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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. A chord of circle of radius 14 cm subtends an angle of 45Â° at the centre. Then the area of the corresponding sector is (1 Mark)****(a) 50 cm ^{2}**

We have, Area Of the sector of Circle of radius 21 cm subtending an angle of 45Â° at the centre

= (11 x 7 ) cm

Here, first term,a = 5

and common difference, d = 2 - 5 = -3

âˆ´ Fifth term, T

= 5 - 12 = -7

[Here, Modal class is 155 - 160 and median class is 160 - 165

Now, required sum= 155 + 165 = 320

(a) 8

(b) 2âˆš7

(c) 10

(d) 6

Where, x

So, distance between P (-6,8) and origin 0 (0,0) is given by,

= âˆš100

= 10

(a) X

(b) x

(c) 2x

(d) 3x

x

x

2x

3x

âˆ´ 2 is the root of the equation if 2x

(a) 1

(b) 2

(c) 3

(d) 0

(a) 0

(b) -1

(c) 1

(d) 2

[Given that sin Î± + sin

â‡’ sin Î± = 1 - sin

â‡’ sin

â‡’ 1 = cos

(a) R

(b) R

(c) R

(d) Nothing definite can be said about the relation among R

Here,

âˆ´

Hence, the given system of linear equation is consistent.

(a) 21 cm

(b) 16 cm

(c) 17 cm

(d) 18 cm

In right angled Î”OPQ, use Pythagoras theorm

Slant height of the frustum

l = âˆšh

= âˆš(16)

= âˆš256 + 144

= âˆš400

= âˆš20 cm

Q. Assertion (A): The discriminant of the quadratic equation (x + 5)

The given equation is

(x + 5)

â‡’ x

â‡’ x

Comparing it with ax

a = 1, b = 0, c = 31

âˆ´ D = b

Solution:

[P (an orange ball =

Number of orange ball = 10

âˆ´ , where x is total number of ball

â‡’

= 1/2 [ x

= 1/2 [0(0-5) + 6(5-0) + 0(0-0)]

= 1/2 [6 x 5] = 15 sq. units

CP = CQ, AR = AQ and BR = BP

Now perimeter of Î”ABC = AB + BC + CA

= (AR + BR) + (BP + CP) + (AQ + CQ)

= (AR + AQ) + (BR + BP) + (CP + CQ)

= (AQ + AQ) + (BR + BR) + (CP + CP)

= 2AQ + 2BR + 2 CP

= 2(4) + 2(6) + 2(5) cm

= (8 + 12 + 10) cm

= 30 cm

In right angled Î”ABC,

â‡’ 6 x AB = 18 x 7 â‡’ AB = =21

Hence, the length of the shadow is 21 m.

Solution:

Î± = 18, d = -2 and S

Now,

â‡’ 18n - n

â‡’ 19n - n

â‡’ n(19 - n) =0

â‡’ n = 0 or n = 19

(a) q

(b) q + 1

(c) 2q

(d) 2q + 1

Explanation : According to Euclid's division lemma, a = bq + r, where o

lf b = 2, then a â€” 2q or a â€” 2q + 1

Since a = 2q + 1 is not divisible by 2, then 2q + 1 is an odd integer.

**Section B**

**Q.21. Points A(3,1), B(5,1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b. (2 Mark)**

The coordinates of the mid-point of diagonal BD are

We know that the diagonals of a parallelogram bisect each other. So, coordinates of the mid-point of diagonal AC are the same as the coordinates of the mid-point of diagonal BD.

âˆ´

â‡’ and

â‡’ 3 + a = 9 and 1 + b = 4

â‡’ a = 9 - 3 and b = 4 - 1

â‡’ **a = 6 and b = 3****Or****Points P and Q trisect the line segment joining the points A(- 2, 0) and B(0, 8) such that P is nearer to A. Find the coordinates Of points P and Q.****Solution:** The given points are A(- 2, 0) and B(0, 8). P arid Q are the points of trisection such that P is nearer to A. Then AP - PQ = QB.

