RRB JE Algebra Questions with Detailed Solutions

Given
m + n + o = 15
mn + no + om = 75
Formula Used
​$$m^3 + n^3 + o^3 - 3mno = (m + n + o)((m + n + o)^2 - 3(mn + no + om))$$​​
Solution
​$$m^3 + n^3 + o^3 - 3mno = 15 × (15^2 - 3 × 75) \\m^3 + n^3 + o^3 - 3mno = 15 × (225 - 225)\\m^3 + n^3 + o^3 - 3mno = 15 × 0 = 0\\$$​​
Final Answer
So the correct answer is (d)

Given:
Equation: $$(5x+1)^3 + (x+3)^3 + 8(3x-4)^3 = 6(5x+1)(x+3)(3x-4)$$​​
Formula Used:
If $$a^3 + b^3 + c^3 = 3abc$$​, then a + b + c = 0 (for distinct real numbers).
Solution:
Rewrite the third cubed term and the right-hand side to match the standard identity form:
​$$8(3x-4)^3 = [2(3x-4)]^3 = (6x-8)^3$$​​
6(5x+1)(x+3)(3x-4) = 3(5x+1)(x+3)[2(3x-4)] = 3(5x+1)(x+3)(6x-8)
Now the equation is in the form:
​$$a^3 + b^3 + c^3 = 3abc$$​​
where:
a = 5x + 1
b = x + 3
c = 6x - 8
Applying the identity condition a + b + c = 0:
(5x + 1) + (x + 3) + (6x - 8) = 0
12x - 4 = 0
12x = 4
x =$$\frac{4}{12} = \frac{1}{3}$$​​
Final Answer
So the correct answer is (c)

​Given

​$$\frac{x}{x^2-4x+1}=\frac{1}{9} \\\ \\$$​​
Formula Used
a³ + b³ = (a + b)³ - 3ab(a + b)
Solution
​$$\frac{x}{x^2-4x+1}=\frac{1}{9} \\\ \\9x=x^2-4x+1 \\\ \\x^2-13x+1=0 \\\ \\\frac{x^2-13x+1}{x}=0 \\\ \\x-13+\frac{1}{x}=0 \\\ \\x+\frac{1}{x}=13 \\\ \\x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=13^3-3(13) \\\ \\=2197-39=2158$$​​
Final Answer
So the correct answer is (d)

Given
Expression: $$\frac{x + 3}{x^2 - 2x} × \frac{2x - 1}{x^2 + 2x + 4} × \frac{x^4 - 8x}{2x^2 + 5x - 3}$$​​
Formula Used
Algebraic identity: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$​​
Solution
Denominator 1:  $$x^2 - 2x = x(x - 2)$$​​
Numerator 3:  $$x^4 - 8x = x(x^3 - 8) = x(x^3 - 2^3) = x(x - 2)(x^2 + 2x + 4)$$​​
Denominator 3: $$2x^2 + 5x - 3= 2x^2 + 6x - x - 3= 2x(x + 3) - 1(x + 3)= (2x - 1)(x + 3)$$​​
Substitute all these factorized forms back into the expression:
​$$= \frac{x + 3}{x(x - 2)} × \frac{2x - 1}{x^2 + 2x + 4} × \frac{x(x - 2)(x^2 + 2x + 4)}{(2x - 1)(x + 3)}$$​​
Now, cancel out the common terms from numerator and denominator:
(x + 3) cancels with (x + 3)
(2x - 1) cancels with (2x - 1)
x(x - 2) cancels with x(x - 2)
$$(x^2 + 2x + 4)$$​ cancels with $$(x^2 + 2x + 4)$$​​
Everything cancels out leaving exactly 1.
Final Answer
So the correct answer is (d)

Given
Equation: $$\frac{4(x^2+1)-8x}{3x} = 6,$$​where x > 0
Formula Used
​$$(a + b)^2 = a^2 + b^2 + 2ab$$​​
Solution
Multiply both sides by 3x: 
​$$\frac{4(x^2+1)-8x}{3x} = 6 \\\ \\4x^2 + 4 - 8x = 18x \\4x^2 - 26x + 4 = 0\\4x - 26 + \frac{4}{x} = 0\\\ \\4(x + \frac{1}{x}) = 26\\\ \\x + \frac{1}{x} = \frac{26}{4} = \frac{13}{2}\\\ \\$$​​
We need to find the value of $$\sqrt x + \frac{1}{\sqrt x}$$​​​
​$$(\sqrt x + \frac{1}{\sqrt x} )^2 = x + \frac{1}{x} + 2 \\\ \\(\sqrt x + \frac{1}{\sqrt x} )^2 = \frac{13}{2} + 2 = \frac{17}{2}$$​​
​$$\sqrt x + \frac{1}{\sqrt x} = \sqrt {\frac{17}{2}}$$​​

