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# Algebra Expressions MCQ Questions

## Algebra Expressions MCQ Questions with details Solution

Q1. If 3x – $\frac{1}{4\mathrm{y}}$ = 6, then the value of 4x – $\frac{1}{3\mathrm{y}}$is

(a) 2

(b) 4

(c) 6

(d) 8

 Answer: (d) 8Solution: 3x – $\frac{1}{4\mathrm{y}}$ = 6⇒ 4x – $\frac{1}{3\mathrm{y}}$ = 8 (multiplying 4/3 both side)

Q2. If   $\mathrm{x}-\frac{1}{\mathrm{x}}$ = 5 then the value of  $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$

(a) 7

(b) 9

(c) 27

(d) 81

 Answer: (c) 27Solution: $\mathrm{x}-\frac{1}{\mathrm{x}}$ = 5⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,-2.\mathrm{x}.\frac{1}{\mathrm{x}}=25$ (squaring both side)⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,27$Using Trick ⇒$\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,5^2+2=27$Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$ = n2 – 2 and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$ = n2 + 2

Q3.  If   $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=7$then the value of $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$

(a) 21

(b) 47

(c) 49

(d) 51

 Answer: (b) 47Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 7⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,+2.\mathrm{x}.\frac{1}{\mathrm{x}}=49$ (squaring both side)⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,47$Using Trick ⇒$\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,7^2-2=47$Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$ = n2 – 2 and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$ = n2 + 2

Q4.  If   $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=2$then the value of $\mathrm{x}^{2020}+\frac{1}{\mathrm{x}^{2020}}$

(a) 0

(b) 2

(c) 4

(d) 2020

 Answer: (b) 2Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=2$⇒ x2 +1 = 2x⇒ x2 – 2x + 1 = 0⇒ (x – 1)2 = 0⇒ x = 1∴ $\mathrm{x}^{2020}+\frac{1}{\mathrm{x}^{2020}}\,\,=\,\,1^2+\frac{1}{1^2}=2$Short Tricks: If   $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=2$ then the value of$\mathrm{x}^{\mathrm{n}}+\frac{1}{\mathrm{x}^{\mathrm{n}}}\,\,=\,\,2$, where n = integer

Q5.  If   $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=5$then the value of $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$

(a) 527

(b) 530

(c) 550

(d) 625

 Answer: (a) 527Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 5⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,+2.\mathrm{x}.\frac{1}{\mathrm{x}}=25$ (squaring both side)⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,23$⇒ $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,+2.\mathrm{x}^2.\frac{1}{\mathrm{x}^2}=529$ (squaring both side)⇒ $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,=\,\,527$Using Trick ⇒$\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,=\,\,\left( 5^2-2 \right) ^2-2=527$Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$ = (n2 – 2)2 – 2 andIf $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$ = (n2 + 2)2 – 2

Q6.  If   $\mathrm{x}-\frac{1}{\mathrm{x}}\,\,=3$ then the value of $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$

(a) 81

(b) 100

(c) 125

(d) 119

 Answer: (d) 119Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 9⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,-2.\mathrm{x}.\frac{1}{\mathrm{x}}=9$ (squaring both side)⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,11$⇒ $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,+2.\mathrm{x}^2.\frac{1}{\mathrm{x}^2}=121$ (squaring both side)⇒ $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,=\,\,119$Using Trick ⇒$\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,=\,\,\left( 3^2+2 \right) ^2-2=119$Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$ = (n2 – 2)2 – 2 andIf $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$ = (n2 + 2)2 – 2

Q7.  If   $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=3$then the value of $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}$

(a) 18

(b) 47

(c) 49

(d) 51

 Answer: (a) 18Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 3⇒ $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,+3.\mathrm{x}.\frac{1}{\mathrm{x}}\left( \mathrm{x}+\frac{1}{\mathrm{x}} \right) =27$ (cubing both side)⇒ $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,=\,\,27 -9=18$Using Trick ⇒$\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,=\,\,3^3-3\times 3=18$Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}$ = n3 – 3.n and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}$ = n3 + 3.n

Q8.  If   $\mathrm{x}-\frac{1}{\mathrm{x}}\,\,=5$then the value of $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}$

(a) 125

(b) 130

(c) 135

(d) 140

 Answer: (d) 140Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 5⇒ $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,-3.\mathrm{x}.\frac{1}{\mathrm{x}}\left( \mathrm{x}-\frac{1}{\mathrm{x}} \right) =125$ (cubing both side)⇒ $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,=\,\,125 +15=140$Using Trick ⇒$\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,=\,\,5^3+3\times 5=140$Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}$ = n3 – 3.n and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}$ = n3 + 3.n

Q9.  If   $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=3$then the value of $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$

(a) 320

(b) 322

(c) 341

(d) 350

 Answer: (b) 322Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 3⇒ $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,+3.\mathrm{x}.\frac{1}{\mathrm{x}}\left( \mathrm{x}+\frac{1}{\mathrm{x}} \right) =27$ (cubing both side)⇒ $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,=\,\,27 -9=18$⇒ $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,+2.\mathrm{x}^3.\frac{1}{\mathrm{x}^3}=\,\,324$(squaring both side)⇒ $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,=\,\,324-2=322$ Using Trick ⇒$\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,=\,\,\left( 3^3-3\times 3 \right) ^2-2=322$Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$ = (n3 – 3.n)2 – 2 andIf $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$ = (n3 + 3.n)2 + 2

Q10.  If   $\mathrm{x}-\frac{1}{\mathrm{x}}\,\,=2$ then the value of $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$

(a) 125

(b) 196

(c) 198

(d) 225

 Answer: (c) 198Solution: $\mathrm{x}-\frac{1}{\mathrm{x}}$ = 2⇒ $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,-3.\mathrm{x}.\frac{1}{\mathrm{x}}\left( \mathrm{x}-\frac{1}{\mathrm{x}} \right) =8$ (cubing both side)⇒ $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,=\,\,8 +6=14$⇒ $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,-2.\mathrm{x}^3.\frac{1}{\mathrm{x}^3}=\,\,196$ (squaring both side)⇒ $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,=\,\,196+2=198$ Using Trick ⇒$\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,=\,\,\left( 2^3+3\times 2 \right) ^2+2=198$Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$ = (n3 – 3.n)2 – 2 andIf $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$ = (n3 + 3.n)2 + 2

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### CTET 2014 September Paper-2 Question with Answer

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