Top 10 Questions Arithmetic Ability for SSC CGL and CHSL
Top 10 Questions (Algebra) for SSC CGL and CHSL
Answers are in bold
1)
If 
,then the value of 
will be
A) 1
B)
C) 2
D)
B)
C) 2
D)
2)
If 
,then the value of 
will be
A) 3
B) 2
C) 1
D) 0
B) 2
C) 1
D) 0
3)
If 
,then the value of 
will be
A) 2
B) 0
C) 1
D) -2
B) 0
C) 1
D) -2
4)
If 
,then the value of x will be
A) 2
B)
C) 1
D) -1
B)
C) 1
D) -1
5)
If 
,then the value of 
will be
A) 
B)
C)
D)
B)
C)
D)
6)
If 
,then the value of 
will be
A) 4
B) 8
C) 9
D) 12
B) 8
C) 9
D) 12
7)
If 
,then the value of 
will be
A) 2
B) -2√2
C) 2√2
D) -√2
B) -2√2
C) 2√2
D) -√2
8)
If 
,then the value of 
will be
A) 4
B) 6
C) 8
D) 10
B) 6
C) 8
D) 10
9)
If 
,then the value of (a+1)will be
A) a-1
B)
C)
D)
B)
C)
D)
10)
For two positive numbers a and b, if 
=24 and ab=6, then
is equal to
A) 96
B) 10
C) 102
D) 108
B) 10
C) 102
D) 108
provide solutions also
ReplyDeletehow explain
ReplyDeletekindly explain question no 2
ReplyDeletex+ 1/x=rt 3 squaring on both side we get x^2 +1/x^2 =1 ......... now let us assume x as A and 1/x as B then A^3 +B^3 = (A+B)(A^2 -AB +B^2) .......... we know the value of A+B =rt 3 and also value of A^2 +B^2 =1 and value of AB=1 putting all the value we get
Delete(rt3) (1-1) =0
for Question NO 1 put the value of a =1 and b=1
ReplyDeletefor question no 3 put the value of x=1
for question no 4 put the value of x=1
Q-6 both side multiply by 3/2
ReplyDeleteQ-8.... x^2+y^2 -2x +6y +10=0........ this can be written as x^2 -2x +1 +y^2 +6y +9=0........
ReplyDelete= (x-1)^2 + (y+3)^2=0 this condition only satisfy when (x-1)=0 and (y+3)=0 from this we can get the value of x and y i.e x=1 and y=-3 ...
now put the value of x and y in x^2 + Y^2 = 1+9+10
correction required 1+9=10
Deletefor ques no 9: answer must be 1/(a-1).
ReplyDelete(a^2) - 1 = 1 => (a+1)(a-1) = 1 => a+1 = 1/(a-1)
correct me if I am wrong
you are correct. if the given answer is correct, it suggests a+1=a-1 meaning 1 equals -1.
Deleteoption d is correct.
ReplyDelete