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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) (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. ** ** (1 Mark)****Ans.** Choice (c) is correct.

We have,**Q.2. When a solid is converted from one shape another, the volume of the new shape (1 Mark)(a) remains the same(b) decrease(c) increases 4 times(d) increases 2 timesAns.** (a) A solidâ€™s volume remains the same

(a) 10

Ans.

[The x-coordinate of the point of intersection of more than type and less than type o give is the required median]

(a) 8

(b) 2âˆš7

(c) 10

(d) 6

Ans.

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

Ans.

First term, a = 18, and common difference, and last term (a

Let number of terms be n.

Hence, the number of terms in the given A.P. are 27.

(a) 4/3

(b) -4/3

(c) 2/3

(d) -2/3

Ans.

Let p(x) = (k - h)x

Since, - 3 is a zero of polynomial,

âˆ´ p(-3) = 0

âˆ´ (k-1) (-3)

=> 9(k-1 )-3k + 1 = 0

=> 9k - 9 - 3k + 1=0

=> 6k - 8 = 0

=>6k = 8

âˆ´

(b) 4

Ans.

Ans. (a) R

(b) R

(c) R

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

Ans.

If tan Î¸ = 3/2, then the value of

Ans.

(b) 60Â°

(c) 65Â°

(d) 55Â°

Ans.

âˆ OBP = 90Â°

âˆ´ âˆ AOB + âˆ APB = 180Â°

â‡’ 115Â° + âˆ APB = 180Â°

â‡’ âˆ APB = (180Â° - 115Â°) = 65Â°

Ans.

Ans.

Slant height of the frustum

(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.

Assertion ( A) : The equation is a quadratic equation.

Reason (R) : Degree of quadratic equation is 2.

Ans.

The given equation is

â‡’ 2x

Ans.

Q.15. Arun has a cubical block with one word written on each face â€˜COME TO LEARN GO TO SERVEâ€™. The block is thrown, the probability of getting â€˜TOâ€™ = ________.

Ans.

[Number of possible outcomes = 6 ;

Number of favourable outcomes = 2

Required probability = 2/6 =1/3]

Ans.

Ans.

âˆ´

Clearly, 2 is a rational number and it lies between

Ans.

Ans.

Thus, two irrational numbers lie between

are 1.4242242224 ... and 1.5050050005.**Q.20. 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 => m = 2

**Section B**

**Q.21. How many multiples of 4 lie between 10 and 205 ? (2 Mark)**

It is an A.P. with first term, a â€” 12, common difference, d = 4 and last term = 204.

Let number of terms be n

Hence, the number of multiples of 4 between 10 and 205 are 49.

Ans.

âˆ´ a

â‡’ {a + (7 - 1)d} - {a + (5 - 1 )d} = 12 [âˆµ a

â‡’ (a + 6d) - (a + 4d) = 12

â‡’ 2d = 12

â‡’ d = 6 ...(1)

and a

â‡’ a + (3-1)d = 16

â‡’ a + 2 x 6 = 16 [using(1)]

â‡’ a = 16 - 12

â‡’ a = 4 ...(2)

The A.P. is 4, 10,16, 22, 28.........

Or

Show that is irrational.

Ans.

p,q are coprimes. ...(i)

But this contradicts the fact that is irrational.

Hence is an irrational number.

Or

Let us assume that 6-2âˆš3 is rational number.

Then, it will be of the form a/b where a, b are coprime integers and b â‰ 0.

...(1)

Since, 6 and a/b are rational. So, their difference will be rational.

âˆ´ 2 and âˆš3 are rational.

As 2 is rational, so it is true.

But we know that, âˆš3 is irrational.

So, this contradicts the fact that âˆš3 is irrational. Therefore, our assumption is wrong.

Hence, 6 - 2âˆš3 is irrational

Ans.

Now, ar (quad. BCED)

= ar(Î”ABC)- ar (Î”ADE) 36 -16 = 20 cm

Or

Through D, draw DG || BF intersecting AC in G.

