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Class - X Maths Sample Paper

SECTION - A

Question numbers 1 to 10 carry 1 mark each. For each of the questions 1 to 10 four alternative choices have been provided of which only one is correct. You have to select the correct choice.
1. Which of the following is not a quadratic equation :
(A) (x-2)² +1 = 2x-3       (B) x(x+1) + 8 = (x+2)(x-2)
(C) x(2x +3) = x²+1        (D) (x+2)  = x -4

2. For what value of p are 2p+1, 13, 5p-3, three consecutive terms of an A.P. ?
(A) 2       (B) - 2        (C) 4          (D) -4

3. A tangent PA is drawn from an external point P to a circle of radius 3√2  cm such that the distance of the point P from O is 6 cm as shown in figure.  The value of ∠APO  is

(A)    30°           (B)    45°          (C)    60°        (D)     50°

4. The maximum number of common tangents that can be drawn to two circles intersecting at two distinct points is
(A)    1         (B)    2          (C)    3             (D)    4

5. In the given figure, O is the centre of the circle.  If PA and PB are tangents from an external point P to the circle, then ∠AQB is equal to

(A) 100°          (B) 80°        (C) 70°          (D) 50°

6. The probability  that a leap year has 53 Sundays is 
(A)          (B)          (C)         (D) 
7. If the angle of depression of an object from a  5 m high tower is 30°, then the distance of the object from the base of tower is
(A) 25√3  m     (B) 50√3  m      (C) 75√3  m     (D) 150m

 8. The perimeter of a quadrant of a circle of radius   cm is
(A) 3.5 cm    (B) 5.5 cm     (C) .5 cm     (D) 12.5 cm

9. If a solid right circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of a sphere, then the radius of the sphere is
(A) 6 cm (B) 4 cm (C) 8 cm (D) 12 cm

10. In the given figure, AA₁=A₁A₂=A₂A =A₃B.  If B₁A₁||CB, then A₁  divides AB in the ratio
 
(A) 1 : 2      (B) 1 : 3       (C) 1 : 4      (D) 1 : 1

SECTION - B

Question numbers 11 to 18 carry 2 marks each.
11. The sum of circumferences of two circles is 132 cm. If the radius of one circle is 14 cm, find the radius of the second circle.

12. For what value(s) of k, the equation x² - 2kx - k= a will  have equal roots ?
13. Find the 10th term from the end of the A.P. 4, 9, 14, ........ 254.

14. A point P is at a distance of √10  from the point (2, 3). Find the coordinates of the point P if its y coordinate is twice its x coordinate.

OR

Find the coordinates of the point B, if the point P(-4, 1) divides the line segment joining the points A(2, -2) and B in the ratio 3 : 5.

15. If the points A (4, 3) and B (x, 5) are on the circle with  centre O(2, 3) ; find  the value of x.

16. A bag contains cards numbered from 2 to 26.  0ne  card is drawn  from the bag at random.  Find the probability  that it has a number divisible by both 2 and 3.
17. How many lead shots, each 0.3 cm in diameter, can be made from a cuboid of dimensions 9 cm x 11 cm x 12 cm.

18. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB+CD=AD+BC. 

SECTION - C

Question numbers 19 to  28 carry 3 marks each.
19. Solve for x :
      ;x  - 1, -2, -4

OR 

Solve for x :
9x² - 3(a+b)x + ab = a

20. Determine the ratio in which the line 3x+y-9=a divides the line segment joining the points (1, 3) and (2, 7)

21. The area of a triangle whose vertices are (-2, -2), (-1, -3) and (x, a) is 3 square units.  Find the value of x.

22. How many terms of the A.P.  8,  1, 64, ...... are needed to give the sum 465 ?  Also find the last term of this A.P.

23.In the given figure, diameter AB is 12 cm long.  AB is trisected at points P and Q. Find the area of the shaded region.

OR 

A chord of a circle of radius 12 cm subtends an angle of 120°at the centre.  Find the  area of the corresponding minor segment of the circle.  (Use π=3.14)
24. Water is flowing  at the rate of 5 km/hour through  a pipe of diameter 14 cm into a rectangular tank, which is 50 m long and 44 m wide.  Determine the time in which the level of water in the tank will  rise by 7cm.

25. In the figure, OP is equal to the diameter of the circle. Prove that ABP is an equilateral triangle.

OR

In the figure, AB and CD are two parallel tangents to a circle with  centre O.  ST is the tangent segment between two parallel tangents touching the circle at Q.  Show that ∠SOT=90°.

