Mathematics – Standard (041)
Sample Question Paper 2019-20
Max. Marks: 80
Duration : 3 hrs
(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.
Q.1. If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers,, then HCF (a,b) is : (1 Mark)
Ans. Correct option : (b)
Explanation : Since a = x3y2 = and b = xy3
Thus, HCF of a and b =
Q.2. If n is any natural number, then 6n - 5n always ends with : (1 Mark)
Ans. (a) 1
[For any n ∈ N, 6n and 5n end with 6 and 5 respectively.
∴ 6n - 5n always ends with 6 - 5 i.e., 1]
Q.3. If A and B are acute angles such that sin A = cos B, then A + B is equal to (1 Mark)
Ans. (d) Given, sin A = cos B
⇒ sinA = sin(90° - B)
⇒ A = 90° - B
⇒ A+ B = 90°
Q.4. The areas of two similar triangles ABC and DEF are 64 cm! and 121 cm2 respectively. If EF = 15.4 cm, then the value of BC will be (1 Mark)
(a) 11.2 cm
(c) 8.14 cm
(d) 11.4 cm
Ans. Choice (a) is correct
Given, ΔABC - ΔDEF, EF - 15.4 cm, ar(ΔABC) = 64 sq. cm and ar(ΔDEF) = 121 sq. cm.
The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides
Q.5. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio : (1 Mark)
(a) 2 : 3
(b) 4 : 9
(c) 81 :1 6
(d) 16 : 81
In the given figure, AT is a tangent to the circle with centre 'O' such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to :
(a) 4 cm
(b) 2 cm
(c) 2√3 cm
(d) 4√3 cm
Ans. Correct option : (d)
Explanation : We know that the ratio of the areas of the triangles will be equal to the square of the ratio of the corresponding sides of the triangles.
Thus, required ratio of the areas of the two
Correct option: (c)
Explanation : Join OA. OA is radius and AT is tangent at contact point A
∴ ∠OAT = 90°,
Given that, OT = 4 cm
Q.6. If x = a secθ cosϕ, y = b sec θ sin ϕ and (1 Mark)
Q.7. The area of a quadrant of a circle whose circumference is 20 cm is (1 Mark)
Ans. (b) Given, the circumference of a quadrant = 20
Now, area of a quadrant
Q.8. If the radius of a circle is 6 cm, then the distance between two parallel tangents is _________. (1 Mark)
Ans. 12 cm
Distance between two parallel tangents = Diameter = 2 x Radius
= 2 x 6 cm
= 12 cm
Q.9. The surface areas of two spheres are in the ratio 16 : 9. The ratio of their volumes is (a) 64 :27 (1 Mark)
(b) 16 : 9
(c) 4 : 3
(d) 163 : 93
Ans. Correct option : (a)
Now, volumes of two spheres,
Q.10. If P(2, a) is the mid-point of the line segment joining the points L(6, -5) and M(-2, 11), then value of a is (1 Mark)
Ans. (d) 3
[Here, P(2, a) is the mid-point of the line segment joining L(6, -5) and M(-2, 1)
Q.11. A line which divides two sides of a triangle in the same ratio is .......... to the third side. (1 Mark)
Q.12. If the sum of first n terms of an A.P. is 2n2 + 6n. Its second term will be 12. (1 Mark)
Let Sn denote the sum of first n terms and an denote nth term; then
⇒ a1 = 8 ...(2)
Putting n = 2 m (1), we get
S2 = 2 x 22+ 6 x 2 = 8 +12 = 20
Q.13. In figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20 √3 m long. Find the Sun's altitude. (1 Mark)
Ans. Let the ∠ACB be θ, ∠B = 90°
⇒ θ = 30°
Thus, the Sun's altitude is 30°.
Q.14. The common difference of an AP, in which a12 - a8 = 24, is ________. (1 Mark)
Q.15. Trigonometric ratio can be determined in _______ angled triangle only. (1 Mark)
Q.16. Find the value of 0 in the equation 2 sin 2θ = 1. (1 Mark)
Ans. We have,
2 sin 2θ = 1
Q.17. Find the sum of first 16 terms of the A.P 10,6,2,........ (1 Mark)
Ans. Here, a = 10, d = 6 - 10 = - 4 and n = 16
Q.18. If 47x + 31y = 18 and 31x+ 47y = 60, then find the value of x + y. (1 Mark)
Ans. Adding the given equations, we have 78x + 78y = 78
⇒ x + y = 1
Q.19. Find the 9th term of the AP whose nth term is given by 3n - 5. (1 Mark)
Ans. Given, Tn = 3n-5
⇒ T9 = 3(9)-5 = 22
Hence, the 9th term of the AP is 22.