In Î”CBF, D is the mid-point of BC and DG || BF

â‡’ G is the mid-point of CF

i.e., CG - GF ...(i)

Again, in Î”ADG, E is the mid-point of AD and EF is parallel to DG

â‡’ F is the mid-point of AG

Ans.

or, âˆ PAB =âˆ PBA = 60Â°

âˆ´ Î”PAB is an equilateral triangle.

Hence, AB - PA - 5 cm.

âˆ´ Total number of possible outcomes in the sample space = 52

Number of spade cards =13 and number of king cards other than spade = 3

âˆ´ Total number of cards which are neither a spade nor a king = 52 - (13 + 3) = 3

Probability of drawing a card which is neither a spade nor a king

Q.26. Calculate the mode for the following frequency distribution

Ans.

**Section C**

**Q.27. Prove that (3+2âˆš5) is an irrational number, given that âˆš5 is an irrational number. (3 Mark)Ans.** Suppose 3+2âˆš5 is a rational number. Then there exists co-prime integers a and b such that

â‡’

â‡’

â‡’ is a rational number,

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

Hence, 3+2 âˆš5 is an irrational number.

Q.28. Read the following passage and answer the question that follows:

A mathematics teacher Ramnivas Mathuriya organised a workshop for Class Xth students in the school auditorium to investigate the practice of students about subject knowledge. In order he write a polynomial x

Ans.

divide x

by x

Here, remainder = (2k - 9)x - (8 - k)k +10

But the remainder is given in the form of x+ a,

âˆ´ (2k - 9)x - (8 - k)k +10 = x + a

On comparing the coefficients of x and constant term both sides, we get

2k - 9 = 1 and - (8 - k)k + 10 = a

â‡’ 2k = 10 and a = - (8 - k) k + 10

â‡’ k = 5 and a = - 3x5 + 10

= -15 + 10 = -5

âˆ´ k = 5 and a = - 5

Or

In a competitive examination, one mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer 120 questions and got 90 marks. How many questions did she answer correctly ?

Ans.

...(ii)

Equations (i) and (ii) can be rewritten as

Hence, the required solution is x = 1 and y =1.

Or

Let x and y be the number of correct answers and wrong answers.

According to the statement of the question, we have

x + y = 120 ...(i)

Hence, total number of questions correctly answered by Jayanti are 100.

Ans.

(Perpendicular drawn from the centre to a chord bisects it.)

OB = 10 cm

OC

=10

OC

OC = 6 cm

âˆ´ Distance of the chord from the centre = 6 cm.

Or

Hence, System has no solution when P = 2.

Ans.

BC

â‡’ BC

â‡’ BC = 10 cm

Let O be the centre of the circle. The circle touches sides AB, BC and CA at E, F and G respectively. As tangent at any point of a circle is perpendicular to the radius through the point.

âˆ´ OE âŠ¥ AB, OF âŠ¥ BC, OG âŠ¥ CA

Join O to A, B and C.

Now,

(1 + cot A + tan A)(sin A - cos A)

sec

= sin

Ans.

Find the area of the shaded region. [Use Ï€ = 3.14]

Ans.

Area of semicircle A = 1/2Ï€(1.5)

Area of semicircle B = 1/2Ï€(1.5)

Area of semicircle C = 1/2Ï€(1.5)

For circle D, we have

diameter = 4.5 cm â‡’ radius = 4.5/2 cm

Area of circle D =

For semicircle E, we have

diameter = 9 cm â‡’ radius = 4.5 cm

Area of semicircle E = 1/2Ï€(4.5)

Now, Area of shaded region

= area of E + area of B - area of A - area of C - area of D

= 1.57 {7.875}

= 12.36375 cm

Ans.

âˆ´

where p and q are co-prime integers and q â‰ 0

On squaring both the sides, we get,

Or

Or p

âˆ´ p

âˆ´ p is divisible by 2. ...(i)

âˆ´ q is divisible by 2 ...(ii)

From (i) and (ii), p and q are divisible by 2, which contradicts the fact that p and q are co-primes. Hence, our assumption is false.

âˆ´ âˆš2 is an irrational number.

