2017 - JAMB Mathematics Past Questions and Answers - page 3
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: 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\(\sqrt{3}\)
Users' Answers & Comments= 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: c
Let the speed of the 1st trip be x miles/hr
and the speed of the 2nd trip be 3x miles/hr
Speed = distance/time
∴ Time taken to cover a distance of 50 miles on the 1st trip
= \(\frac{50}{xhr}\)
time taken to cover a distance of 300 miles on the next trip
= \(\frac{300}{3xhr}\)
= \(\frac{100}{xhr}\)
∴the new time compared with the old time is twice as much
Users' Answers & Commentsand the speed of the 2nd trip be 3x miles/hr
Speed = distance/time
∴ Time taken to cover a distance of 50 miles on the 1st trip
= \(\frac{50}{xhr}\)
time taken to cover a distance of 300 miles on the next trip
= \(\frac{300}{3xhr}\)
= \(\frac{100}{xhr}\)
∴the new time compared with the old time is twice as much
23
A
40o
B
55o
C
50o
D
60o
correct option: a
Sum of angle at a point = 360o
2x + 3x + 4x = 360
9x = 360
x = \(\frac{360}{9}\)
x = 40o
Users' Answers & Comments2x + 3x + 4x = 360
9x = 360
x = \(\frac{360}{9}\)
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: b
Users' Answers & Comments25
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: 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%
Users' Answers & Comments= 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: b
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
Users' Answers & CommentsIt 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
Evaluate \(\frac{0.00000231}{0.007}\) and leave the answer in standard form
A
3.3 x 104
B
3.3 x 10-3
C
3.3 x 10-4
D
3.3 x 10-8
correct option: c
\(\frac{0.00000231}{0.007}\) to standard form
= \(\frac{231 \times 10^{-8}}{7 \times 10^{-3}}\)
= 33 × 10\(^{-8 − (−3)}\)
= 33 × 10\(^{− 8 + 3}\)
= 33 × 10-5
Users' Answers & Comments= \(\frac{231 \times 10^{-8}}{7 \times 10^{-3}}\)
= 33 × 10\(^{-8 − (−3)}\)
= 33 × 10\(^{− 8 + 3}\)
= 33 × 10-5
28
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: a
Actual measurement = 10cm
approximated 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 %
Users' Answers & Commentsapproximated 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 %
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: 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
Users' Answers & CommentsAmount = 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: d
Total angle at a point = 3600
∴ 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