CBSE Mathematics sample paper for class 10 solved for sa 2
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January 23, 2016 |
1 Comment |
Sample Paper
Subject: Mathematics
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
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 = p, b = −2p, c = 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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