26. The shadow of a tower standing on a level ground is found to be 4a m longer when the sun's altitude is 30° than when it is 60°.  Find the height of the tower.

27. Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6 cm from its centre O.

28. All the three face cards of spades are removed from a well - shuffled pack of 52 cards.A card is then drawn  at random from the remaining pack.  Find the probability  of getting
(a) a black face card
(b) a queen
(c) a black card 

SECTION - D

Question numbers  29 to 34  carry 4 marks each.
29. Prove that the lengths of tangents drawn from an external point to a circle are equal.

30. A person on tour has Rs. 360 for his daily expenses. If he extends his tour for four days, he has to cut down his daily expenses by Rs. 3. Find the original duration of the tour.

31. 200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row ?

32. An aircraft is flying  at a constant height with  a speed of 360 km/hour.  From a point on the ground, the angle of elevation at an instant was observed to be 45°.   After 20 seconds, the angle of elevation was observed to be 30°.   Determine the height at which the aircraft is flying.   (use  √3 = 1.732 )

OR

A boy 2 m tall is standing at some distance from a 30 m tall building. The angle of elevation from his eyes of the top of the building  increases from 30° to 60° as he walks towards the building.   Find the distance he walked towards the building.

33. The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle.Find the area of the shaded region.  (use π= 3.14 and √3 = 1.73205)
 
34. The internal radii of the ends of a bucket, full of milk and of internal height 16 cm, are14 cm and 7cm. If this milk is poured into a hemispherical vessel, the vessel is completely filled.  Find the internal diameter of the hemispherical vessel.

OR

A right  triangle, whose sides are 6 cm and 8 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.

Class - XI Maths Sample Paper

Instructions 

1. Questions 1 to 10 carry 1 mark each.
2. Questions 11 to 22 carry 4 marks each.
3. Questions 23 to 29 carry 6 marks each.

1.  Write the following set in the set-builder form: {5, 25, 125, 625}.

Sol. It can be seen that 5 = 5¹, 25 = 5², 125 = 5³, and 625 = 5⁴.

∴ {5, 25, 125, 625} = {x: x = 5ⁿ, n ∈ N and 1 ≤ n ≤ 4}

2. Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements. 

Sol.. It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.

We know that A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B.

∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B.

Since n(A) = 3 and n(B) = 2, it is clear that A = {x, y, z} and B = {1, 2}.

3. Let f be the subset of Z ×Z defined by f={(ab,a+b):a,b∈Z} Is f a function from Z to Z? Justify your answer.

Sol. We observe that          
         1 x 6 = 6 and 2 x 3 = 6
⇒      (1×6,1+6) ∈ f and (2×3,2+3) ∈ f
⇒      (6,7) ∈ f and (6,5) ∈ f.
We observe that an element 6 have appeared more than once as the first component of the ordered pairs in f. So f is not a function for Z to Z.

4. Express the given complex number in the form a + ib: (1 – i) – (–1 + i6).

Sol. (1 – i) – (–1 + i6) = 1 – i + 1 – 6i = 2 – 7i

5. If 2x+ i (x - y) = 5, where x and y are real numbers, find the values of x and y.

Sol. We have 2x+ i(x - y) = 5
or                  2x+ i(x - y) = 5 + 0.i
        
Comparing the real and imaginary parts,we get

2x = 5 and x - y = 0
⇒ x = 5/2 and x = y

Thus x = y = 5/2.

6. Solve the given inequality for real x: 4x + 3 < 5x + 7.

Sol. 4x + 3 < 5x + 7

⇒ 4x + 3 – 7 < 5x + 7 – 7

⇒ 4x – 4 < 5x

⇒ 4x – 4 – 4x < 5x – 4x

⇒ –4 < x

Thus, all real numbers x, which are greater than –4, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–4, ∞).

7. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Sol. There will be as many ways as there are ways of filling 3 vacant places

    In succession by the given six digits. In this case, the units place can be filled by 2 or 4 or 6 only i.e., the units place can be filled in 3 ways. The tens place can be filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled by any of the 6 digits in 6 different ways, as the digits can be repeated.

    Therefore, by multiplication principle, the required number of three digit even numbers is 3 × 6 × 6 = 108.

8. Find the sum of odd integer from 1 to 21.

Sol. The odd integer from 1 to 21 are 11, namely 1, 3, 5, 7, 9, 11,13,15,17,19, 21.

9. Write the equations for the x and y-axes.

Sol. The y-coordinate of every point on the x-axis is 0.

Therefore, the equation of the x-axis is y = 0.

The x-coordinate of every point on the y-axis is 0.

