Practice Simplifying roots of roots - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) Find the value of $√{10+√{ 25+√{ 108+√{ 154+√{ 225}}}}}$.
(a)
(b)
(c)
(d)
Expression
=$√{10+√{ 25+√{ 108+√{ 154+√{ 225}}}}}$
=$√{10+√{ 25+√{ 108+√{ 154+15}}}}$
=$√{10+√{ 25+√{ 108+√{169}}}}$
=$√{10+√{ 25+√{ 108+13}}}$
=$√{10+√{ 25+√{121}}}$
=$√{10+√{ 25+11}}$
=$√{10+6}=√16=4$
Q-2) $√{6+√{6+√{6 +...}}}$ is equal to
(a)
(b)
(c)
(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)
Q-3) $√{12+√{12+√{12 +...}}}$ is equal to
(a)
(b)
(c)
(d)
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 25If $√{x+√{x+√{x +...∞}}}$ where, x=n(n + 1)then $√{x+√{x+√{x +...∞}}}$ = (n + 1)
$√{12+√{12+√{12}}}$=4
It is because
12 = 3 × 4 = n (n + 1)
Q-4) The value of the following is : $√{12+√{12+√{12 +...}}}$
(a)
(b)
(c)
(d)
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
Q-5) ${√{10+√{ 25+√{ 108+√{ 154+√{ 225}}}}}}/√^3{8}$= ?
(a)
(b)
(c)
(d)
${√{10+√{ 25+√{ 108+√{ 154+√{ 225}}}}}}/√^3{8}$= ?
?=${√{10+√{ 25+√{ 108+√{ 154+15}}}}}/√^3{2×2×2}$
=${√{10+√{ 25+√{ 108+√{169}}}}}/2$
=${√{10+√{ 25+√{ 108+13}}}}/2$
=${√{10+√{ 25+√{121}}}}/2$
=${√{10+√{ 25+11}}}/2$
=${√{10+√{36}}}/2={√{10+6}}/2$
=${√{16}}/2=4/2=2$
Q-6) $√{2+√{2+√{2 +...}}}$ is equal to
(a)
(b)
(c)
(d)
Let x = $√{2+√{2+√{2 +...}}}$
On squaring both sides
$x^2 = 2+√{2+√{2+√{2 +...}}}$
$x^2$ = 2 + x
$x^2$ - x - 2 = 0
$x^2$ - 2x + x - 2 = 0
x (x - 2) + 1 (x - 2) = 0
(x - 2) (x + 1) = 0
x = 2 or - 1
But sum of positive numbers can't be negative.
x = 2
Q-7) Find the value of $√{30+√{30+√{30+...}}}$
(a)
(b)
(c)
(d)
Let x = $√{30+√{30+√{30+...}}}$
On squaring both sides,
$x^2 = 30+√{30+√{30+√{30+...}}}$
$x^2$ = 30 + x
$x^2$ - x - 30 = 0
$x^2$ - 6x + 5x - 30 = 0
x (x - 6) + 5 (x - 6) = 0
(x - 6) (x + 5) = 0
x = 6 because x ≠ - 5
Using Rule 25
$√{30+√{30+√{30+...}}}$ = 6
It is because
30 = 5 × 6 = n (n +1)
Q-8) The value of the expression $√{6+√{6+√{6 +...+upto∞}}}$ is
(a)
(b)
(c)
(d)
Let, x = $√{6+√{6+√{6 +...∞}}}$
On squaring,
$x^2 = 6 + √{6+√{6+√{6 +...∞}}}$
$x^2$ = 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
Using Rule 25
$√{6+√{6+√{6 +...}}}$ = 3
It is because, 6 = 2 × 3 =n (n+1)
Q-9) The value of $√{72+√{72+√{72 +...}}}$ is
(a)
(b)
(c)
(d)
x = $√{72+√{72+√{72 +...}}}$
On squaring both sides,
$x^2 = 72+√{72+√{72+√{72 +...}}}$
$x^2$ = 72 + x
$x^2$ - x - 72 = 0
$x^2$ - 9x + 8x - 72 = 0
x (x - 9) + 8 (x - 9) = 0
(x + 8) (x - 9) = 0
x = 9 because x ≠ - 8
Using Rule 25
$√{72+√{72+√{72 +...}}}$ = 9
It is because 72 = 8×9 = n (n + 1)
Q-10) The value of $√{2^3√{4√{2^3√{4√{2^3√{4…}}}}}}$ is
(a)
(b)
(c)
(d)
x=$√{2^3√{4√{2^3√{4√{2^3√{4…}}}}}}$
On squaring
$x^2 = 2 √{2^3√{4√{2^3√{4√{2^3√{4…}}}}}}$
On cubing,
$x^6$= 8 × 4x
$x^5 = 32 = 2^5$ ⇒ x = 2