Algebra - Study Mode
[#146] If a = 7, b = 5 and c = 3, then the value of a 2 + b 2 + c 2 - ab - bc - ca is?
Correct Answer
(A) 12
Explanation
Solution: $$eqalign{
& {a^2} + {b^2} + {c^2} - ab - bc - ca cr
& = frac{1}{2}left[ {{{left( {a - b}
ight)}^2} + {{left( {b - c}
ight)}^2} + {{left( {c - a}
ight)}^2}}
ight] cr
& = frac{1}{2}left[ {{{left( {7 - 5}
ight)}^2} + {{left( {5 - 3}
ight)}^2} + {{left( {3 - 7}
ight)}^2}}
ight] cr
& = frac{1}{2}left( {4 + 4 + 16}
ight) cr
& = frac{{24}}{2} cr
& = 12 cr} $$
[#147] If $${{ ext{7}}^x}{ ext{ = }}frac{1}{{343}}{ ext{,}}$$ xa0 then the value of x is?
Correct Answer
(B) -3
Explanation
Solution: $$eqalign{
& {{ ext{7}}^x}{ ext{ = }}frac{1}{{343}} cr
& Rightarrow {{ ext{7}}^x}{ ext{ = }}frac{1}{{{7^3}}} cr
& Rightarrow {{ ext{7}}^x}{ ext{ = }}{{ ext{7}}^{ - 3}} cr
& Rightarrow x = - 3 cr} $$ (If bases are equal then their power are also equal)
[#148] $$frac{{frac{1}{3}.frac{1}{3}.frac{1}{3} + frac{1}{4}.frac{1}{4}.frac{1}{4} - 3.frac{1}{3}.frac{1}{4}.frac{1}{5} + frac{1}{5}.frac{1}{5}.frac{1}{5}}}{{frac{1}{3}.frac{1}{3} + frac{1}{4}.frac{1}{4} + frac{1}{5}.frac{1}{5} - left( {frac{1}{3}.frac{1}{4} + frac{1}{4}.frac{1}{5} + frac{1}{5}.frac{1}{3}}
ight)}}{ ext{ is?}}$$
Correct Answer
(C) $$frac{{47}}{{60}}$$
Explanation
Solution: $$frac{{frac{1}{3}.frac{1}{3}.frac{1}{3} + frac{1}{4}.frac{1}{4}.frac{1}{4} - 3.frac{1}{3}.frac{1}{4}.frac{1}{5} + frac{1}{5}.frac{1}{5}.frac{1}{5}}}{{frac{1}{3}.frac{1}{3} + frac{1}{4}.frac{1}{4} + frac{1}{5}.frac{1}{5} - left( {frac{1}{3}.frac{1}{4} + frac{1}{4}.frac{1}{5} + frac{1}{5}.frac{1}{3}}
ight)}}$$ A 3 + B 3 + C 3 - 3ABC = (A + B + C)(A 2 + B 2 + C 2 - AB - BC - CA) $$ herefore frac{{{{left( {frac{1}{3}}
ight)}^3} + {{left( {frac{1}{4}}
ight)}^3} - 3.frac{1}{3}.frac{1}{4}.frac{1}{5} + {{left( {frac{1}{5}}
ight)}^3}}}{{{{left( {frac{1}{3}}
ight)}^2} + {{left( {frac{1}{4}}
ight)}^2} + {{left( {frac{1}{5}}
ight)}^2} - frac{1}{3}.frac{1}{4} - frac{1}{4}.frac{1}{5} - frac{1}{5}.frac{1}{3}}}$$ $$ = frac{{left( {frac{1}{3} + frac{1}{4} + frac{1}{5}}
ight)left[ {{{left( {frac{1}{3}}
ight)}^2} + {{left( {frac{1}{4}}
ight)}^2} + {{left( {frac{1}{5}}
ight)}^2} - frac{1}{3}.frac{1}{4} - frac{1}{4}.frac{1}{5} - frac{1}{5}.frac{1}{3}}
ight]}}{{left[ {{{left( {frac{1}{3}}
ight)}^2} + {{left( {frac{1}{4}}
ight)}^2} + {{left( {frac{1}{5}}
ight)}^2} - frac{1}{3}.frac{1}{4} - frac{1}{4}.frac{1}{5} - frac{1}{5}.frac{1}{3}}
ight]}}$$ $$eqalign{
& = frac{{20 + 15 + 12}}{{60}} cr
& = frac{{47}}{{60}} cr} $$
[#149] If 0.13 × p 2 = 13, then p is equal to?
Correct Answer
(A) 10
Explanation
Solution: $$eqalign{
& { ext{0}}{ ext{.13}} imes {{ ext{p}}^2} = 13 cr
& Rightarrow {{ ext{p}}^2} = frac{{13}}{{0.13}} cr
& Rightarrow {{ ext{p}}^2} = frac{{13}}{{13}} imes 100 cr
& Rightarrow {{ ext{p}}^2} = 100 cr
& Rightarrow { ext{p}} = 10 cr} $$
[#150] If $$x = 7 - 4sqrt 3 { ext{,}}$$ xa0xa0 then $$sqrt x { ext{ + }}frac{1}{{sqrt x }}$$ xa0 is equal to?
Correct Answer
(D) 4
Explanation
Solution: $$eqalign{
& x = 7 - 4sqrt 3 cr
& Rightarrow x = 4 + 3 - 4sqrt 3 cr
& Rightarrow x = {left( 2
ight)^2} + {left( {sqrt 3 }
ight)^2} - 2 imes 2sqrt 3 cr
& Rightarrow {left( {2 - sqrt 3 }
ight)^2} cr
& herefore left[ {{a^2} + {b^2} - 2ab = {{left( {a - b}
ight)}^2}}
ight] cr
& Rightarrow x = {left( {2 - sqrt 3 }
ight)^2} cr
& Rightarrow sqrt x = 2 - sqrt 3 cr
& Rightarrow frac{1}{{sqrt x }} = frac{1}{{2 - sqrt 3 }} imes frac{{2 + sqrt 3 }}{{2 + sqrt 3 }} cr
& Rightarrow 2 + sqrt 3 cr
& herefore sqrt x { ext{ + }}frac{1}{{sqrt x }}{ ext{ }} cr
& = 2 - sqrt 3 + 2 + sqrt 3 cr
& = 4 cr} $$