Algebra | Free Online Test and Practice

Algebra | Free Online Test and Practice

“Algebra” is a common chapter in many competitive exams such as SSC, Bank, UPSC  and other exams. Many students practice a lot of Algebra Formula and equations to make his and her calculation strong. Sometimes, A student gets confused at the time of choosing the right answer among the given options.

Basic Algebra formula to improve accuracy. 

  • a2 + b2 = (a + b)2 – 2ab
  • a2 – b2 = (a – b)(a + b)
  • (a+b)2 = a2 + 2ab + b2
  • (a – b)2 = a2 – 2ab + b2
  • (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
  • (a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca
  • a3 – b3 = (a – b)(a2 + ab + b2)
  • a3 + b3 = (a + b)(a2 – ab + b2)
  • (a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)
  • (a – b)3 = a3 – 3a2b + 3ab2 – b3
  • a4 – b4 = (a – b)(a + b)(a2 + b2)
  • (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
  • (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4
  • a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)

Click Here to Download Algebra Formula PDF

Let’s have some warm-up questions before taking the test…

1. If x = 2 then the value of x3 + 27x2 + 243x + 631 is…..?
A. 1213
B. 1233
C. 1336
D. 1356
Ans: B
Solution
Given equation,
f(x) = x3 + 27x2 + 243x + 631
⇒ x(x2 + 27x+ 243) + 631
Now, put the value of x = 2
⇒ 2(22 + 27 × 2 + 243) + 631 ⇒ 2 (4 + 54 + 243) + 631
⇒ 2(301) + 631 = 602 + 631 = 1233.

2. The expression x4 – 2x2 + k will be a perfect square when the value of k is….?
A. 1
B. 0
C. -1
D. 2
Ans: A
Solution
Let’s assume x2 = m, therefore the given eq. will be:
m2 – 2m + k which is a quadratic equation (ax2 + bx + c).
Now we know that a quadratic eqn. is a perfect square if its discriminant (b2 – 4ac) is equal to zero.
In the eq. a = 1, b = – 2, c = k
∴    (– 2)2 – 4 (1).k = 0
–4k = –4
k = 1

3. If x2 + y2 + 1 = 2x, then the value of x3 + y5 is…..?
A. 1
B. -2
C. 2
D. -1
Ans: A
Solution
x2 + y2 + 1 = 2x
⇒ x2 + y2 + 1 – 2x = 0 ⇒ x2 – 2x + 1 + y2 = 0
⇒ (x – 1)2 + y2 = 0
In the above eq. the L.H.S. only becomes zero when the base of terms; (x – 1) and y becomes zero because for any other value the sum of their squares shall always be a positive integer.
Taking (x – 1) = 0
and y = 0
Therefore, x = 1 and y = 0
∴     x3 + y5 = 1 + 0 = 1.

You are doing well, Let’s have a few more question…

4. If x = 2015, y = 2014 and z = 2013, then value of x2 + y2 + z2 – xy – yz – zx is…?
A. 12
B. 9
C. 3
D. 6
Ans: C
Solution
x – y = 2015 – 2014 = 1
y – z = 2014 – 2013 = 1
z – x = 2013 – 2015 = –2
∴ x2 + y2 + z2 – xy – yz – zx
Numerator & denominator multiplied by 2
= 1/2 (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx) ⇒ 1/2 (x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx)
= 1/2 [(x– y)2 + (y – z)2 + (z – x)2] ⇒ 1/2 [1 + 1 + 4] = 1/2 × 6 = 3

5. Consider the following statements
I. x + 3 is the factor of x3 + 2x2 + 3x + 8.
II. x – 2 is the factor of x3 + 2x2 + 3x + 8.
Which of the statements given above is/are correct?

A. Only I
B. Only II
C. Neither I nor II
D. Both I and II
Ans:C
Solution
Put x = –3 in equation x3 + 2x2 + 3x + 8
= (–3)3 + 2(–3)2 + 3(–3) + 8
= –10 ≠ 0
So, (x + 3) is not the factor of x3 + 2x2 + 3x + 8
Similarly, put x = 2 in above equation
= (2)3 + 2(2)2 + 3(2) + 8
= 30 ≠ 0
So, (x – 2) is also not the factor of x3 + 2x2 + 3x + 8.

Algebra | Free Online Test Details

Algebra Free Online Test Details

Exam Practice NameAlgebra
Number of Sets2
Number of Questions20 X 2 = 40
Type of QuestionsMULTIPLE CHOICE
LanguagesEnglish / Hindi

Are you ready for this exam practice?

 Free Online Test Set-1

Instructions:

Each question carry 1 mark, no negative marks.
Click the ‘Finish quiz’ button to Submit your answers.
Don’t refresh the page or click back button of browser. your work will be lost.
If you want to give the answer to the question later, click the Review Question button.

 Free Online Test Set-2

Instructions:

Each question carry 1 mark, no negative marks.
Click the ‘Finish quiz’ button to Submit your answers.
Don’t refresh the page or click back button of browser. your work will be lost.
If you want to give the answer to the question later, click the Review Question button.

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