Algebra Questions and Answers

Answer: D

The value of \((p + q + r)\) is 30.

Add A, B and C term by term, then subtract D:

\[ \begin{aligned} A + B + C &= -3x^2 + 4x + 18 \\ (A + B + C) - D &= -4x^2 + 5x + 29 \end{aligned} \]

So p = -4, q = 5, r = 29, and p + q + r = 30. Forgetting to change the signs of D's terms when subtracting gives \(-2x^2 + 3x + 7\), a total of 8, and none of 27, 28 or 31 follows from these polynomials.

Answer: A

The coefficient of xy is -6.

Only the cross terms of each product contain xy:

\[ \begin{aligned} (5x - 3y)(6x + 5y) &: 25xy - 18xy = 7xy \\ (3x + 7y)(4x - 5y) &: -15xy + 28xy = 13xy \\ 7xy - 13xy &= -6xy \end{aligned} \]

The second product is subtracted, so its xy term changes sign. Adding the two products instead would give 20, and the option 13 is the second product's coefficient on its own; forgetting the subtraction is the usual trap here.

Answer: C

One factor is 7x - 2y.

The first three terms form a perfect square, so the expression is a difference of two squares:

\[ \begin{aligned} &(9x - 7y)^{2} - (5x + 3y)^{2} \\ &= (9x - 7y - 5x - 3y)(9x - 7y + 5x + 3y) \\ &= (4x - 10y)(14x - 4y) \\ &= 4(2x - 5y)(7x - 2y) \end{aligned} \]

So the factors are 2x - 5y and 7x - 2y. Options A and B have the right numbers with the wrong sign (7x + 2y, 2x + 5y), and 5x - 7y does not appear at all.

Answer: B

The value is 3/2.

Simplify the bracket first, then clear the fractions:

\[ \begin{aligned} 2x - \frac{1 - x}{2} &= \frac{5x - 1}{2} \\ \frac{7x - 1}{4} - \frac{5x - 1}{14} &= 4 \\ 7(7x - 1) - 2(5x - 1) &= 112 \\ 39x - 5 &= 112,\ x = 3 \end{aligned} \]

Then

\[ \frac{5x - 3}{3x - 1} = \frac{12}{8} = \frac{3}{2} \]

Check: with x = 3 the left side is 20/4 - (1/7)(6 + 1) = 5 - 1 = 4. Since x is fixed, the expression has only one value, so 1, 7/5 and 11/8 cannot be right.

Answer: A

She has 18 and 7 coins of ₹ 5 and ₹ 10 respectively.

Let x be the number of ₹ 5 coins; then 25 - x are ₹ 10 coins:

\[ \begin{aligned} 5x + 10(25 - x) &= 160 \\ 250 - 5x &= 160 \\ x &= 18 \end{aligned} \]

Check: 18 × 5 + 7 × 10 = 90 + 70 = 160. The other pairs also total 25 coins but give the wrong amount: 10 and 15 make 200, 15 and 10 make 175, and 20 and 5 make 150.

Answer: D

The value of \(x^{2} - 2x + 1\) is \(\frac{25}{9}\).

Solve the equation first:

\[ \begin{aligned} 15x + 20 - 48x - 56 &= 9x - 8 \\ -33x - 36 &= 9x - 8 \\ -42x &= 28 \\ x &= -\tfrac{2}{3} \end{aligned} \]

Then \(x^{2} - 2x + 1 = (x - 1)^{2} = \left(-\tfrac{5}{3}\right)^{2} = \tfrac{25}{9}\).

\(\tfrac{2}{3}\) is the magnitude of x itself, \(\tfrac{4}{9}\) is \(x^{2}\), and \(\tfrac{5}{3}\) is \(|x - 1|\) before squaring.

Answer: B

The value is 154.

With a = 1, b = −3, c = −2, note that \(b^{2} = 9\) (the square of a negative number is positive):

\[ \begin{aligned} a(a + b^{2} + c) &= 1(1 + 9 - 2) = 8 \\ b^{2}(a^{2} + b^{2} + c^{2}) &= 9(1 + 9 + 4) = 126 \\ -c(a + b^{2}) &= 2(1 + 9) = 20 \end{aligned} \]

The total is 8 + 126 + 20 = 154. Treating the last term as \(-20\) (forgetting that \(-c = +2\)) gives 114, and 138 and 176 come from other sign slips.

Answer: C

The expression equals \(x^{3} + y^{3} + xy(y - x)\).

Use the identities \((x - y)(x^{2} + xy + y^{2}) = x^{3} - y^{3}\) and \((x + y)(x^{2} - xy + y^{2}) = x^{3} + y^{3}\):

\[ \begin{aligned} &(x^{3} - y^{3}) + (x^{3} + y^{3}) - (x + y)(x^{2} - y^{2}) \\ &= 2x^{3} - (x^{3} + x^{2}y - xy^{2} - y^{3}) \\ &= x^{3} + y^{3} - x^{2}y + xy^{2} \\ &= x^{3} + y^{3} + xy(y - x) \end{aligned} \]

The option with \(xy(x - y)\) has the sign of the last term reversed. Check with x = 2, y = 1: the expression is 7 + 9 − 9 = 7, and \(8 + 1 + 2(1 - 2) = 7\).

Answer: A

The value is \(-\dfrac{7}{4}\).

First find \(x\) by cross-multiplying:

\[ \begin{aligned} 7(5 - 7x) &= -40(8 + 3x) \\ 35 - 49x &= -320 - 120x \\ 71x &= -355 \\ x &= -5 \end{aligned} \]

Then substitute:

\[ \frac{2x + 3}{x + 9} = \frac{-10 + 3}{-5 + 9} = -\frac{7}{4} \]

Check: with \(x = -5\), \(\frac{5 + 35}{8 - 15} = \frac{40}{-7}\), as given. Once \(x\) is fixed the fraction has only one value, so the other three options cannot be right; a common slip is losing the minus sign on \(\frac{40}{7}\), which gives a different \(x\).

Answer: A

The factorisation is (3x - 2y + 3)(3x - 2y + 4).

Group the terms around \(t = 3x - 2y\):

\[ \begin{aligned} 9x^{2} - 12xy + 4y^{2} &= (3x - 2y)^{2} = t^{2} \\ 21x - 14y &= 7(3x - 2y) = 7t \end{aligned} \]

So the expression is a quadratic in \(t\):

\[ \begin{aligned} t^{2} + 7t + 12 &= (t + 3)(t + 4) \\ &= (3x - 2y + 3)(3x - 2y + 4) \end{aligned} \]

Options with \(3x + 2y\) would give \(+12xy\) and \(+14y\), the wrong signs. (t + 2)(t + 6) multiplies out to \(t^{2} + 8t + 12\), not \(t^{2} + 7t + 12\).