model 5 simplifying roots of roots Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 5 EXERCISES

Top 10,000+ Aptitude Memory Based Exercises

The following question based on power, indices and surds topic of quantitative aptitude

Questions : $√{6+√{6+√{6 +...}}}$ is equal to

(a) 5

(b) 4

(c) 6

(d) 3

The correct answers to the above question in:

Answer: (d)

Let x = $√{6+√{6+√{6 +...}}}$

Squaring on both sides,

$x^2 = 6+√{6+√{6+√{6 +...}}}$

$x^2$ = 6 + x

$x^2$ - x - 6 = 0

$x^2$ - 3x + 2x - 6 = 0

x (x - 3) + 2 (x - 3) = 0

(x + 2) (x - 3) = 0

x = 3 because x ≠ - 2

$√{6+√{6+√{6 +...}}}$ = 3

It is because

6 = 2 × 3 = n (n + 1)

Practice power, indices and surds (model 5 simplifying roots of roots) Online Quiz

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Read more simplifying roots of roots Based Quantitative Aptitude Questions and Answers

Question : 1

$√{1+√{1+√{1 +...}}}$

a) lies between 1 and 2

b) lies between 0 and 1

c) is greater than 2

d) equals 1

Answer: (a)

Let x = $√{1+√{1+√{1 +...}}}$

On squaring both sides

$x^2 =1+√{1+√{1+√{1 +...}}}$

$x^2$ = 1 + x

$x^2$ - x - 1 = 0

$x = {+1± √{1+4}}/2 = {+1 ±√ 5}/2$

But sum of + ve numbers can't be negative.

$x={1+ √5}/2 ={1+ 2.236}/2$

= ${3.236}/2 =1.618$

Thus 1 < 1.618 < 2

Question : 2

$√{12+√{12+√{12 +...}}}$ is equal to

a) 6

b) 4

c) 2

d) 3

Answer: (b)

Let x = $√{12+√{12+√{12 +...}}}$

On squaring both sides,

$x^2 =12+√{12+√{12+√{12 +...}}}$

$x^2$ = 12 + x

$x^2$ - x - 12 = 0

$x^2$ - 4x + 3x - 12 = 0

x (x - 4) + 3 (x - 4) = 0

(x - 4) (x + 3) = 0

x = 4, - 3

The given expression is positive.

x = 4

Using Rule 25
If $√{x+√{x+√{x +...∞}}}$ where, x=n(n + 1)
then $√{x+√{x+√{x +...∞}}}$ = (n + 1)

$√{12+√{12+√{12}}}$=4

It is because

12 = 3 × 4 = n (n + 1)

Question : 3

The value of the following is : $√{12+√{12+√{12 +...}}}$

a) 2

b) $2√3$

c) 4

d) $2√2$

Answer: (c)

x = $√{12+√{12+√{12 +...}}}$

On squaring both sides,

$x^2 = 12 + √{12+√{12+√{12 +...}}}$

$x^2$ = 12 + x

$x^2$ - x - 12 = 0

$x^2$ - 4x + 3x - 12 = 0

x (x - 4) + 3 (x - 4) = 0

(x - 4) (x + 3) = 0

x = 4 because x ≠ - 3

Question : 4

Find the value of $√{10+√{ 25+√{ 108+√{ 154+√{ 225}}}}}$.

a) 8

b) 10

c) 4

d) 6

Answer: (c)

Expression

=$√{10+√{ 25+√{ 108+√{ 154+√{ 225}}}}}$

=$√{10+√{ 25+√{ 108+√{ 154+15}}}}$

=$√{10+√{ 25+√{ 108+√{169}}}}$

=$√{10+√{ 25+√{ 108+13}}}$

=$√{10+√{ 25+√{121}}}$

=$√{10+√{ 25+11}}$

=$√{10+6}=√16=4$

Question : 5

$√{3√{3√{3...}}}$ is equal to

a) $2√3$

b) 3

c) $3√3$

d) $√3$

Answer: (b)

Let x = $√{3√{3√{3...}}}$

Squaring both sides,

$x^2 = 3√{3√{3√{3...}}}$ = 3x

$x^2$ - 3x = 0

x (x - 3) = 0

x = 3 because x ≠ 0

Using Rule 23
$√{x√{x√{x...n times}}}= x^(1-1/{x^n})$

$√{3+√{3+√{3+...∞}}}$ = 3

It is because, here

n = ∞ and x =3

$√{3+√{3+√{3+...∞}}}=3^(1-1/{3∞})$

= $3^(1 - 0)$ [${something}/∞ = 0$] = 3

Question : 6

If m = $√{5+√{5+√{5 +...}}}$ and n = $√{5-√{5-√{5-...}}}$, then among the following the relation between m and n holds is

a) m + n + 1 = 0

b) m + n - 1 = 0

c) m - n - 1 = 0

d) m - n + 1 = 0

Answer: (c)

m = $√{5+√{5+√{5 +...}}}$

On squaring both sides,

$m^2 = 5 + m ⇒ m^2$ - m = 5 ....(i)

Again,

n = $√{5-√{5-√{5-...}}}$

On squaring both sides,

$n^2$ = 5 - n

$n^2$ + n = 5 .........(ii)

$m^2$ - m = $n^2$ + n

$(m^2 - n^2)$ = m + n

(m + n) (m - n) - (m + n) = 0

(m + n) (m - n - 1) = 0

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