Final Answer
So the correct answer is (d)

Given:
$$x + \frac{1}{17x} = 3$$​ 
Formula Used:
​$$(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$$​​
Solution:
We need to find the value of $$17x^{3} + \frac{1}{289x^{3}}$$​ 

Cubing the given equation $$x + \frac{1}{17x} = 3$$ on both sides:

​$$(x + \frac{1}{17x})^{3} = 3^{3} \\ \ \\x^{3} + \frac{1}{(17x)^{3}} + 3 × x × \frac{1}{17x} × (x + \frac{1}{17x}) = 27 \\ \ \\x^{3} + \frac{1}{4913x^{3}} + \frac{3}{17} × 3 = 27 \\ \ \\x^{3} + \frac{1}{4913x^{3}} = 27 - \frac{9}{17}$$​​
Now, multiply the entire equation by 17:

​$$17 × (x^{3} + \frac{1}{4913x^{3}}) = 17 × (27 - \frac{9}{17})$$​​

​$$17x^{3} + \frac{1}{289x^{3}} = 459 - 9 = 450$$​​
Final Answer
So the correct answer is (a)

Given
Equations: x + y - 2z = 10; 3x + y - z = 12; x + 2y - z = 5
Solution
Let's verify option B where x = 3, y = -1, z = -4.
Substitute in the first equation: 3 + (-1) - 2(-4) = 3 - 1 + 8 = 10. (True)
Substitute in the second equation: 3(3) + (-1) - (-4) = 9 - 1 + 4 = 12. (True)
Substitute in the third equation: 3 + 2(-1) - (-4) = 3 - 2 + 4 = 5. (True)
All conditions are perfectly satisfied.
Final Answer
So the correct answer is (b)

Given
Polynomial: P(x) = $$x^2 - 4x + 7$$​​
Divisor: (x - 2)
Formula Used
Remainder Theorem: If a polynomial P(x) is divided by (x - a), the remainder is P(a).
Solution
Find the value of a from the divisor x - 2 = 0 $$\Rightarrow x = 2.$$​​
Substitute x = 2 into the polynomial P(x).
P(2) = $$(2)^2 - 4(2) + 7$$​​
P(2) = 4 - 8 + 7
P(2) = 3
Final Answer
So the correct answer is (a)

Given: ​

(5a + 4b + 9c$$)^2$$ 

Formula Used: ​

​$$(a+b+c)^2 = a^2 +b^2 +c^2+2ab +2bc+2ac$$ 

Solution: 

From the formula:

(5a + 4b + 9c$$)^2$$ 

​$$=(5a)^2+(4b)^2+(9c)^2 +2(5a)(4b)+2(4b)(9c)+2(9c)(5a) \\ \ \\ = 25a^2+16b^2+81c^2+40ab+72bc+90ac$$​​

Given
Equation 1: 2x + 3y - 5z = 4
Equation 2: x + 3y - 2z = 12
Equation 3: 2x + y - 3z = 5
Solution
Subtract Equation 3 from Equation 1 to eliminate x:
(2x + 3y - 5z) - (2x + y - 3z) = 4 - 5
2y - 2z = -1
y - z = -0.5
z = y + 0.5 ... (Equation 4)
Multiply Equation 2 by 2:
2x + 6y - 4z = 24 ... (Equation 5)
Subtract Equation 5 from Equation 1 to eliminate x:
(2x + 3y - 5z) - (2x + 6y - 4z) = 4 - 24
-3y - z = -20
3y + z = 20
Substitute Equation 4 (z = y + 0.5) into this result:
3y + (y + 0.5) = 20
4y + 0.5 = 20
4y = 19.5
y = $$\frac{19.5}{4} = \frac{39}{8}$$​​
Now find z using Equation 4:
z  $$= \frac{39}{8} + \frac{1}{2} = \frac{39}{8} + \frac{4}{8} = \frac{43}{8}$$​​
Substitute y and z into Equation 2 to find x:
 $$x + 3(\frac{39}{8}) - 2(\frac{43}{8}) = 12 \\\ \\x + \frac{117}{8} - \frac{86}{8} = 12\\\ \\x + \frac{31}{8} = \frac{96}{8}\\\ \\x = \frac{96 - 31}{8} = \frac{65}{8} \\\ \\$$​​
Final Answer
So the correct answer is (b)

