Mathbench

OMPT-G mixed paper

30 questions drawn from every sheet on the OMPT-G syllabus, and only from the topics OMPT-G itself examines.

All OMPT worksheets


Questions

Sheet length
1

Give the equation of the line through (−1,0)(-1, 0) and (−2,−2),(-2, -2)\mathclose{,} in the form y=mx+c.y = mx + c\mathclose{.}

Answer:

2

Solve for x:x\mathclose{:} x7=x+63\dfrac{x}{7} = \dfrac{x + 6}{3}

Answer:

3

Solve: 6x+7≥23−2x6 x + 7 \ge 23 - 2 x

Answer:

4

Solve the system and give the value of x:x\mathclose{:} 5x+y=315 x + y = 31 and 5x−2y=135 x - 2 y = 13

Answer:

5

The system x−3y=0x - 3 y = 0 and 3x−9y=03 x - 9 y = 0 has infinitely many solutions. Write yy in terms of x.x\mathclose{.}

Answer:

6

What is the maximum value of f(x)=−x2+4x+1?f(x) = - x^{2} + 4 x + 1\mathclose{?}

Answer:

7

Solve x2+7x+12=0x^{2} + 7 x + 12 = 0 and give the LARGER solution.

Answer:

8

Solve exactly and give the LARGER solution: 3x2+4x−1=03 x^{2} + 4 x - 1 = 0

Answer:

9

Give the equation of the VERTICAL asymptote of f(x)=x+6x+2.f(x) = \dfrac{x + 6}{x + 2}\mathclose{.}

Answer:

10

Solve for x:x\mathclose{:} 4xx+7=3\dfrac{4 x}{x + 7} = 3

Answer:

11

Divide x3−3x2−x+8x^{3} - 3 x^{2} - x + 8 by x−2.x - 2\mathclose{.} Give the remainder.

Answer:

12

The graph of y=x3y = x^{3} is shifted 1 unit to the right and 5 units up. Give the equation of the new graph.

Answer:

13

Give the inverse function f−1(x)f^{-1}(x) of f(x)=x−5+1,f(x) = \sqrt{x - 5} + 1\mathclose{,} and the values of xx it is defined for.

Answer:

14

What is the degree of the polynomial (x2−5)2(2x3−2)2?\left(x^{2} - 5\right)^{2}\left(2 x^{3} - 2\right)^{2}\mathclose{?}

Answer:

15

Write as a single logarithm: 3log⁡10(x)−log⁡10(4).3\log_{10}(x) - \log_{10}(4)\mathclose{.} Give only the ARGUMENT of the resulting logarithm.

Answer:

16

Evaluate exactly: log⁡9(729)\log_{9}\left(729\right)

Answer:

17

Two angles of a triangle are 96∘96^\circ and 27∘.27^\circ\mathclose{.} How many degrees is the third?

Answer:

18

Give the exact value: cos⁡(2π3)\cos\left(\frac{2\pi}{3}\right)

Answer:

19

In triangle ABC,ABC\mathclose{,} a=7,a = 7\mathclose{,} b=8b = 8 and the angle between them is C=120∘.C = 120^\circ\mathclose{.} Calculate the length c.c\mathclose{.}

Answer:

20

Give the amplitude of f(x)=−5cos⁡(3x)+2.f(x) = -5\cos\left(3 x\right) + 2\mathclose{.}

Answer:

21

Give the exact value of arcsin⁡(sin⁡(−2π3)).\arcsin\left(\sin\left(-\frac{2\pi}{3}\right)\right)\mathclose{.}

Answer:

22

Solve 6sin⁡(x)=−36\sin(x) = -3 for 0≤x<2π.0 \le x < 2\pi\mathclose{.} Give ALL the solutions.

Answer:

23

Solve 4sin⁡(x)=−224\sin(x) = - 2 \sqrt{2} for 0≤x<2π.0 \le x < 2\pi\mathclose{.} Give ALL the solutions.