Therefore, the equation of the y-axis is y = 0.

10. A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?

Sol. When a die is rolled, the sample space is given by

S = {1, 2, 3, 4, 5, 6}

Accordingly, E = {4} and F = {2, 4, 6}

It is observed that E ∩ F = {4} ≠ Φ

Therefore, E and F are not mutually exclusive events.

11. A wheel makes 500 revolutions per minute. How many radians it turns in one second?

Sol. In one revolution angle covered is 2 π radians
In one minute, number of revolutions = 500
In one second, number of revolutions = 500/60 = 50/6 

∴ In one second angle covered = 50/6 X 2 π radians

or  In one second angle covered =  π radians

12. Write the value of tan 75°.

Sol. tan 75°

= tan(45°+30°)

13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Sol. Let U be the set of all students who took part in the survey.

Let T be the set of students taking tea.

Let C be the set of students taking coffee.

Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100

To find: Number of student taking neither tea nor coffee i.e., we have to find n(T' ∩ C').

   n(T' ∩ C') = n(T ∪ C)'

= n(U) – n(T ∪ C)

= n(U) – [n(T) + n(C) – n(T ∩ C)]

= 600 – [150 + 225 – 100]

= 600 – 275

= 325

Hence, 325 students were taking neither tea nor coffee.

14. Find the value of .

Sol.

15. Solve the quadratic equation 25x² – 30x + 11 = 0.

Sol. 25x² – 30x + 11 = 0

Using formula for the roots of quadratic equations, we get

Hence, the roots are

16. In how many of the distinct permutations of the letters in MISSISSIPPI do the four 1’s not come together?

Sol. Total number of letters in “MISSISSIPPI” = 11
Out of these I is repeated four times
                  S is repeated four times
                  P is repeated two times
Thus, the total number of distinct permutations
                   
There are 4 I’s viz, I, I, I, I. Considering these four I’s as one letter we have 8 letters (MSSSSPP and one letter obtained by combining all 1’s), out of which
                   S is repeated four times
                   P is repeated two times
                   and rest are different
Thus, the total number of distinct permutations in which 4 I’s are together
                   
Hence, the total number of distinct permutations of the letters in MISSISSIPPI do the four I’s not come together
                           

                   = 7×6×5×161

                   = 210 x 161

                   = 33810                  

17. Let the sum of n, 2n, 3n terms of an A.P. be S₁,S₂and S₃ respectively, show that : S₃ = 3(S₂ - S₁).

Sol.

        

        

        

18. Find coordinates of the foot of  perpendicular from the point (3,-4) to line 4x – 15y + 17 = 0.

Sol. The equation of the given line is 4x – 15y + 17 =   0                      … (i)

The equation of a line perpendicular to the given line is 15x + 4y – k = 0, where k is a constant.

If this line passes through the point (3, -4), then

4 x 3 – 15 x (-4) – k = 0
            12 + 60 – k = 0
                           k = 72

Therefore the equation of a line passing through the point (3, -4) and perpendicular to the given line is

15x + 4y – 72 = 0                                            … (ii)

 The required foot of the perpendicular is the point of intersection of lines (i) and (ii).

Solving equation (i) and (ii), we get

Therefore the required point is

19. Find the equation of the circle which passes though the points (3,7), (5,5) and has its centre on the line x - 4y = 1.

Sol. Let the equation of the required circle be x² + y² + 2gx + 2fy + C = 0 ..(i)

It passes through (3,7) and (5,5)

9 + 49 + 6g + 14f + C = 0      ...(ii)
25 + 25 + 10g + 10f + C = 0  ...(iii) 

The centre (-g, -f) lies on x - 4y = 1

-g + 4f = 1                          ...(iv)

Subtracting (ii) from (iii), we get 

4g - 4f - 8 = 0.                    ...(v)
or g - f = 2                         ...(vi)

Solving equation (iv) and (vi)

3f = 3 ,  g - 1 = 2
  f = 1 ,      g = 3.

Substituting the values of g and f in (ii) 

We get
9 + 49 + 18 + 14 + C = 0.
C = -90

Hence the equation of the circle 

x² + y² + 6x + 2y - 90 = 0.

20. Find the equation of the circle which passes through the points (2, –2), and  (3, 4) and whose centre lies on the line x + y = 2.

Sol. Let the equation of the circle be (x - h)² + (y - k)² = r² 
Since the circle passes through (2, – 2) and (3, 4), we have

(2 - h)² + (-2 - k)² = r² ...(1) and
(3 - h)² + (4 - k)² = r²  ...(2)

Also, the centre lies on the line x + y = 2, we have

h + k = 2                   ...(3)

Solving the equations (1), (2) and (3), we get

h = 0.7, k = 1.3 and r² = 12.58

Hence, the equation of the required circle is

(x - 0.7)² + (y - 1.3)² = 12.58.