Therefore, point P divides AB internally in the ratio 1 : 2,

Using section formula, the coordinates of point P are

Also, point Q divides AB internally in the ratio 2:1,

Using section formula, the coordinates of point Q are

Hence, the points of trisection are P **Q.22. In the given figure, if DE â•‘ BC, then find the value of x. (2 Mark)**

â‡’ ...... (1)

[by basic proportionality theorem]

â‡’

â‡’

âˆ´.....(1)

Prove that: AC

or

In a Î”PQR, N is a point on PR such that QN âŠ¥ PR. If PN.NR = QN

â‡’ BC = 2BD

Now, in right AABD, we have

AD

BD

In right AABC, we have

AC

= 4(AD

= 4AD

= 4AD

Or

Here, QN

â‡’ âˆ l = âˆ 2 â€” 90Â°

Also, PN.NR = QN.QN

â‡’

âˆ´ Î”PNQ - Î”QNR

â‡’ âˆ 3 = âˆ 4 and âˆ 5 = âˆ 6

[âˆµ corresponding angles of similar Î”s]

Now, in APQR, we have

âˆ 3 + âˆ 4 + âˆ 5 + âˆ 6 = 180Â°

âˆ 4 + âˆ 4 + âˆ 5 + âˆ 5 = 180

[âˆµ âˆ 3 = âˆ 4 and âˆ 6 = âˆ 5]

â‡’ 2(âˆ 4 + âˆ 5) = 180

â‡’ âˆ 4 + âˆ 5 = 90Â°

â‡’âˆ PQR = 90Â°

âˆ´ PQ = OP = OQ

âˆ´ âˆ POQ = 60Â°

(angle of equilateral Î”.) 1/2

âˆ P = âˆ Q = 90Â°

(radius âŠ¥ tangent)

âˆ´ âˆ PTQ + 90Â° + 90Â° + 60Â° = 360Â°

(angle sum property) 1/2

âˆ PTQ = 120Â°

x

Comparing it with ax

a - 1, b = -8, c = 18

D = b

= (- 8)

=64 - 72

= - 8 < 0

Here, D < 0.

âˆ´ The given equation has no real roots.

Thus, the given equation has

â‡’ 4600 + 50p = 5160 + 30p

â‡’ 50p - 30p = 5160 - 4600 â‡’ 20p = 560

âˆ´ p = 28....(i)

Or

Here, the ciass 30-35 has maximum frequency. So, it is the modal class

âˆ´ l = 30, h = 5, f

Mode = .......(1)

=

=

= 30 + 0.625 = 30.625.........(i)

**Section C**

**Q.27. The perpendicular from A on side BC of a Î”ABC meets BC at D such that DB **=** 3CD. Prove that 2AB ^{2} = 2AC^{2} + BC^{2}. (3 Mark)**

We have,

DB = 3 CD

âˆµ BC = BD + DC

â‡’ BC = 3CD + CD

â‡’ BC = 4CD

â‡’

âˆ´

In right triangle ABD, right angled at D, we have

AB

Also, in a right triangle ACD, right angled at D), we have

AC

Subtracting (2) from (1), we get

AB

â‡’

â‡’

â‡’

â‡’

â‡’ 2(AB

â‡’

Given : Two triangles ABC and FOR such that Î”ABC

âˆ´ BC = 2BD and QR = 2QM

Now, Î”ABC

âˆ´ and âˆ B = âˆ Q

â‡’

Thus, in Î”ADB and Î”PMQ, we have

By SAS-criterion of similarity, we have

â‡’

â‡’

â‡’

â‡’

â‡’ - 30 = x

On comparing the standard quadratic equation ax

a = 1, p = - 3 and c = 2

By using quadratic formula, we get

Hence, the roots of the given equation are 2 and 1. ......(1)

or

Let .....(i)

Then, (1/2)

Therefore, the given equation reduces to

â‡’ 2y

[by factorisation method]

â‡’ 2y (y -5 ) + 5(y-5) = 0 â‡’ (y-5)(2y + 5) = 0

â‡’ ......(i)

Now, putting y = 5 in Eq. (i), we get

â‡’ 5x - 15 = 2x + 3

â‡’ 3x = 18 â‡’ x = 6.....(1/2)

Again, putting y = in Eq. (i), we get

â‡’ - 5x +15 = 4x + 6

âˆ´ 9 x = 9 â‡’ x = 1

Hence, the values of x are 1 and 6.....(i)

Or

At present Asha's age (in years) is 2 more than the square of her daughter Nishaâ€™s age. When Nisha grows to her motherâ€™s present age. Ashaâ€™s age would be one year less than 10 times the present ag e o f Nisha. Find the present ages of both Asha and Nisha.