Answer:

24

An isosceles triangle has a top angle of 100∘100^\circ between its two equal sides. How many degrees is each of the other two angles?

Answer:

25

A right-angled triangle has legs of length 77 and 24.24\mathclose{.} How long is its hypotenuse?

Answer:

26

How far apart are the points (7,2)(7, 2) and (11,5)?(11, 5)\mathclose{?}

Answer:

27

What angle does the line y=1−xy = 1 - x make with the positive xx-axis? Give it in degrees, between 0∘0^\circ and 180∘.180^\circ\mathclose{.}

Answer:

28

The circle (x+3)2+(y+5)2=100\left(x + 3\right)^{2} + \left(y + 5\right)^{2} = 100 passes through P(−11,1).P(-11, 1)\mathclose{.} Give the equation of the tangent to the circle at P,P\mathclose{,} in the form y=mx+c.y = mx + c\mathclose{.}

Answer:

29

The arithmetic sequence 5,1,−3,…5, 1, -3, \ldots continues. Which term of it is equal to −63?-63\mathclose{?}

Answer:

30

What is the 4th term of the geometric sequence 6,−12,24,…?6, -12, 24, \ldots\mathclose{?}

Answer:

Answers

Every answer below was re-derived independently before this page was built.

1

Answer: y=2x+2y = 2 x + 2

The slope is m=−2−0−2−(−1)=2.m = \dfrac{-2 - 0}{-2 - (-1)} = 2\mathclose{.} The point (−1,0)(-1, 0) lies on the line, so 0=2⋅(−1)+c,0 = 2 \cdot (-1) + c\mathclose{,} which gives c=2.c = 2\mathclose{.} The line is y=2x+2.y = 2 x + 2\mathclose{.}

Common mistakes, and the answer each one gives:

  • Inverted the slope (run over rise) → y=x2+12y = \frac{x}{2} + \frac{1}{2}
  • Found the slope but left out the intercept → y=2xy = 2 x
  • Got the sign of the intercept wrong → y=2x−2y = 2 x - 2
2

Answer: −212-\dfrac{21}{2}

Multiply both sides by 2121 to clear the denominators: 3x=7(x+6).3x = 7\left(x + 6\right)\mathclose{.} Expand: 3x=7x+42.3 x = 7 x + 42\mathclose{.} Collect: −4x=42,- 4 x = 42\mathclose{,} so x=−212.x = -\dfrac{21}{2}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Multiplied out 7(x+6)7\left(x + 6\right) as 7x+6,7 x + 6\mathclose{,} multiplying only the xx → −32- \frac{3}{2}
  • Cross-multiplied the wrong way, writing 7(x)=3(x+6)7\left(x\right) = 3\left(x + 6\right) → 92\frac{9}{2}
3

Answer: x≥2x \ge 2

Collect the xx terms on the left and the numbers on the right: 8x≥16.8 x \ge 16\mathclose{.} Divide by 8,8\mathclose{,} a positive number, so the sign stays: x≥2.x \ge 2\mathclose{.}

Common mistakes, and the answer each one gives:

  • Forgot to divide the number side by 88 → x≥16x \ge 16
  • Moved the number across without changing its sign → x≥154x \ge \frac{15}{4}
4

Answer: 55

Multiply the first by 2,2\mathclose{,} so the yy terms become 2y2 y and −2y,- 2 y\mathclose{,} then add: 15x=75,15 x = 75\mathclose{,} so x=5.x = 5\mathclose{.} Put that into the first equation: y+25=31,y + 25 = 31\mathclose{,} so y=6.y = 6\mathclose{.}

Common mistakes, and the answer each one gives:

  • Gave the value of yy instead of xx → 66
  • Multiplied only the left-hand sides when scaling the equations → 4415\frac{44}{15}
5

Answer: y=x3y = \frac{x}{3}

The second equation is 33 times the first, so only one condition remains: x−3y=0,x - 3 y = 0\mathclose{,} which gives y=x3y = \frac{x}{3} — one solution for every x.x\mathclose{.}

