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. 40o
- B. 55o
- C. 50o
- D. 60o
Correct Option: Answer is A
Sum of angle at a point = 360o
2x + 3x + 4x = 360
9x = 360
x = 360/9
x = 40o
24. Divide 4x³ – 3x + 1 by 2x – 1
- A. 2x2 -x + 1
- B. 2x2 – x -1
- C. 2x2 + x + 1
- D. 2x2 + 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