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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 sin x + cos y = 1; x = 30 ^{0 }and y is an acute angle, find the value of y. (1 Mark)(a) 60^{0}(b) 70^{0}(c) 56^{0}(d) 50^{0}Ans.** Choice (a) is correct.

Given,

sin x + cos y = 1

â‡’

â‡’ cos y = cos 60

â‡’ y = 60

(a) 2 cm

(b) 2âˆš2 cm

(c) 2 + âˆš2 cm

(d) 4 cm

Ans.

We have, AC = CB = 2 cm

By Pythagoras theorem,

(a)

(b)

(c)

(d)

Ans.

[Here, new mean

(a) 60

(b)45

(c) 30

(d) 90

OR

The angle of depression of a car parked on the road from the top of 150 m high tower is 30

(a) 50âˆš3

(a) 150âˆš3

(a) 150âˆš2

(a) 75

Ans.

OR

(b) Explanation: In Î”ABC, âˆ B = 90

(a) 50âˆš3 m

(b) 150âˆš3 m

(c) 150âˆš2 m

(d) 75 m

Ans.

â‡’

â‡’ x = 150âˆš3 m

(a) 590

(b) 620

(c) 640

(d) 660

Ans.

Now, sum of first 20 terms

= 640

(a) 1

(b) -1

(c) 0

(d) 1+âˆš3

Ans.

(a) 5

(b) 20

(c) 25

(d) 30

Ans.

Explanation: In the given A.P., d = 5 Thus,

a

Ans.

We have,

cos 3Î¸ = sin 45

(a) -1/3

(b) Other than -1/2

(c) -1/4

(d) Other than -1/4

Ans.

x + ky = 0 and 2x - y = 0

On comparing these equations with

Condition for unique solution,

â‡’

â‡’

Ans.

[Here, radius of the base

= height of the cone = r

âˆ´ Volume of the cone

Ans.

âˆš2 = 1.414

and âˆš3 = 1.732

Now, we can write 'n' rational number between those e.g., just greater than 1.414 and less than 1.732 and it should be terminating or not

e.g.,: 1.415659,1.416893,1.715644, ...

Therefore, one rational number between âˆš2 and âˆš3 is 1.416893.

Ans.

Ans.

Ans

[Average of given ten numbers

Required probability = 7/10]

Ans.

d = âˆš6-âˆš3

= âˆš3(âˆš2-1)

Again, d = âˆš9-âˆš6

= 3-âˆš6

= âˆš12-âˆš9 = 2âˆš3-3

As common difference are not equal.

Hence, the given series is not in A.P.

Marks obtained | Less than 10 | Less than 20 | Less than 30 | Less than 40 | Less than 50 |

Number of students | 4 | 7 | 18 | 33 | 40 |

**Change the above data into a continuous grouped frequency distribution.Ans.** Changing given cumulative frequency distribution into continuous grouped frequency distribution, we get

Class interval | Frequency |

0-10 | 4 |

10-20 | 7-4 = 3 |

20-30 | 18-7 = 11 |

30-40 | 33-18 = 15 |

40-50 | 40-33 = 7 |

**Q.18. Which term of the AP: 21, 18,15,... is -81? (1 Mark)Ans. **Given, AP is 21, 18, 15, ...

Here, a = 21 and d = 18 - 21 = -3

Let nth term of given AP be -81.

Then, a

â‡’ a + [n-1)d = -81 [âˆ´a

On putting the values of a and d in Eq. (i), we get

21 + (n -1) x (-3)= -81

Hence, 35th term of given AP is -81.

Ans.

Hence, the required constant added and subtracted is 1/64.

Ans.

Or, sin A = sin (90

or, A = 90

âˆ´ A + B = 90

**SECTION B**

**Q.21. If S _{n}, the sum of the first n terms of an A.P. is given by S_{n} = 2n^{2} + n, then find its nth term.**

â‡’ a

OR

Let a

a

â‡’ 7d = 7

â‡’ d = 1

Hence, the common difference, d is 1.

Ans.

On putting k = -1 in the equation 2x

we get,

Hence, the other root is - 3/2.

OR

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Ans.

âˆ l = âˆ 2 [âˆ´ AB || DC || EF âˆ´âˆ and âˆ 2 are corresponding angles]

âˆ FDG = âˆ ADB [common]

So, by AA criterion of similarity, we have

Î”DFG âˆ¼ Î”DAB

â‡’ ...(i)

In trapezium ABCD, we have EF || AB || DC

In Î”BEG and Î”BCD, we have

Adding (iii) and (iv), we have

Ans.

i.e., 184,........ 13,10,7.

Hence, 166 is the 7

Ans.

It can be drawn in 46 different ways.

âˆ´ Total number of possible outcomes = 46

(i) Let E be the event that card drawn from the box has a prime number less than 10.

The prime numbers less than 10 are 5, 7.

These are 2 numbers.

âˆ´ Number of cases in favour of event E = 2

Probability of getting a card having a prime number less than 10

(ii) Let F be the event that card drawn from the box has a perfect square.

