Algebra Questions for RRB NTPC Set-3 PDF

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Algebra Questions for RRB NTPC Set-3 PDF
Algebra Questions for RRB NTPC Set-3 PDF

Algebra Questions for RRB NTPC Set-3 PDF

Download RRB NTPC Algebra Questions Set-3 PDF. Top 10 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Download Algebra Questions for RRB NTPC Set-3 PDF

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Question 1: 8+57+38+108+169

a) 4

b) 6

c) 8

d) 10

Question 2: If a * b = 2a + 3b – ab, then the value of (3 * 5 + 5 * 3) is

a) 10

b) 6

c) 4

d) 2

Question 3: If a * b = ab, then the value of 5 * 3 is

a) 125

b) 243

c) 53

d) 15

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Question 4: If x=1+2+3 , then the value of (2x48x35x2+26x28) is __?

a) 66

b) 0

c) 36

d) 26

Question 5: If a2+b2+c2=2(a2bc3) then the value of a+b+c is

a) 3

b) 0

c) 2

d) 4

 

Question 6: Find the simplest value of 250+1872 is __? (2=1.414).

a) 9.898

b) 10.312

c) 8.484

d) 4.242

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Question 7: If a3b3c3=0 then the value of a9b9c93a3b3c3 is

a) 1

b) 2

c) 0

d) -1

Question 8: If x + y + z = 6 and x2+y2+z2=20 then the value of x3+y3+z3-3xyz is

a) 64

b) 70

c) 72

d) 76

Question 9: If a and b are the roots of the equation x2+4x2=0, find the equation, which will have the roots, ab and ba.

a) x2+5x+6=0

b) x210x+5=0

c) x2+10x+1=0

d) 2x2+5x+1=0

Question 10: Find x in the equation 810.7565610.25=3x

a) 6

b) 8

c) 2

d) 5

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Answers & Solutions:

1) Answer (A)

Start from the root of 169 then second root will reduce to 11, thrid root will reduce to 7, fourth root will reduce to 8, and finally it reduce to value 4

2) Answer (A)

For 3*5 put a=3 and b=5 in given equation
and for 5*3 put a=5 and b=3 in equation
now add both values

3) Answer (A)

Put a=5 and b=3 in given equation
hence it will be 53 = 125

4) Answer (A)

x = 1+ 2+3
(x1)2 = (2+3)2
x2+12x=5+26
x22x=4+26 ( eq. (1) )
(x22x)2=x4+4x24x3=40+166 eq (2)
Now in 2x48x35x2+26x28
or 2(x44x3)5x2+26x28 ( putting values from eq (1) and eq (2) )
After solving we will get it reduced to 66

5) Answer (B)

Given a2+b2+c2=2(a2bc3),

So, (a1)2+(b+2)2+(c1)2=0

Hence, a=1, b=-2 and c=1

So, the sum of the equation is

6) Answer (A)

Given equation can be reduced in the form of 102+3262=72
Hence  72 will be around 9.898

7) Answer (C)

shortcut :

put c = 0 in  a3b3c3=0  a3=b3

a9b9(0)93a3b3(0)3a9b9(a3)3(b3)3 =  (a)3(a)3 = 0  ( a3=b3 )

so the answer is option C.

normal method :

a3b3c3=0

a3=b3+c3

cubing on both sides,

(a3)3=(b3+c3)3

a9=b9+c9+3b3c3(b3+c3)

a9=b9+c9+3b3c3(a3)

a9b9c93a3b3c3=0

so the answer is option C.

8) Answer (C)

We know that x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzxz)
x3+y3+z33xyz=(6)(20xyyzxz)
Hence the solution must be a multiple of 6.
Out of the given options only Option C is a multiple of 6.
Hence Option C is the correct answer.

9) Answer (C)

Sum of the roots of new equation = ab+ba = a2+b2ab = (a+b)22abab = (4)22(2)(2) =-10
Product of the roots = abba = 1
Therefore, the required equation is x2+10x+1=0.

10) Answer (D)

810.75=27
65610.25=9
==> 810.7565610.25=35
==> x = 5.

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We hope this Algebra Questions Set-3 pdf  for RRB NTPC Exam will be highly useful for your Preparation.

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