Algebra Expressions MCQ Questions

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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) 8

Solution: 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) 27

Solution: $\mathrm{x}-\frac{1}{\mathrm{x}}$ = 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) 47

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 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) 2

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=2$

⇒ 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) 527

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 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 and

If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ 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) 119

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 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 and

If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ 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) 18

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 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) 140

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 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) 322

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 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 and

If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ 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) 198

Solution: $\mathrm{x}-\frac{1}{\mathrm{x}}$ = 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 and

If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$ = (n3 + 3.n)2 + 2

 

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