2001 - JAMB Mathematics Past Questions and Answers - page 1
1
Evaluate 21.05347 - 1.6324 x 0.43 to 3 decimal places
A
20.980
B
20.351
C
20.981
D
20.352
correct option: d
Hint: Use BODMAS, in other words, do multiplication of the second and the last first before subtracting value obtained from the first.
Users' Answers & Comments3
Given that \(p = 1 + \sqrt{2}\hspace{1mm}and\hspace{1mm}q = 1 -\sqrt{2},\hspace{1mm}evaluate\hspace{1mm}\frac{(p^{2} - q^{2})}{2pq}\)
A
2(2+√2)
B
-2(2+√2)
C
2√2
D
-2√2
correct option: d
HINT:
\(\frac{(p^{2} - q^{2})}{2pq}=\frac{(1+\sqrt{2})^{2}}{2((1+\sqrt{2})(1-\sqrt{2}))}\)
Use difference of two squares to expand the bracket and reduce to \(\frac{2(2\sqrt{2})}{-2} = -2\sqrt{2}\)
Users' Answers & Comments\(\frac{(p^{2} - q^{2})}{2pq}=\frac{(1+\sqrt{2})^{2}}{2((1+\sqrt{2})(1-\sqrt{2}))}\)
Use difference of two squares to expand the bracket and reduce to \(\frac{2(2\sqrt{2})}{-2} = -2\sqrt{2}\)
4
A car dealer bought a second-hand car for N250,000 and spent N70,000 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 = N(250,000 + 70,000) = N320,000
Selling price = N400,000 (given)
Gain = SP - CP = N(400,000 - 320,000) = N80,000
Gain % = gain/CP x 100 = (80,000/320,000) x 100
Gain % = 25%
Users' Answers & CommentsSelling price = N400,000 (given)
Gain = SP - CP = N(400,000 - 320,000) = N80,000
Gain % = gain/CP x 100 = (80,000/320,000) x 100
Gain % = 25%
5
If \(x = \frac{y}{2}\),evaluate\(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right)\)
A
5/8
B
5/2
C
5/4
D
5/16
correct option: b
Users' Answers & Comments6
Find the principal which amounts to N5,500 at a simple interest in 5 years at 2% per annum.
A
N4,900
B
N5,000
C
N4,700
D
N4,800
correct option: b
Principal, P = Amount, A - Interest, I.
A = P + I
I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10
But A = P + I,
=> 5500 = P + (P/10)
=> 55000 = 10P + P
=> 55000 = 11P
Thus P = 55000/11 = N5,000
Users' Answers & CommentsA = P + I
I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10
But A = P + I,
=> 5500 = P + (P/10)
=> 55000 = 10P + P
=> 55000 = 11P
Thus P = 55000/11 = N5,000
7
Evaluate (0.142 x 0.275) 7(0.02) to 3 decimal places.
A
0.039
B
0.358
C
0.033
D
0.308
correct option: a
(0.142 x 0.275) 7(0.02) = (0.14 x 0.14 x 0.275)/(7 x 0.02) = (0.14 x 0.14 x 0.275)/(0.14)
=0.14 x 0.275 = 0.0385 Approx. 0.039
Users' Answers & Comments=0.14 x 0.275 = 0.0385 Approx. 0.039
8
Divide: ax3x - 26x2x + 156ax - 216 by a2x - 24ax + 108
A
ax - 2
B
ax + 2
C
ax - 8
D
ax - 6
correct option: a
Users' Answers & Comments9
If two graphs y = px2 + q and y = 2x2 -1 intersect at x = 2, find the value of p in terms q.
A
\(\frac{q-8}{7}\)
B
\(\frac{7-q}{4}\)
C
\(\frac{8-q}{2}\)
D
\(\frac{7+q}{8}\)
correct option: b
\(y = px^{2}+q\y = 2x^{2}-1\\Rightarrow px^{2}+q=2x^{2}-1\px^{2}=2x^{2}-1-q\p=\frac{(2x^{2}-q-1)}{x^{2}}\At\hspace{1mm}x=2\p=\frac{2(2^{2})-1-q)}{2^{2}}\p=\frac{(8-1-q)}{4}\=\frac{7-q}{4}\)
Users' Answers & Comments10
Find the integral values of x and y satisfying the inequality 3y + 5x \(\leq\) 15, given that y > 0, y < 3 and x > 0.
A
(1,1), (1,2), (1,3)
B
(1,1), (2,1), (1,3)
C
(1,1), (3,1), (2,2)
D
(1,1), (1,2), (2,1)
correct option: d
Hint: Sketch the inequality graph for the 3 conditions given and read out your points from the co-ordinates.
Users' Answers & Comments