OMPT-B mixed paper

Answer: $\frac{5}{8}$

Write both over the common denominator $8$, combine the numerators, then cancel: $\frac{5}{8}$.

Common mistakes, and the answer each one gives:

• Added (or subtracted) numerators and denominators separately → $\frac{1}{3}$

Answer: $\frac{25}{9}$

A negative exponent turns the fraction over, and the fraction $\frac{2}{3}$ means the cube root raised to the power $2$: $\left(\dfrac{125}{27}\right)^{2/3} = \frac{25}{9}$.

Common mistakes, and the answer each one gives:

• Ignored the minus sign, so the fraction was never turned over → $\frac{9}{25}$
• Took the root but never applied the power → $\frac{5}{3}$
• Turned the fraction over and stopped there → $\frac{125}{27}$

Answer: $\frac{2}{3}$

The bars mean the inside is $12$ away from zero in EITHER direction, so there are two equations: $6 x + 8 = 12$ and $6 x + 8 = -12$. They give $\frac{2}{3}$ and $- \frac{10}{3}$; the larger is $\frac{2}{3}$.

Common mistakes, and the answer each one gives:

• Gave the smaller of the two solutions → $- \frac{10}{3}$
• Solved the positive branch, then negated the answer instead of solving the second branch → $- \frac{2}{3}$

Answer: $\frac{46}{105}$

Multiplication comes before addition, so work out $\dfrac{1}{3} \times \dfrac{5}{7} = \frac{5}{21}$ first, then add $\dfrac{1}{5}$: $\frac{46}{105}$.

Common mistakes, and the answer each one gives:

• Worked strictly left to right, adding before multiplying → $\frac{8}{21}$

Answer: $\frac{18 \left(x + 4\right)}{\left(x + 3\right) \left(x + 5\right)}$

The common denominator is $(x+5)(x+3)$. $\dfrac{9(x+3) + 9(x+5)}{(x+5)(x+3)} = \frac{18 \left(x + 4\right)}{\left(x + 3\right) \left(x + 5\right)}$.

Common mistakes, and the answer each one gives:

• Added numerators and denominators separately → $\frac{18}{2 x + 8}$
• Used $(x+5+3)$ as the common denominator → $\frac{18}{x + 8}$

Answer: $25 x^{2} - 16$

$(p \pm q)^2 = p^2 \pm 2pq + q^2$ and $(p+q)(p-q) = p^2 - q^2$. Here that gives $25 x^{2} - 16$.

Common mistakes, and the answer each one gives:

• Kept a middle term; in this product the two cancel → $25 x^{2} + 40 x - 16$
• Added the squares instead of subtracting → $25 x^{2} + 16$
• Squared only the first term → $25 x^{2} - 4$

Answer: $\left(x + 2\right) \left(x + 8\right)$

Find two numbers whose PRODUCT is $16$ and whose SUM is $-10$: they are $-8$ and $-2$. So the factorisation is $\left(x + 2\right) \left(x + 8\right)$.

Common mistakes, and the answer each one gives:

• Used the numbers with their signs unchanged in the brackets → $x^{2} - 10 x + 16$
• Matched the sum to the constant and the product to the middle term → $x^{2} + 9 x - 10$

Answer: $x^{6}$

Multiplying adds the exponents and dividing subtracts them: $3 + 8 - 5 = 6$, so the answer is $x^{6}$.

Common mistakes, and the answer each one gives:

• Multiplied the exponents instead of adding them → $x^{19}$
• Divided the exponents instead of subtracting → $x^{\frac{11}{5}}$

Answer: $\frac{x - 3}{x - 5}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x - 3}{x + 6} \times \dfrac{x + 6}{x - 5}$. The $x + 6$ then cancels, leaving $\frac{x - 3}{x - 5}$.

Common mistakes, and the answer each one gives:

• Multiplied straight across without flipping the second fraction → $\frac{x^{2} - 8 x + 15}{x^{2} + 12 x + 36}$
• Flipped the FIRST fraction instead of the second → $\frac{x^{2} + 12 x + 36}{x^{2} - 8 x + 15}$

Answer: $\frac{7}{2}$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{3 - (-4)}{-2 - (-4)} = \frac{7}{2}$.