Given
First root = $$\frac{1}{\sqrt{2}}$$​​
Second root = $$\frac{1}{\sqrt{2}}$$​​
Formula Used
Quadratic equation: $$x^2$$​ - (Sum of roots)x + (Product of roots) = 0
Solution
Sum = $$\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$​​

Product = $$\frac{1}{\sqrt{2}} × \frac{1}{\sqrt{2}} = \frac{1}{2}$$​​
Substitute the sum and product into the quadratic equation formula:
​$$x^2 - (\sqrt{2})x + \frac{1}{2} = 0 \\2x^2 - 2\sqrt{2}x + 1 = 0$$​​
Final Answer
So the correct answer is (a)

Given
Quadratic equation: 15x^2 - x - 2 = 0
Formula Used
Splitting the middle term method for  $$ax^2 + bx + c = 0$$​​
Solution
We need two numbers that multiply to (15 × -2 = -30) and add to -1.
The numbers are -6 and 5.
Rewrite the middle term:
​$$15x^2 - 6x + 5x - 2 = 0\\3x(5x - 2) + 1(5x - 2) = 0\\(3x + 1)(5x - 2) = 0\\3x + 1 = 0 => x = \frac{-1}{3} \\\ \\5x - 2 = 0 => x = \frac{2}{5}\\\ \\$$​​
Final Answer
So the correct answer is (c)

Given
Equation: $$ax^3 - 24x^2 + 188x - 480 = 0$$​​
Roots are three consecutive even natural numbers.
Formula Used
Sum of roots = $$\frac{-b}{a}$$​​
Product of roots = $$\frac{-d}{a}$$​​
Solution
Let the three consecutive even natural numbers be k-2, k, and k+2.
Sum of roots = (k-2) + k + (k+2) = 3k
From the equation, Sum of roots = $$\frac{24}{a}$$​​
Therefore, 3k =  $$\frac{24}{a} => ak = 8$$​​
Since the roots are positive even natural numbers and a is a positive integer (from options), k must be an even integer. 
If a = 1, then k = 8. The roots are 6, 8, and 10.
Let us verify the product of roots: 6 × 8 × 10 = 480.
From the equation, Product of roots = $$\frac{480}{a} = \frac{480}{1} = 480.$$​​
The condition is perfectly satisfied for a = 1.
Final Answer
So the correct answer is (b)

Given:
Roots of equation $$x^{2} - 7x + c = 0$$​ are a and b.
Roots of equation $$x^{2} - dx + 216 = 0$$ are $$a^{2}b$$​ and $$b^{2}a.$$​​
Formula Used:
For a quadratic equation $$x^{2} - Sx + P = 0:$$​​
Sum of roots = α + β
Product of roots = αβ
Solution:
From the first equation $$x^{2} - 7x + c = 0:$$​​
Sum of roots a + b = 7
Product of roots ab = c
From the second equation $$x^{2} - dx + 216 = 0:$$​​
Sum of roots $$a^{2}b + b^{2}a = d$$​​
Factoring ab from the sum of roots:
ab(a + b) = d
Substituting ab = c and a + b = 7:
c × 7 = d
d = 7c
Now, using the product of roots from the second equation:
​$$(a^{2}b) × (b^{2}a) = 216$$​​
​$$a^{3}b^{3} = 216$$​​
​$$(ab)^{3} = 216$$​​
​$$c^{3} = 216$$​​
​$$c = \sqrt[3]{216} = 6$$​​
Substitute c = 6 into d = 7c:
d = 7 × 6 = 42.
Therefore, the value of d is 42.
Final Answer
So the correct answer is (b)

Given
Equation 1: $$x^2 - kx - 49$$​= 0
Equation 2: $$x^2 - 5kx + 11$$​= 0
Condition: k > 0
Formula Used
Substitute the common root $$\alpha$$​ into both equations and subtract to eliminate $$\alpha^2.$$​​
Solution
Let $$\alpha$$​ be the common root.
​$$\alpha^2 - k\alpha - 49 = 0\\\alpha^2 - 5k\alpha + 11 = 0$$​​
Subtracting the second equation from the first:
​$$4k\alpha - 60 = 0\\4k\alpha = 60$$​​
​$$k\alpha = 15 \Rightarrow \alpha = \frac{15}{k}$$​​
Substitute \alpha back into the first equation:
​$$(\frac{15}{k})^2 - k(\frac{15}{k}) - 49 = 0 \\\frac{225}{k^2} - 15 - 49 = 0\\\frac{225}{k^2} - 64 = 0\\\frac{225}{k^2} = 64\\k^2 = \frac{225}{64}\\$$​​
Since k > 0, taking the positive square root:
​$$k = \frac{15}{8}$$​​
Final Answer
So the correct answer is (d)