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

**Sample Paper**

Subject: Mathematics

Subject: Mathematics

**Class 10**

^{th}

**(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**

*a*_{20}−*a*_{15}is**Solution:**

Let
the first term of the A.P. be

*a*.
a

_{n}= a + (n-1)d
a

_{20}– a_{15 }_{ }_{}_{ }

_{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 = π

*r*
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

∴

*px*^{2}− 2*px*+ 6 = 0
Since
the roots of the given quadratic equation are equal, its discriminant is equal
to 0.

⇒ D = 0

⇒

*b*^{2}− 4*ac*= 0
⇒ (− 2

*p*)^{2}− 4 ×*p*× 6 = 0 [*a*=*p*,*b*= −2*p*,*c*= 6]
⇒ 4

*p*^{2}− 24*p*= 0
⇒ 4

*p*(p − 6) = 0
⇒ 4

*p*= 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

*n*^{th }term of an AP is given by,*an = a*+ (*n*− 1)*d*
49
= 4 + (10 − 1)

*d*
49
− 4 = 9

*d*
45
= 9

*d**d*

*= 5*

Thus,
the common difference of the AP is 5.

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