Sample Question Paper Mathematics

First Term (SA - I) Class X

Time: 3 to 3 ½ hours M.M.:90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D. Section A

comprises of 8 questions of 1 mark each, section B comprises of 6 questions of 2 marks each, section C comprises of 10 questions of 3 marks each and section D comprises of 10 questions of 4 marks each.

(iii) Question numbers 1 to 8 in section A are multiple choice questions where you have to select one

correct option out of the given four. (iv) There is no overall choice. However, internal choice has been provided in 1 question of two

marks, 3 questions of three marks each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions.

Section A Question numbers 1 to 8 carry 1 mark each. For each question, four alternative choices have been provided of which only one is correct. You have to select the correct choice. Q. 1 Give that HCF (2520, 6600) = 40, LCM (2520, 6600) = 252 × k, then the value of k is: (A) 1650 (B) 1600 (C) 165 (D) 1625 Solution:

Q. 2 If are the zeroes of f(x) = px2 – 2x + 3p and , then the value of p is:

Solution:

Q. 3

In the given figure

Solution:

Q. 4

Solution:

Q. 5 In the given figure, secA – cosec C is equal to:

Solution:

Q. 6

Solution:

Q. 7 From the following, the rational number whose decimal expression is terminating is:

Solution:

The prime factorization of q, i.e. 5 is of the form 2m × 5n. Where m and n are whole numbers. has terminating decimal expression.

Ans: (A) Q. 8 For a given data with 70 observations, the ‘less than ogive’ and ‘more than ogive’ intersect at (20.5, 35) The median of the data is: (A) 20 (B) 35 (C) 70 (D) 20.5 Solution: The median of grouped data is the x-coordinate of the point of intersection of the two ogives. Here, the ‘less than ogive’ and ‘more than ogive’ intersect at (20.5, 35). Median = 20.5 Ans: (D) Section B Question numbers 9 to 14 carry 2 marks each. Q. 9 In the given factor tree, find the numbers m and n:

Solution: y = 3 × 3 = 9 n = 2 × 9 = 18

x = 2 × 18 = 36 m = 2 × 36 = 72

Q. 10 Find a quadratic polynomial whose zeroes are Solution: Zeroes of quadratic polynomial are (3 + ) and (3 – )

Sum = (3 + ) (3 – ) = 3 + + 3 – = 6 Product = (3 + ) (3 – ) = 32 – ( )2 = 9 – 5 = 4 Required polynomial = x2 – (Sum of zeroes) x + Product of zeroes = x2 – 6x + 4 Q. 11 What type of solution does the pair of equations

Solution:

The given system of equations has a unique solution.

Q. 12

OR

If sin (A + B) = cos (A – B) = and A, B (A > B) are acute angles, find the values of A and

B. Solution:

OR

Q. 13 In a given figure, find ‘x’ in terms of p, q and r

Solution:

Q. 14 For what value of x, is the mean of the given observations 2x – 5 , x + 3, 7 – x, 5 – x and x + 9 with frequencies 2, 3, 4, 6 and 1 respectively 4? Solution:

2x – 5 2 4x – 10 x + 3 3 3x + 9 7 – x 4 28 – 4x 5 – x 6 30 – 6x x + 9 1 x + 9 Total 16 66 – 2x

Section C Question numbers 15 to 24 carry 3 Marks each. Q. 15 Prove that is an irrational number.

OR

Prove that is irrational.

Solution: Let us assume, to the contrary, that 12 – 41 is rational.

OR

Q. 16 Solve the given pair of linear equations by the cross multiplication method: ax + by = a – b bx – ay = a + b

OR

Solve : mx – ny = m2 + n2 x – y = 2n Solution:

ax + by = a – b or ax + by = – (a – b) = 0 …(1) bx – ay = a + b or bx – ay = – (a + b) = 0 …(2)

OR

y = n + m – 2n y = m – n

Hence, the solution of the given system of equations is x = n + m, y = m – n.

Q. 17

Prove that: (cosec A – sin A) (sec A – cos A) =

Solution:

Q. 18 If A + B = , then prove that :

Solution:

Q. 19 In figure, are on the same base BC. If AD intersects BC at O, then prove that:

Solution: Draw AM and DN perpendicular to BC

Q. 20 Calculate mode for the following distribution:

Age Below 30 Below 40 Below 50 Below 60 Below 70 No. of persons 12 22 47 62 70

OR

The distribution below gives the weight of 30 students of a class. Find the median weight of the students.

Weight ( in kg) 40 – 45 45 - 50 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 No. of students 2 2 8 6 6 3 2

Solution:

Age 20 – 30 30 – 40 40 – 50 50 – 60 60 - 70 No. of persons 12 10 25 15 8

Class 40 – 50 has maximum frequency, i.e. 25

OR

Weight (in kg) No. of students Cumulative frequency 40 – 45 2 2 45 – 50 3 5 50 – 55 8 13 55 – 60 6 19 60 – 65 6 25 65 – 70 3 28 70 - 75 2 30

Total number of students (n) = 30

Cumulative frequency greater than 15 is 19 and corresponding class is 55 – 60.

