SCC is a place where u can get free study material for academic & competitive exams like CTET,SSC,IIT,NDA,Medical exams etc,

Saturday, 23 January 2016

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.
To read complete paper Download file............. 
 Download file from here
Share on Google Plus Share on whatsapp

Search

Popular Posts

Facebook

Blogger Tips and TricksLatest Tips For BloggersBlogger Tricks
SCC Education © 2017. Powered by Blogger.

Total Pageviews