Common mistakes, and the answer each one gives:

  • Lost the sign of the xx term moving it across → y=−x3y = - \frac{x}{3}
6

Answer: 55

The turning point of ax2+bx+cax^2+bx+c sits at x=−b2a.x = -\dfrac{b}{2a}\mathclose{.} Here a=−1a = -1 and b=4,b = 4\mathclose{,} so x=−42⋅(−1)=2.x = -\dfrac{4}{2 \cdot (-1)} = 2\mathclose{.} The maximum value is f(2)=5.f(2) = 5\mathclose{.} (Completing the square gives f(x)=−(x−2)2+5,f(x) = -\left(x - 2\right)^{2} + 5\mathclose{,} which shows the same thing.)

Common mistakes, and the answer each one gives:

  • Gave the xx-coordinate of the turning point instead of the value → 22
  • Substituted x=b2a,x = \dfrac{b}{2a}\mathclose{,} without the minus sign → −11-11
7

Answer: −3-3

Factorise: x2+7x+12=(x+4)(x+3)=0,x^{2} + 7 x + 12 = \left(x + 4\right)\left(x + 3\right) = 0\mathclose{,} so x=−4x = -4 or x=−3.x = -3\mathclose{.} The larger is −3.-3\mathclose{.}

Common mistakes, and the answer each one gives:

  • Gave the smaller solution → −4-4
  • Read the solutions off the brackets with the signs unchanged → 44
8

Answer: −2+73\dfrac{-2 + \sqrt{7}}{3}

Here a=3,a = 3\mathclose{,} b=4b = 4 and c=−1,c = -1\mathclose{,} so the discriminant is b2−4ac=16−(−12)=28b^2 - 4ac = 16 - (-12) = 28 — positive but not a perfect square, so the formula is needed: x=−b±b2−4ac2a=−4±286.x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-4 \pm \sqrt{28}}{6}\mathclose{.} Since 28=27,\sqrt{28} = 2 \sqrt{7}\mathclose{,} that is −4±276=−2±73.\dfrac{-4 \pm 2 \sqrt{7}}{6} = \dfrac{-2 \pm \sqrt{7}}{3}\mathclose{.} The larger solution is −2+73.\dfrac{-2 + \sqrt{7}}{3}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used +b+b in the numerator instead of −b-b → 2+73\dfrac{2 + \sqrt{7}}{3}
  • Divided by aa instead of 2a2a → −4+273\dfrac{-4 + 2\sqrt{7}}{3}
  • Gave the smaller solution → −2−73\dfrac{-2 - \sqrt{7}}{3}
9

Answer: x=−2x = -2

The vertical asymptote sits where the denominator is zero: x+2=0,x + 2 = 0\mathclose{,} so x=−2.x = -2\mathclose{.} (The horizontal one is y=1,y = 1\mathclose{,} the ratio of the leading coefficients.)

Common mistakes, and the answer each one gives:

  • Set the NUMERATOR to zero (that gives the xx-intercept) → x=−6x = -6
  • Gave the horizontal asymptote instead → y=1y = 1
  • Lost the minus sign when solving the denominator → x=2x = 2
10

Answer: 2121

Multiply both sides by x+7:x + 7\mathclose{:} 4x=3(x+7),4 x = 3\left(x + 7\right)\mathclose{,} which is 4x=3x+21.4 x = 3 x + 21\mathclose{.} Collect: x=21.x = 21\mathclose{.} That is not x=−7,x = -7\mathclose{,} the value the equation forbids, so it stands.