Q. 21 Using trigonometric identities, write the given expression as an integer:

Solution:

Q. 22 If one zero of the quadratic polynomial 2x2 – (3k+1)x – 9 is negative of the other, find the value of k. Solution:

Q. 23 ABC is a triangle in which AB = AC and D is a point on AC such that = AC × CD. Prove that BD = BC Solution:

Q. 24 The median class of a frequency distribution is 125 – 145. The frequency of this class and cumulative frequency of the class preceding to the median class are 20 and 22 respectively. Find the sum of the frequencies, if the median is 135 Solution: Median class = 125 – 145 Frequency of median class (f) = 20 Cumulative frequency of the class preceding to the median class (cf) = 22 Median = 135 Class size (h) = 145 – 125 = 20 Lower Limit of median class (l) = 125 Let the sum of the frequencies be n

Section D Question numbers 25 to 34 carry 4 marks each. Q. 25 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

OR Prove that in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

Solution:

OR

Construction : Construct a right triangle PQR, right angled at Q such that PQ = AB and QR = BC Proof : In right triangle PQR,

In ∆ ABC and ∆ PQR AB = PQ [Given] BC = QR [Given] AC = PR [Proved in (3)]

∆ ABC ∆ PQR [SSS congruency condition]

Q. 26 Prove that :

OR

If tan A = , show that sin A cos A =

Solution:

OR

Q. 27 Evaluate:

Solution:

Q. 28 Solve the system of equations graphically x + 2y = 5; 2x – 3y = – 4. Also find the points where the lines meet the x-axis. Solution: x + 2y = 5 …(1)

x 3 1 3 y 4 2 1

x 2 1 4 y 0 2 4

The two lines intersect each other at the point (1, 2) Hence, x = 1, y = 2 is the solution of the system.

The line x + 2y = 5 meets the x-axis at (5, 0). The line 2x – 3y = – 4 meets the x-axis at ( 2, 0) Q. 29 Draw less than and more than ogive for the following distribution and hence obtain the median.

Marks 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 - 100 No. of students 14 6 10 20 30 8 12

Solution:

Marks No. of student Cumulative Frequency ‘less than’

Cumulative Frequency ‘More than’

30 – 40 14 Less than 40 14 More than or equal to 30

100

40 – 50 6 Less than 50 14+6 = 20 More than or equal to 40

100 – 14 = 86

50 – 60 10 Less than 60 20 + 10 = 30 More than or equal to 50

86 – 6 = 80

60 – 70 20 Less than 70 30 + 20 = 50 More than or equal to 60

80 – 10 = 70

70 – 80 30 Less than 80 50 + 30 = 80 More than or equal to 70

70 – 20 = 50

80 – 90 8 Less than 90 80 + 8 = 88 More than or equal to 80

50 – 30 = 20

90 - 100 12 Less than 100 88 + 12 = 100 More than or equal to 90

20 – 8 = 12

From the graph, Median = 70. Q. 30

In an equilateral triangle ABC, D is a point on side BC such that .

Prove that 9AD2 = 7AB2

Solution: Given : An equilateral triangle ABC and D is a point on BC such that BD =

To prove : Construction : Draw AE BC. Join AD.

Q. 31 Show that any positive odd integer is of the form 8q + 1 or 8q + 3 or 8q + 5 or 8q + 7 Solution: Let a be any positive integer and b= 8 Then, by Euclid’s algorithm, a = 8q + r, for some integer q ≥ 0 and 0 ≤ r <8 i.e., the possible remainders are 0, 1, 2, 3, 4, 5, 6, 7. If r = 0, 2, 4, 6; a is even If r = 1, 3, 5, 7 ; a is odd. a is an odd positive integer in the form 8q +1, 8q + 3, 8q + 5, 8q + 7. Q. 32 Students of a class are made to stand in rows. If one student is extra in a row, there would be 2 rows less. If one student is less in a row, there would be 3 rows more. Find the number of students in the class. Solution: Let number of students in each row be x and number of rows be y.

Total number of students = xy Case I : If one student is extra in a row, number of students in each row = x + 1, and number of rows = y – 2

Total number of students = (x + 1) (y – 2) xy = (x + 1) ( y – 2) xy = xy – 2x + y – 2 2x – y = – 2 …(1)

Case II : If one student is less in a row, number of students in each row = x – 1 and number of rows = y + 3

Total number of students = (x – 1) (y + 3) xy = xy + 3x – y – 3 3x – y = 3 …(2)

Subtracting (1) from (2), we get

Substituting the value of x in eq (1), we get 2 × 5 – y = – 2

– y = – 2 – 10 y = 12

Total number of students in the class = 5 × 12 = 60.

Q. 33 If the remainder on division of x3 – kx2 + 13x – 21 by 2x – 1 is – 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x3 – kx2 + 13x. Solution:

Q. 34 Find the mean marks of students for the following distribution by step deviation method.

Marks Number of students 0 and above 80 10 and above 77 20 and above 72 30 and above 65 40 and above 55 50 and above 43 60 and above 28 70 and above 16 80 and above 10 90 and above 8 100 and above 0

Solution:

Marks No. of students

0 – 10 80 – 77 = 3 5 4 12 10 – 20 77 – 72 = 5 15 3 15 20 – 30 72 – 65 = 7 25 2 14 30 – 40 65 – 55 = 10 35 1 10 40 – 50 55 – 43 = 12 45 = a 0 0 50 – 60 43 – 28 = 15 55 1 15 60 – 70 28 – 16 = 12 65 2 24 70 – 80 16 – 10 = 6 75 3 18 80 – 90 10 – 8 = 2 85 4 8 90 – 100 8 95 5 4

Total 80 54