= (x - y) km/h

Effective speed in downward journey

= (x + y) km/h

According to the statement of the question, we obtain

and

Put we obtain

30a + 44b = 10 .....(i)

40a + 55b = 13 .....(ii)

By using cross-multiplication, we have

â‡’

Again, putting the values of a and b, we obtain

x - y= 5 and x + y =11

Adding the two equations, we have

2x = 16 â‡’ x = 8

And x + y =11

â‡’ y = 11 - 8 = 3

Hence, the speed of the boat in still water is 8 km/h.

Or

Let present age of daughter Nisha be x years

âˆ´ Present age of Asha = (x

Then, her motherâ€™s age would be

x

Now, as per statement of the question, we obtain

â‡’ 2x

â‡’ 2x

â‡’ (x - 5) (2x - 1) = 0

â‡’ Either x - 5 = 0 or 2x - 1 = 0

â‡’ x = 5 or x = 1/2 (Rejecting)

Hence, present age of Nisha is 5 years and her motherâ€™s age 5

AP + PB + DR + CR = AS + BQ + DS + CQ

or, AB +C D = AD + BC

From (i), 2AB = 2AD or AB = AD

or, ABCD is a rhombus

OR

We have, for equation

(2m - l) x + 3y -5 = 0 ...(i)

a

and for equation

3x + (n - 1)y - 2 = 0 .....(ii)

a

For a pair of linear equations to have infinite number of solutions

or

or 2(2m -1) = 15 and 5(n-1) = 6

Hence,

(i) sin = cos A/2

(ii) If âˆ A = 90Â°, then find the value of tan

Solution:

âˆ´A + B + C = 180Â° ....(i)

â‡’ B + C = 180Â° - A

â‡’

â‡’

(ii) When A = 90Â°, from (1), we get

90Â° + B + C = 180Â°

â‡’ B + C = 180Â° - 90Â° = 90Â°

â‡’

tan (A + B) = 1

â‡’ tan (A + B) = tan 45Â°

â‡’ A + B = 45Â° ..........(i)

and

Adding and subtracting (1) and (2), we get

2A = 75Â° and 2B =15Â°

â‡’ and

â‡’ and

Now, the perimeter of the shaded region = AB + PB + AP

Solution:

âˆ´

â‡’

â‡’

âˆš5 is irrational and is rational (as assumed).

But rational number cannot be equal to an irrational number

âˆ´3 + âˆš5 is an irrational number.

**Section D**

**Q.35. A pole has to be erected at a point on the boundary of a circular park of diameter 13 m in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 m. Is it possible to do so ? If yes, at what distances from the two gates should the pole be erected ? (4 Mark)**

Let A and B be positions of fixed gates which are diametrically opposite in a circular park of diameter 13 m. Let P be the position of the pole such that AP = x m and BP = y m; x > y.

According to the given condition, we have

x - y = 7

â‡’ x = y + 7 ...(1)

Since angle in a semi-circle is 90Â°

â‡’ âˆ APB = 90Â°

In right triangle ABP, we have

AP

â‡’ x

â‡’ (y + 7)

â‡’ y = 5 [ âˆµ y cannot be negative]

âˆ´ x =y + 7 = 5 + 7 = 12

Hence, distances of the pole from the gates are

Solution:

and 3 x - y = 3 ..(ii)

Table for 5x - y = 5 or y = 5x - 5 is

Plot the points 4(1, 0) and B( 0, - 5) on a graph paper and join these points to form a line AB. (Table for 3x- y = 3 or v = 3x - 3 is

Plot the points A (1,0) and C (0, -3) on the same graph paper and join these points to form a line AC.

Hence, the triangle formed by given lines is Î”ABC whose vertices are A(1,0),B(0, - 5) and C(0, -3).

Or

in the return journey. If he returned at a speed of 10 km/hour more than the speed while going, find the speed per hour in each direction.

By the given conditions, we have

x

and y

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

x

â‡’ x

Which is a quadratic equation.

Here, a = 1, b = - 8 , c = - 180

By the quadratic formula, we have

x= 18; x = - 10, it is not possible.

