# Mathematics 2017 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.

Given T = { even numbers from 1 to 12 }

N = {common factors of 6, 8 and 12}

Find T ∩ N**A.**{2, 3}**B.**{2, 3, 4}**C.**{3, 4, 6}**D.**{2}

**Correct Option: Answer is D**

T = {evenn numbers from 1 to 12}

N = {common factors of 6,8 and 12}

Find T ∩ N

T = {2, 4, 6, 8, 10, 12}

N = {2}

T ∩ N = {2} i.e value common to T & N

**2. **What is the next number in the series 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\)…

**A.**\(\frac{1}{3}\)**B.**\(\frac{2}{8}\)**C.**\(\frac{3}{7}\)**D.**\(\frac{1}{8}\)

**Correct Option: Answer is D**

2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\)…..

There are 4 terms in the series

Therefore the next number will be the 5th term

T_{n} = ar\(^{n − 1}\) (formular for geometric series)

a = first term = 2

r = common rate = \(\frac{\text{next term}}{\text{previous term}}\) = \(\frac{1}{2}\)

n = number of terms

T_{5} = 5th term = ?

T_{5} = ar\(^{5 – 1}\)

= ar\(^4\)

= 2 × (ar\(^{n − 1}\))^{4}

= 2 × \(\frac{1}{16}\)

= \(\frac{1}{8}\)

**3. **If U = {x : x is an integer and 1 ≤ x ≤ 20 }

E1 = {x: x is a multiple of 3}

E2 = {x: x is a multiple of 4} and an integer is picked at random from U, find the probability that it is not in E2

**A.**3/4**B.**$\frac{\mathrm{3/}}{10}$**C.**$\frac{\mathrm{1/}}{4}$**D.**1/20

**Correct Option: Answer is A**

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

E1 = {3, 6, 9, 12, 15, 18}

E2 = {4, 8, 12, 16, 20}

Probability of E2 = \(\frac{5}{20}\) i.e \(\frac{\text{Total number in}E_2}{\text{Entire number in set}}\)

Probability of set E2 = 1 − \(\frac{5}{20}\)

= \(\frac{15}{20}\)

= \(\frac{3}{4}\)

**4. **The curved surface area of a cylinder 5cm high is 110cm^{2}. Find the radius of its base

π =

**A.**2.6cm**B.**3.5cm**C.**3.6cm**D.**7.0cm

**Correct Option: Answer is B**

Curved surface area of cylinder = 2πrh

110 = 2 × \(\frac{22}{7}\) × r × 5

r = \(\frac{110 \times 7}{44 \times 5}\)

= 3.5cm

**5. **If two graphs Y = px^{2} + q and y = 2x^{2} − 1 intersect at x =2, find the value of p in terms of q

**A.**q − \(\frac{8}{7}\)**B.**7 − \(\frac{q}{4}\)**C.**8 − \(\frac{q}{2}\)**D.**7 + \(\frac{q}{8}\)

**Correct Option: Answer is B**

Y = Px^{2} + q

Y = 2x^{2} – 1

Px^{2} + q = 2x^{2} – 1

Px^{2} = 2x^{2} – 1 – q

p = \(\frac{2x^2 – 1 – q}{x^2}\)

at x = 2

P = \(\frac{2(2)^2 – 1 – q}{2^2}\)

= \(\frac{2(4) – 1 -q}{4}\)

= \(\frac{8 – 1 – q}{4}\)

P = \(\frac{7 – q}{4}\)

**6. **

Evaluate (\(\sin\)45^{o} + \(\sin\)3^{o} ) in surd form

**A.**\(\frac{\sqrt{3}}{2\sqrt{2}}\)**B.**√3 − \(\frac{1}{2}\)**C.**\(\frac{1}{2}\)√2**D.**1 + \(\frac{\sqrt{2}}{2}\)

**Correct Option: Answer is D**

hypotenuse

sin = \(\frac{1}{2}\)

\(\sin45 = \frac{1}{\sqrt{2}}\)

= \(\frac{2}{2}\)

∴ (sin45 + sin30)

= \(\frac{1}{\sqrt{2}} + \frac{1}{2}\)

= \(\frac{\sqrt{2}}{2}\) + \(\frac{1}{2}\)

= \(\frac{\sqrt{2} + 1}{2}\)

= \(\frac{1 + \sqrt{2}}{2}\)

**7. **If y = x Sin x, find \(\frac{dy}{dx}\) when x = \(\frac{\pi}{2}\)

**A.**\(\frac{- \pi}{2}\)**B.**-1**C.**1**D.**\(\frac{ \pi}{2}\)

**Correct Option: Answer is C**

y = xsinx

\(\frac{dy}{dx}\) = \(1 \sin x + x \cos x\)

= \(\sin x + x \cos x\)

At x = \(\frac{\pi}{2}\)

= sin\(\frac{\pi}{r}\) + \(\frac{\pi}{2} \cos {\frac{\pi}{2}}\)

= 1 + \(\frac{\pi}{2}\) × 10

= 1

**8. **If temperature t is directly proportional to heat h, and when t = 20^{o}C, h = 50 J, find t when h = 60J

**A.**24^{o}C**B.**20^{o}C**C.**34^{o}C**D.**30^{o}C

**Correct Option: Answer is A**

t ∝ h, t = 20, h

t = ? h = 60

t = kh where k is constant

20 = 50k

k = \(\frac{20}{50}\)

k = \(\frac{2}{5}\)

when h = 60, t = ?

t = \(\frac{2}{5}\) × 60

t = 24^{o}C

**9. **Evaluate 1 – (\(\frac{1}{5}\) x \(\frac{2}{3}\)) + ( 5 + \(\frac{2}{3}\))

**A.**4**B.**3**C.**2\(\frac{2}{3}\)**D.**3\(\frac{2}{3}\)

**Correct Option: Answer is D**

**10. **Given m = N\(\sqrt{\frac{SL}{T}}\) make T the subject of the formula

**A.**\(\frac{\text{NSL}}{M}\)**B.**\(\frac{N^2SL}{M^2}\)**C.**\(\frac{N^2SL}{M}\)**D.**\(\frac{NSL}{M^2}\)

**Correct Option: Answer is B**

M = N \(\sqrt{\frac{SL}{T}}\),

make T subject of formula square both sides

M\(^{2}\) = \(\frac{N^2SL}{T}\)

TM\(^{2}\) = N\(^{2}\)SL

T = \(\frac{N^2SL}{M^2}\)</p