Algebra - Study Mode

[#221] When a number x is divided by a divisor it is seen that the divisor = 4 times the quotient = double of remainder. If the remainder is 80, then the value of x is?
Correct Answer

(A) 6480

Explanation

Solution: According to the question, Divisor = 2 × remainder = 2 × 80 = 160 Again, 4 × quotient = 160 ⇒ Quotient = $$frac{{160}}{4}$$ = 40 ∴ x = Divisor × Quotient + Remainder x = 160 × 40 + 80 = 6480

[#222] If p = 99, then the value of p(p 2 + 3p + 3) is?
Correct Answer

(B) 999999

Explanation

Solution: $$eqalign{
& x08ecause p = 99 cr
& pleft( {{p^2} + 3p + 3}
ight) cr
& = {p^3} + 3{p^2} + 3p + 1 - 1 cr
& = {left( {p + 1}
ight)^3} - 1 cr
& = {left( {99 + 1}
ight)^3} - 1 cr
& = {left( {100}
ight)^3} - 1 cr
& = 1000000 - 1 cr
& = 999999 cr} $$

[#223] If the sum of square of two real numbers is 41, and their sum is 9. Then the sum of cubes of these two numbers is ?
Correct Answer

(C) 189

Explanation

Solution: $$eqalign{
& { ext{Let the numbers are a, b}} cr
& {a^2} + {b^2} = 41 cr
& a + b = 9 cr
& Rightarrow {left( {a + b}
ight)^2} = {a^2} + {b^2} + 2ab cr
& Rightarrow {9^2} = 41 + 2ab cr
& Rightarrow 81 - 41 = 2ab cr
& Rightarrow ab = 20 cr
& { ext{Take }} cr
& a = 5 cr
& b = 4 cr
& Rightarrow {a^3} + {b^3} = {5^3} + {4^3} cr
& Rightarrow {a^3} + {b^3} = 125 + 64 cr
& Rightarrow {a^3} + {b^3} = 189 cr} $$

[#224] A complete factorisation of x 4 + 64 is?
Correct Answer

(D) (x 2 + 4x + 8)(x 2 - 4x + 8)

Explanation

Solution: $$eqalign{
& left( {{x^4} + 64}
ight) cr
& = {x^4} + {8^2} + 2.{x^2}.8 - 2.{x^2}.8 cr
& = {left( {{x^2} + 8}
ight)^2} - left( {16{x^2}}
ight) cr
& = {left( {{x^2} + 8}
ight)^2} - {left( {4x}
ight)^2} cr
& = left( {{x^2} + 8 + 4x}
ight)left( {{x^2} + 8 - 4x}
ight) cr
& = left( {{x^2} + 4x + 8}
ight)left( {{x^2} - 4x + 8}
ight) cr} $$

[#225] The simplified value of $$left( {1 - frac{{2xy}}{{{x^2} + {y^2}}}}
ight)$$ xa0 $$ ÷ $$ $$left( {frac{{{x^3} - {y^3}}}{{x - y}} - 3xy}
ight)$$ xa0xa0 is?
Correct Answer

(B) $$frac{1}{{{x^2} + {y^2}}}$$

Explanation

Solution: $$left( {1 - frac{{2xy}}{{{x^2} + {y^2}}}}
ight) div left( {frac{{{x^3} - {y^3}}}{{x - y}} - 3xy}
ight)$$ $$ = left( {frac{{{x^2} + {y^2} - 2xy}}{{{x^2} + {y^2}}}}
ight) div $$ xa0xa0 $$left( {frac{{{x^3} - {y^3} - 3xyleft( {x - y}
ight)}}{{x - y}}}
ight)$$ $$eqalign{
& = frac{{{{left( {x - y}
ight)}^2}}}{{{x^2} + {y^2}}} div frac{{{{left( {x - y}
ight)}^3}}}{{x - y}} cr
& = frac{{{{left( {x - y}
ight)}^2}}}{{{x^2} + {y^2}}} div {left( {x - y}
ight)^2} cr
& = frac{{{{left( {x - y}
ight)}^2}}}{{{x^2} + {y^2}}} imes frac{1}{{{{left( {x - y}
ight)}^2}}} cr
& = frac{1}{{{x^2} + {y^2}}} cr} $$