Algebra Questions and Answers

Answer: C

The solution set is shown in (C): all x from a point just to the right of 0 onwards.

Multiply both sides by 3 and collect terms:

\[ \begin{aligned} 3(2 - 5x) &\le -(6x - 5) \\ 6 - 15x &\le -6x + 5 \\ 1 &\le 9x \\ x &\ge \frac{1}{9} \end{aligned} \]

So the bold part starts at a filled dot at the small positive value 1/9 and runs to the right, which is line (C).

Line (A) shows x at most a positive value, the inequality reversed. Line (B) starts the set at a negative value, which no correct rearrangement gives, since 1/9 is positive. Line (D) shows two separate rays, the shape of a quadratic inequality, not a linear one.

Answer: A

The correct option is f(2,2) = g(2,2).

f(p,q) is p raised to the power q, and g(p,q) is a tower of q copies of p. With two terms each:

\[ \begin{aligned} f(2,2) &= 2^{2} = 4 \\ g(2,2) &= 2^{2} = 4 \end{aligned} \]

The other options are false. f(g(2,2),2) = f(4,2) = 16 and f(2,g(2,2)) = f(2,4) = 16, so neither is less than the other. g(2,1) = 2 = f(2,1) by definition, so they are equal. f(3,2) = 9 while g(3,2) = \(3^{3}\) = 27, so f(3,2) is smaller, not greater.

Answer: C

The larger number is 7.

Let the numbers be x and y:

\[ \begin{aligned} 2x + 3y &= 25 \\ 3x + 2y &= 20 \end{aligned} \]

Adding gives \(5x + 5y = 45\), so \(x + y = 9\); subtracting the first from the second gives \(x - y = -5\). Hence \(x = 2\) and \(y = 7\).

Check: \(4 + 21 = 25\) and \(6 + 14 = 20\). The larger of 2 and 7 is 7; 5 is their difference, and 10 and 8 do not satisfy the equations.

Answer: D

The tank holds 35 buckets.

Let the tank hold T buckets. Taking out 22 and pouring in 11 is a net removal of 11 buckets, which lowered the level from \(\frac{5}{7}\) to \(\frac{2}{5}\):

\[ \begin{aligned} \tfrac{5}{7}T - \tfrac{2}{5}T &= 11 \\ \tfrac{25 - 14}{35}T &= 11 \\ T &= 35 \end{aligned} \]

Check: \(\frac{5}{7}\) of 35 is 25, and \(25 - 11 = 14\), which is \(\frac{2}{5}\) of 35. The other capacities do not divide evenly by both 7 and 5.

Answer: C

The value of (a - b) is -9.

Expand the right-hand side and compare coefficients:

\[ \begin{aligned} (x + a)(bx + 2y) &= bx^{2} + 2xy + abx + 2ay \\ b &= 5,\quad 2a = -8 \Rightarrow a = -4 \\ ab &= -20\ \text{(checks with } -20x) \\ a - b &= -4 - 5 = -9 \end{aligned} \]

9 is b - a, the subtraction done the wrong way round, and -1 and 1 come from taking a = 4 or mixing up the signs.

Answer: D

The value of (5p + q - r) is 8.

\[ \begin{aligned} A &= x^{2} - 3x^{2}y^{2} \\ B &= x^{2}y^{2} - 2x^{2} \\ C &= 3y^{2} - 12x^{2}y^{2} \\ A + B - C &= -x^{2} + 10x^{2}y^{2} - 3y^{2} \end{aligned} \]

So p = -1, q = 10, r = -3, and 5p + q - r = -5 + 10 + 3 = 8.

10 is q alone. 12 and 18 come from sign slips; the step that most often goes wrong is subtracting C, which turns its -12x²y² into +12x²y² and its 3y² into -3y².

Answer: C

The equation with the same solution is 3 (2x + 3) = 5 (x + 1).

Solve the given equation by cross-multiplying:

\[ \begin{aligned} 23(-x - 3) &= 3 - 5x \\ -23x - 69 &= 3 - 5x \\ -18x &= 72 \Rightarrow x = -4 \end{aligned} \]

Substitute x = -4 in each option: 3(-8 + 3) = -15 and 5(-4 + 1) = -15, so option 3 holds. The others fail: 5(-7) = -35 against 3(-3) = -9; 3(-5) = -15 against 2(-1) = -2; and 2(-11) = -22 against 3(-3) = -9.

Answer: D

The value is \(\frac{5}{8}\).

Multiply the equation by 21 to clear the fractions:

\[ \begin{aligned} 3(x - 4) - 21 &= 7(5 - x) + 21x \\ 3x - 33 &= 35 + 14x \\ -11x &= 68 \end{aligned} \]

So 11k = -68, and

\[ \begin{aligned} \frac{11k + 18}{11k - 12} &= \frac{-68 + 18}{-68 - 12} \\ &= \frac{-50}{-80} = \frac{5}{8} \end{aligned} \]

Answer: D

The ascending order will now be changed to descending order.

Multiplying by a negative number reverses an inequality. The numbers become -1, -3.5, -4.5 and -6, so

\[ -1 \gt -3.5 \gt -4.5 \gt -6 \]

which is descending order. The number of terms stays four, the ascending order does not survive, and the mean changes from 7.5 to -3.75, so the other three statements are false.

Answer: C

One factor is 2q - 3r.

Take out p, then group the terms:

\[ \begin{aligned} &p(20q^2 - 60qr + 45r^2 + 4q - 6r) \\ &= p[5(2q - 3r)^2 + 2(2q - 3r)] \\ &= p(2q - 3r)(10q - 15r + 2) \end{aligned} \]

So the factors are p, 2q - 3r and 10q - 15r + 2. Option 2, 2r - 3q, swaps the letters, and options 1 and 4 have p in place of q.