Practice MCQ Questions and Answer on Algebra

441.

If x is real, $$x+\frac{1}{x}\ne 0$$   and $$x^3\text{ + }\frac{1}{x^3}=0\text{,}$$   then the value of $${\left(x+\frac{1}{x}\right)}^4\text{is?}$$

• (A) 4
• (B) 9
• (C) 16
• (D) 25

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 $$\eqalign{ & {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 0{\text{ }} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = 3\left( {x + \frac{1}{x}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr & \,\,\,\,\left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {\left[ {{{\left( {x + \frac{1}{x}} \right)}^2}} \right]^2} = {\left( 3 \right)^2} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$

442.

If $$a^3+\frac{1}{a^3}=2\text{,}$$   then the value of $$\frac{a^2+1}{a}$$  is (a positive number) ?

• (A) 1
• (B) 2
• (C) 3
• (D) 4

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 $$\eqalign{ & {\text{But }}a = 1 \cr & {a^3} + \frac{1}{{{a^3}}} = 2 \cr & \Rightarrow {1^3} + \frac{1}{{{1^3}}} = 2 \cr & \Rightarrow 2 = 2{\text{ }}\left( {{\text{Satisfy}}} \right) \cr & {\text{So, }}\frac{{{a^2} + 1}}{a} \cr & = \frac{{{1^2} + 1}}{1} \cr & = 2 \cr} $$

443.

If $$\frac{3a+5b}{3a-5b}=5,$$   then a : b is equal to?

• (A) 2 : 1
• (B) 2 : 3
• (C) 1 : 3
• (D) 5 : 2

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 $$\eqalign{ & \frac{{3a + 5b}}{{3a - 5b}} = 5 \cr & \Rightarrow 3a + 5b = 15a - 25b \cr & \Rightarrow 12a = 30b \cr & \Rightarrow 2a = 5b \cr & a:b \cr & 5:2 \cr} $$

444.

If x² - 16x + 59 = 0, then what is the value of $${\left(x-6\right)}^2+\frac{1}{{\left(x-6\right)}^2}?$$

• (A) 14
• (B) 18
• (C) 16
• (D) 20

Show Answer

 $$\eqalign{ & {x^2} - 16x + 59 = 0 \cr & {\left( {x - 6} \right)^2} + \frac{1}{{{{\left( {x - 6} \right)}^2}}} = ? \cr & {\text{Let }}x - 6 = t \cr & x = t + 6 \cr & {t^2} + \frac{1}{{{t^2}}} = ? \cr & {\left( {t + 6} \right)^2} - 16\left( {t + 6} \right) + 59 = 0 \cr & {t^2} + 12t - 36 - 16t - 96 + 59 = 0 \cr & {t^2} - 4t - 1 = 0 \cr & {\text{Dividing by }}t,{\text{ we get}} \cr & t - 4 - \frac{1}{t} = 0 \cr & t - \frac{1}{t} = 4 \cr & {\text{By squaring, we get}} \cr & {t^2} + \frac{1}{{{t^2}}} - 2 = 16 \cr & {t^2} + \frac{1}{{{t^2}}} = 18 \cr} $$

445.

If $${\text{7}}^x\text{ = }\frac{1}{343}\text{,}$$   then the value of x is?

• (A) 3
• (B) -3
• (C) $$\frac{1}{3}$$
• (D) $$\frac{1}{7}$$

Show Answer

 $$\eqalign{ & {{\text{7}}^x}{\text{ = }}\frac{1}{{343}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}\frac{1}{{{7^3}}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}{{\text{7}}^{ - 3}} \cr & \Rightarrow x = - 3 \cr} $$ (If bases are equal then their power are also equal)

446.

When a number x is divided by a divisor it is seen that the divisor = 4 times the quotient = double of remainder. If the remainder is 80, then the value of x is?

• (A) 6480
• (B) 9680
• (C) 8460
• (D) 4680

Show Answer

 According to the question, Divisor = 2 × remainder = 2 × 80 = 160 Again, 4 × quotient = 160 ⇒ Quotient = $$\frac{{160}}{4}$$ = 40 ∴ x = Divisor × Quotient + Remainder x = 160 × 40 + 80 = 6480

447.

If $$\frac{144}{0.144}=\frac{14.4}{x}\text{,}$$    then the value of x is?

• (A) 144
• (B) 14.4
• (C) 1.44
• (D) 0.0144

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 $$\eqalign{ & \frac{{144}}{{0.144}} = \frac{{14.4}}{x} \cr & \Rightarrow \frac{{144 \times 1000}}{{144}} = \frac{{144}}{{x \times 10}} \cr & \Rightarrow 1000 = \frac{{144}}{{10x}} \cr & \Rightarrow x = \frac{{144}}{{1000 \times 10}} \cr & \Rightarrow x = \frac{{144}}{{10000}} \cr & \Rightarrow x = 0.0144 \cr} $$

448.

If $$x=3+2\sqrt{2}\text{,}$$   then the value of $$x^2\text{ + }\frac{1}{x^2}\text{ is?}$$

• (A) 36
• (B) 30
• (C) 32
• (D) 34

Show Answer

 $$\eqalign{ & {\text{ }}x = 3 + 2\sqrt 2 \cr & \Rightarrow {x^2} = {\left( {3 + 2\sqrt 2 } \right)^2} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2} = 9 + 8 + 12\sqrt 2 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr & \Rightarrow \frac{1}{{{x^2}}} = \frac{1}{{17 + 12\sqrt 2 }} \times \frac{{17 - 12\sqrt 2 }}{{17 - 12\sqrt 2 }} \cr & \Rightarrow \frac{1}{{{x^2}}} = 17 - 12\sqrt 2 \cr & \therefore {\text{ }}{x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr & = 17 + 12\sqrt 2 + 17 - 12\sqrt 2 \cr & = 34 \cr} $$

449.

If x + y + z = 22 and xy + yz + zx = 35, then what is the value of (x - y)² + (y - z)² + (z - x)²?

• (A) 793
• (B) 681
• (C) 758
• (D) 715

Show Answer

 x + y + z = 22 xy + yz + zx = 35 (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) (22)2 = x2 + y2 + z2 + 2 × 35 484 - 70 = x2 + y2 + z2 x2 + y2 + z2 = 414 (x - y)2 + (y - z)2 + (z - x)2 = 2(x2 + y2 + z2 - xy - yz - zx) = 2(414 - 35) = 2 × 379 = 758

450.

If $$\sqrt{x}=\sqrt{3}-\sqrt{5}\text{,}$$    then the value of x² - 16x + 6 is?

• (A) 0
• (B) -2
• (C) 2
• (D) 4

Show Answer

 $$\eqalign{ & \sqrt x = \sqrt 3 - \sqrt 5 \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow x = 3 + 5 - 2.\sqrt {3.} \sqrt 5 \cr & \Rightarrow x = 8 - 2\sqrt {15} \cr & \Rightarrow x - 8 = - 2\sqrt {15} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2} + 64 - 16x = 60 \cr & \Rightarrow {x^2} + 4 - 16x = 0 \cr & \Rightarrow {x^2} + 6 - 16x = 2 \cr & \Rightarrow {x^2} - 16x + 6 = 2 \cr} $$