Algebra - Study Mode

[#166] If x varies inversely as (y 2 - 1) and x is equal to 24 when y = 10, then the value of x when y = 5 is?
Correct Answer

(A) 99

Explanation

Solution: $$eqalign{
& x propto frac{1}{{{y^2} - 1}}{ ext{ }}left( {{ ext{Given}}}
ight) cr
& x = k imes frac{1}{{{y^2} - 1}}left( {k{ ext{ is constant}}}
ight) cr
& { ext{Now }}x = 24{ ext{ when }}y = 10{ ext{ given}} cr
& Rightarrow 24 = k imes frac{1}{{{{left( {10}
ight)}^2} - 1}} cr
& Rightarrow 24 = frac{k}{{99}} cr
& Rightarrow k = 24 imes 99 cr
& x = ? cr
& y = 5 cr
& Rightarrow x = 24 imes 99 imes frac{1}{{25 - 1}} cr
& Rightarrow x = 24 imes 99 imes frac{1}{{24}} cr
& Rightarrow x = 99 cr} $$

[#167] If x 2 + y 2 + 2x + 1 = 0, then the value of x 31 + y 35 is?
Correct Answer

(A) -1

Explanation

Solution: $$eqalign{
& {x^2} + {y^2} + 2x + 1 = 0 cr
& Rightarrow {x^2} + 2x + 1 + {y^2} = 0 cr
& Rightarrow {left( {x + 1}
ight)^2} + {y^2} = 0 cr} $$ Hence both terms are squares and there addition is zero. So, it can be possible only when both terms are zeros. $$eqalign{
& herefore x + 1 = 0 cr
& Rightarrow x = - 1 cr
& y = 0 cr
& herefore {x^{31}} + {y^{35}} cr
& Rightarrow {left( { - 1}
ight)^{31}} + {left( 0
ight)^{35}} cr
& Rightarrow - 1 cr} $$

[#168] If a, b, c are real and a 2 + b 2 + c 2 = 2(a - b - c) -3, then the value of 2a - 3b + 4c is?
Correct Answer

(C) 1

Explanation

Solution: a 2 + b 2 + c 2 = 2(a - b - c) - 3 ⇒ a 2 + b 2 + c 2 = 2a - 2b - 2c - 3 ⇒ a 2 + b 2 + c 2 - 2a + 2b +2c + 1 + 1 + 1 = 0 ⇒ (a 2 - 2a + 1) + (b 2 + 2b + 1) + (c 2 + 2c + 1) = 0 ⇒ (a - 1) 2 + (b + 1) 2 + (c + 1) 2 = 0 a = 1 b = -1 c = -1 ∴ 2a - 3b + 4c = 2 × 1 - 3 × (-1) + 4 × (-1) = 2 + 3 - 4 = 1

[#169] If (3a + 1) 2 + (b - 1) 2 + (2c - 3) 2 = 0 then the value of (3a + b + 2c) is equal to?
Correct Answer

(A) 3

Explanation

Solution: $$eqalign{
& {left( {3a + 1}
ight)^2} + {left( {b - 1}
ight)^2} + {left( {2c - 3}
ight)^2} = 0 cr
& {left( {3a + 1}
ight)^2} = 0 cr
& Rightarrow 3a = - 1 cr
& Rightarrow a = - frac{1}{3} cr
& {left( {b - 1}
ight)^2} = 0 cr
& Rightarrow b - 1 = 0 cr
& Rightarrow b = 1 cr
& {left( {2c - 3}
ight)^2} = 0 cr
& Rightarrow c = frac{3}{2} cr
& herefore 3a + b + 2c cr
& = 3 imes - frac{1}{3} + 1 + frac{3}{2} imes 2 cr
& = - 1 + 1 + 3 cr
& = 3 cr} $$

[#170] The value of the expression $$frac{{{{left( {a - b}
ight)}^2}}}{{left( {b - c}
ight)left( {c - a}
ight)}} + $$ xa0 $$frac{{{{left( {b - c}
ight)}^2}}}{{left( {a - b}
ight)left( {c - a}
ight)}} + $$ xa0xa0 $$frac{{{{left( {c - a}
ight)}^2}}}{{left( {a - b}
ight)left( {b - c}
ight)}}$$ xa0 = ?
Correct Answer

(B) 3

Explanation

Solution: $$frac{{{{left( {a - b}
ight)}^2}}}{{left( {b - c}
ight)left( {c - a}
ight)}} + $$ xa0 $$frac{{{{left( {b - c}
ight)}^2}}}{{left( {a - b}
ight)left( {c - a}
ight)}} + $$ xa0xa0 $$frac{{{{left( {c - a}
ight)}^2}}}{{left( {a - b}
ight)left( {b - c}
ight)}}$$ Now, $$ Rightarrow frac{{{{left( {a - b}
ight)}^2}}}{{left( {b - c}
ight)left( {c - a}
ight)}} imes frac{{left( {a - b}
ight)}}{{a - b}}$$ Multiply divide by (a - b) in 1 st term Multiply divide by (b - c) in 2 nd term Multiply divide by (c - a) in 3 rd term $$ Rightarrow frac{{{{left( {a - b}
ight)}^2}left( {a - b}
ight)}}{{left( {b - c}
ight)left( {c - a}
ight)left( {a - b}
ight)}} + $$ xa0 xa0 $$frac{{{{left( {b - c}
ight)}^2}left( {b - c}
ight)}}{{left( {a - b}
ight)left( {b - c}
ight)left( {c - a}
ight)}} + $$ xa0 xa0 $$frac{{{{left( {c - a}
ight)}^2}left( {c - a}
ight)}}{{left( {a - b}
ight)left( {b - c}
ight)left( {c - a}
ight)}}$$ $$eqalign{
& { ext{Let, }}a - b = x cr
& b - c = y cr
& c - a = z cr
& herefore x + y + z = 0 cr
& herefore {x^3} + {y^3} + {z^3} = 3xyz cr
& herefore {left( {a - b}
ight)^2} + {left( {b - c}
ight)^2} + {left( {c - a}
ight)^2} cr
& = 3left( {a - b}
ight)left( {b - c}
ight)left( {c - a}
ight) cr
& herefore frac{{3left( {a - b}
ight)left( {b - c}
ight)left( {c - a}
ight)}}{{left( {a - b}
ight)left( {b - c}
ight)left( {c - a}
ight)}} cr
& = 3 cr} $$