2015 - JAMB Mathematics Past Questions and Answers - page 7
61
Find the distance between the points (-2,-3) and (-2,4)
A
3m
B
2.4m
C
3.2m
D
5.1m
correct option: d
Users' Answers & Comments62
Determine the third term of a geometrical progression whose first and second term are 2 and 14 respectively
A
1458
B
1485
C
1345
D
1258
correct option: a
1st G.P. = a =2
2nd G.P. = ar − 1 = 54
2(r) = 54
r = 54/2 = 27
r = 27
3rd term = ar2 = (2) (27)2
2 × 27 × 27
= 1458
Users' Answers & Comments2nd G.P. = ar − 1 = 54
2(r) = 54
r = 54/2 = 27
r = 27
3rd term = ar2 = (2) (27)2
2 × 27 × 27
= 1458
63
Given that S and T are sets of real numbers such that S = {x : 0 \(\leq\) x \(\leq\) 5} and T = {x : − 2 < x < 3} Find S \(\cup\) T
A
−3 < x < \(\leq\)3
B
−2< x < \(\leq\)5
C
2< x < \(\geq\) −5
D
−1< x \(\geq\) \(\leq\)2
correct option: b
S = {0, 1, 2, 3, 4, 5}
T = {− 1, 0, 1, 2}
S \(\cup\) T = {− 1, 0, 1, 2, 3, 4, 5 }
⇒ − 2 < x ≤ 5
Users' Answers & CommentsT = {− 1, 0, 1, 2}
S \(\cup\) T = {− 1, 0, 1, 2, 3, 4, 5 }
⇒ − 2 < x ≤ 5
64
If an investor invest N450,000 in a certain organization in order to yield X as a return of N25,000. Find the return on an investment of N700,000 by Y in the same organization.
A
N14,950.50K
B
N25,150.30K
C
N15,000.00K
D
N38,888.90K
correct option: d
[Return ÷ Investment] as a ratio ;
i.e The Ratio is Return : Investmen
[(Return1÷ Investment1 ) = (Return2÷ Investment2)]
R1 = N 25000
R2 =?
I1 = N450,000,
I1 = N 700000
(25000 ÷ 450000) = (R2 ÷ 700000)
R2 = [(25000 × 700000 ) ÷ 450000]
= N38,888.90K
∴ The return on a investment of Y = N38888.90K
Users' Answers & Commentsi.e The Ratio is Return : Investmen
[(Return1÷ Investment1 ) = (Return2÷ Investment2)]
R1 = N 25000
R2 =?
I1 = N450,000,
I1 = N 700000
(25000 ÷ 450000) = (R2 ÷ 700000)
R2 = [(25000 × 700000 ) ÷ 450000]
= N38,888.90K
∴ The return on a investment of Y = N38888.90K
65
Evaluate log717
A
1.35
B
1.353
C
1.455
D
0.455
correct option: c
log717
= [log 17 ÷ log7]
= [1.2304 ÷ 0.8451]
[100.0899 ÷ 101.9270]
= 1.455(antilog)
Users' Answers & Comments= [log 17 ÷ log7]
= [1.2304 ÷ 0.8451]
[100.0899 ÷ 101.9270]
= 1.455(antilog)
66
100112 + *****2 + 111002 + 1012 = 10011112
A
11112
B
110112
C
101112
D
110012
correct option: b
Convert the binary to base 10 and they convert back to base two
100112 + xxxxx2 + 111002 + 1012 = 10011112
(1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20) + xxxxx2 +(1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20) + (1 × 22 + 0 × 21 + 1 × 20)
= (16 + 0 + 0 + 2 + 1) + xxxxx2 + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)
=(64 + 0 + 0 + 8 + 4 + 2 + 1)
19 + xxxxx2 + 33 = 79
xxxxx2 + 52 = 79
xxxxx2 = 79 − 52
xxxxx2 = 2710
\( \begin{array}{c|c}
2 & 27 \\ \hline
2 & 13 \text{ rem 1}\\ 2 & 6 \text{ rem 1}\\ 2 & 3 \text{ rem 0}\\ 2 & 1 \text{ rem 1}\\ & 0 \text{ rem 1}\\ \end{array}\uparrow \)
2710 = 110112
Therefore xxxxx2 = 2710 = 110112
Users' Answers & Comments100112 + xxxxx2 + 111002 + 1012 = 10011112
(1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20) + xxxxx2 +(1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20) + (1 × 22 + 0 × 21 + 1 × 20)
= (16 + 0 + 0 + 2 + 1) + xxxxx2 + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)
=(64 + 0 + 0 + 8 + 4 + 2 + 1)
19 + xxxxx2 + 33 = 79
xxxxx2 + 52 = 79
xxxxx2 = 79 − 52
xxxxx2 = 2710
\( \begin{array}{c|c}
2 & 27 \\ \hline
2 & 13 \text{ rem 1}\\ 2 & 6 \text{ rem 1}\\ 2 & 3 \text{ rem 0}\\ 2 & 1 \text{ rem 1}\\ & 0 \text{ rem 1}\\ \end{array}\uparrow \)
2710 = 110112
Therefore xxxxx2 = 2710 = 110112
67
Evaluate log5(\( y^2x^5 ÷ 125b½) \)
A
2 log5y + 5log5 y2 − 3
B
log5 y2 + 5log5 x + 3
C
25logy 5 + 3
D
2log5y + 5log5x − ½ log5b −3
correct option: d
Users' Answers & Comments