Common mistakes, and the answer each one gives:

• Inverted the fraction (run over rise) → $\frac{2}{7}$
• Subtracted the coordinates in opposite orders → $- \frac{7}{2}$

Answer: $-2$

Expand the bracket: $3 x - 15 = 7 x - 7$. Collect the $x$ terms on one side and the numbers on the other, then divide: $x = -2$.

Common mistakes, and the answer each one gives:

• Multiplied only the $x$ by the number outside the bracket → $\frac{1}{2}$
• Moved a term across without changing its sign → $- \frac{29}{4}$

Answer: $\frac{17}{2}$

Cross-multiply to clear both denominators: $6\left(x - 7\right) = 2\left(x - 4\right)$. Expand, collect the $x$ terms and divide: $x = \frac{17}{2}$.

Common mistakes, and the answer each one gives:

• Multiplied only the numerators by the other denominator → $\frac{19}{2}$
• Added the fractions instead of cross-multiplying → $\frac{25}{4}$

Answer: $0$

Eliminate $y$ by scaling and adding, or substitute. The solution is $x = 0$, $y = 5$.

Common mistakes, and the answer each one gives:

• Solved for y instead of x → $5$

Answer: $4$

The first equation already gives $y$, so put it straight into the second: $2x + 5\left(6 - 3 x\right) = -22$. That leaves one unknown, and $x = 4$.

Common mistakes, and the answer each one gives:

• Gave the value of $y$ rather than of $x$ → $-6$
• Substituted into the equation it came from, which says nothing → $9$

Answer: $5$

$\left(x - 5\right) \left(x + 1\right) = 0$, so $x = 5$ or $x = -1$. The larger is $5$.

Common mistakes, and the answer each one gives:

• Gave the smaller root → $-1$
• Read the roots off with the signs unchanged → $1$

Answer: $- \frac{5}{6} + \frac{\sqrt{133}}{6}$

The discriminant is $b^2 - 4ac = 133$, which is positive but not a perfect square, so the roots are irrational and the formula is needed: $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$. The larger root is $- \frac{5}{6} + \frac{\sqrt{133}}{6}$.

Common mistakes, and the answer each one gives:

• Used $+b$ in the numerator instead of $-b$ → $\frac{5}{6} + \frac{\sqrt{133}}{6}$
• Divided by $a$ instead of $2a$ → $- \frac{5}{3} + \frac{\sqrt{133}}{3}$
• Gave the smaller root → $- \frac{\sqrt{133}}{6} - \frac{5}{6}$

Answer: $6$

The turning point of $ax^2+bx+c$ sits at $x = -\dfrac{b}{2a}$. Here $a = -2$ and $b = 24$, so $x = -\dfrac{24}{2(-2)} = 6$. (Completing the square gives $-2\left(x - 6\right)^2$, which shows the same thing.)

Common mistakes, and the answer each one gives:

• Used $\dfrac{b}{2a}$ without the minus sign → $-6$
• Gave the MAXIMUM VALUE instead of where it happens → $0$

Answer: $7$

Where the graphs meet, the two expressions are equal. Bring everything to one side: $x^{2} - 6 x - 7 = 0$, which factorises as $\left(x - 7\right) \left(x + 1\right)$. The roots are $-1$ and $7$.

Common mistakes, and the answer each one gives:

• Gave the smaller root → $-1$
• Read the roots straight off the brackets with the signs unchanged → $1$

Answer: $-8$

The vertical asymptote sits where the denominator is zero: $x + 8 = 0$, so $x = -8$. (The horizontal one is $y = -2$, the ratio of the leading coefficients.)

Common mistakes, and the answer each one gives:

• Set the NUMERATOR to zero (that gives the x-intercept) → $1$
• Gave the horizontal asymptote instead → $-2$
• Forgot the sign when solving → $8$

Answer: $\frac{x - 1}{x + 6}$

Factorise both: $\dfrac{\left(x - 6\right) \left(x - 1\right)}{\left(x - 6\right) \left(x + 6\right)}$. The factor $x - 6$ is common to both and cancels, leaving $\frac{x - 1}{x + 6}$.

Common mistakes, and the answer each one gives:

• Cancelled the $x^2$ terms, which are terms and not factors → $\frac{7 x}{36} - \frac{1}{6}$
• Cancelled the wrong bracket → $\frac{x - 6}{x + 6}$

Answer: $45 x^{2} - 120 x + 84$

Square the bracket FIRST, then multiply by $5$, then add the constant — the order matters. $\left(3 x - 4\right)^2 = 9 x^{2} - 24 x + 16$, and the whole thing comes to $45 x^{2} - 120 x + 84$.

