Algebra MCQ Quiz - Objective Question with Answer for Algebra - Download Free PDF
Last updated on Jul 17, 2025
Latest Algebra MCQ Objective Questions
Algebra Question 1:
Simplify the following.
\(\frac{0.01 \times 0.01 \times 0.01 +0.003 \times 0.003 \times 0.003}{0.05 \times 0.05 - 0.015 \times 0.05+0.015 \times 0.015}\)
Answer (Detailed Solution Below)
Algebra Question 1 Detailed Solution
Given:
\(\frac{0.01 × 0.01 × 0.01 +0.003 × 0.003 × 0.003}{0.05 × 0.05 - 0.015 × 0.05+0.015 × 0.015}\)
Formula used:
a3 + b3 = (a + b)(a2 - ab + b2)
Calculation:
\(\frac{0.01 × 0.01 × 0.01 +0.003 × 0.003 × 0.003}{0.05 × 0.05 - 0.015 × 0.05+0.015 × 0.015}\)
⇒ (0.013 + 0.0033)/(5 × 0.01 × 5 × 0.01 - 5 × 0.01 × 5 × 0.003 + 5 × 0.003 × 5 × 0.003)
⇒ (0.013 + 0.0033)/(25 × 0.01 × 0.01 - 25 × 0.01 × 0.003 + 25 × 0.003 × 0.003)
⇒ (0.013 + 0.0033)/25(0.012 - 0.01 × 0.003 + 0.0032)
⇒ (0.01 + 0.003)/25
⇒ 13/25 × 10- 3
∴ The value is 13/25 × 10-3
Algebra Question 2:
If \( \left({x^2} + {{ 1} \over x^2}\right) = 6\), and 0 < x < 1, what is the value of \( {x^4} - {{ 1} \over x^4}\)?
Answer (Detailed Solution Below)
Algebra Question 2 Detailed Solution
Concept used:
If \( \left({x^2} + {{ 1} \over x^2}\right) = a\), then \( \left({x^2} - {{ 1} \over x^2}\right)\) = - √(a2 - 4) (if 0 < x < 1)
Calculation:
\( \left({x^2} + {{ 1} \over x^2}\right) = 6\) , then \( \left({x^2} - {{ 1} \over x^2}\right)\) = - √(62 - 4) = - √32
Now, \( {x^4} - {{ 1} \over x^4}\)
⇒ \( {(x^2)^2} - ({{ 1} \over x^2})^2\)
⇒ \( \left({x^2} + {{ 1} \over x^2}\right)\)\( \left({x^2} - {{ 1} \over x^2}\right)\)
⇒ 6 × (-√32)
⇒ - 24√2
∴ The correct answer is - 24√2
Mistake Points Please note that
if 0 < x < 1
then
1/x > 1
so, 1/x4 > 1
and 0 < x4 < 1
so,
x4 - 1/x4 < 0
so the answer will be negative.
Algebra Question 3:
The square of the difference between two given natural numbers is 324, while the product of these two given numbers is 144. Find the positive difference between the squares of these two given numbers.
Answer (Detailed Solution Below)
Algebra Question 3 Detailed Solution
Given:
The square of the difference between two given natural numbers is 324, while the product of these two given numbers is 144.
Calculation:
Let the numbers are x and y
(x - y)2 = 324
So, x - y = 18, xy = 144
(x + y)2 = (18)2 + 4× 144
⇒ 900
⇒ x + y = 30
Then, x is (30 + 18) / 2 = 24 and y = 6
So , x2 - y2 = 242 - 62
⇒ 576 - 36 = 540
∴ The correct option is 2
Algebra Question 4:
What is the value of a3 + b3 + c3 if (a + b + c) = 0?
Answer (Detailed Solution Below)
Algebra Question 4 Detailed Solution
Given Data:
a + b + c = 0
Formula Used:
a3+ b3 + c3 – 3abc = (a+b+c)( a2+ b2+c2 −ab−bc–ca)
Calculation:
According to the formula,
a3+ b3 + c3 – 3abc = (a + b + c)( a2 + b2 + c2 − ab − bc – ca)
But (a+b+c) = 0
⇒ a3+ b3 + c3 – 3abc = 0
⇒ a3+ b3 + c3 = 3abc
∴ a3 + b3 + c3 is 3abc.
