Algebra - Study Mode

[#396] If $$2x + frac{1}{{3x}} = 5{ ext{,}}$$ xa0 then the value of $$frac{{5x}}{{6{x^2} + 20x + 1}}$$ xa0 is?
Correct Answer

(D) $$frac{1}{7}$$

Explanation

Solution: $$eqalign{
& 2x + frac{1}{{3x}} = 5 cr
& 6{x^2}{ ext{ + 1 = 15x}},......{ ext{(i)}} cr
& { ext{Now,}}frac{{5x}}{{6{x^2} + 20x + 1}} cr
& = frac{{5x}}{{6{x^2} + 1 + 20x}} cr
& left[ {{ ext{From equation (i)}}}
ight] cr
& = frac{{5x}}{{15x + 20x}}{ ext{ }} cr
& = frac{{5x}}{{35x}} cr
& = frac{1}{7} cr} $$

[#397] If a + b = 10 and ab = 21, then the value of (a - b) 2 is?
Correct Answer

(B) 16

Explanation

Solution: $$eqalign{
& a + b = 10{ ext{ and }}ab = 21 cr
& left( {a + b}
ight) = 10 cr
& Rightarrow {a^2} + {b^2} + 2ab = 100 cr
& Rightarrow {a^2} + {b^2} = 100 - 2ab cr
& Rightarrow {a^2} + {b^2} = 100 - 2 imes 21 cr
& Rightarrow {a^2} + {b^2} = 100 - 42 cr
& {a^2} + {b^2} = 58,.........(i) cr
& {left( {a - b}
ight)^2} = {a^2} + {b^2} - 2ab cr
& {left( {a - b}
ight)^2} = 58 - 2 imes 21 cr
& left[ {{ ext{from equation (i)}}}
ight] cr
& = { ext{58}} - { ext{42}} cr
& {left( {a - b}
ight)^2} = 16 cr} $$

[#398] If x : y = 4 : 15 then the value of $$left( {frac{{x - y}}{{x + y}}}
ight)$$ xa0 is?
Correct Answer

(A) $$frac{{11}}{{19}}$$

Explanation

Solution: $$eqalign{
& y:x = 4:15 cr
& herefore frac{y}{x} = frac{4}{{15}} cr
& herefore frac{{x - y}}{{x + y}} cr
& Rightarrow frac{{xleft( {1 - frac{y}{x}}
ight)}}{{xleft( {1 + frac{y}{x}}
ight)}} cr
& { ext{Taking }}x{ ext{ common}} cr
& Rightarrow frac{{1 - frac{4}{{15}}}}{{1 + frac{4}{{15}}}} cr
& Rightarrow frac{{11}}{{15}} imes frac{{15}}{{19}} cr
& Rightarrow frac{{11}}{{19}} cr} $$

[#399] If (x - 3) 2 + (y - 5) 2 + (z - 4) 2 = 0, then the value of $$frac{{{x^2}}}{9}{ ext{ + }}frac{{{y^2}}}{{25}}{ ext{ + }}frac{{{z^2}}}{{16}}$$ xa0xa0 is?
Correct Answer

(C) 3

Explanation

Solution: $$eqalign{
& {left( {x - 3}
ight)^2}{ ext{ + }}{left( {y - 5}
ight)^2}{ ext{ + }}{left( {z - 4}
ight)^2} = 0 cr
& herefore {left( {x - 3}
ight)^2} = 0{ ext{ }}x = 3 cr
& {left( {y - 5}
ight)^2} = 0{ ext{ }}y = 5 cr
& {left( {z - 4}
ight)^2} = 0{ ext{ }}z = 4{ ext{ }} cr
& herefore frac{{{x^2}}}{9}{ ext{ + }}frac{{{y^2}}}{{25}}{ ext{ + }}frac{{{z^2}}}{{16}} cr
& Rightarrow frac{9}{9} + frac{{25}}{{25}} + frac{{16}}{{16}} cr
& Rightarrow 3 cr} $$

[#400] x varies inversely as square of y. Given that y = 2 for x = 1, the value of x for y = 6 will equal to?
Correct Answer

(D) $$frac{1}{9}$$

Explanation

Solution: $$eqalign{
& x propto frac{1}{{{y^2}}} cr
& left( {{ ext{Inversely proportional}}}
ight) cr
& x = frac{k}{{{y^2}}} cr
& left( {{ ext{Given}}}
ight), cr
& left( {y = 2}
ight){ ext{ for }}left( {x = 1}
ight) cr
& herefore 1 = frac{k}{{{{left( 2
ight)}^2}}} cr
& Rightarrow 1 = frac{k}{4} cr
& Rightarrow k = 4 cr
& herefore { ext{For }}y = 6 cr
& x = frac{4}{{{{left( 6
ight)}^2}}} cr
& x = frac{4}{{36}} cr
& x = frac{1}{9} cr} $$