# SCC Education

## CBSE Mathematics sample paper for class 10 solved for sa 2

Sample Paper
Subject: Mathematics
Class 10th

(EXPECTED QUESTIONS)
Time allowed: 3hrs                                                                                                                                                                               Marks: 90
Instructions:-
1.All questions are compulsory. 2. This Q.P consists of 31 questions divided into four sections A,B,C,D. 3. section A comprises of 4 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 ,section D comprises of 11 questions of  4 marks each.

Section A( 1 X 4 =4 )

Q1. If the common differences of an A.P. is 3, then what is a20 − a15 is
Solution:
Let the first term of the A.P. be a
an = a + (n-1)d
a20 – a15
ans

Q2. A tower stands vertically on the ground. From a point on the ground which is 25 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 45°. Then the height (in meters) of the tower is
Solution:
Let AB be the tower and C be the point on the ground 25 m away from the foot of the
tower such that ACB = 45°.
In right ÄABC:
AB = 25 m
Thus, the height of the tower is 25 m.

Q3.  If the diameter of a semicircular protractor is 14 cm, then find its perimeter.
Solution:
Diameter = 14 cm
Radius =
Length of the semicircular part = πr
Total perimeter = Length of semicircular part + Diameter
= 22 cm + 14 cm
= 36 cm
Thus, the perimeter of the protractor is 36 cm.

Q4. What is the distance between the points A(c, 0) and B(0, −c)?

Solution:
Using distance formula, the distance between the points A(c, 0) and B (0, −c) is given by:
AB =
Thus, the distance between the given is  .

Section B( 2 X 6 =12)
Q5. Find the value of p for which the roots of the equation px (x − 2) + 6 = 0, are equal.

Solution:
The given quadratic equation is px(x − 2) + 6 = 0.
Let us factorize the given quadratic equation.
px(x − 2) + 6 = 0
px2 − 2px + 6 = 0
Since the roots of the given quadratic equation are equal, its discriminant is equal to 0.
D = 0
b2 − 4ac = 0
(− 2p)2 − 4 × p × 6 = 0 [a = pb = −2pc = 6]
4p2 − 24p = 0
4p (p − 6) = 0
4p = 0 or p − 6 = 0
p = 0 or p = 6

Q6. Find the common differnece of an A.P. whose first term in 4, the last  term is 49 and the sum of all its terms is 265.

Solution:
It is known that the sum of n terms of an AP whose first term is a and last term is l is given as
Here, first term, a = 4
Last term, l = 49
Sum of n terms (Sn) = 265
Therefore, the series has 10 terms.
It is known that the nth term of an AP is given by, an = a + (n − 1) d
49 = 4 + (10 − 1) d
49 − 4 = 9d
45 = 9d
d = 5
Thus, the common difference of the AP is 5.
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