Algebra Questions for RRB NTPC Set-2 PDF

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Algebra Set-2 Questions for RRB NTPC PDF
Algebra Set-2 Questions for RRB NTPC PDF

Algebra Questions for RRB NTPC Set-2 PDF

Download RRB NTPC Algebra Set-2 Questions and Answers PDF. Top 15 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

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Question 1: What is the minimum number that must be added to 9, 23, 41, 83 in order to make them proportional?

a) 3

b) 7

c) 5

d) 9

Question 2: 10 years ago, the average age of Saurav and Sachin was 20 years. 5 years from hence, the average age of Saurav, Sachin and Viru will be 40 years. What is the present age of Viru?

a) 35 years

b) 45 years

c) 20 years

d) Cannot be determined

Question 3: A man has a certain number of chocolates with him. He gave one-fifth of the chocolates to his son, one-fourth of them to his daughter and one-third of them to his nephew. If he has 65 chocolates remaining with him, what is the number of chocolates that the man had initially?

a) 400

b) 320

c) 300

d) 200

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Question 4: The smallest positive integer n with 24 divisors considering 1 and n as divisors is

a) 420

b) 240

c) 360

d) 480

Question 5: The sum of two numbers is equal to 27 and their product is equal to 182. What are the two numbers?

a) 15, 12

b) 11, 16

c) 9, 18

d) 13, 14

e) 19, 8

Question 6: If a = 6, b = 2 and c = -8, then what is the value of $a^3 + b^3 + c^3$?

a) 0

b) -2146

c) -286

d) -288

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Question 7: If $x + \dfrac{1}{x} = 3$, find the value of $x^3 + \dfrac{1}{x^3}.

a) 15

b) 18

c) 12

d) 9

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Question 8: Sum of three consecutive even numbers is 48. what is the square of largest number ?

a) 196

b) 256

c) 324

d) 400

Question 9: What is the minimum value of the expression $x^{2}+12x+42$ ?

a) 3

b) 4

c) 6

d) 5

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Question 10: Find the remainder when $2x^{4}-x^{3}+5x^{2}-2x+6$=0 is divided by x+1.

a) 15

b) 14

c) 13

d) 16

Question 11: The difference between the square and half of the natural number is 33 then what is the cube of the number.

a) 512

b) 216

c) 343

d) 125

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Question 12: What is the remainder when $7x^{8}-5$ is divided by (x+1) ?

a) 2

b) 1

c) 0

d) 3

Question 13: Arithmetic mean of two natural numbers is 15 and what is the minimum value of the product of two numbers ?

a) 225

b) 220

c) 210

d) 200

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Question 14: Sum of the reciprocals of two consecutive natural numbers is 11/30.what is the sum of squares of those numbers ?

a) 51

b) 61

c) 113

d) 41

Question 15: $x^{3}-4x^{2}+12x-k$ is perfectly divisible by (x-3) then what is the value of k ?

a) -27

b) 27

c) -23

d) 23

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Answers & Solutions:

1) Answer (B)

Let the number be x.
=> (9+x)(83+x) = (23+x)(41+x)
=> $747 + 92x + x^2 = 943 + 64x + x^2$
=> 28x = 196
=> x = 7

2) Answer (B)

Total age of Saurav and Sachin 10 years ago = 20 * 2 = 40 years
So, total age of Sachin and Saurav 5 years hence = 40 + 10*2 + 5*2 = 70 years
Total age of all three 5 years hence = 40*3 = 120 years
So, Viru’s age 5 years hence = 120 – 70 = 50
Viru’s present age = 50 – 5 = 45 years

3) Answer (C)

Let the number of chocolates with the man be N.
So, chocolates remaining with him after the distribution = N – (N/5 + N/4 + N/3) = N – 47N/60 = 13N/60 = 65
So, N = 65 * 60/13 = 300

4) Answer (C)

For any given number, that can be represented as $ A^{x} \times B ^ {y} $, etc

The number of factors is denoted by (x+1) x ( y+1), etc

360 = $ 2 ^ {3} \times 3 ^ {2} \times 5^ {1}$

So the number of factors = (3 +1) x (2+ 1) (1+ 1) = 4x3x2 = 24

For 240, it is $ 2 ^ {4} \times 3 ^ {1} \times 5^ {1}$

Number of factors = 5 x 2 x 2 = 20 only

5) Answer (D)

Only in option D the product of numbers is 182.∴option D is the right choice.

6) Answer (D)

Given, a = 6, b = 2, c = -8
Here, a+b+c = 0.
We know that if a+b+c = 0, then $a^3+b^3+c^3 = 3abc$
Therefore, $a^3 + b^3 + c^3 = 3 \times 6 \times 2 \times (-8) = -288$

7) Answer (B)

Given, $x + \dfrac{1}{x} = 3$
Cubing on both sides,
$(x + \dfrac{1}{x})^3 = 27$
$x^3 + \dfrac{1}{x^3} + 3 \times x \times \dfrac{1}{x} (x + \dfrac{1}{x}) = 27$
$x^3 + \dfrac{1}{x^3} + 3 \times 3 = 27$
Therefore, $x^3 + \dfrac{1}{x^3} = 27-9 = 18$

8) Answer (C)

Let the three numbers be 2n-2, 2n,2n+2
sum=6n
We have 6n=48
n=8
Substituting n=8 we have 14,16,18
Square of 18 is 324.

9) Answer (C)

By expressing the given value in terms of square we have $(x+6)^{2}+6$
Minimum value of the term in square is zero
And so minimum value of the expression is 6.

10) Answer (D)

From the remainder theorem we have if f(x) is divided by (x-a) then f(a) is the remainder and so by substituting -1 in f(x) we have
=$2*(-1)^{4}-(-1)^{3}+5*(-1)^{2}-2*(-1)+6$
=2+1+5+2+6
=16

11) Answer (B)

let the number be x
Given $x^{2}-(x/2)$=33
$2x^{2}-x$=66
$2x^{2}-x-66$=0
$2x^{2}-12x+11x-66$=0
2x(x-6)+11(x-6)=0
(2x+11)(x-6)=0
x=6
Cube of 6=216

12) Answer (A)

From the remainder theorem when f(x) is divided by (x-a) then f(a) will be the remainder and so substituting x=-1 in the equation we have
=7-5
=2

13) Answer (A)

We know that AM$\geq$GM
15$\geq$GM
Let the numbers be a and b
$\sqrt{ab}\leq15$
ab$\geq$225

14) Answer (B)

Let the numbers be n and n+1
Then we have (1/(n+1))+(1/(n))=11/30
(2n+1)/(n)(n+1) =11/30
60n+30=$11n^{2}$+11n
$11n^{2}-49n-30$=0
$11n^{2}-55n+6n-30$=0
11n(n-5)+6(n-5)=0
n=5 and so n+1=6
Sum of squares=25+36
=61

15) Answer (B)

From the remainder theorem we know that if f(x) is divisible by (x-a) then f(a)=0 an so by substituting x=3 in the expression we have
$3^{3}-4*3^{2}+12*3-k$=0
27-36+36-k=0
27-k=0
k=27

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