21. A group of two persons is to be selected from two boys and two girls. What is the probability that the group will have

(a) two boys           (b) one boy          (c) only girls

Sol. Total number of persons = 2 + 2 = 4. Two persons can be selected in ⁴C₂ ways. 

(a) Two boys can be selected in ²C₂ ways.

Hence, probability = ²C₂ /⁴C₂ = 1/6.

(b) It means that there is one boy and one girl in the group. One boy out of 2 boys can be selected in ²C₁ ways. One girl out of 2 girls can be selected in ²C₁ ways.

Hence, probability = ²C₁ x ²C₁/⁴C₂ = 2 x 2/2 x 3 = 2/3.

(c) Two girls can be selected from 2 girls in ²C₂ ways.

Hence, probability = ²C₂/⁴C₂ = 1/6.

22. Find the point in XY-plane which is equidistant from three points A(2,0,3), B(0,3,2) and C(0,0,1).

Sol. The z-coordinate = 0 on xy-plane.

Let P (x, y, 0)be a point on xy-plane such that PA = PB = PC

Now, PA = PB

⇒ PA² = PB²

⇒ (x - 2)² + (y - 0)² + (0 - 3)² = (x -0)² + (y -3)² + (0 - 2)²

⇒ 4x - 6y = 0 or 2x - 3y = 0 ...(1)

    PB = PC

    PB² = PC²

⇒ (x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 0)² + (0 - 1)²

⇒ - 6y + 12= 0 ory =2 ...(2)

By (1), we get x = 3

Hence, the required point is (3, 2, 0).

23. A market research group conducted a survey of 1000 consumers and reported that 720 consumers like product A and 450 consumers like product B. What is the leastnumber that must have liked both products?

Sol. Let U be the set of all consumers who were questioned, PA be the set of consumers who liked product A and PB be the set of consumers who liked product B.
It is given that :
    n(U) = 1000, n(P_A) = 720, n(P_B) = 450
We know that :
    n(P_A∪P_B) = n(P_A)+n(P_B) - n(P_A∩P_B)
⇒ n(P_A∪P_B) = 720 + 450 - n(P_A∩P_B)
⇒ n(P_A∪P_B) = 1170 - n(P_A∪P_B)
⇒ n(P_A∪P_B) = 1170 - n(P_A∩P_B)   ....(1)
Since P_A∪P_B⊂U, therefore
             n(P_A∪P_B) ≤ n(U)
⇒          -n(P_A∪P_B) ≥ n(U)
⇒ 1170 - n(P_A∪P_B) ≥ 1170 - n(U) .....(2)
From (1) and (2), we have
       n(P_A∩P_B) ≥ 1170 - n(U)              But P_A∪P_B⊂U
⇒    n(P_A∩P_B) ≥ 1170 - 1000             ⇒    n(P_A∪P_B) ≤ n(U) = 1000
⇒    n(P_A ∩ P_B) ≥ 170                       ⇒ Maximum value of n(P_A∪P_B) = 1000
Thus, the least value of n(P_A∩P_B) is 170 and the maximum value of n(P_A∩P_B) is 1000.

24. Solve : 2 cos² x+3 sin x = 0.

Sol. We have

                       2 cos² x+3 sin x = 0
⇒                 2(1-sin² x)+3 sin x = 0                                [sin² x+cos² x=1]
⇒                  2-2 sin² x+3 sin x = 0
⇒                   2 sin² x-3 sin x-2 = 0
⇒          2 sin² x+sin x-4 sin x-2 = 0
⇒ sin x(2 sin x + 1)-2(2 sin x+1) = 0
⇒                 (2 sin x+1)(sin x-2) = 0
⇒   Either 2 sin x+1=0 or sin x-2 = 0
But sin x = 2 is not possible as sin x cannot be greater than 1.
∴                               2 sin x+1 = 0
⇒      
⇒                                    
Hence, the solution is given by
                                             

25. Find  centroid of a triangle, mid-points of whose sides are (1, 2, - 3), (2, 0, 1) and (- 1, 1, - 4).

Sol. Let A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) are vertices of triangle ABC.

P(1, 2, -3), Q(3, 0, 1) and R(- 1, 1, - 4) are mid-points of sides AB, BC and AC respectively.

