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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 two solid hemispheres of the same base radius r are joined together along their bases, then volume of this new solid is (1 Mark)(a) 4/3Ï€r ^{3} (b) 1/3Ï€r^{3} (c) 2/3Ï€r^{3} (d) 4Ï€r^{2}Ans: ** (a)

When two solid hemispheres of the same base radius r are joined together along their bases, then new solid formed is a sphere of radius r.

âˆ´ Volume of this new solid formed

(b) -2

(c)-6

(d) -4

Ans: (a)

Since, Î± and Î² are zeroes of p( x) = x

(a) 23

(b) 24

(c) 27

(d ) 25

Ans:

(a) âˆš3

(b) 1/ âˆš3

(c) âˆš3/2

(d) 1

Ans:

Given, sin A = 1/2

(a) 32 cm

(b) 16 cm

(c) 4âˆš2 cm

(d) 16âˆš2 cm

Ans:

Given, AC = 4 cm and ABC is an isosceles triangle right angled at C.

CB = AC = 4 cm

By Pythagoras Theorem in right triangle, right angled at C,

(a) 6 units

(b) 5 units

(c) 4 units

(d) 2 units

Ans:

(a) 0Â°

(b) 30Â°

(c) 60Â°

(d) 90Â°

Ans:

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than 2 real roots

Ans: (c)

Therefore, the equation has no real roots

Radius of hemisphere, r = 7 cm

âˆ´ Curved surface area of the hemisphere =

(a) 3

(b) 4

(c) 2

(d) 5

Ans:

[âˆ´ dividing numerator and denominator by cos Î¸ ]

Ans:

Ans:

Here, roots of the quadratic Polynomial are given as - 3 and 4.

âˆ´ Sum of the roots = - 3 + 4 = 1

and Product of the roots = - 3 X 4 = - 1 2

âˆ´ The quadratic polynomial is

Hence the required polynomial is

(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.

Reason (R):

Ans:

Ans:

**Q 15. If a number x is chosen from the numbers 1, 2, 3 and a number y is selected from the numbers 1, 4, 9, then P(xy < 9) =_______. (1 Mark)Ans:** 5/9

[Sample space = 3 x 3 = 9 and xy < 9

are (1, 1), (1, 4), (2, 1), (2, 4), (3, 1) i.e., 5 cases

Hence

Ans

Ans:

Given sin A = 3/4

we know that

sin

cos

or

How many terms are there in the following AP?

25, 25, 30, .....100

Ans:

we have

[âˆµ putting n = 15, given]

or

Let the number of terms be n.

We have, a

âˆ´ Using the formufa, a

= 10 = 20+ (n-1) (5)

n - 1 = 16

n = 17

Ans:

OR

In the given figure, PQ and PR are tangents to the circle with centre O such that âˆ QPR = 50Â°, then find âˆ OQR.

or

**Section B**

**Q 21. Write the smallest number which is divisible by both 306 and 657. (2 Mark)Ans: **22338

The smallest number divisible by 306 and 657 is their LCM.

we have

306 = 2 x 153 = 2 x 3 x 51 = 2 x 3 x 3 x 17 = 2 x 3

and 657 = 3 x 219 = 3 x 3 x 73 = 3

âˆ´ LCM = 2

Let one number be x, then the other number will be (9-x)

According to the question

sum of reciprocals = 1/2

ACS = AB

In the given figure, ABC and DBC are two triangles o n the sam e base B C . If AD intersects BC at O, show that:

In Î”ABC, ADâŠ¥ BC

Now, in Î”ADC, âˆ D = 90Â°

âˆ´ AC

or

Draw AP

[by AA similarity criterion]

Ans:

Let A(3,0), B(6,4) and C(-l, 3)

AB

âˆ´ Triangle is isosceles.

25 + 25 = 50

Also, AB

Since, Pythagoras theorem is verified, therefore triangle is a right angled triangle.

**Q 25. The larger of two supplementary angles exceeds the smaller by 18Â°. Find the angles. (2 Mark)Ans**: 99Â° and 81Â° or Sumit is 45

The larger of two supplementary angles be x.

âˆ´ Other angle = (180Â° - x)

According to given condition, we have

x - (180Â° - x ) = 18Â°

â‡’ x - 180Â° + x - 18Â°

â‡’ 2x = 198Â°

â‡’ x = 99Â°

âˆ´ 180Â° - x = 180Â° - 9 9 Â° = 81Â°

Hence, the angles are 99Â° and 81Â°.

Or**Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present ?****Solution**: Let the present age of Sumit be x years and the present age of his son be y years.

Using the given information, we have

x = 3y ...(1)

Five years later, Sumit's age = (x + 5) years and his sonâ€™s age = (y + 5) years

Using the given information, we have

...(2)

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

2 x 3y - 5y = 15 â‡’ y â€” 15

From (1), we have

x â€” 3y â€” 3 x 15 â€” 45

Hence, the present age of father , sumit is 45 years.**Q 26. Compute the Arithmetic Mean for the following data (2 Mark)Ans:** 28.6

Here, h = 10

Let assumed mean (A) = 3

**Section C**

**Q 27. Prove that 2 + 5 âˆš3 is an irrational number, given that âˆš3 is an irrational number. (3 Mark)**

Suppose, 2 + 5 âˆš3

This contradicts the given fact that âˆš3 is an irrational number. So, our supposition is wrong.