Common mistakes, and the answer each one gives:

  • Ignored the denominator and solved 4x=34 x = 3 → 34\frac{3}{4}
  • Multiplied out 3(x+7)3\left(x + 7\right) as 3x+7,3 x + 7\mathclose{,} multiplying only the xx → 77
11

Answer: 22

x3÷x=x2;x^{3} \div x = x^{2}\mathclose{;} subtract x3−2x2,x^{3} - 2 x^{2}\mathclose{,} leaving −x2−x+8.- x^{2} - x + 8\mathclose{.} −x2÷x=−x;- x^{2} \div x = - x\mathclose{;} subtract −x2+2x,- x^{2} + 2 x\mathclose{,} leaving 8−3x.8 - 3 x\mathclose{.} −3x÷x=−3;- 3 x \div x = -3\mathclose{;} subtract 6−3x,6 - 3 x\mathclose{,} leaving 2:2\mathclose{:} the quotient is x2−x−3x^{2} - x - 3 and the remainder 2.2\mathclose{.} Check it with the remainder theorem: the remainder on dividing by x−2x - 2 is the value at x=2,x = 2\mathclose{,} and f(2)=2.f(2) = 2\mathclose{.}

Common mistakes, and the answer each one gives:

  • Put in x=−2x = -2 instead of x=2x = 2 → −10-10
  • Gave the constant term of the quotient → −3-3
12

Answer: y=(x−1)3+5y = \left(x - 1\right)^{3} + 5

Shifting 1 unit to the right replaces xx by x−1;x - 1\mathclose{;} shifting 5 units up adds 5:5\mathclose{:} y=(x−1)3+5.y = \left(x - 1\right)^{3} + 5\mathclose{.}

Common mistakes, and the answer each one gives:

  • Shifted the wrong way horizontally → y=(x+1)3+5y = \left(x + 1\right)^{3} + 5
  • Shifted the wrong way vertically → y=(x−1)3−5y = \left(x - 1\right)^{3} - 5
13

Answer: (x−1)2+5 for x≥1\left(x - 1\right)^{2} + 5 \text{ for } x \ge 1

Write y=x−5+1y = \sqrt{x - 5} + 1 and undo the steps in reverse order: take 11 off, y−1=x−5;y - 1 = \sqrt{x - 5}\mathclose{;} square, (y−1)2=x−5;\left(y - 1\right)^{2} = x - 5\mathclose{;} then x=(y−1)2+5.x = \left(y - 1\right)^{2} + 5\mathclose{.} Swap the letters: f−1(x)=(x−1)2+5.f^{-1}(x) = \left(x - 1\right)^{2} + 5\mathclose{.} A square root is never negative, so ff only takes values y≥1y \ge 1 — and those are the xx the inverse is defined for: x≥1.x \ge 1\mathclose{.}

Common mistakes, and the answer each one gives:

  • Squared both sides before taking the 11 off → x2+4x^{2} + 4
  • Undid the steps in the original order instead of the reverse order → (x+5)2−1\left(x + 5\right)^{2} - 1
14

Answer: 1010

The highest power in the product comes from the highest powers of the factors: (x2)2⋅(2x3)2\left(x^{2}\right)^{2} \cdot \left(2x^{3}\right)^{2} has degree 4+6=10.4 + 6 = 10\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added only the exponents outside the brackets → 44
  • Multiplied the degrees instead of adding them → 2424
15

Answer: x34\frac{x^{3}}{4}

The power rule turns 3log⁡10(x)3\log_{10}(x) into log⁡10(x3),\log_{10}(x^{3})\mathclose{,} and the quotient rule combines the two into log⁡10(x34).\log_{10}\left(\frac{x^{3}}{4}\right)\mathclose{.}

Common mistakes, and the answer each one gives:

  • Subtracted the arguments instead of dividing them → x3−4x^{3} - 4
  • Divided the wrong way round → 4x3\frac{4}{x^{3}}
  • Multiplied by the power instead of raising to it → 3x4\frac{3 x}{4}
16

Answer: 33

Write both numbers as powers of 3:3\mathclose{:} 9=329 = 3^{2} and 729=36.729 = 3^{6}\mathclose{.} Then log⁡9(729)=62=3,\log_{9}(729) = \dfrac{6}{2} = 3\mathclose{,} because 99 raised to that power is 36=729.3^{6} = 729\mathclose{.}

