1990 - JAMB Mathematics Past Questions and Answers - page 4
31
A cylinder pipe, made of metal is 3cm thick.If the internal radius of the pope is 10cm.Find the volume of metal used in making 3m of the pipe.
A
153\(\pi\)cm3
B
207\(\pi\)cm3
C
15 300\(\pi\)cm3
D
20 700\(\pi\)cm3
correct option: d
Users' Answers & Comments32
If the heights of two circular cylinder are in the ratio 2 : 3 and their volumes?
A
27 : 32
B
27 : 23
C
23 : 32
D
27 : 23
correct option: a
\(\frac{h_1}{h_2}\) = \(\frac{2}{3}\)
h2 = \(\frac{2h_1}{3}\)
\(\frac{r_1}{r_2}\) = \(\frac{9}{8}\)
r2 = \(\frac{9r_1}{8}\)
v1 = \(\pi\)(\(\frac{9r_1}{8}\))2(\(\frac{2h_1}{3}\))
= \(\pi\)r1 2h1 x \(\frac{27}{32}\)
v = \(\frac{\pi r_1 2h_1 \times \frac{27}{32}}{\pi r_1 2h_1}\) = \(\frac{27}{32}\)
v2 : v1 = 27 : 32
Users' Answers & Commentsh2 = \(\frac{2h_1}{3}\)
\(\frac{r_1}{r_2}\) = \(\frac{9}{8}\)
r2 = \(\frac{9r_1}{8}\)
v1 = \(\pi\)(\(\frac{9r_1}{8}\))2(\(\frac{2h_1}{3}\))
= \(\pi\)r1 2h1 x \(\frac{27}{32}\)
v = \(\frac{\pi r_1 2h_1 \times \frac{27}{32}}{\pi r_1 2h_1}\) = \(\frac{27}{32}\)
v2 : v1 = 27 : 32
33
The locus of a point which moves so that it is equidistant from two intersecting straight lines is the
A
perpendicular bisector of the two lines
B
angle bisector of the two lines
C
bisector of the two lines
D
line parallel to the two lines
34
4, 16, 30, 20, 10, 14 and 26 are represented on a pie chart. Find the sum of the angles of the bisectors representing all numbers equals to or greater than 16
A
48o
B
84o
C
92o
D
276o
correct option: d
Given that 4, 16, 30, 20, 10, 14 and 26
Adding up = 120
nos \(\geq\) 16 are 16 + 30 + 20 + 26 = 92
The requires sum of angles = \(\frac{92}{120}\) x \(\frac{360^o}{1}\)
= 276o
Users' Answers & CommentsAdding up = 120
nos \(\geq\) 16 are 16 + 30 + 20 + 26 = 92
The requires sum of angles = \(\frac{92}{120}\) x \(\frac{360^o}{1}\)
= 276o
35
The mean of ten positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number
A
3
B
16
C
18
D
38
correct option: d
Mean of 10 numbers = 16
The total sum of numbers = 16 x 10 = 160
Mean of 11 numbers = 18
Total sum of numbers = 11 x 18
= 198
The 11th no. = 198 - 160
= 38
Users' Answers & CommentsThe total sum of numbers = 16 x 10 = 160
Mean of 11 numbers = 18
Total sum of numbers = 11 x 18
= 198
The 11th no. = 198 - 160
= 38
36
Below are the scores of a group of students in a test
\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}\)
If the average scores is 3.5, find the value of x
\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}\)
If the average scores is 3.5, find the value of x
A
1
B
2
C
3
D
4
correct option: b
\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}\)
Average = 3.5
3.5 = \(\frac{(1 \times 1) + (2 \times 4) + (3 \times 5) + (4 \times 6) + 5x + (6 \times 2)}{1 + 4 + 5 + 6 + x + 2}\)
\(\frac{3.5}{1}\) = \(\frac{1 + 8 + 15 + 24 + 5x + 12}{18 + x}\)
\(\frac{3.5}{1}\) = \(\frac{60 + 5x}{18 + x}\)
60 + 5x = 3.5(18 \(\div\) x)
60 + 5x = = 63 + 1.5x
5x - 1.5x = 63 - 60
1.5x = 3
x = \(\frac{3}{1.5}\)
\(\frac{30}{15}\) = 2
Users' Answers & CommentsAverage = 3.5
3.5 = \(\frac{(1 \times 1) + (2 \times 4) + (3 \times 5) + (4 \times 6) + 5x + (6 \times 2)}{1 + 4 + 5 + 6 + x + 2}\)
\(\frac{3.5}{1}\) = \(\frac{1 + 8 + 15 + 24 + 5x + 12}{18 + x}\)
\(\frac{3.5}{1}\) = \(\frac{60 + 5x}{18 + x}\)
60 + 5x = 3.5(18 \(\div\) x)
60 + 5x = = 63 + 1.5x
5x - 1.5x = 63 - 60
1.5x = 3
x = \(\frac{3}{1.5}\)
\(\frac{30}{15}\) = 2
37
Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?
