Practice MCQ Questions and Answer on Algebra

31.

If [8(x + y)³ - 27(x - y)³] ÷ (5y - x) = Ax² + Bxy + Cy², then the value of (A + B + C) is:

• (A) 26
• (B) 19
• (C) 16
• (D) 13

Show Answer

 [8(x + y)3 - 27(x - y)3] ÷ (5y - x) = Ax2 + Bxy + Cy2 Let, x = 1, y = 1 [8(1 + 1)3 - 27 × 0] ÷ (5 - 1) = (A + B + C) $$\frac{{8 \times 8}}{4}$$  = A + B + C 16 = A + B + C

32.

If $${x^2} + 5x + 6 = 0{\text{,}}$$    then the value of $$\frac{{2x}}{{{x^2} - 7x + 6}}$$   is?

• (A) $$\frac{1}{6}$$
• (B) $$\frac{1}{3}$$
• (C) $$ - \frac{1}{6}$$
• (D) $$ - \frac{1}{3}$$

Show Answer

 $$\eqalign{ & {\text{If }}{x^2} + 5x + 6 = 0 \cr & {\text{then, }}{x^2} + 6 = - 5x \cr & {\text{So,}}\frac{{2x}}{{{x^2} + 6 - 7x}} \cr & = \frac{{2x}}{{ - 5x - 7x}} \cr & = \frac{{2x}}{{ - 12x}} \cr & = - \frac{1}{6}{\text{ }} \cr} $$

33.

If (5x + 1)³ + (x - 3)³ + 8(3x - 4)³ = 6(5x + 1)(x - 3)(3x - 4), then x is equal to:

• (A) $$\frac{5}{6}$$
• (B) $$\frac{1}{3}$$
• (C) $$\frac{2}{3}$$
• (D) $$\frac{3}{4}$$

Show Answer

 a3 + b3 + c3 = 3abc If a + b + c = 0 (5x + 1) + (x - 3) + (6x - 8) = 0 12x - 10 = 0 12x = 10 x = $$\frac{5}{6}$$

34.

Simplify $$\frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}}{\text{if }}x + \frac{{{y^2}}}{x} = 5.$$

• (A) $$\frac{5}{{{y^2}}}$$
• (B) $$\frac{7}{{{y^2}}}$$
• (C) $$ - \frac{5}{{{y^2}}}$$
• (D) $$ - \frac{7}{{{y^2}}}$$

Show Answer

 $$\eqalign{ & {\text{Given:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {\text{Put }}y = 2,\,x = 1 \cr & {\text{Now,}} \cr & \frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{{{\left( 1 \right)}^2} + 2 \times 1 + {{\left( 2 \right)}^2}}}{{{{\left( 1 \right)}^3} - 5{{\left( 1 \right)}^2}}} \cr & = \frac{7}{{ - 4}} \cr & {\text{Put }}y = 2{\text{ in option and option D will satisfy the condition}} \cr & \cr & {\bf{Alternate \,solution:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {x^2} + {y^2} = 5x \cr & {x^2} - 5x = - {y^2} \cr & {\text{Now, }}\frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{5x + 2x}}{{x\left( {{x^2} - 5x} \right)}} \cr & = \frac{{7x}}{{x\left( { - {y^2}} \right)}} \cr & = - \frac{7}{{{y^2}}} \cr} $$

35.

If P = 7 + 4√3 and PQ = 1, then what is the value of $$\frac{1}{{{P^2}}} + \frac{1}{{{Q^2}}}?$$

• (A) 196
• (B) 194
• (C) 206
• (D) 182

Show Answer

 $$\eqalign{ & P = 7 + 4\sqrt 3 \cr & PQ = 1,\,Q = \frac{1}{P} \cr & Q = \frac{1}{{7 + 4\sqrt 3 }} = 7 - 4\sqrt 3 \cr & P + Q \cr & = 7 + 4\sqrt 3 + 7 - 4\sqrt 3 \cr & = 14 \cr & \Rightarrow \frac{1}{{{P^2}}} + \frac{1}{{{Q^2}}} \cr & = \frac{{{P^2} + {Q^2}}}{{{{\left( {PQ} \right)}^2}}} \cr & = \frac{{{{\left( {P + Q} \right)}^2} - 2PQ}}{{{{\left( {PQ} \right)}^2}}} \cr & = {\left( {14} \right)^2} - 2 \cr & = 196 - 2 \cr & = 194 \cr} $$

36.

