Algebra - Study Mode
[#211] If a = 4.36, b = 2.39 and c = 1.97, then the value of a 3 - b 3 - c 3 - 3abc is?
Correct Answer
(C) 0
Explanation
Solution: $$eqalign{
& a = 4.36 cr
& b = 2.39 cr
& c = 1.97 cr
& { ext{ }}a - b - c cr
& = 4.36 - 2.39 - 1.97 cr
& = 0 cr
& { ext{ }}{{ ext{a}}^3} - {b^3} - {c^3} - 3abc cr
& = frac{1}{2}left( {a - b - c}
ight)left[ {{{left( {a - b}
ight)}^2} + {{left( {b - c}
ight)}^2} + {{left( {c - a}
ight)}^2}}
ight] cr
& = 0 cr} $$
[#212] If $$frac{{3a + 5b}}{{3a - 5b}} = 5,$$ xa0 then a : b is equal to?
Correct Answer
(D) 5 : 2
Explanation
Solution: $$eqalign{
& frac{{3a + 5b}}{{3a - 5b}} = 5 cr
& Rightarrow 3a + 5b = 15a - 25b cr
& Rightarrow 12a = 30b cr
& Rightarrow 2a = 5b cr
& a:b cr
& 5:2 cr} $$
[#213] If x : y = 3 : 4, then (7x + 3y) : (7x - 3y) is equal to?
Correct Answer
(C) 11 : 3
Explanation
Solution: $$eqalign{
& x:y = 3:4 cr
& frac{{left( {7x + 3y}
ight)}}{{left( {7x - 3y}
ight)}} cr
& = frac{y}{y}left( {frac{{7frac{x}{y} + 3}}{{7frac{x}{y} - 3}}}
ight) cr
& = frac{{7 imes frac{3}{4} + 3}}{{7 imes frac{3}{4} - 3}} cr
& = frac{{frac{{21}}{4} + 3}}{{frac{{21}}{4} - 3}} cr
& = frac{{frac{{21 + 12}}{4}}}{{frac{{21 - 12}}{4}}} cr
& = frac{{11}}{3} ,,or,, 11 : 3 cr} $$
[#214] For what value (s) of a is $$x + frac{1}{4}sqrt x + {a^2}$$ xa0xa0 a perfect square?
Correct Answer
(B) $$frac{1}{8}$$
Explanation
Solution: $$eqalign{
& x + frac{1}{4}sqrt x + {a^2} cr
& = {left( {sqrt x }
ight)^2} + 2 imes frac{1}{8} imes sqrt x + {a^2} cr
& left[ {left( {{{ ext{A}}^2} + { ext{2AB}} + {{ ext{B}}^2}}
ight) = {{left( {{ ext{A}} + { ext{B}}}
ight)}^2}}
ight] cr
& { ext{Here, A}} = sqrt x { ext{ and }} cr
& { ext{B}} = a cr
& { ext{B}} = frac{1}{8} cr
& herefore a = frac{1}{8} cr} $$
[#215] If x, y are two positive real number and $${x^{frac{1}{3}}} = {y^{frac{1}{4}}},$$ xa0 then which of the following relations is true?
Correct Answer
(D) x 20 = y 15
Explanation
Solution: $$eqalign{
& { ext{ }}{x^{frac{1}{3}}} = {y^{frac{1}{4}}}, cr
& Rightarrow { ext{LCM of 3, 4}} = 12{ ext{ }} cr
& herefore { ext{ }}{left( {{x^{frac{1}{3}}}}
ight)^{12}} = {left( {{y^{frac{1}{4}}}}
ight)^{12}} cr
& Rightarrow {x^4} = {y^3} cr
& { ext{take power '5' on both sides}} cr
& Rightarrow {left( {{x^4}}
ight)^5} = {left( {{y^3}}
ight)^5} cr
& Rightarrow {x^{20}} = {y^{15}} cr} $$