Algebra Question 5:
If 2x + 3y = 16 and xy = 9, then find 8x3 + 27y3
Answer (Detailed Solution Below)
Algebra Question 5 Detailed Solution
Given:-
2x + 3y = 16 and xy = 9
Concept used:-
(a + b)3 = a3 + b3 + 3ab(a + b)
Calculation:-
a3 + b3 = (a + b)3 - 3ab(a + b)
8x3 + 27y3
⇒ (2x)3 + (3y)3
⇒ (2x + 3y)3 - 3 × 2x × 3y (2x + 3y)
⇒ (2x + 3y)3 - 18xy (2x + 3y)
⇒ 163 - 18 × 9 × 16
⇒ 4096 - 2592
⇒ 1504
∴ The required answer is 1504.
Top Algebra MCQ Objective Questions
If \(\rm x-\frac{1}{x}=-6\), what will be the value of \(\rm x^5-\frac{1}{x^5}\)?
Answer (Detailed Solution Below)
Algebra Question 6 Detailed Solution
Download Solution PDFGiven:
x - (1/x) = (- 6)
Formula used:
If x - (1/x) = P, then
x + (1/x) = √(P2 + 4)
If x + (1/x) = P, then
x3 + (1/x3) = (P3 - 3P)
x5 - (1/x5) = {x3 + (1/x3)} × {x2 - 1/x2} + {x - (1/x)}
Calculation:
x - (1/x) = (- 6)
x + (1/x) = √{(- 6)2 + 4} = √40 = 2√10
So, x2 - 1/x2 = (x + 1/x) (x - 1/x) = 2√10 × (-6) = -12√10
and x3 + (1/x3) = (√40)3 - 3√40
⇒ 40√40 - 3√40 = 37 × 2√10 = 74√10
Now,
x5 - (1/x5) = {x3 + (1/x3)} × {x2 - 1/x2} + {x - (1/x)}
⇒ {74√10 × (-12√10)} + (- 6)
⇒ - 74 × 12 × (√10 × √10) - 6
⇒ (- 8880) - 6 = - 8886
∴ The correct answer is - 8886.
If \(a + \frac{1}{a} = 7\), then \(a^5 + \frac{1}{a^5} \)is equal to:
Answer (Detailed Solution Below)
Algebra Question 7 Detailed Solution
Download Solution PDFGiven:
\(a + \frac{1}{a} = 7\)
Formula used:
(a + 1/a) = P ; then
(a2 + 1/a2) = P2 - 2
(a3 + 1/a3) = P3 - 3P
\(a^5 + \frac{1}{a^5} \) = (a2 + 1/a2) × (a3 + 1/a3) - (a + 1/a)
Calculation:
a + (1/a) = 7
⇒ (a2 + 1/a2) = (7)2 - 2 = 49 - 2 = 47
⇒ (a3 + 1/a3) = (7)3 - (3 × 7) = 343 - 21 = 322
a5 + (1/a5) = (a2 + 1/a2) × (a3 + 1/a3) - (a + 1/a)
⇒ 47 × 322 - 7
⇒ 15134 - 7 = 15127
∴ The correct answer is 15127.
If (a + b + c) = 19 and (a2 + b2 + c2) = 155, find the value of (a - b)2 + (b - c)2 + (c - a)2.
Answer (Detailed Solution Below)
Algebra Question 8 Detailed Solution
Download Solution PDFGiven:
(a + b + c) = 19
(a2 + b2 + c2) = 155
Formula used:
a2 + b2 + c2 - (ab + bc + ca) = (1/2) × [(a - b)2 + (b - c)2 + (c - a)2]
Calculation:
a + b + c = 19
Squaring both sides
⇒ (a + b + c)2 = (19)2
⇒ a2 + b2 + c2 + 2 × (ab + bc + ca) = 361
⇒ 155 + 2 × (ab + bc + ca) = 361
⇒ 2 × (ab + bc + ca) = (361 - 155)
⇒ (ab + bc + ca) = 206/2 = 103
Now,
a2 + b2 + c2 - (ab + bc + ca) = (1/2) × [(a - b)2 + (b - c)2 + (c - a)2]
⇒ 2 × (155 - 103) = (a - b)2 + (b - c)2 + (c - a)2
⇒ (a - b)2 + (b - c)2 + (c - a)2 = 104
∴ The correct answer is 104.
