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
Solution:
, then the value of p is:
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 +
Sum = (3 +
) (3 –
)
=3+
+3–
=6
Product = (3 +
= 32 – (
=9–5
=4
) (3 –
)2
) and (3 –
)
)
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) =
B.
Solution:
OR
and A, B (A > B) are acute angles, find the values of A and
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
x+3
7–x
5–x
x+9
Total
2
3
4
6
1
16
4x – 10
3x + 9
28 – 4x
30 – 6x
x+9
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:
or
or
ax + by = a – b
ax + by = – (a – b) = 0
bx – ay = a + b
bx – ay = – (a + b) = 0
…(1)
…(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 =
Solution:
, then prove that :
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
No. of persons
Below 30
12
Below 40
22
Below 50
47
Below 60
62
Below 70
70
OR
The distribution below gives the weight of 30 students of a class. Find the median weight of the
students.
Weight ( in kg)
No. of students
40 – 45
2
45 - 50
2
20 – 30
12
30 – 40
10
50 – 55
8
55 – 60
6
60 – 65
6
Solution:
Age
No. of persons
Class 40 – 50 has maximum frequency, i.e. 25
40 – 50
25
50 – 60
15
60 - 70
8
65 – 70
3
70 – 75
2
OR
Weight (in kg)
40 – 45
45 – 50
50 – 55
55 – 60
60 – 65
65 – 70
70 - 75
No. of students
2
3
8
6
6
3
2
Cumulative frequency
2
5
13
19
25
28
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
Prove that BD = BC
Solution:
= AC × CD.
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
BC = QR
AC = PR
∆ ABC  ∆ PQR
[Given]
[Given]
[Proved in (3)]
[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
x
y
…(1)
3
4
1
2
3
1
x
y
2
0
1
2
4
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
No. of students
30 – 40
14
40 – 50
6
50 – 60
10
60 – 70
20
70 – 80
30
80 – 90
8
90 - 100
12
Solution:
Marks
No. of student
30 – 40
14
Cumulative Frequency
‘less than’
Less than 40
14
40 – 50
6
Less than 50
14+6 = 20
50 – 60
10
Less than 60
20 + 10 = 30
60 – 70
20
Less than 70
30 + 20 = 50
70 – 80
30
Less than 80
50 + 30 = 80
80 – 90
8
Less than 90
80 + 8 = 88
90 - 100
12
Less than 100
88 + 12 = 100
Cumulative Frequency
‘More than’
More than or
100
equal to 30
More than or
100 – 14 = 86
equal to 40
More than or
86 – 6 = 80
equal to 50
More than or
80 – 10 = 70
equal to 60
More than or
70 – 20 = 50
equal to 70
More than or
50 – 30 = 20
equal to 80
More than or
20 – 8 = 12
equal to 90
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
0 and above
10 and above
20 and above
30 and above
40 and above
50 and above
60 and above
70 and above
80 and above
90 and above
100 and above
Number of students
80
77
72
65
55
43
28
16
10
8
0
Solution:
Marks
No. of students
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
60 – 70
70 – 80
80 – 90
90 – 100
Total
80 – 77 = 3
77 – 72 = 5
72 – 65 = 7
65 – 55 = 10
55 – 43 = 12
43 – 28 = 15
28 – 16 = 12
16 – 10 = 6
10 – 8 = 2
8
80
5
15
25
35
45 = a
55
65
75
85
95
4
3
2
1
0
1
2
3
4
5
 12
 15
 14
 10
0
15
24
18
8
4
54