Common mistakes, and the answer each one gives:

• Multiplied the bracket by the outside number before squaring → $225 x^{2} - 600 x + 404$
• Squared each term inside the bracket separately → $45 x^{2} + 84$

Answer: $\frac{48}{5}$

Multiply both sides by $x - 6$: $3 x = 8\left(x - 6\right)$. Expand, collect and divide: $x = \frac{48}{5}$. (Check it is not $x = 6$, which the original forbids.)

Common mistakes, and the answer each one gives:

• Multiplied only the numerator by the right-hand side → $\frac{8}{3}$
• Set the denominator to zero, which is the EXCLUDED value → $6$

Answer: $-5$

A cube is undone by a cube root, and unlike a square root it keeps the sign: $\sqrt[3]{-64} = -4$. So $x + 1 = -4$ and $x = -5$.

Common mistakes, and the answer each one gives:

• Divided by 3 instead of taking the cube root → $- \frac{67}{3}$
• Took the cube root but lost the sign → $3$

Answer: $11 x^{2}$

The power rule turns $2\log_{10}(x)$ into $\log_{10}(x^{2})$, and the product rule combines the two into $\log_{10}(11 x^{2})$.

Common mistakes, and the answer each one gives:

• Added the arguments instead of multiplying them → $x^{2} + 11$
• Multiplied by the power instead of raising to it → $22 x$

Answer: $6$

Equal logarithms with the same base have equal arguments, so $x + 2 = 8$ and $x = 6$. Check the domain: $x + 2 > 0$ holds.

Common mistakes, and the answer each one gives:

• Subtracted the shift from the wrong side → $10$
• Treated the log as a multiplier and divided → $4$

Answer: $1$

Write the right-hand side as a power of $2$: $2 = 2^{1}$. With the same base on both sides the exponents must be equal, so $4 x - 3 = 1$ and $x = 1$.

Common mistakes, and the answer each one gives:

• Set the exponent equal to 2 instead of to 1 → $\frac{5}{4}$
• Forgot to move the constant in the exponent across → $\frac{1}{4}$

Answer: $\frac{4}{3}$

Write both numbers as powers of $2$: $8 = 2^{3}$ and $16 = 2^{4}$. Then $\log_{8}16 = \dfrac{4}{3} = \frac{4}{3}$, because the base has to be raised to that power to reach the argument.

Common mistakes, and the answer each one gives:

• Divided the two numbers → $2$
• Turned the fraction of exponents upside down → $\frac{3}{4}$

Answer: $12 x \left(2 x^{2} + 1\right)^{2}$

Outer power 3, inner $2x^2 + 1$, so $f' = 3(2x^2+1)^{2}\cdot 4x$. This gives $f'(x) = 12 x \left(2 x^{2} + 1\right)^{2}$.

Common mistakes, and the answer each one gives:

• Differentiated each part separately and combined them → $3 \left(2 x^{2} + 1\right)^{2}$

Answer: $-4$

$f'(x) = 9 x^{2} - 144 = 0$ at $x = -4$ and $x = 4$. $f''(x) = 18 x$, and $f''(-4) = -72 < 0$, so $x = -4$ is the maximum.

Common mistakes, and the answer each one gives:

• Gave the local MINIMUM instead → $4$
• Gave the y-value rather than the x-coordinate → $384$
• Solved f(x) = 0 instead of f'(x) = 0 → $0$

Answer: $\frac{- 4 x^{2} + 4 x + 4}{x^{4} + 2 x^{2} + 1}$

The quotient rule is $\dfrac{u'v - uv'}{v^2}$ — and the ORDER of the two products matters, unlike the product rule. With $u = 4 x - 2$ and $v = x^{2} + 1$: $u' = 4$ and $v' = 2x$, giving $\frac{- 4 x^{2} + 4 x + 4}{x^{4} + 2 x^{2} + 1}$.

Common mistakes, and the answer each one gives:

• Wrote $uv' - u'v$ in the numerator, the two products swapped → $\frac{4 x^{2} - 4 x - 4}{x^{4} + 2 x^{2} + 1}$
• Differentiated top and bottom separately → $\frac{2}{x}$