The perfect square numbers from 5 to 50 are 9, 16, 25, 36, 49.

These are 5 numbers.

âˆ´ Number of cases in favour of event F = 5

Probability of getting a card having a perfect square number from 5 to 50.

OR

Find the coordinates of points on the X-axis which are at the distance of 17 units from the point (11, - 8).

Ans.

Let the ratio be k : 1, then by section formula, the coordinates of the point which divides AB in the ratio k : 1 are

Thus, point lies on the X-axis, and we know that on the X-axis, the ordinate is zero.

On squaring both sides, we get

(11- x)

**SECTION C**

**Q.27. Point P divides the line segment joining the points A(2, 1) and P(5, -8) such that If P lies on the line 2x - y + k = 0, find the value of k.****OR****For what value of p, are the points (2,1), (p,-1) and (-1,3) collinear? (3 Mark)**

It is given that

â‡’ 3AP=AP

â‡’ 3AP=AP + PB

â‡’ 2AP = PB

â‡’

â‡’ AP : PB = 1 : 2

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

Using section formula, the coordinates of point P are

Hence, the coordinates of P are (3, -2).

Point P(3, -2) lies on the line given by the equation

2x-y+k = 0

âˆ´ The coordinates of point P satisfies above equation.

Let the given points be A(2,1), B(p, -1) and C(-1, 3). Since the points A(2, 1), B(p, -1) and C(-1, 3) are collinear, therefore, the area of the triangle formed by them is zero.

âˆ´ Area of Î”ABC = 0

â‡’

â‡’ |2p - 10| = 0

â‡’ 2p - 10 = 0

â‡’ p = 5

Hence, the given points are collinear when p = 5.

Ans.

n = 4q +r, 0 â‰¤ r < 4

â‡’ n = 4q + r, where r can have one and only one value of 0, 1, 2 or 3.

â‡’ n = 4q or 4q + 1, or 4q + 2, or 4q + 3 and one and only one of these above possibilities can happen.

(i) If n = 4q, then n is divisible by 4.

(ii) If n = 4q + 1, then n + 3 = 4q + 4 = 4(q +1), therefore n + 3 is divisible by 4.

(iii) If n = 4q + 2, then n + 6 = 4q + 8 = 4(q + 2), therefore n + 6 is divisible by 4.

(iv) If n = 4q + 3, then n + 9 = 4q + 12 = 4(q + 3), therefore n + 9 is divisible 4.

A boat goes 24 km upstream and 28 km downstream in 6 hours. If goes 30 km upstream and 21 km downstream in hours. Find the speed of boat in still water and also speed of the stream.

Ans.

âˆ´ Given pair of equations can be rewritten as

...(i)

Adding these two equations, we have

6x = 6 â‡’ x = 1

And 3(1) + y = 4 â‡’ y = 1

Hence, the values of x and y are x = 1, y = 1.

Let speed of the boat in still wafer be x km/h and speed of the stream be y km/h. In case of downstream, effective speed of the boat = (x + y) km/h.

In case of upstream, effective speed of the boat = (x - y) km/h

According to the statement of the question, we have

and

Multiplying equation (i) by 5, we have

Solving (iii) and (iv), we have

Hence, speed of boat in still water = 10 km/h and speed of the stream

= 4 km/h.

OR

An electric pole is 12 m high. A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole upright. If the wire makes an angle of 60

Ans.

â‡’ tanÎ¸ = 3/4

â‡’ sinÎ¸ = 3/5 and cosÎ¸ = 4/5

= 13/11

OR

Let OA be the electric pole and B be the point on the ground to fix the wire.

Let BA be x m.

In Î”ABO, AO/AB = sin60

â‡’12/x = âˆš3/2

â‡’

Hence, the length of wire = 13.84 m.

Ans.

BD = CD = BC/2

â‡’ BC = 2 CD ...(1)

In right triangle ABC, right angled at B, we have

AC

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

AD

Subtracting (3) from (2), we have

It is given that AABC is an isosceles with AB - AC.

âˆ´ âˆ C = âˆ B

Now, in triangles ABD and ECF, we have

âˆ ABD = âˆ ECF [âˆ´ âˆ B - âˆ C]

âˆ ADB = âˆ EFC = 90

So, by AA-criterion of similarity, we have

Î”ABD ~ Î”ECF

OR

Ans.

OR

= RHS

Ans.

= 28âˆš2 m [âˆµ Diagonal of square = âˆš2 x side]

Also, radius of sector COD = 28âˆš2m

âˆ´ Area of (sector AOB + sector COD)

OR

The largest possible sphere is cut out from a wooden solid cube of side 7 cm. Find the volume of the wood left.

Ans.

Base of triangle = diameter of semicircle

= 42 cm

and its height - radius of semicircle

= 42/2 = 21 cm

Area of shaded portion = Area of semicircle - area of Î”ABC

Hence, the area of shaded portion = 252 cm

OR

Given, the side of cube a = 7 cm

Since, the diameter of the largest possible sphere = Side of the cube

Hence, the radius of sphere = 7/2 cm.