From equation (ii), we have

y

â‡’ y

Hence, required numbers are 18, 12 and 18 ,- 1 2 .

Or

Let the speed of onward journey be x km/h

âˆ´ Speed of return journey â€” (x + 10) km/h

Total distance = 150 km

cording to the statement of the equation, we obtain

Rejecting x = - 30, because speed cannot be negative, we have x = 20

Hence, speed of onward journey is 20 km/h and speed of return journey is 20 + 10

i.e., 30 km/h.

OM âŠ¥ PQ, bisects PQ.

PM = 8 cm (as PQ = 16 cm.)

or, TP = 80/6

or, TP = 40/3

Hence length of TP = 40/3 cm.

OR

Steps of Construction:

1. Draw a line segment BC of length 5 cm.

2. Draw the angles of 60Â° and 30Â° on the points B and C respectively which intersect each other at A.

3. Î”ABC is the given triangle.

4. Draw a ray BX making an acute angle with BC.

5. Locate three points B

6. Join B

7. Draw 'B

8. Through C draw a line parallel to AC inlersectiug extended line segment BA at A'. Î”A'BC'is the required triangle.

To prove:

âˆ´

and

â‡’ ...(3) [using (1) and (2)]

Similarly, ....(4)

Note that Î”BDE and Î”DEC are on the same base DE and between the same parallels BC and DE

âˆ´ ar(Î”BDE) = ar(Î”DEC) ....(5)

From (4) and (5), we have

.....(6)

Again from (3) and (6), we have

Hence,

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

i.e., AC

In Î”ADB and Î”ABC

âˆ ADB = âˆ ABC

and âˆ A = âˆ A

âˆ´ . By AA-similaritv criterion, we have

So, [In similar triangles, corresponding sides are proportional]

â‡’ AD.AC =AB

In Î”BDC and Î”ABC

âˆ CDB = âˆ ABC [Each = 90Â°]

and âˆ C = âˆ C

âˆ´ ABDC ~ AABC [By AA-similarity]

So, [In similar triangles, corresponding sides are proportional]

â‡’ CD.AC = BC

Adding (1) and (2), we have

AD.AC + CD. AC = AB

â‡’ (AD + CD) AC = AB

â‡’AC.AC = AB

Hence, AC

What is the probability that it will point at

(i) 8?

(i) 4 or 5.

âˆµ Total number of points on the circle = 8

(i) Let E

âˆ´ Number of outcomes favourable to E

Probability that arrow comes at number 8,

(ii) Let E

Here, odd numbers are 1,3,5 and 7.

âˆ´ Number of outcomes favourable to E

Probability that arrow comes at an odd number,

(iii) Let E

âˆ´ Number of outcomes favourable to E

(iv) Let E

âˆ´ Number of outcomes favourable to E

Probability that arrow comes at a number less than 9,

Or

Total number of possible outcomes on rolling two dice together = 36

(i) To get the sum of numbers, 4 or 5, favourable outcomes are (1,3), (3, 1), (2, 2), (1,4), (4, 1), (2, 3) and (3, 2).

âˆ´ Number of favourable outcomes = 7

Now, P (getting a sum 4 or 5) = 7/36

(ii) To get the sum of numbers, 7, 8 or 9, favourable outcomes are (1,6), (6,1), (2,5), (5, 2), (3, 4), (4, 3), {2, 6), (6, 2), (3, 5), (5, 3), (4, 4), {3, 6), (6, 3), (4, 5) and (5, 4).

âˆ´ Number of favourable outcomes = 15

Now, P (getting a sum 7, 8 or 9) =

(iii) To get the sum of numbers, between 5 and 8, i.e. getting sum 6 or 7, favourable outcomes are (1, 5), (5,1), (2, 4), (4, 2), (3,3), (1,6), (6,1), {2, 5), (5, 2), (3, 4) and (4, 3).

Number of favourable outcomes = 11 Now. P faettina a sum between 5 and 8)

= 11/36

(iv)To get the sum of numbers more than 10, i.e. getting sum 11 or 12, favourable outcomes are (5, 6), (6, 5) and (6, 6).

âˆ´ Number of favourable outcomes = 3

Now, P (getting a sum more than 10)

=

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