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Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

Algebra MCQ Questions and answers with easy and logical explanations. Arithmetic Ability provides you all type of quantitative and competitive aptitude MCQ questions on Algebra with easy and logical explanations. Algebra MCQ is important for exams like Banking exams, IBPS, SCC, CAT, XAT, MAT, etc. Page-1 section-2

Each section contains 100 MCQs on Algebra .
1
If $${64^{x + 1}} = \frac{{64}}{{{4^x}}}{\text{,}}$$   then the value of x is?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {64^{x + 1}} = \frac{{64}}{{{4^x}}} \cr & \Rightarrow {\left( {{4^3}} \right)^{x + 1}} - \frac{{{4^3}}}{{{4^x}}} \cr & \Rightarrow {4^{3x + 3}} = {4^{3 - x}} \cr & \Rightarrow 3x + 3 = 3 - x \cr & \Rightarrow 4x = 0 \cr & \Rightarrow x = 0 \cr} $$
2
If ax2 + bx + c = a(x - p)2, then the relation among a, b, c would be?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & a{x^2} + bx + c = a{\left( {x - p} \right)^2} \cr & \Rightarrow a{x^2} + bx + c = a{\left( {{x^2} + {p^2} - 2px} \right)^2} \cr & \Rightarrow a{x^2} + bx + c = a{x^2} + a{p^2} - 2apx \cr & {\text{Comparing confficients of }}{x^2}{\text{and }}x \cr & \Rightarrow b = - 2ap \cr & \Rightarrow p = - \frac{b}{{2a}}\,.......(1) \cr & and{\text{ }}c = a{p^2} \cr & \Rightarrow c = a \times \frac{{{b^2}}}{{4{a^2}}}\left[ {{\text{From (i)}}} \right] \cr & \Rightarrow 4ac = {b^2} \cr} $$
3
If a2 + b2 + c2 + 3 = 2(a + b + c) then the value of (a + b + c) is?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {a^2} + {b^2} + {c^2} + 3 = 2\left( {a + b + c} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 3 = 2a + 2b + 2c \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} - 2b + 1 + {c^2} - 2c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {c - 1} \right)^2} = 0 \cr & a = 1 \cr & b = 1 \cr & c = 1 \cr & \therefore \left( {a + b + c} \right) \cr & = 1 + 1 + 1 \cr & = 3 \cr} $$
4
If $$x - \frac{1}{x} = 5{\text{,}}$$   then $${x^2}{\text{ + }}\frac{1}{{{x^2}}}$$   is?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & x - \frac{1}{x} = 5 \cr & \left[ {{\text{Squaring both sides}}} \right] \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} - 2 = 25 \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = 27 \cr} $$
5
If $$n = 7 + 4\sqrt 3 {\text{,}}$$   then the value of $$\left( {\sqrt n + \frac{1}{{\sqrt n }}} \right)$$   is:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & n = 7 + 4\sqrt 3 \cr & \Rightarrow n = 4 + 3 + 4\sqrt 3 \cr & \Rightarrow n = {\left( 2 \right)^2} + {\left( {\sqrt 3 } \right)^2} + 2 \times 2 \times \sqrt 3 \cr & \Rightarrow n = {\left( {2 + \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt n = 2 + \sqrt 3 \cr & \Rightarrow \frac{1}{{\sqrt n }} = 2 - \sqrt 3 \cr & \therefore \sqrt n + \frac{1}{{\sqrt n }} \cr & = 2 + \sqrt 3 + 2 - \sqrt 3 \cr & = 4 \cr} $$
6
If $$x = \sqrt 3 + \sqrt 2 {\text{,}}$$    then the value of $$\left( {x + \frac{1}{x}} \right)\,{\text{is?}}$$
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{ }}x = \sqrt 3 + \sqrt 2 \cr & \frac{1}{x} = \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \frac{1}{x} = \sqrt 3 - \sqrt 2 \cr & \therefore x + \frac{1}{x} \cr & = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 \cr & = 2\sqrt 3 \cr} $$
7
If p + q = 10 and pq = 5, then the numerical value of $$\frac{p}{q}{\text{ + }}\frac{q}{p}$$   will be?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & p + q = 10\,.........{\text{(i)}} \cr & pq = 5 \cr & {\text{Squaring both sides of equation (i)}} \cr & {\left( {p + q} \right)^2} = {\left( {10} \right)^2} \cr & {p^2} + {q^2} + 2pq = 100 \cr & {p^2} + {q^2} + 2 \times 5 = 100 \cr & {p^2} + {q^2} = 90 \cr & {\text{Now,}} \cr & \therefore \frac{p}{q}{\text{ + }}\frac{q}{p} \cr & = \frac{{{p^2} + {q^2}}}{{pq}} \cr & = \frac{{90}}{5} \cr & = 18 \cr} $$
8
If x = 3 + 2$$\sqrt 2 $$ and xy = 1, then the value of $$\frac{{{x^2} + 3xy + {y^2}}}{{{x^2} - 3xy + {y^2}}}$$   is?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & x = 3 + 2\sqrt 2 {\text{ and }}xy = 1 \cr & {y^2} = \frac{1}{{{x^2}}} \cr & y = \frac{1}{x} = \frac{1}{{3 + 2\sqrt 2 }} = 3 - 2\sqrt 2 \cr & \therefore x + \frac{1}{x} = 3 + 2\sqrt 2 + 3 - 2\sqrt 2 = 6 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} = 36 - 2 = 34 \cr & \frac{{{x^2} + 3xy + {y^2}}}{{{x^2} - 3xy + {y^2}}} \cr & = \frac{{{x^2} + \frac{1}{{{x^2}}} + 3}}{{{x^2} + \frac{1}{{{x^2}}} - 3}} \cr & = \frac{{34 + 3}}{{34 - 3}} \cr & = \frac{{37}}{{31}} \cr} $$
9
If x - y = 2, xy = 24, then the value of (x2 + y2) is?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Given,}} \cr & x - y = 2{\text{ and }}xy = 24 \cr & {\text{By squaring}} \cr & \Rightarrow {x^2} + {y^2} - 2xy = 4 \cr & \Rightarrow {x^2} + {y^2} - 2 \times 24 = 4 \cr & \Rightarrow {x^2} + {y^2} = 4 + 48 \cr & \Rightarrow {x^2} + {y^2} = 52 \cr} $$
10
If the expression $$\frac{{{x^2}}}{{{y^2}}} + tx + \frac{{{y^2}}}{4}$$   is a perfect square, then the value of t is?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\frac{{{x^2}}}{{{y^2}}} + tx + \frac{{{y^2}}}{4}\left( {{\text{ Given}}} \right)$$
To make it a perfect square it should be in the form
$$\eqalign{ & {{\text{A}}^2} \pm 2{\text{AB}} + {{\text{B}}^2} = {\left( {{\text{A}} \pm {\text{B}}} \right)^2} \cr & = {\left( {\frac{x}{y}} \right)^2} \pm tx + {\left( {\frac{y}{2}} \right)^2} \cr & = {{\text{A}}^2} \pm 2{\text{AB}} + {{\text{B}}^2} \cr & {\text{A}} = \frac{x}{y}{\text{, B}} = \frac{y}{2}\,\,\& \,\,{\text{2AB}} = tx \cr & {\text{So, }}tx = \pm 2 \times \frac{x}{y} \times \frac{y}{2} \cr & \Rightarrow tx = \pm x \cr & \Rightarrow t = \pm 1 \cr} $$