Common mistakes, and the answer each one gives:

  • Divided the two numbers → 8181
  • Turned the fraction of exponents upside down → 13\frac{1}{3}
17

Answer: 5757

The three angles of any triangle add to 180∘,180^\circ\mathclose{,} so the third is 180−96−27=57.180 - 96 - 27 = 57\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used 360∘,360^\circ\mathclose{,} the angles round a POINT → 237237
  • Subtracted only one of the two given angles → 8484
18

Answer: −12- \frac{1}{2}

2π3\frac{2\pi}{3} is a special angle, so the value is exact. Its reference value is 12,\frac{1}{2}\mathclose{,} and the quadrant fixes the sign: cos⁡(2π3)=−12.\cos\left(\frac{2\pi}{3}\right) = - \frac{1}{2}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used the right size but the wrong sign for that quadrant → 12\frac{1}{2}
  • Gave the sine instead → 32\frac{\sqrt{3}}{2}
19

Answer: 1313

The cosine rule: c2=a2+b2−2abcos⁡C,c^2 = a^2 + b^2 - 2ab\cos C\mathclose{,} and cos⁡120∘=−12.\cos 120^\circ = -\dfrac{1}{2}\mathclose{.} So c2=49+64+56=169,c^2 = 49 + 64 + 56 = 169\mathclose{,} and c=169=13.c = \sqrt{169} = 13\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used +2abcos⁡C+2ab\cos C instead of −2abcos⁡C-2ab\cos C → 57\sqrt{57}
  • Forgot to take the square root → 169169
20

Answer: 55

The amplitude is the size of the number in front, ∣−5∣=5:|-5| = 5\mathclose{:} the graph swings 55 above and below its middle line.

Common mistakes, and the answer each one gives:

  • Gave the vertical shift instead → 22
  • Gave the coefficient of xx instead → 33
  • Kept the minus sign; an amplitude is a distance → −5-5
21

Answer: −π3-\frac{\pi}{3}

sin⁡(−2π3)=−32,\sin\left(-\frac{2\pi}{3}\right) = - \frac{\sqrt{3}}{2}\mathclose{,} and arcsin⁡\arcsin gives back the ONE angle in [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] with that value: −π3.-\frac{\pi}{3}\mathclose{.} That is not −2π3,-\frac{2\pi}{3}\mathclose{,} which lies outside the range.

Common mistakes, and the answer each one gives:

  • Cancelled arcsin⁡\arcsin against sin⁡\sin → −2π3-\frac{2\pi}{3}
  • Gave the value of sin⁡,\sin\mathclose{,} not an angle → −32- \frac{\sqrt{3}}{2}
22

Answer: x=7π6 or x=11π6x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}

Divide by 6:6\mathclose{:} sin⁡(x)=−12.\sin(x) = - \frac{1}{2}\mathclose{.} One angle is α=arcsin⁡(−12)=−π6;\alpha = \arcsin\left(- \frac{1}{2}\right) = -\frac{\pi}{6}\mathclose{;} sine is also equal at π−α,\pi - \alpha\mathclose{,} and both are brought into [0,2π).[0, 2\pi)\mathclose{.} So x=7π6 or x=11π6.x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Took the second solution as π+α\pi + \alpha instead of π−α\pi - \alpha → x=5π6 or x=11π6x = \frac{5\pi}{6} \text{ or } x = \frac{11\pi}{6}
  • Gave only the first solution → x=7π6x = \frac{7\pi}{6}
23