P is the mid-point of AB, therefore

(x₁ + x₂)/2 = 1  ...(1)
(y₁ + y₂)/2 =2   ...(2)
(z₁ + z₂)/2 =- 3 ...(3)

Q is the mid-point of BC, therefore

(x₂ + x₃)/2 =3  ...(4)
(y₂ + y₃)/2 =0  ...(5)
(z₂ + z₃)/2 =1  ...(6)

R is the mid-point of AC, therefore

(x₁ + x₃)/2 =- 1 ...(7)
(y₁ + y₃)/2 =1   ...(8)
(z₁ + z₃)/2 =- 4 ...(9)

Adding (1), (4) and (7), we get

(x₁ + x₂)/2 + (x₂ + x₃)/2 + (x₁ + x₃)/2 = 1 +3-1

⇒ (x₁ + x₂ + x₃) =3

Adding (2), (5) and (8), we get

(y₁ + y₂)/2 + (y₂ + y₃)/2 + (y₁ + y₃)/2 = 2 + 0 + 1

⇒ (y₁ + y₂ + y₃) =3

Adding (3), (6) and (9), we get

(z₁ + z₂)/2 + (z₂ + z₃)/2 + (z₁ + z₃)/2 = - 3 + 1 -4

⇒ (z₁ + z₂ + z₃) =- 6

We have centroid O of triangle is given by

{(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3, (z₁ + z₂ + z₃)/3}= {3/3, 3/3, -6/3}
= (1, 1, - 2)

Therefore, centroid is (1, 1, - 2).

26. If the ratio of the coefficients of 3rd and 4th terms in the expansion of  is 1:2 then find the value of n.

Sol. Given expansion is

T₃ = ⁿC₂ x^n-2

T₄ = ⁿC₃ x^n-3

But we are given,

⇒ -n + 2 = 12

⇒ n = -10

27. Find the value of n so that  may be the geometric mean between a and b.

Sol. G. M. of a and b is

By the given condition,

Squaring both sides, we obtain

⇒ a²ⁿ⁺² + 2aⁿ⁺¹ bⁿ⁺¹ + b²ⁿ⁺² = (ab)(a²ⁿ + 2aⁿbⁿ + b²ⁿ)

⇒ a²ⁿ⁺² + 2aⁿ⁺¹ bⁿ⁺¹ + b²ⁿ⁺² = (a²ⁿ⁺¹ b + 2aⁿ⁺¹ bⁿ⁺¹ +ab²ⁿ⁺¹)

⇒ a²ⁿ⁺² + b²ⁿ⁺² = a²ⁿ⁺¹ b + ab²ⁿ⁺¹

⇒ a²ⁿ⁺² - a²ⁿ⁺¹ b = ab²ⁿ⁺¹ - b²ⁿ⁺²

⇒ a²ⁿ⁺¹(a –b) = b²ⁿ⁺¹ (a – b)

⇒ 2n + 1 = 0

28. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Sol. Let the observations be x₁, x₂, x₃, x₄, x₅, and x₆.

It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are y_i, then

y_i = 3x, i.e., , for i = 1 to 6

From (1) and (2), it can be observed that,

Substituting the values of x_i and in (2), we obtain

Therefore, variance of new observations

Hence, the standard deviation of new observations is

29. P,Q,R shot to hit a target. If P hits it 3 times in 4 trials, Q hits it 2 times in 3 trials and R hits it 4 times in 5 trials, what is the probability that the  target is hit by at least two persons?

Sol. Let E₁, E₂ and E₃ be independent events that P, Q and R hit the target respectively. Then
P(E₁) = 3/4, P(E₂) = 2/3and P(E₃) = 4/5
P(E₁^c) = 1 - 3/4 = 1/4, P(E₂^c) = 1 - 2/3 = 1/3and P(E₃^c) = 1 - 4/5 = 1/5 .

The target is hit by atleast two persons in the following mutually exclusive ways:

(a) P hits, Q hits and R does not hit i.e., E₁ ∩ E₂ ∩ E₃^c
(b) P hits, Q does not hit and R hits i.e., E₁ ∩ E₂^c ∩ E₃
(c) P does not hit, Q hits and R hits i.e., E₁^c ∩ E₂ ∩ E₃
(d) P hits, Q hits and R hits i.e., E₁ ∩ E₂ ∩ E₃
Required Probability = P[(a) U (b) U (c) U (d)] = P(a) + P(b) + P(c) + P(d)
                              = (3/4 x 2/3 x 1/5) + (1/4 x 2/3 x 4/5) + (3/4 x 2/3 x 4/5)
                              = 25/30 = 5/6.