Practice Simplification problems - 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) $8^{2/3}$ is equal to :
(a)
(b)
(c)
(d)
$8^{2/3} = (2^3)^{2/3}$
= $2^{3×2/3} = 2^2 = 4$
Q-7) The value of $(√^3{3.5} + √^3{2.5})((√^3{3.5})^2 - √^3{8.75} + (√^3{2.5})^2)$ is :
(a)
(b)
(c)
(d)
$(√^3{3.5} + √^3{2.5})((√^3{3.5})^2 - √^3{8.75} + (√^3{2.5})^2)$
Let $√^3{3.5} = a$ and $√^3{2.5} = b$
Expression
= (a + b)$(a^2 - ab + b^2) = a^3 + b^3$
= $(√^3{3.5})^3 + (√^3{2.5})^3$
= 3.5 + 2.5 = 6
Q-8) $(16^{0.16} × 2^{0.36})$ is equal to
(a)
(b)
(c)
(d)
$(16^{0.16} × 2^{0.36})$
= $(16^{16/100} × 2^{36/100})$
= $(2^{4×16/100} × 2^{36/100})$
= $(2^{64/100 + 36/100}) = (2^{100/100}) = 2$
Q-9) The value of $(256)^{0.16} × (16)^{0.18}$ is :
(a)
(b)
(c)
(d)
Expression
= $(256)^{0.16} × (16)^{0.18}$
= $(4)^{4×0.16} × (4)^{2×0.18}$
= $(4)^{0.64} × (4)^{0.36}$
= $(4)^{0.64+0.36} = (4)^1$ = 4
Q-10) The value of $√{2}^4 + √^3{64} + √^4{2^8}$ is :
(a)
(b)
(c)
(d)
? = $√{2}^4 + √^3{64} + √^4{2^8}$
= $2^{4×1/2} + 4^{3×1/3} + 2^{8×1/4}$
= $2^2 + 4 + 2^2$
= 4 + 4 + 4 = 12