Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

4(4x + 5) > 2x - 1 > 4x - 3
16x + 20 > 2x - 1
14x > -21
x > $$\frac{{ - 3}}{2}$$
2x - 1 > 4x - 3
2x < 2
x < 1
$$\frac{{ - 3}}{2}$$ < x < 1
Only 0 lies between (-1.5, 1)
Go through option D
x = 0 satisfied

$$\eqalign{ & \frac{{4\left[ {{{\left( {17} \right)}^3} - {{\left( 7 \right)}^3}} \right]}}{{\left( {{{17}^2} + {7^2} + p} \right)}} = 40 \cr & {\text{We know,}} \cr & \left[ {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)} \right] \cr & \frac{{4\left[ {\left( {17 - 7} \right)\left( {{{17}^2} + {7^2} + 17 \times 7} \right)} \right]}}{{{{17}^2} + {7^2} + p}} = 40 \cr & {17^2} + {7^2} + 119 = {17^2} + {7^2} + p \cr & p = 119 \cr} $$

$$\eqalign{ & \frac{{3\left( {{x^2} + 1} \right) - 7x}}{{3x}} = 6 \cr & 3{x^2} + 3 - 7x = 18x \cr & 3{x^2} + 3 = 25x \cr & 3\left( {{x^2} + 1} \right) = 25x \cr & {\text{divided by }}'x' \cr & 3\left( {x + \frac{1}{x}} \right) = 25 \cr & x + \frac{1}{x} = 25 \cr & x + \frac{1}{x} + 2 = \frac{{25}}{3} \cr & {\left( {x + \frac{1}{x}} \right)^2} = \frac{{25}}{3} + 2 \cr & {\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)^2} = \frac{{31}}{3} \cr & \sqrt x + \frac{1}{{\sqrt x }} = \sqrt {\frac{{31}}{3}} \cr} $$

$$\eqalign{ & {x^3} + {y^3} + {z^3} - 3xyz \cr & = \frac{{\left( {x + y + z} \right)}}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr & = \frac{{255 + 256 + 257}}{2}\left[ {{1^2} + {1^2} + {2^2}} \right] \cr & = \frac{{768 \times 6}}{2} \cr & = \frac{{4608}}{2} \cr & = 2304 \cr} $$

$$\eqalign{ & \sqrt {\left( {1 - {p^2}} \right)\left( {1 - {q^2}} \right)} = \frac{{\sqrt 3 }}{2}\,........\,\left( {\text{i}} \right) \cr & {\text{Put value of }}p{\text{ and }}q \cr & p = 0,\,q = \frac{1}{2} \cr & {\text{Equation }}\left( {\text{i}} \right){\text{ is satisfying}} \cr & {\text{Then, }} \cr & \sqrt {2{p^2} + 2{q^2} + 2pq} + \sqrt {2{p^2} + 2{q^2} - 2pq} \cr & = \sqrt {0 + \frac{2}{4} + 0} + \sqrt {0 + \frac{2}{4} - 0} \cr & = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} \cr & = \frac{2}{{\sqrt 2 }} \cr & = \sqrt 2 \cr} $$

(4x + 2y)³ + (4x - 2y)³ = 16(Ax³ + Bxy²)
⇒ (4x)³ + (2y)³ + 3 × 4x × 2y(4x + 2y) + (4x)³ - (2y)³ - 3 × 4x × 2y(4x - 2y) = 16(Ax³ + Bxy²)
⇒ 2(4x)³ + 96x²y + 48xy² - 96x²y + 48xy² = 16(Ax³ + Bxy²)
⇒ 128x³ + 96xy² = 16(Ax³ + Bxy²)
⇒ 16(8x³ + 6xy²) = 16(Ax³ + Bxy²)
On comparing on both side-
A = 8 & B = 6
Then,
$$\eqalign{ & \frac{1}{2}\left( {\sqrt {{A^2} + {B^2}} } \right) \cr & = \frac{1}{2}\left( {\sqrt {64 + 36} } \right) \cr & = \frac{1}{2} \times 10 \cr & = 5{\text{ Answer}} \cr} $$

$$\eqalign{ & x + \frac{4}{x} = 4 \cr & x = 2 \cr & {\text{So, }}{x^2} - 4 = ? \cr & {x^2} - 4 = 4 - 4 = 0 \cr} $$

$$\eqalign{ & 4{x^2} - 6x + 1 = 0 \cr & 2x\left( {2x - 3 + \frac{1}{{2x}}} \right) = 0 \cr & 2x + \frac{1}{{2x}} = 3 \cr & 8{x^3} + \frac{1}{{8{x^3}}} = 27 - 3 \times 3 \cr & 8{x^3} + \frac{1}{{8{x^3}}} = 18 \cr} $$