AQ = k : 1, then coordinates of

But A lies on the X-axis, therefore its y-coordinate is 0.

â‡’

â‡’ 4k - 2 = 0

â‡’ k = 1/2

âˆ´ The ratio is 1:2.

The coordinates of the point of division A is

Breadth of rectangle = 6 cm

âˆ´ The diagonal of a rectangle

Since the rectangle is inscribed in a circle, therefore the diagonal of a rectangle is the diameter of a circle.

âˆ´ Diameter of a circle = 10 cm

Radius of a circle = 5 cm

From the figure,

Area of the shaded portion

= Area of the circle of radius 5 cm - Area of a rectangle

**Section D**

**Q.35. ****Draw the graphs of the following equations : 2x - y = 1 and x + 2y = 13 Find the solution of the equations from the graph and shade the triangular region formed by the lines and the Y-axis.ORThe sum of the squares of two consecutive odd numbers is 394. Find the numbers.**Given, 2x - y = 1

â‡’ y = 2x - 1

â‡’

Plotting the above points and drawing the lines joining them, we get the graph of above equations.

Two obtained lines intersect at point A(3,5).

Hence, x = 3 and y - 5.

ABC is the triangular shaded region formed by the obtained lines with the Y-axis.

OR

Let the first odd no. be 2x + 1

âˆ´ Second consecutive odd number = 2x + 1 + 2

= 2x +3

Now, according to question,

(2x + 1)

â‡’ 4x

â‡’ 8x

â‡’ x

â‡’ x

â‡’ x(x + 8) -6(x + 8) = 0

x = - 8 or 6

âˆ´ First number = 2 x 6 + 1

= 13

and second odd number = 15

Ans.

Const.: Join BE and CD.

Draw EL âŠ¥ AB and DM âŠ¥ AC.

Since Î”BDE and Î”CDE are triangles on the same base DE and between the same, parallels DE and BC.

âˆ´ Area of Î”BDE = Area of Î”CDE ... (v)

From (iii), (iv) and (v), we have

AD/DB = AE/EC**Q.37. In the given figure, PQL and PRM are tangents to the circle with centre O at the points Q and R, respectively. If S is a point on the circle such that âˆ SQL = 50Â° and âˆ SRM = 60Â°, then find the value of âˆ QSR. (4 Mark)**

Therefore, âˆ OQL = âˆ ORM = 90Â°

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

Now, âˆ OQS = âˆ OQL - âˆ SQL = 90Â°-50Â° = 40Â°

But âˆ OOS = âˆ OSQ

[âˆµ angles opposite to equal sides are equal] âˆ OSQ = 40Â°

Similarly, âˆ OSR = âˆ ORS = âˆ ORM - âˆ SRM

= 90Â°-60Â° = 30Â°

Now, âˆ QSR = âˆ OSQ + âˆ OSR = 40Â° + 30Â° = 70Â°

Then, volume of the bucket, i.e., capacity of a bucket.

.......(1)

It is given that the capacity of a open bucket

= 12308.8 cm

From (1) and (2), we have

â‡’

â‡’ h = 15 cm

Slant height of a open bucket (l)

Total area of metal sheet used in the bucket

Area of semi-circle =1386/2 = 693 cm

= 1386/2 = 693 cm

Area of shaded region = 693 - 441

= 252 sq cm

Or

Since âˆ C = 90Âº

AC = BC = x(say)

AB

AB

2x

or, x = 2 and r =

âˆ´ Slant height of conical portion = 2 m

Total surface area of toy = 2Ï€rh Ï€r

Let assumed mean (A) = 170

Modal class is 160-180

âˆ´l = 160, h = 20, f

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