Q.20. In the given figure, O is the centre of the circle. PT and PQ are tangents to the circle from an external point P. If (1 Mark)
Ans. Join OT and OQ.
Since OTPQ is a cyclic quadrilateral
∴ ∠QOT + ∠QPT = 180°
⇒ ∠QOT + 70° = 180°
⇒ ∠QOT= 180° -70°
⇒ ∠QOT = 110° ...(1)
Q.21. Find the smallest natural number by which 1,200 should be multiplied so that the square root of the product is a rational number. (2 Mark)
Ans. Since, 1,200 = 4 x 3 x (2 x 5)2
= 24 x 3 x 52
Hence, the required smallest natural number is 3.
Q.22. Lets denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F respectively, prove that BD = s - b. (2 Mark)
Ans. We know that, tangents from an external point to a circle are of equal length.
Q.23. Find the type of triangle formed by the points P (-5,6) Q (-4,-2) and R (7,5).
A parallelogram has vertices P(1,4), Q(7,11), R(a,4) and S(1,-3). Find the value of a. (2 Mark)
Ans. Given, vertices of triangle are P (-5,6), Q (-4,-2) and R (7,5).
Hence, the given triangle is a scalene triangle.
Let P(1,4), Q(7,11), R(a,4) and S(1,-3) be the vertices of a parallelogram POPS, respectively. Join PR and QS. Let PR and QS intersect at the point T.
We know that, the diagonals of a parallelogram bisect each other.
So, T is the mid-point of PR as well as that of QS. Mid-point of PR = Mid-point of QS
On comparing x-coordinate from both sides, we get
Hence, the value of a is 7.
Q.24. Three different coins are tossed simultaneously. Find the probability of getting exactly one head. (2 Mark)
Ans. When three different coins are tossed simultaneously, the possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
∴ Total number of possible outcomes in the sample space = 8.
Let A be the event of getting exactly one head.
Then, cases favourable to the event A are HTT, THT and TTH.
Number of cases in favour of event A = 3.
Probability of getting exactly one head - P(A)
Q.25. If the perimeter of a protractor is 72 cm, calculate its area,
If the total surface area of a solid hemisphere is 462 cm2 , find its volume (2 Mark)
Ans. Perimeter of semi-circle,
Q.26. Two types of water tankers are available in a shop at the same cost One is in a conical form of diameter 1 m and height 2 m and another is in the form of a cylinder of diameter 1 m and height 1.5 m. The shopkeeper advises to purchase cylindrical tanker. Out of the two tankers, whose volume or quantity is more and by how much ? (2 Mark)
Shopkeeper is telling truth.
Difference in the quantity of the two tankers = 1.18 - 0.52 = 0.66 m3
Q.27. A bag contains cards numbered 1 to 49. Find the probability that the number on the drawn card is : (3 Mark)
(i) an odd number
(ii) a multiple of 5
(iii) Even prime
Ans. Total cards = 49
Odd numbers from 1 to 49 are 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47 and 49.
No. of favourable outcomes = 25
(ii) Multiple of 5 are 5,10, 15, 20, 25, 30, 35, 40 and 45.
No. of favourable outcomes = 9
(ii) P(multiple of 5) = 9/49
Only even prime number is 2 i.e., P(even prime) 1. D
(iii) P(even prime)= 1/49
Q.28. The sums of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The first term of each AP is 1 and their common differences are 1, 2 and 3 respectively. Prove that : S1 + S3 = 2S2 (3 Mark)
Ans. Given that, S1, S2 and S3 are the sums of first n terms of three APs, whose first term of each AP is 1 and common differences are 1, 2 and 3 respectively.
Q.29. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 h to complete the total journey, then find its original average speed. (3 Mark)
Ans. Let the original average speed of train be x km/h.
Then, the speed for the journey of 72 km will be (x + 6) km/h.
[by factorisation method]
Since, x is the average speed of the train, so x cannot be negative.
