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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 an event that cannot occur, then its probability is : (1 Mark)****(a) 1****(b) 3/4****(c) ****1/2****(d) 0****Ans.**(d)

An event that cannot occur has 0 probability, such an event is called impossible event.**Q.2. The smallest rational number by which 1/7 should be multiplied so that its decimal expansion terminates after one place of decimal is: (1 Mark)**

(b) 7/10

(c) 7/100

(d) 1/10

[Here, 1/7 is non - terminating decimal expansion

which terminates after one place of decimal]

(a) 8, 5

(b) - 8, - 5

(c) 8, -5

(d) -8, 5

Ans.

The zeroes of the given polynomial are given by the equation

x^{2} - 13x + 40 = 0

â‡’ x^{2} - 5x - 8x + 40 - 0

â‡’ x(x - 5) - 8(x - 5) = 0

â‡’ (x - 5)(x - 8) = 0

â‡’ x = 5 = 0 or x - 8 = 0

â‡’ x = 5 or x = 8

Hence, the zeroes of the Polynomial are 5 and 8.**Q.5. (sec A + tan A) (1 - sin A) = (1 Mark)**

(sec A + tan A) (1 - sin A)

[Here, p sin (90Â° - Î¸) cot (90Â° - Î¸)

= cos (90Â° - Î¸) â‡’ cos Î¸.tan Î¸ = sin Î¸

Splitting the middle term we get,

2x

â‡’ 2x (x - 3) + 1 (x - 3) = 0

â‡’

Hence, x = 3 is the solution.

b

[Here, PA = PB, where P(x, y) is any point on the perpendicular bisector of AB.

Î¸ = Ï† = Ï€/4

Total number of outcomes = 52

Number of favourable events = 3 as there are only 3 face cards of spade i.e., J, Q, K.

âˆ´ P(Face cards of space) = 3/52.

âˆ´

â‡’ k = 7

10 - x = 3x + 2 - x - 10

â‡’ 3x = 18 â‡’ x = 6]

Comparing it with the standard equation ax

a = 3, b = k, c = 3

The roots will be real and equal if discriminant,

D = b

i.e., k

â‡’ k

â‡’ k = Â± 6

Hence, the roots of the equation are real and equal for k = Â± 6.

âˆ´

252, 255, 258, ..., 999

Using a + (n - 1)d = a

252 + (n - 1)3 = 999

(n - 1)3 = 999 - 252

= 747

= n - 1 = 747/3 = 249

âˆ´ n = 249 + 1 = 250

Now, required sum of all such numbers

Or

Find the values of t, so that (t + 4)x

Ans. Given, 3a

a = 3,b = 6 andc = 1

Discriminant, D = b

= (6)

= 36 - 12 = 24 >0

Hence, the given equation has two distinct real roots.

Or

Discriminant = b

= (t + 1)

= t

For equal roots, discriminant = 0

âˆ´ t

â‡’ t

â‡’ t = 5 or t = - 3

Hence, the value of t is 5, - 3.

The coordinates of the point satisfy the above equation.

2 x 3 - 3 x a = 5

â‡’ - 3a = - 1

â‡’ a = 1/3

**Section B**

**Q.21. Explain whether 3 x 12 x 101 + 4 is a prime number or a composite number. (2 Mark)**

= 4(909 + 1) ................

= 4(910)

= 2 x 2 x 2 x 5 x 7 x 13

= a composite number

[âˆµ Product of more than two prime factors]

Since OP is the radius of the circle and 1 is tangent.

âˆ´ OP âŠ¥ l

Similarly, OQ âŠ¥ m

Since l and m are two parallel tangents.

âˆ´ Perpendicular at P and Q passes through the centre of the circle.

By using prime factorisation method,

Now, 15 = 3 x 5

and 20 = 2 x 2 x 5 = 2 2 x 5

âˆ´ LCM (15, 20) = 2

In every 60 min, tour leaves at the same time.

âˆ´ HCF = 2

Or

Let a be any odd positive integer.

When a is divided by 6, remainder will be one of 0, 1, 2, 3, 4, 5.

a = 6m + 0 i.e., 6m or 6m +1 or 6m + 2 or 6m + 3 or 6m + 4 or 6m + 5

When a = 6m, which is even positive integer as it is divisible by 2.

When a = 6m + 2 = 2(3m + 1), which is even positive integer as it is divisible by 2.

When a = 6m + 4 = 2(3m + 2), which is even positive integer as it is divisible by 2.

Hence, every odd positive integer is of the form 6m + 1 or 6m + 3 or 6m + 5.

Therefor

Hence, required ratio is 5 :1

Hence, point on y- axis is

= 2.8 m

Height of cylindrical portion = 3,5 m

Height of conical portion = 2.1 m

âˆ´ Slant height of conical portion

Area of canvas used to make the tent

=C.S.A. of cylindrical portion + C.S.A. of conical portion

**Section C**

**Q.27. Find the roots of the quadratic equation : (3 Mark)**

â‡’ b

â‡’ (b

Now, length of first semicircle (l

= Perimeter of first semicircle

Now, length of second semicircle (l

= Perimeter of second semicircle

Similarly, we can get the length of other

and so on upto 13 semicircles.

Let S be total length of all semicircles.

â‡’ x

â‡’ -1(p + 1)+2(-1-3) + 5(3-p) = 0

â‡’ - p - 1 - 8 + 15 - 5p = 0

â‡’ - 6p + 6 = 0 â‡’ p = 1

Thus, for p = 1, the given points are coliinear.

Now, let S(2,p) = B(2,1) divides the line segment joining the points A and C internally in the ratio k:1.