**Section D**

**Q.35. Solve for x : (4 Mark)**

[On dividing by (b + 2a)]

Cross-multiplying, we get

â‡’ x + a =0 or 2x + b = 0

â‡’ x= - a or x = - b/2

Hence, the roots of the equation be - a and -b/2.

Or

Ans.

sides of two squares be x m and y m such that x > y.

Sum of their areas =(x2 + y2) m

and the difference of their perimeters = (4x - 4y) m

According to the given condition, we have

x

and 4x - 4y = 64

â‡’ x - y = 16 ...(2)

From (2), we get

y = x - 16 ...(3)

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

â‡’ x = 24 [âˆ´ Side (x) cannot be negative]

From (3), we get y = x - 16 = 24 - 16 = 8.

Hence, the sides of two squares are 24 m and 8 m.

Or

Find The centre of a circle passing through the points (6,-6), (3,-7) and (3,3).

Ans.

Now, coordinates of D =

and length of median AD is given by

Let C (x, y) be the centre of the circle passing through the points P(6,-6), Q(3,-7) and R(3,3)

Or

Two water.taps together can fill a tank in hours. The tap of smaller diameter takes 5 hours more than the larger one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Ans.

Real and unequal roots.

By using quadratic formula, we have

Hence, roots of the given quadratic

Time taken by two taps together to fill the tank

Let V be the volume of tank and x the number of hours required by the larger tap to fill the tank.

âˆ´ Number of hours required by the smaller tap to fill the tank = (x + 5)

âˆ´ Portion of the tank filled by larger tap in 1 hour = v/x

Portion of the tank filled by smaller tap in

Now, we have a = 9, b = -155, c = - 500 By using quadratic formula, we obtain

But -25/9 is not possible because time cannot be negative.

Hence, time required by the larger tap is x = 20 hours and time required by the smaller tap is

x + 5 = 20 + 5 = 25 hours.

Ans.

Let circle touches CB at M, CA at N and AB at P.

Now OM âŠ¥ CB and ON âŠ¥ AC (radius âŠ¥ tangent)

OM = ON (radii)

CM = CN (Tangents)

âˆ´ OMCN is a square.

Let OM = r = CM = CN

AN = AP, CN = CM and BM = BP (tangent from external point)

AN = AP

â‡’ AC -CN = AB-BP

b-r = c-BM

b-r = c-(a-r)

b-r = c-a+r

2r - a+b-c

Or

Steps of construction:

1. Draw a line segment BC = 7 cm

2. At point B draw a line BY making an angle of 60Â°.

3. With centre B mark an arc A of length 6 cm.

4. Join CA,

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

6. Locate four points B

7. Join B

8. Through C' draw a line parallel to AC intersecting line segment AB at A'.

Hence Î”A'BC' is the required triangle.

Ans.

Radius of the lower end, r - 8 cm; radius of the upper end, R - 20 cm and height of the container, h - 16 cm.

Let V be the volume of the container. Then

Volume of the container =

Let S be the surface area of the metal sheet used to make the container. Surface area of the metal sheet used to make the container is given by

Cost of metal sheet = Rs 10 per 100 cm

Draw 'less than ogive' and â€˜more than ogive'. Also, find the median.

Or

The mean of the following table is 50. The frequencies f

Ans.

Now, we mark the upper limits along X-axis and cumulative frequencies along Y-axis, on the graph paper. Then, plot the points (7,26), (14, 57), (21,92), (28, 134), (35, 216), (42, 287), (49, 341) and (56, 360). Join all these points by a freehand smooth curve to obtain an ogive of less than type,

Now, let us form the cumulative frequency distribution of more than type, as shown below

Now, we plot the points (0, 360), (7, 334), (14, 303), (21,268), (28, 226), (35,144), (42,73) and (49,19) on the same graph paper by choosing a suitable scale. Join all these points by a freehand smooth curve to obtain an ogive of more than type,

The two ogives intersect at point A. Now, we draw a perpendicular line from A to the X-axis, the intersection point on X-axis is 31.9. Thus, the required median is 31.9

or

Let us assumed mean be A = 50 and h = 20.

Table for step deviation is given below

We have, [given]

On solving Eqs. (i) and (ii), we get

f

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