Hence, the original average speed of the train is 42 km/h.
Q.30. In what ratio does the point P(-4, y) divide the line segment joining the points A(-6,10) and B(3, - 8) ? Hence, find the value of y.
Find the value of p for which the points (- 5,1), (1, p) and (4, - 2) are collinear. (3 Mark)
Ans. Given points are A(- 6, 10) and B(3, - 8).
Let the point P(- 4, y) divides the line segment joining the points A(- 6,10) and 5(3, - 8) in the ratio k: 1.
Using section formula, the coordinates of point P are
It is given that point P is (-4,y).
∴ The point P divides AB in the ratio k : 1 i.e., 2/7: 1 i.e., 2 : 7 and y = 6.
Let the given points be (-5,1),B(1,p) and C(4,-2). Since the points A(- 5, 1),B(1,p) and C(4,-2) are collinear, therefore, the area of the triangle formed by them is zero.
∴ Area of ΔABC = 0
Hence, the points are collinear when p = - l .
Q.31. In the given figure, ABC is a right angled triangle at ∠B = 90°. D is the mid-point of BC. Show that AC2 = AD2 + 3CD2. (3 Mark)
Using Pythagoras theorem in the right AABC, we have
Q.32. Prove that: (3 Mark)
Q.33. The New Delhi Municipal Corporation decided to planting trees in between the vacant area of two buildings. If the angle of depression of the top and bottom of 8 m tall building from the top of a multistoried building are 30° and 45° respectively, then find the height of the multistoried building and the distance between the two buildings. (3 Mark)
Ans. Let AB be the building of height 8 m and CD be a multistoried building.
∴ Height of multistoried building,
DC = 10.92 + 8 = 18.92 m
Hence, the distance between the two buildings is 18.92 m.
Q.34. Calculate the mode of the following distribution : (3 Mark)
Ans. Here, the maximum frequency is 20 and the class corresponding to this frequency is 20 - 25. So, modal class is 20 - 25.
Lower limit of modal class, I = 20; Class size, h = 5.
Frequency of modal class, f1 = 20; Frequency of class preceding modal class, f0 = 7;
Frequency of class succeeding the modal class, f2 - 8.
Using the formula :
Q.35. Solve graphically the pair of linear equations :
3x - 4y + 3 = 0 and 3x + 4y - 21 = 0.
Find the co-ordinates of the vertices of the triangular region formed by these lines and x-axis. Also, calculate the area of this triangle. (4 Mark)
(i) These lines intersect each other at point (3, 3).
Hence, x = 3 and y = 3.
(ii) The vertices of triangular region are (3, 3), (-1, 0) and (7, 0).
Q.36. Sides AB and BC and median AD of a triangle ABC are respectively proportional to the sides PQ and QR and median PM of APQR, Show that ΔABC ~ ΔPQR. (4 Mark)
Ans. Given : AD and PM are the medians of ΔABC and ΔPQR respectively, such that
To Prove : ΔABC - ΔPQR Const. Produce AD to E, such that AD = DE.
Produce PM to N, such that PM = MN.
Join BE and QN.
Proof : In ΔADC and ΔEDB
Q.37. In ΔABC, points P and Q on sides AB and AC, a re such that is extended to T such that PT - BC and PB - TC, then find the value of x. (4 Mark)
In ΔABC, it is given that
∴ PTCB is a parallelogram
⇒ ∠1= ∠2
On applying angle sum property in ΔAPQ, we get
Q.38. A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/min.
There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole. (4 Mark)
Ans. Let B be the top of the cliff AB of height 150 metres. Let C and D be initial and final positions of the boat moving away from the cliff. Let BX be the horizontal line through B. It is given that angles of depression of the boat at C and D as observed from B are 60° and 45° respectively. Let AC = x metres.
In ΔADB, we have
Time taken by the boat in moving C to D = 2 minutes
Let AS and CD be two poles on either sides of river just opposite to each other such that AS = 60 metres; CD = h metres and AC = x metres. Let BX be horizontal line through B. It is given that the angles of depression of the top D and the foot C of the pole CD are 30° and 60° respectively. Draw DE ⊥ from D on AB.
In ΔACB, we have
Putting the value of x from (2) in (1), we get
Q.39. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building. (4 Mark)
⇒ h = 20 m.
Q.40. Find the mean of the following distribution (4 Mark)