If \((x^2+\frac{1}{x^2})=7\), and 0 < x < 1, find the value of \(x^2-\frac{1}{x^2} \).
Answer (Detailed Solution Below)
Algebra Question 9 Detailed Solution
Download Solution PDFGiven:
x2 + (1/x2) = 7
Formula used:
x2 + (1/x2) = P
then x + (1/x) = √(P + 2)
and x - (1/x) = √(P - 2)
⇒ x2 - (1/x2) = {x + (1/x)} × {x - (1/x)}
Calculation:
x2 + (1/x2) = 7
⇒ x + (1/x) = √(7 + 2) = √9
⇒ x + (1/x) = 3
⇒ x - (1/x) = -√(7 - 2)
⇒ x - (1/x) = - √5 {0 < x < 1}
x2 - (1/x2) = {x + (1/x)} × {x - (1/x)}
⇒ 3 × (- √5)
∴ The correct answer is - 3√5.
Mistake Points
Please note that
0 < x < 1
so
1/x > 1
so
x + 1/x > 1
and
x - 1/x < 0 (because 0 < x < 1 and 1/x > 1 so x - 1/x < 0)
so
(x - 1/x)(x + 1/x) < 0
If \(7 b-\frac{1}{4 b}=7\), then what is the value of \(16 b^2+\frac{1}{49 b^2}\) ?
Answer (Detailed Solution Below)
Algebra Question 10 Detailed Solution
Download Solution PDFFormula used
(a - b)2 = a2 + b2 - 2ab
Calculation
Multiplying the expression by 4/7.
⇒ 4/7 × (7b - 1/4b) = 7 × 4/7
⇒ 4b - 1/7b = 4
Squaring both sides:
⇒ (4b - 1/7b)2 = 42
⇒ \(16 b^2+\frac{1}{49 b^2}\)- 2 × 4 × 1/7 = 16
⇒ \(16 b^2+\frac{1}{49 b^2}\) = 16 + 8/7
⇒ \(16 b^2+\frac{1}{49 b^2}\) = 120/7
The value is 120/7.
If \((x - \frac{1}{x})\)= √6, and x > 1, what is the value of \((x^8 - \frac{1}{x^8})\)?
Answer (Detailed Solution Below)
Algebra Question 11 Detailed Solution
Download Solution PDFGiven:
x - (1/x) = √6
Formula used:
x8 - (1/x8) = {x4 + (1/x4)} × {x2 + (1/x2)} × {x + (1/x)} × {x - (1/x)}
If x - (1/x) = a, then x + (1/x) = √(a2 + 4)
Calculation:
x - (1/x) = √6
x2 + (1/x2) = (√6)2 + 2 = 8
Square both sides:
x4 + (1/x4) = (8)2 - 2 = 62
Using, If x - (1/x) = a, then x + (1/x) = √{(√a)2 + 4} .
x + (1/x) = √{(√6)2 + 4} = √10
Substituting the values, we get
x8 - (1/x8) = {x4 + (1/x4)} × {x2 + (1/x2)} × {x + (1/x)} × {x - (1/x)}
⇒ 62 × 8 × √10 × √6 = 496 × 2 × √15 = 992√15
∴ The correct answer is 992√15.