A
\(\frac{2}{3}\)
B
\(\frac{2}{15}\)
C
\(\frac{1}{2}\)
D
\(\frac{1}{4}\)
correct option: c
\(\begin{array}{c|c} 1 & 2 & 3 & 4\\hline 1(1, 1) & (1, 2) & (1, 3) & (1, 4)\ \hline 2(2, 1) & (2 , 2) & (2, 3) & (2, 4) \ \hline 3(3, 1) & (3, 2) & (3, 3) & (3, 4)\ \hline 4(4, 1) & (4, 2) & (4, 3) & (4, 4)\end{array}\)
sample space = 16
sum of nos. removed are (2), 3, (4), 5
3, (4), 5, (6)
(4), 5, (6), 7
(5), 6, 7, (8)
Even nos. = 8 of them
Pr(even sum) = \(\frac{8}{16}\)
= \(\frac{1}{2}\)
Users' Answers & Commentssample space = 16
sum of nos. removed are (2), 3, (4), 5
3, (4), 5, (6)
(4), 5, (6), 7
(5), 6, 7, (8)
Even nos. = 8 of them
Pr(even sum) = \(\frac{8}{16}\)
= \(\frac{1}{2}\)
38
Find the probability that a number selected at random from 41 to 56 is a multiply of 9
A
\(\frac{1}{8}\)
B
\(\frac{2}{15}\)
C
\(\frac{3}{16}\)
D
\(\frac{7}{8}\)
correct option: a
Given from 41 to 56
41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56
The nos multiple of 9 are: 45, 54
P(multiple of 9) = \(\frac{2}{16}\)
= \(\frac{1}{8}\)
Users' Answers & Comments41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56
The nos multiple of 9 are: 45, 54
P(multiple of 9) = \(\frac{2}{16}\)
= \(\frac{1}{8}\)
39
A
y = x2 + x - 2
B
y = -x2 - x + 2
C
y = x2 - x - 2
D
y = -x2 - x + 2
correct option: c
The required equation is y = x2 - x - 2
i.e. B where the graph touches the graph touches the x-axis y = 0
x2 - x - 2 = 0 = (x + 1)(x - 2) = 0
Hence roots of the equation are -1 and 2 as shown in the graph
Users' Answers & Commentsi.e. B where the graph touches the graph touches the x-axis y = 0
x2 - x - 2 = 0 = (x + 1)(x - 2) = 0
Hence roots of the equation are -1 and 2 as shown in the graph
40
A
40o
B
50o
C
80o
D
100o
correct option: b
Since PS = QS = RS
S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius
QPR \(\pm\) \(\frac{100^o}{2}\) = 50o
Users' Answers & CommentsS is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius
QPR \(\pm\) \(\frac{100^o}{2}\) = 50o