By using internal section formula, we get

â‡’

[comparing x-coordinate from both sides]

â‡’ 5k - 1 = 2k + 2

â‡’ 5k - 2k = 2 + 1

â‡’ 3k = 3 â‡’ k = 1

Hence, the required ratio is 1:1.

Or

Let PQRS be a square with opposite vertices P (- 1,2) and R (3,2). Again, iet the coordinates of other vertices Q be (x, y).

âˆµ All sides of a squareâ€™are equal,

âˆ´ PQ = QR

â‡’ PQ

Hence, other two vertices of the square are (1,0) and (1, 4).

âˆ´ Total charges for 25 days = Fixed hostel charges + Food charges for 25 days

= Rs. (x + 25y) = Rs. 4500 (Given)

âˆ´ x+ 25y = 4500 ...(1)

Similarly, x + 30y = 5200 ...(2)

Subtracting (1) from (2), we get

5y = 70

â‡’ y = 140

Putting the value of y from (3) in (1), we get ...(3)

x + 25 x 140 = 4500

â‡’ x + 3500 = 4500

â‡’ x = 4500 - 3500 = 1000

âˆ´ Fixed hostel charges = Rs. 11000 and cost of food per day = Rs. 140.

Diagonals of parallelogram bisect each other

âˆ´ Radius of semi-circle, ACQA = 7âˆš2 cm

Now, area of shaded portion = Area of APCQA

= Area of semi-circle ACQA - (Area of quadrant BAPC - Area of Î”ABC)

= 154 - 154 + 98

= 98 cm

Or

Given, length of rope = 5 m

Radius of arc = 5 m

(i) Area of the field the horse can graze

(ii) If length of rope = 10 m = r

Then, area of the field graze by the horse,

âˆ´ Required increase in the grazing area

= A

= 58.875 m

âˆ´ Volume of the sand in the cylindrical bucket = Ï€r

Height of conical heap, H = 24 cm

Let R be the radius of the conical heap.

Volume of the cylindrical bucket = Volume of the conical heap

âˆ´

â‡’ R

â‡’ R = 36 cm

Slant height l of the conical heap is given by

Hence, the slant height of the heap = 43.2 cm

**Section D**

**Q.35. The following table gives the weight of 120 articles :****Change the distribution to a 'more than type' distribution and draw its give. (4 Mark)**

âˆ A = âˆ P and âˆ C = âˆ Q

In Î”BPQ and Î”BAC [âˆµ corr. angles]

âˆ B = âˆ B [common]

âˆ´ Î”BPQ ~ Î”BAC

[by AAA similarity rule]

To prove Î”AFE, Î”FBD, Î”EDC and Î”DEF are similar to Î”ABC.

Proof In Î”ABC, F and E are the mid-points of sides AB and AC, respectively

FE â•‘ BC

[segment joining the mid-points of any two sides of triangles is parallel to the third side]

Similarly, DF â•‘ CA and DE â•‘ BA

In Î”AFE and Î”ABC,

âˆ A = âˆ A [common angle]

âˆ AFE = âˆ B [corresponding angles]

âˆ´ Î”AFE âˆ¼ âˆ ABC ...(i)

[by AA similarity criterion]

Similarly, we can prove

Î”FBD âˆ¼ Î”ABC ...(ii)

and Î”EDC âˆ¼ Î”ABC ...(iii).

Now, in Î”DEF and Î”ABC,

âˆ D = âˆ A

[opposite angles of parallelogram AFDE]

âˆ DEF = âˆ B

[opposite angles of parallelogram BDEF]

âˆ´ Î”DEF - Î”ABC ......(iv)

[by AA similarity criterion]

From Eqs. (i), (ii), (iii) and (iv), we conclude that each of Î”AFE, Î”FBD, Î”FDC and Î”DEF is similar to Î”ABC. Hence proved.

Or

In right angled Î”PSQ and Î”PSR,

(PQ)

and (PR)

On adding Eqs, (i) and (ii), we get (PQ)

Hence, Î”PQR is a right angled triangle, right angled at P.

The angle of elevation of an aeroplane from a point A on the ground is 60Â°. After a flight of 30 seconds, the angle of elevation changes to 30Â°. If the plane is flying at a constant height of 3600âˆš3 metres, find the speed of the aeroplane.

AC = 100 m, âˆ ACB = 30Â° and âˆ AEF = 45Â°

From right Î”ACB, we have

â‡’ 1/2 = AB/100

Hence, distance of the bird from the girl = 42.42 m.

Or

Let P and Q be initial and final positions of the aeroplane. Let ABC be the horizontal line through point of observation A It is given that the angles of elevation of the plane in two positions P and Q from the point A are 60Â° and 30Â° respectively. Draw Pi? and QC perpendiculars from P and Q on the horizontal. The plane is flying at a constant height 3600âˆš3m.

Distance travelled by the plane = PQ = BC = AC -AB = (10800- 3600) m = 7200 m

Time taken by the plane in travelling from P to Q = 30 seconds.**Q.39. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is 29/20. Find the original fraction. (4 Mark)**

So, the fraction = x-3/x

By the given condition, new fraction

= x-1/x+2

â‡’ 20[(x-3) (x + 2) + x (x - 1)] - 29(x

â‡’ 20(x

â‡’ 11x

â‡’ 11x

(11x + 12) (x - 10) = 0 or x = 10

The fraction is 7/10

Median : We have

Mode : We have

Modal class = 20 - 30 [âˆµ Max. frequency 36 belongs to 20 - 30]

l = 20, h = 10, f

Now

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