Upper Sec Mixed Check #01

1. Step 1Expand both brackets

   3(4a + 6b) − 2(2a + 3b) = 12a + 18b − 4a − 6b

2. Step 2Collect like terms

   = 8a + 12b

3. Step 3Factorise fully

   The greatest common factor is 4. 8a + 12b = 4(2a + 3b)

Answer: 4(2a + 3b)

If this broke, look here first: check whether your expanded and simplified expression is correct before trying to factorise it.

1. Step 1Identify the sides relative to 38°

   ST = 7 cm is opposite angle R. RT is the hypotenuse.

2. Step 2Choose the ratio

   sin 38° = opposite / hypotenuse. sin 38° = 7 / RT.

3. Step 3Rearrange and calculate

   RT = 7 / sin 38° = 11.370… cm

Answer: 11.4 cm (3 s.f.)

If this broke, look here first: identifying the opposite side and hypotenuse before choosing the ratio.

1. Step 1Add all seven marks

   6 + 13 + 16 + 32 + 33 + 34 + 36 = 170

2. Step 2Divide by the number of students

   Mean = 170 / 7 = 24.285714…

3. Step 3Round only at the end

   To 2 decimal places, 24.285714… becomes 24.29.

Answer: 24.29 (2 d.p.)

If this broke, look here first: checking that all seven marks are included in the total before dividing.

1. Step 1Find the greatest common factor

   HCF(8, 12) = 4. Both terms also contain m. The greatest common monomial factor is 4m.

2. Step 2Divide each term

   8mn ÷ 4m = 2n; −12m ÷ 4m = −3. 8mn − 12m = 4m(2n − 3).

Answer: 4m(2n − 3)

If this broke, look here first: including m in the common factor and dividing both terms.

1. Step 1Find BC in triangle ABC

   ABC is right-angled at B; AC = 13 cm is the hypotenuse. BC² = AC² − AB² = 13² − 5². BC² = 169 − 25 = 144, so BC = 12 cm.

2. Step 2Move to triangle BCD

   For angle CBD, opposite = CD = 9 cm. Adjacent = BC = 12 cm.

3. Step 3Choose the ratio

   tan(CBD) = 9 / 12

4. Step 4Find the angle

   CBD = tan⁻¹(9/12) = 36.869…°

Answer: 36.9° (1 d.p.)

If this broke, look here first: finding the shared side with Pythagoras before selecting the angle ratio.

1. Step 1Count all possible tokens

   The endpoints are included: 47 − 18 + 1 = 30.

2. Step 2List valid multiples of 4

   20, 24, 28, 32, 36, 40, 44. There are 7 favourable tokens.

3. Step 3Form the probability

   P(multiple of 4) = 7/30

Answer: 7307/30

If this broke, look here first: counting both endpoints in the total, then listing multiples inside the range.

1. Step 1Take out the greatest common factor

   6p² − 24q² = 6(p² − 4q²)

2. Step 2Recognise the difference of squares

   p² − 4q² = p² − (2q)²

3. Step 3Factorise the bracket

   p² − (2q)² = (p − 2q)(p + 2q)

Answer: 6(p − 2q)(p + 2q)

If this broke, look here first: taking out the common factor before checking the bracket for two squares.

1. Step 1Find two terms for the middle

   Product = 3 × (−4) = −12; sum = −11. Use −12 and +1.

2. Step 2Split the middle term

   3x² − 11x − 4 = 3x² − 12x + x − 4

3. Step 3Factor by grouping

   = 3x(x − 4) + 1(x − 4)

4. Step 4Take out the common bracket

   = (3x + 1)(x − 4)

Answer: (3x + 1)(x − 4)

If this broke, look here first: finding two numbers with product −12 and sum −11 before splitting the middle term.

1. Step 1Count the ratio parts

   2 + 3 + 4 = 9 parts

2. Step 2Find one part

   81 ÷ 9 = 9 cards per part

3. Step 3Count favourable cards

   Red: 2 × 9 = 18; green: 4 × 9 = 36. Red or green: 18 + 36 = 54 cards.

4. Step 4Form the probability

   P(red or green) = 54/81 = 2/3

Answer: 232/3

If this broke, look here first: using the ordered ratio to identify the two favourable colours.

1. Step 1Translate the wording

   “The excess of A over B” means A − B. Write 6(2t − 1) − (3t + 8).

2. Step 2Expand carefully

   = 12t − 6 − 3t − 8

3. Step 3Collect like terms

   = 9t − 14

Answer: 9t − 14

If this broke, look here first: translating “excess of A over B” as A − B, with brackets around B.

1. Step 1Identify the student’s error

   m is in the original denominator. It must stay in the denominator when n is isolated.

2. Step 2Clear the fraction

   E = 12/(mn). Multiply both sides by mn: Emn = 12.

3. Step 3Make n the subject

   Divide both sides by Em: n = 12/(Em).

4. Step 4State the restriction on E

   Since m and n are nonzero, 12/(mn) cannot be 0. Therefore E ≠ 0.

Answer: n = 12/(Em), E ≠ 0

If this broke, look here first: checking where m belongs when isolating n from Emn = 12.

1. Step 1Find the mean

   Total = 2 + 4 + 4 + 6 + 9 + 11 = 36. Mean = 36 ÷ 6 = 6.

2. Step 2Apply the condition

   Only 9 and 11 are greater than 6. The result equal to 6 does not qualify.

3. Step 3Form the probability

   2 favourable cards out of 6: 2/6 = 1/3

Answer: 131/3

If this broke, look here first: finding the mean first, then treating “greater than” as a strict condition.