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

• (A) 0
• (B) 5
• (C) 1
• (D) 4

Show Answer

 $$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & {\text{Put }} \cr & a = 0 \cr & b = 0 \cr & c = 2 \cr & d = 2 \cr & \therefore a + b + c + d = 4 \cr & \Rightarrow 0 + 0 + 2 + 2 = 4 \cr & \Rightarrow 4 = 4\left( {{\text{ satisfy}}} \right) \cr & \frac{1}{{\left( {1 - 0} \right)\left( {1 - 0} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 0} \right)\left( {1 - 2} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 2} \right)\left( {1 - 0} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 0} \right)\left( {1 - 0} \right)}} \cr & = \frac{1}{{ - 1}} + \left( {\frac{1}{{ + 1}}} \right) + \frac{1}{{ - 1 \times - 1}} + \frac{1}{{ - 1}} \cr & = - 1 + 1 + 1 - 1 \cr & = 0 \cr} $$

37.

If x = 3 + 2$$\sqrt 2 $$ and xy = 1, then the value of $$\frac{{{x^2} + 3xy + {y^2}}}{{{x^2} - 3xy + {y^2}}}$$   is?

• (A) $$\frac{{30}}{{31}}$$
• (B) $$\frac{{70}}{{31}}$$
• (C) $$\frac{{35}}{{31}}$$
• (D) $$\frac{{37}}{{31}}$$

Show Answer

 $$\eqalign{ & x = 3 + 2\sqrt 2 {\text{ and }}xy = 1 \cr & {y^2} = \frac{1}{{{x^2}}} \cr & y = \frac{1}{x} = \frac{1}{{3 + 2\sqrt 2 }} = 3 - 2\sqrt 2 \cr & \therefore x + \frac{1}{x} = 3 + 2\sqrt 2 + 3 - 2\sqrt 2 = 6 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} = 36 - 2 = 34 \cr & \frac{{{x^2} + 3xy + {y^2}}}{{{x^2} - 3xy + {y^2}}} \cr & = \frac{{{x^2} + \frac{1}{{{x^2}}} + 3}}{{{x^2} + \frac{1}{{{x^2}}} - 3}} \cr & = \frac{{34 + 3}}{{34 - 3}} \cr & = \frac{{37}}{{31}} \cr} $$

38.

The lines 2x + y = 5 and x + 2y = 4 intersect at the point?

• (A) (1, 2)
• (B) (2, 1)
• (C) (2, 0)
• (D) (0, 2)

Show Answer

 $$\eqalign{ & 2x + y = 5..............(i) \cr & x + 2y = 4..............(ii) \cr & {\text{Multiply equation (ii) by 2}} \cr & 2x + 4y = 8............(iii) \cr} $$ Now subtracting equation (i) from (iii) $$\eqalign{ & {\text{ }}2x + 4y = 8 \cr & \mathop {}\limits_ - 2x\mathop + \limits_ - \,\,y\, = 5 \cr & \overline {\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3y = 3{\text{ }}} } \cr & y = 1 \cr & x = 2 \cr & \therefore {\text{Intersection point = }}\left( {2,1} \right) \cr} $$

39.

$${\text{If }}\frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }} = a + b\sqrt 6 {\text{,}}$$       then the value of a and b are respectively?

• (A) $$\frac{9}{{15}}, - \frac{4}{{15}}$$
• (B) $$\frac{3}{{11}},\frac{4}{{33}}$$
• (C) $$\frac{9}{{10}},\frac{2}{5}$$
• (D) $$\frac{3}{5},\frac{4}{{15}}$$

Show Answer

 $$\eqalign{ & \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {16 \times 3} + \sqrt {9 \times 2} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} \times \frac{{4\sqrt 3 - 3\sqrt 2 }}{{4\sqrt 3 - 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{\left( {4\sqrt 3 + 5\sqrt 2 } \right)\left( {4\sqrt 3 - 3\sqrt 2 } \right)}}{{48 - 18}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 + 18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 }}{{30}} + \frac{{18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{4}{{15}}\sqrt 6 + \frac{3}{5} = a + b\sqrt 6 \cr & \Rightarrow \frac{3}{5} + \frac{4}{{15}}\sqrt 6 = a + b\sqrt 6 \cr} $$ By comparing coefficients of rational and irrational parts. $$\eqalign{ & \Rightarrow a = \frac{3}{5}{\text{ , }}b = \frac{4}{{15}} \cr & \therefore \left( {\frac{3}{5},\frac{4}{{15}}} \right) \cr} $$

40.

If $$\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}}$$   then the value of $$\frac{{x - 1}}{x}$$  is?

• (A) -2
• (B) 0
• (C) 1
• (D) -1

Show Answer

 $$\eqalign{ & {x^2} + \frac{{{\text{ }}1}}{{{x^2}}} = 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} + 2.x.\frac{1}{x} = 2 \cr & \Rightarrow x - \frac{1}{x} = 2 - 2 \cr & \Rightarrow x - \frac{1}{x} = 0 \cr} $$