Answer: x=5π4 or x=7π4x = \frac{5\pi}{4} \text{ or } x = \frac{7\pi}{4}

Divide by 4:4\mathclose{:} sin⁡(x)=−22.\sin(x) = - \frac{\sqrt{2}}{2}\mathclose{.} One angle is α=arcsin⁡(−22)=−π4;\alpha = \arcsin\left(- \frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}\mathclose{;} sine is also equal at π−α,\pi - \alpha\mathclose{,} and both are brought into [0,2π).[0, 2\pi)\mathclose{.} So x=5π4 or x=7π4.x = \frac{5\pi}{4} \text{ or } x = \frac{7\pi}{4}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Took the second solution as π+α\pi + \alpha instead of π−α\pi - \alpha → x=3π4 or x=7π4x = \frac{3\pi}{4} \text{ or } x = \frac{7\pi}{4}
  • Gave only the first solution → x=5π4x = \frac{5\pi}{4}
24

Answer: 4040

The other two angles are equal and share what is left of 180∘:180^\circ\mathclose{:} 180−1002=40.\dfrac{180 - 100}{2} = 40\mathclose{.}

Common mistakes, and the answer each one gives:

  • Did not halve what was left → 8080
  • Halved the top angle instead → 5050
25

Answer: 2525

Pythagoras: the square on the hypotenuse equals the sum of the squares on the legs. 72+242=625,7^2 + 24^2 = 625\mathclose{,} and 625=25.\sqrt{625} = 25\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the legs instead of their squares → 3131
  • Subtracted the squares, which finds a LEG, not the hypotenuse → 527\sqrt{527}
26

Answer: 55

The horizontal gap is ∣11−7∣=4|11 - 7| = 4 and the vertical gap is ∣5−2∣=3,|5 - 2| = 3\mathclose{,} so Pythagoras gives 42+32=25=5.\sqrt{4^2 + 3^2} = \sqrt{25} = 5\mathclose{.}

Common mistakes, and the answer each one gives:

  • Added the two gaps instead of using Pythagoras → 77
  • Forgot the square root → 2525
27

Answer: 135135

The slope is the tangent of that angle: tan⁡θ=−1,\tan\theta = -1\mathclose{,} so θ=135∘.\theta = 135^\circ\mathclose{.}

Common mistakes, and the answer each one gives:

  • Gave the angle in radians → 3π4\frac{3\pi}{4}
  • Gave the acute angle with the axis → 4545
28

Answer: y=4x3+473y = \frac{4 x}{3} + \frac{47}{3}

The radius to PP runs from the centre (−3,−5)(-3, -5) to (−11,1),(-11, 1)\mathclose{,} so its slope is 6−8=−34.\dfrac{6}{-8} = -\dfrac{3}{4}\mathclose{.} The tangent is perpendicular to it, with slope 43.\dfrac{4}{3}\mathclose{.} Through P:P\mathclose{:} y=43(x+11)+1,y = \dfrac{4}{3}\left(x + 11\right) + 1\mathclose{,} which is y=4x3+473.y = \frac{4 x}{3} + \frac{47}{3}\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used the slope of the radius itself → y=−3x4−294y = - \frac{3 x}{4} - \frac{29}{4}
  • Turned the radius's slope over but kept its sign → y=−4x3−413y = - \frac{4 x}{3} - \frac{41}{3}
29

Answer: 1818

The common difference is d=−4,d = -4\mathclose{,} and term number nn is 5+(n−1)×(−4).5 + (n - 1) \times (-4)\mathclose{.} Setting that equal to −63:-63\mathclose{:} (n−1)×(−4)=−68,(n - 1) \times (-4) = -68\mathclose{,} so n−1=17n - 1 = 17 and n=18.n = 18\mathclose{.}

Common mistakes, and the answer each one gives:

  • Forgot to add 1 at the end → 1717
30

Answer: −48-48

The common ratio is r=−12÷6=−2.r = -12 \div 6 = -2\mathclose{.} The nnth term is arn−1,ar^{n-1}\mathclose{,} so the ratio is applied 33 times: 6×(−2)3=−48.6 \times \left(-2\right)^{3} = -48\mathclose{.}

Common mistakes, and the answer each one gives:

  • Used arn,ar^n\mathclose{,} applying the ratio one time too many → 9696
  • Added the ratio each time instead of multiplying → 00