Shortcut Trick
If a - 1/a = m,
• Then a2 +1/a2 = m2 + 2
• Then a4 + 1/a4 = m4 + 4m2 + 2
if a2 +1/a2 = m , then a +1/a = √(m+2)
⇒ x - 1/x =√6
⇒ x2+1/x2 = (√6)2 +2 = 8
⇒ x +1/x = √10
⇒ x4+ 1/x4 = (√6)4+4(√6)2 +2 = 36+24+2 = 62
⇒ x8 - (1/x8) = {x4 + (1/x4)} × {x2 + (1/x2)} × {x + (1/x)} × {x - (1/x)}
⇒ 62 × 8 × √10 × √6 = 496 × 2 × √15
⇒ 992√15
If (a + b + c) = 12, and (a2 + b2 + c2) = 50, find the value of (a3 + b3 + c3 - 3abc)
Answer (Detailed Solution Below)
Algebra Question 12 Detailed Solution
Download Solution PDFGiven :
(a + b + c) = 12, (a2 + b2 + c2) = 50
Formula Used :
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc +ac)
(a3 + b3 + c3 - 3abc) = (a2 + b2 + c2 - ab - bc - ca)(a + b + c)
Calculation :
⇒ 144 = 50 + 2(ab + bc +ac)
⇒ (ab + bc +ac) = 94/2 = 47
Now,
⇒ (a3 + b3 + c3 - 3abc)
⇒ (a2 + b2 + c2 - ab - bc - ca)(a + b + c) = (50 - 47)(12)
⇒ 3 × 12 = 36
∴ The correct answer is 36.
If x2 - 1/x2 = 4 \(\sqrt2\), what is the value of x4 - 1/x4?
Answer (Detailed Solution Below)
Algebra Question 13 Detailed Solution
Download Solution PDFx2 -1/x2 = 4√2
Formula used:-
(A + B)2 = A2 + B2 + 2AB
(A2 - B2) = (A+ B) (A - B)
Calculation:-
Square both side
⇒ (x2 -1/x2)2 = (4√2 )2
⇒ x4 + 1/x4 - 2 = 32
⇒ x4 + 1/x4 = 34
Add 2 on both sides
⇒ x4 + 1/x4 + 2 = 34 +2
⇒ (x2 + 1/x2)2 = 62
⇒ (x2 + 1/x2) = 6 ....(1)
According to the question,
⇒ x4 - 1/x4 = (x2 + 1/x2) (x2 -1/x2)
⇒ (4√2) × 6 = 24√ 2
∴ The required answer is 24√ 2.
A and B have some toffees. If A gives one toffee to B, then they have equal number of toffees. If B gives one toffee to A, then the toffees with A are double with B. The total number of toffees with A and B are __________.
Answer (Detailed Solution Below)
Algebra Question 14 Detailed Solution
Download Solution PDFCalculation
let the number of toffee with A be x and with B be y.
If A gives one toffee to B, then:
⇒ x - 1 = y + 1
⇒ x = y + 2 .........(1)
Now when B gives one toffee to A, then the toffees with A are double with B:
⇒ x + 1 = 2 (y - 1) ......(2)
Putting the value of eq.(1) in eq. (2).
⇒ y + 3 = 2y - 2
⇒ y = 5
If y = 5 then x = 7.
⇒ x + y = 12
The total number of toffees with A and B are 12.
The square of the sum of two given natural numbers is 784, while the product of the two given numbers is 192. Find the positive difference between the squares of these two given numbers.
Answer (Detailed Solution Below)
Algebra Question 15 Detailed Solution
Download Solution PDFLet the numbers are X and Y
Given:
(X + Y)2 = 784 and XY = 192
Calculation:
(X + Y)2 = 784 ⇒ (X + Y) = 28
⇒ X2 + Y2 + 2XY = 784
⇒ X2 + Y2 + 2 × 192 = 784
⇒ X2 + Y2 = 400
So,
⇒ X2 + Y2 - 2XY = 400 - 2 × 192
⇒ X2 + Y2 - 2XY = 16
⇒ (X - Y)2 = 16
⇒ X - Y = 4
Now,
X2 - Y2 = (X + Y)(X - Y)
⇒ 28 × 4 = 112
∴ The correct option is 4
Alternate MethodAccording to the question,
(x + y)2 = 784
⇒ x + y = √784 = 28
xy = 192
x - y = √{(x + y)2 - 4xy}
⇒ √{282 - 4 × 192} = √16 = 4
Now, x2 - y2 = (x + y)(x - y) = 28 × 4 = 112