Volume of the wood left = Volume of cube - Volume of sphere

Hence, Volume of wood left = 163.3 cm

**SECTION D**

**Q.35. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 h less for the same journey. Find the speed of the train. (4 Mark)**

Time taken to cover 360 km with usual speed = 360/x h

If the speed had been increased by 5 km/h, then increased speed = (x + 5) km/h

Time taken to cover 360 km with increased speed

âˆ´ According to the given condition in the question, we have

Hence, the usual speed of the train is 40 km/h.

The given equation is

[On dividing by (a + b)]

Cross-multiplying, we get

x(x + a + b) = - ab

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

Ans. Here, logs stacked in each row form a series

21 + 20 + 19 + 18 + 17 + ...

which is in AP with first term, a = 21 and common difference, d = 20 - 21 = -1.

Let the number of rows be n, then S

and for n = 27,

T

which is not possible.

[âˆ´ number of logs in any row cannot be negative]

So, the number of rows is 16 and number of logs in the top row = 6.

OR

Sum of the areas of two squares is 468m

Ans. Let B takes x days to finish the work and A takes (x - 5) days to finish the same work.

A's one day work + Bâ€™s one day work

But (A + B)â€™s one day work = 1/6

But x can rtot be less than 6. So, B alone can finish the work in 15 days.

OR

Let the side of first square be x m and the side of second square be y m.

Then, the area of first square = x

and the area of second square = y

Perimeter of the first square = 4x m

and perimeter of the second square = 4y m Now, according to the given statement, we have

x

and 4x - 4y = 24

â‡’ x - y = 6 ...(ii)

From equation (ii), we have x = 6 + y ...(iii)

Putting this value of x in (i), we have

which is a quadratic equation in y.

We can solve this equation by quadratic formula.

Here, a = 1, b = 6, c = - 216

âˆ´ D = b

But the side of a square cannot be -ve.

âˆ´ y = 12

Putting this value of y in (iii), we have x = 6 + 12 = 18

Hence, the side of the first square is 18 m and the side of the second square = 12 m.

OR

Prove that:

Ans. Let the speed of car by x m/minutes

In Î”ABC,

h/y = tan 45

â‡’ h = y

Hence, time taken from C to B = 6(âˆš3+1) minutes

OR**Q.39. Find the values of frequecies x and y in the following frequency distribution table, if N = 100 and median is 32. (4 Mark)**

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | Total |

No. of students | 10 | x | 25 | 30 | y | 10 | 100 |

OR

For the following frequency distribution, draw a cumulative frequency curve (ogive) of â€˜more than typeâ€™ and hence obtain the median value.

Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |

Frequency | 5 | 15 | 20 | 23 | 17 | 11 | 9 |

**Ans. **We have,

Median = 32 and N= 100

We prepare cumulative frequency table to calculate median:

We have, N = 75 + x + y = 100

â‡’ x + y = 25 ...(1)

Given, median = 32, so that median class is 30-40.

l = lower limit of the median class = 30, f = frequency of the median, class = 30,

h = width of the median class = 10, cf = cumulative frequency o f the class just preceding the median class = 35 + x.

Using the formula:

OR

We prepare the cumulative frequency table by more than type method as given below:

Here, 0, 10, 20, 30, 40, 50, 60 are the lower limits of the respective class interval 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60, 60 - 70. To represent the data in the table graphically, we mark the lower limits of the class interval on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis (y-axis), by choosing a convenient scale. Other than the given class intervals, we assume the class interval 70 - 80 succeeding to the last class mark 60 - 70 with zero frequency. Now, plot the points (0, 100), (10, 95), (20, 80), (30, 60), (40, 37), (50, 20), (60, 9) and (70, 0) on a graph paper and join them by a free hand smooth curve. This curve is called an ogive more than type (see figure).

Mark N/2 = 100/2 = 50 on the y-axis.

From the point, draw a line parallel to x-axis cutting the curve at a point P. From this point P, draw a perpendicular to the x-axis. The point of intersection M of this perpendicular with the x-axis determines the median of the given data.**Q.40. From a well-shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is(i) a queen of black colour.(ii) a card with number 5 or 6.(iii) a card with number less than 8.(iv) a card with number between 2 and 9.**

Ans. Since, one card is drawn from 52 well-shuffled cards,

âˆ´ Total number of possible outcomes = 52

(i) Since, there are 2 queens of black colour.

âˆ´ P (getting a queen of black colour)

= 2/52 = 1/26

(ii) In each suit, there are 2 cards with number 5 and 6. So, total such ca rd s are 4 tim es 2 = 8

âˆ´ P (getting a card with number 5 or 6)

= 8/52 = 2/13

(iii) In each suit, there are 6 cards with number less than 8, namely 2, 3, 4, 5, 6 and 7

âˆ´ P (getting a card with number less than 8)

(iv) In each suit, there are 6 cards with number between 2 and 9, namely 3, 4, 5, 6, 7,8.

âˆ´ P( getting a card with number between 2 and 9)

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