Practice Power indices and surds - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) The value of : $√{ - √{3} + √{3 + 8√{7 + 4√{3}}}}$ is
(a)
(b)
(c)
(d)
$√{ - √{3} + √{3 + 8√{7 + 4√{3}}}}$
= $√{ - √{3} + √{3 + 8√{4 + 3 + 2 × 2 × √{3}}}}$
= $√{ - √{3} + √{3 + 8√{(2)^2 + (√3)^2 + 2 × 2 × √{3}}}}$
= $√{ - √{3} + √{3 + 8√{(2 × √{3})^2}}}$
= $√{ - √{3} + √{3 + 8(2 × √{3})}}$
= $√{ - √{3} + √{3 + 16 +8√{3}}}$
= $√{-√{3} + √{(√3)^2 + (4)^2 + 2 × 4 √{3}}}$
=$√{-√{3} + √{(4 + √3)^2}}$
= $√{-√{3} + 4 +√{3}} = √4$ = 2
Q-2) $(16)^{0.16} × (16)^{0.04} × (2)^{0.2}$ is equal to :
(a)
(b)
(c)
(d)
$(16)^{0.16} × (16)^{0.04} × (2)^{0.2}$
= $(2^4)^{0.16} × (2^4)^{0.04} × (2)^{0.2}$
= $2^{0.64} × 2^{0.16} × 2^{0.2}$
= $(2)^{0.64+0.16+0.2}$ = 2
Q-3) $(√8 - √4 - √2)$ equals :
(a)
(b)
(c)
(d)
? = $(√8 - √4 - √2)$
= $(2√2 - 2 - √2) = √2 - 2$
Q-4) The simplified form of $(16^{3/2} + 16^{-3/2})$ is :
(a)
(b)
(c)
(d)
$(16^{3/2} + 16^{-3/2})$
= $(4^2)^{3/2} + 1/(16)^{3/2}$
= $4^{2×3/2} + 1/4^{2×3/2} = 4^3 + 1/4^3$
= $64 + 1/64 = {4096 + 1}/64 = 4097/64$
Q-5) The simplified form of $2/{√7 + √5} + 7/{√{12} - √5} - 5/{√{12} - √7}$ is :
(a)
(b)
(c)
(d)
$2/{√7 + √5} + 7/{√{12} - √5} - 5/{√{12} - √7}$
First term = $2/{√7 + √5}$
= ${2 ×(√7 - √5)}/{(√7 + √5)(√7 - √5)}$
= ${2(√7 - √5)}/{7-5} = √7 -√5$
Second term = $7/{√{12} - √5}$
= ${7(√{12}+√5)}/{(√{12}-√{5})(√{12}+√5)}$
=${7(√{12}+√5)}/{12-5}$
= ${7(√{12} +√{5})}/{7} = √{12}+√5$
Third term = $5/{√{12}-√7}$
= ${5(√{12}+√7)}/{(√{12}-√7)(√{12}+√7)}$
= ${5(√{12}+√7)}/{12- 7} = √{12} +√7$
Expression
= $(√7 - √5) + (√{12} +√5) - (√{12}+√7)$
= $√7 -√5 +√{12} +√5 - √{12} -√7 = 0$
Q-6) 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-7) ${√{10+√{ 25+√{ 108+√{ 154+√{ 225}}}}}}/√^3{8}$= ?
(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-8) $√{2+√{2+√{2 +...}}}$ 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-9) 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-10) 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-11) If $√3$ = 1.732, then what is the value of ${4 +3√3}/{√{7 + 4 √3}}$ upto three places of decimal ?
(a)
(b)
(c)
(d)
${4 +3√3}/{√{7 + 4 √3}}$
${4 +3√3}/{√{7 + 4 √3}}={4+3√3}/{√{4+3+2×2×√3}}$
=${4+3√3}/{√(2+√3)^2}={4+3√3}/{2+√3}$
${(4+3√3)(2-√3)}/{(2+√3)(2-√3)}$
=$8-4√3+6√3-9$
=$2√3-1=2×1.732-1$
= 3.464 - 1 = 2.464
Q-12) Given that $√5$ = 2.236 and $√3$ = 1.732; the value of $1/{√5 +√3}$ is
(a)
(b)
(c)
(d)
Expression=$1/{√5 +√3}$
=$1/{√5 +√3}×{√5 -√3}/{√5 -√3}$
(Rationalising the denominator)
=${√5 -√3}/{5-3}={2.236-1.732}/2$
=$0.504/2=0.252$
Q-13) Given that $√2$ = 1.414; the value of $1/{√2 + 1}$ is
(a)
(b)
(c)
(d)
Expression
=$1/{√2 + 1}={√2 - 1}/{(√2 + 1)(√2 - 1)}$
=${√2-1}/{2-1}=√2-1$
= 1.414 - 1 = 0.414
Q-14) Given that $√3$ = 1.732, the value of ${3 +√6}/{5√3 - 2√12- √32+√ 50}$ is
(a)
(b)
(c)
(d)
Expression
=${3 +√6}/{5√3 - 2√12- √32+√ 50}$
=${3 +√6}/{5√3 - 2√{2×2×3}- √{2×2×2×2×2}+√{2×5×5}}$
=${3 +√6}/{5√3 - 4√{3}- 4√{2}+5√{2}}$
=${3 +√6}/{√3 +√2}={(3+√6)(√3-√2)}/{(√3+√2)(√3-√2)}$
[On rationalising the denominator]
=${3√3+√18-3√2-√12}/{3-2}$
=$3√3+3√2-3√2-2√3$
=$√3=1.732$
Q-15) If $√33$ = 5.745, then the value of $√{3/11}$ is approximately
(a)
(b)
(c)
(d)
$√33$ = 5.745(Given)
$√{3/11}=√{{3×11}/{11×11}}=√33/11$
=${5.745}/11 ≈0.5223$
Q-16) Which is greater $√^3{2}$ or $√3$ ?
(a)
(b)
(c)
(d)
$√^3{2}$ or $√3$
$√^3{2} =2^{1/3}=2^{2/6}=√^6{4}$
$√{3} =3^{1/2}=3^{3/6}=√^6{27}$
Q-17) $16^{3/4}$ is equal to :
(a)
(b)
(c)
(d)
$16^{3/4} = (4^2)^{3/4}$
= $4^{2×3/4} = 4^{3/2} = 2^{2×3/2} = 2^3 = 8$
Q-18) $√{3√{3√{3...}}}$ is equal to
(a)
(b)
(c)
(d)
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
Q-19) The value of $√{9 + 2√{ 16} +√^3{ 512}}$ is :
(a)
(b)
(c)
(d)
Expression
=$√{9 + 2√{16} +√^3{ 512}}$
=$√{9 + 2√{4×4} +√^3{8×8×8}}$
=$√{9 + 2×4+8}$
=$√25$ =5
Q-20) The greatest among the numbers $3√2, 3√7, 6√5, 2√20$ is
(a)
(b)
(c)
(d)
$3√2, 3√7, 6√5, 2√20$
$3√2=3×1.4=4.2$
$3√7=3×2.6=7.8$
$6√5=6×2.2=13.2$
$2√20=2×4.5=9$