Hence, 2 + 5 âˆš3

Or

Given integers are 2048 and 960.

Clearly 2048 > 960.

Applying Euclidâ€™s division algorithm, we get

2048 = 960 x 2 + 128

960 = 128 x 7 + 64

128 = 64 x 2 + 0

In the last equation, the remainder is zero and divisor in the last equation is 64.

Hence, HCF of 2048 and 960 = 64.

Ans:

Let p(x) - x

q(x) = x - 2 and

r(x ) = - 2x + 4

By using division algorithm,

we have Dividends (Divisor x Quotient) + Remainder

â‡’ p(x) = g(x) x q(x) + r(x)

On putting the values of p(x), q(x) and r(x), we get

Now, using long division method, divide

x

Thus, we get quotient = x

Hence, g (x ) = x

Effective speed of the boat during downstream = (5 + x) km/h

Effective speed of the boat during upstream = (5 - x) km/h

Distance covered â€” 12 km

According to the statement of the question, we have

â‡’ 24x =25x

â‡’ x

â‡’ (x+ 25) (x-1) = 0

x = -25, x = 1

Since the speed of the stream cannot be negative, therefore, negative value is not admissible

Thus, x = 1

Hence, the speed of thehstream of water is 1 km/h.

Or

Let the number of books bought be x

Now, when number of books bought is x + 4

then, cost of each book =

According to statement

â‡’ x = 16 [ âˆµ x cannot be negative]

Hence, the number of books bought by the shopkeeper is 16.

. LHS - cosec

Ans:

Here OABC is a square inscribed in a quadrant OPBQ.

As all sides of a square are equal.

OA = AB = BC = CO = 15 cm

By Pythagoras theorem in Î”OAB,

âˆ´ Radius of the quadrant = OB = 15 âˆš2 cm

Area of the shaded region - Area of a quadrant of a circle of radius OB (15âˆš2 cm) - Area of a square of side OA = 15 cm

or

Given that ABCD is a square with side 2âˆš2 cm and inscribed in a circle. Join AC or BD. From right triangle ABC right angled at B, diagonal AC is given by

Also, BD = AC = 4 cm [Since diagonals of a square are equal]

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

Diameter of the circle = 4 cm

Radius of the circle = 2 cm

From the figure, Area of the shaded region = Area of the circle of radius 2 cm - Area of a square of side AB â€” 2âˆš2 cm

Ans:

Now, area which can be grazed by three horses = Area of sectors with central angles

âˆ A, âˆ B and âˆ C, radius 7 m

Area of the plot

âˆ´ Area of the plot which remains ungrazed = 336 - 77 = 259 m

Ans:

324 = 252 x 1 + 72

252 = 72 x 3 + 36

72 = 36 x 2 + 0

âˆ´ HCF(324,252) - 36

180 =36 x 5 + 0

âˆ´ HCF(36f 180) =36

âˆ´ HCF of 180,252 and 324 is 36.

**Section D**

**Q 35. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio. (4 Mark)**

Given : A triangle ABC in which a line parallel to BC intersects other two sides AB and AC at D and E respectively.

..(1)

and

....(2)

..(3) using 1 and 2

...(4)

Note that ABDE and ADEC are on thesame base DE and between the same parallel 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

Ans:

Given, a = 1 and sum of first four terms = 1/3( sum of next four term)

Hence, the required common difference is 2

Ans:

Aâ€™s one day work + Bâ€™s one day work

12x - 30 - x2 - 5x

â‡’ x

â‡’ x

â‡’ x

â‡’ x(x- 15) -2(x- 15) = 0

â‡’ (x - 1 5) (x - 2 ) = 0

x = 15 or x = 2

But x cannot be less than 6. So, B alone can finish the work in 15 days.

or

OR

Ans:

Hence, the distance between two ships = 5475 m

or

â‡’ h = âˆš3x .(1)

20(3 +1.73) = 20 X 4.73

Hence, the height of tower is 94.6 m.

First term , a = - 7. and common difference, d = - 12 - ( - 7 ) = - 12 + 7 = - 5

Let - 82 be its nth term.

Hence, - 82 is the 16th term of the given A.P.

Let - 100 be its nth term.

n=98/5 , which is a fraction.

Since number of terms cannot be in fraction.

Hence, - 100 is not any term of the given A.P.**or****How many terms of the Arithmetic Progression 45, 39, 33,.... must be taken so that their sum is 180 ? Explain the double answer.****Ans:** The given A.P. is 45, 39, 33,.....

First term, a = 4 5 and common difference,

d = 39 - 45 = 6

Let 180 be the sum of its n terms.

Sum of first 6 terms = Sum of first 10 terms = 180

Explanation of double answer

Here, common difference is negative.

7th term of the A.P. is given by

Sum of next 4 terms just after first 6 terms = a_{7} + a_{8} + a_{9} + a_{10} = 9 + 3 + (- 3) + (- 9) = 0.

Therefore, sum of first 6 terms is the same as the sum of first 10 terms.**Q 40. Find the median for the following data. (4 Mark)**

The cumulative frequency just greater than 50 is 59 and its corresponding class is 159.5-169.5.

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