## Mathematics 2018 Past Questions | JAMB

Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.

In a class of 40 students, 32 offer Mathematics, 24 offer Physics and 4 offer neither Mathematics nor Physics. How many offer both Mathematics and Physics?

**A.**4**B.**8**C.**16**D.**20

**Correct Option: Answer is D**

Let the number of people that offer both Mathematics and Physics = y

Then, (32−y)+y+(24−y)+4=40

60−y=40⟹y=20

$\therefore $ 20 students offer both Mathematics and Physics.

**2. **Find the values of x for which (x + 2) / 4 -(2x -3)/3 < 4

**A.**x < 8**B.**x > -6**C.**x < 4**D.**x > -3

**Correct Option: Answer is B**

\(\frac {x+2}{4}\) – \(\frac{2x – 3}{3} < 4\)

\(\frac{3(x + 2) – 4(2x – 3)}{12} < 4\)

\(3x + 6 – 8x + 12 < 48 \)

\(18 – 5x < 48 \implies -5x < 30\)

\(\therefore x > -6\)

**3.**

The pie chart shows the monthly expenditure of a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is N7200?

**A.**1000**B.**2000**C.**3000**D.**4000

**Correct Option: Answer is B**

Let the angle for school fees = x°

Then Housing = 2x°

120° + 90° + x° + 2x° = 360°

3x° = 150° , x° = 50°.

Amount spent on housing = 100/360×7200

= N2000.

**4. **A trader realises 10x – x\(^2\) Naira profit from the sale of x bags of corn. How many bags will give him the maximum profit?

**A.**7**B.**6**C.**5**D.**4

**Correct Option: Answer is C**

Profit (P) = 10\(_x\) − \(_x\)2

Maximum profit can be achieved when the differential of profit with respect to number of bags(x) is 0

i.e. \(\frac{dp}{dx}\) = 0

\(\frac{dp}{dx}\) = 10 – 2x = 0

10 = 2x

Then x = \(\frac{10}{2}\) = 5

Answer is C

**5. **If y = 23\(_{5}\) + 101\(_{3}\) , find y, leaving your answer in base two

**A.**1110**B.**10111**C.**11101**D.**111100

**Correct Option: Answer is B**

y = 23\(_{five}\) + 101\(_{three}\)

23\(_{five}\) = \(2 \times 5^1 + 3 \times 5^0\)

= 13\(_{ten}\)

101\(_{three}\) = \(1 \times 3^2 + 0 \times 3^1 + 1 \times 3^0\)

= 10\(_{ten}\)

y\(_{ten}\) = 13\(_{ten}\) + 10\(_{ten}\)

= 23\(_{ten}\)

= 10111\(_{two}\)

**6. **

Find the value of x in the diagram

**A.**10°**B.**28°**C.**36°**D.**40°

**Correct Option: Answer is D**

The diagram shows angles at a point, the total angle at a point is 360

x – 10 + 4x – 50 + 2x + 3x + 20 = 360

10x – 40 = 360

10x = 360 + 40

10x = 400

x = 400/10

x = 40

**7. **Solve for t in the equation \(\frac{3}{4}\)t + \(\frac{1}{3}\)(21 – t) = 11

**A.**\(\frac{9}{13}\)**B.**\(\frac{7}{13}\)**C.**5**D.**9\(\frac{3}{5}\)

**Correct Option: Answer is D**

\(\frac{3}{4}\) t + \(\frac{1}{3}\) (21 – t) = 11

Multiply through by the LCM of 4 and 3 which is 12

12 x(\(\frac{3}{4}\) t) + 12 x (\(\frac{1}{3}\) (21 – t)) = (11 x 12)

9t + 4(21 – t) = 132

9t + 84 – 4t = 132

5t + 84 = 132

5t = 132 – 84 = 48

t = \(\frac{48}{5}\)

t = 9 \(\frac{3}{5}\)

Answer is D

**8. **A school girl spends \(\frac{1}{4}\) of her pocket money on books and \(\frac{1}{3}\) on dress. What fraction remains?

**A.**\(\frac{5}{6}\)**B.**\(\frac{7}{12}\)**C.**\(\frac{5}{12}\)**D.**\(\frac{1}{6}\)

**Correct Option: Answer is C**

**9. **If \(\frac{x}{a + 1}\) + \(\frac{y}{b}\) = 1. Make y the subject of the relation.

**A.**\(\frac{b(a + 1 – x)}{a + 1}\)**B.**\(\frac{a + 1}{b(a – x + 1)}\)**C.**\(\frac{a(b – x + 1)}{b + 1}\)**D.**\(\frac{b}{a(b – x + 1)}\)

**Correct Option: Answer is A**

**10. **Calculate the total surface area of a cupboard which measures 12cm by 10cm by 8cm

**A.**1920cm²**B.**592cm²**C.**296cm²**D.**148cm²

**Correct Option: Answer is B**

Total surface area of a cupboard is given by the equation A = 2(lb + bh + lh) L = 12, b = 10, h = 8

A = 2((12 x 10) + (10 x 8) + (12 x 8))

A = 2(120 + 80 + 96)

A = 2 x 296

A = 592cm2