## 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.

**21. **Simplify 4\(\sqrt{27}\) + 5\(\sqrt{12}\) − 3\(\sqrt{75}\)

**A.**7**B.**− 7**C.**− 7\(\sqrt{3}\)**D.**7\(\sqrt{3}\)

**Correct Option: Answer is D**

4\(\sqrt{27}\) + 5\(\sqrt{12}\) − 3\(\sqrt{75}\)

= 4\(\sqrt{3}\) × 9 + 5\(\sqrt{3}\) × 4 − 3\(\sqrt{3}\) × 25

= 4 × 3\(\sqrt{3}\) + 5 × 2\(\sqrt{3}\) − 3 × 5\(\sqrt{3}\)

= 12\(\sqrt{3}\) + 10\(\sqrt{3}\) − 15\(\sqrt{3}\)

= (12 + 10 − 15)\(\sqrt{3}\)

= 7\(\sqrt{3}\)

**22. **A man covered a distance of 50 miles on his first trip, on a later trip he traveled 300 miles while going 3 times as fast. His new time compared with the old distance was?

**A.**three times as much**B.**the same**C.**twice as much**D.**half as much

**Correct Option: Answer is C**

**23.**

**A.**40^{o}**B.**55^{o}**C.**50^{o}**D.**60^{o}

**Correct Option: Answer is A**

Sum of angle at a point = 360^{o}

2x + 3x + 4x = 360

9x = 360

x = 360/9

x = 40^{o}

**24. **Divide 4x³ – 3x + 1 by 2x – 1

**A.**2x^{2}-x + 1**B.**2x^{2}– x -1**C.**2x^{2}+ x + 1**D.**2x^{2}+ x -1

**Correct Option: Answer is D**

**25. **A car dealer bought a second-hand car for of 250,000 and spent N 70,000 refurbishing it. He then sold the car for N400,000. What is the percentage gain?

**A.**60%**B.**32%**C.**25%**D.**20%

**Correct Option: Answer is C**

Total Cost Price = N(250,000 + 70,000)

= N 32,000

Selling Price = N 400,000(Given)

Gain = Selling Price – Cost Price

= 400,000 – 300,000

= 80,000

% gain = \(\frac{\text{Gain}}{\text{Cost Price}}\) × 100

= \(\frac{80,000}{320,000}\) × 100

Gain % = 25%

**26. **Find the number of ways that the letters of the word EXCELLENCE be arranged

**A.**\(\frac{10!}{2!2!2!}\)**B.**\(\frac{10!}{4!2!}\)**C.**\(\frac{10!}{4!2!2!}\)**D.**\(\frac{10!}{2!2!}\)

**Correct Option: Answer is C**

EXCELLENCE

It is a ten letter word = 10!

Since we have repeating letters, we have to divide to remove the duplicates accordingly. There are 4 Es, 2 Cs, 2 Ls

∴ there are

\(\frac{10!}{4!2!2!}\) ways to arrange

**27. **If a rod 10cm in length was measured as 10.5cm, calculate the percentage error

**A.**5%**B.**10%**C.**8%**D.**7%

**Correct Option: Answer is A**

Actual measurement = 10cmapproximated value of measurement = 10.5cm

% error = \(\frac{\text{Actual measurement − Approximated}}{\text{Actual measure}}\) × 100

= \(\frac{10 − 10.5}{10}\) × 100

= \(\frac{-0.5}{10}\) × 100

ignore -sign i.e take absolute value

= \(\frac{0.5}{10}\) × 100

= 5 %

**28. **Evaluate \(\frac{0.00000231}{0.007}\) and leave the answer in standard form

**A.**3.3 x 10^{-4}**B.**3.3 x 10^{-3}**C.**3.3 x 10^{-5}**D.**3.3 x 10^{-8}

**Correct Option: Answer is A**

\(\frac{0.00000231}{0.007}\) to standard form

= \(\frac{231 \times 10^{-8}}{7 \times 10^{-3}}\)

= 33 × 10\(^{-8 − (−3)}\)

= 33 × 10\(^{− 8 + 3}\)

= 33 × 10\(^{-5}\)

= 3.3 x 10\(^{-4}\)

**29. **Find the principal which amounts to ₦ 5,500 at a simple interest in 5 years at 2% per annum

**A.**₦ 4,900**B.**₦ 5,000**C.**₦ 4,700**D.**₦ 4,000

**Correct Option: Answer is B**

Principal = P, Simple Interest = I, Amount = A

Amount = Principal + Simple Interest

I = \(\frac{PRT}{100}\)

R = rate, T = time

I = \(\frac{P \times 5 \times 2}{100}\)

I = \(\frac{10P}{100}\)

I = \(\frac{P}{10}\)

Amount A = P + I

5500 = P + \(\frac{P}{10}\)

Multiply through by 100

5500 = 10P + P

5500 = 11P

p = \(\frac{5500}{11}\)

p = ₦5000

**30. **

The pie chart shows the allocation of money to each sector in a farm. The total amount allocated to the farm is ₦ 80 000. Find the amount allocated to fertilizer

**A.**₦ 35, 000**B.**₦ 40,000**C.**₦ 25,000**D.**₦ 20,000

**Correct Option: Answer is D**

Total angle at a point = 360º

∴ To get the angle occupied by fertilizer we have,

40 + 50 + 80 + 70 + 30 + fertilizer(x) = 360

270 + x = 360

x = 360 – 270

x = 90

Total amount allocated to the farm

= ₦ 80,000

∴Amount allocated to the fertilizer

= \(\frac{\text{fertilizer (angle) × Total amount}}{\text{total angle}}\)

= \(\frac{90}{360}\) × 80,000

= ₦20,000