1991 - JAMB Mathematics Past Questions and Answers - page 1
1
A
40o
B
30o
C
25o
D
15o
correct option: d
PQM = 110o(Ext. < of a cyclic quad)
MPQ = 180o - (110 + 55)
= 180o - 165o
= 15o
Users' Answers & CommentsMPQ = 180o - (110 + 55)
= 180o - 165o
= 15o
2
A
28o
B
122o
C
150o
D
152o
correct option: b
ADB = 180o - 152o = 28o
28o + y + 30o = 180o
y = 180o - 58o
= 122o
Users' Answers & Comments28o + y + 30o = 180o
y = 180o - 58o
= 122o
3
A
136o
B
124o
C
12o
D
68o
correct option: a
Since PQRS is quadrilateral 2y + 2x = QPS = 360o
i.e. 2(y + x) + QPS = 360o
QPS = 360o - 2(y + x)
But x + y + 68o = 180o
x + y = 180o - 68o = 180o
x + y = 180o - 68o
= 112o
QPS = 360o - 2(112o)
360o - 224o = 136o
Users' Answers & Commentsi.e. 2(y + x) + QPS = 360o
QPS = 360o - 2(y + x)
But x + y + 68o = 180o
x + y = 180o - 68o = 180o
x + y = 180o - 68o
= 112o
QPS = 360o - 2(112o)
360o - 224o = 136o
4
A
64 sq. cm
B
40 sq.cm
C
50 sq.cm
D
10 sq.cm
correct option: d
Users' Answers & Comments5
A
\(\frac{a \pi^2}{3} (x + 3y) \)
B
a\(\pi ^2\)y
C
\(\frac{a \pi ^2}{3}\)(y + x)
D
(\(\frac{1}{3} a \pi ^2 x + y\))
correct option: a
Users' Answers & Comments6
Simplify 3\(\frac{1}{3}\) - 1\(\frac{1}{4}\) x \(\frac{2}{3}\) + 1\(\frac{2}{5}\)
A
2
B
3
C
4
D
5
correct option: c
3 - \(\frac{1}{3}\) - (\(\frac{5}{4}\) x \(\frac{2}{3}\)) + 1\(\frac{2}{5}\)
= \(\frac{10}{3}\) - \(\frac{5}{6}\) + \(\frac{7}{5}\)
= \(\frac{100 - 25 + 42}{30}\)
= \(\frac{117}{30}\)
= 3.9
\(\approx\) 4
Users' Answers & Comments= \(\frac{10}{3}\) - \(\frac{5}{6}\) + \(\frac{7}{5}\)
= \(\frac{100 - 25 + 42}{30}\)
= \(\frac{117}{30}\)
= 3.9
\(\approx\) 4
7
If 2257 is the result of subtracting 4577 from 7056 in base n, find n
A
8\(\frac{17}{30}\)
B
9\(\frac{9}{10}\)
C
10\(\frac{1}{10}\)
D
11\(\frac{11}{36}\)
correct option: a
\(\begin{array}{c|c} 7056 \ \text{- 4577} \\hline 2257 \end{array}\)
By trial and error method
Let the base to 8
i.e. Let n = 8 and it is easily verified that the subtraction holds.
The subtraction does not hold when other values of n are tried
n = 8
Users' Answers & CommentsBy trial and error method
Let the base to 8
i.e. Let n = 8 and it is easily verified that the subtraction holds.
The subtraction does not hold when other values of n are tried
n = 8
8
Find correct to 3 decimal places (\(\frac{1}{0.05}\) + \(\frac{1}{5.005}\)) - (0.05 x 2.05)
A
99.998
B
98.999
C
89.899
D
9.998
correct option: a
(\(\frac{1}{0.05}\) + \(\frac{1}{5.005}\)) - (0.05 x 2.05)
(\(\frac{1}{0.05}\) x 5.005) - (0.05 x 2.05)
\(\frac{500.5}{0.05}\) - 1025
100.1 - 0.1025
= 99.9975
99.998 to 3 sig. fiq.
Users' Answers & Comments(\(\frac{1}{0.05}\) x 5.005) - (0.05 x 2.05)
\(\frac{500.5}{0.05}\) - 1025
100.1 - 0.1025
= 99.9975
99.998 to 3 sig. fiq.
9
express 62 \(\div\) 3 as a decimal correct to 3 significant figures
A
20.6
B
20.667
C
20.67
D
20.7
10
Factory P produces 20,000 bags of cement per day while factory Q produces 15,000 bags per day. If P reduces production by 5% and Q increases production by 5%, determine the effective loss in the number of bags produced per day by the two factories
A
250
B
750
C
1000
D
1250
correct option: a
P produces 20,000 bags
P reduces production by 5%
i.e. reduction = \(\frac{5}{100}\) x \(\frac{20000}{1}\)
= 1000
Q produces 15000 bags
Q increases production by 5%
i.e. increase = \(\frac{5}{100}\) x \(\frac{15000}{1}\)
= 750
Effective loss by the two = 1000 - 750
= 250 bags
Users' Answers & CommentsP reduces production by 5%
i.e. reduction = \(\frac{5}{100}\) x \(\frac{20000}{1}\)
= 1000
Q produces 15000 bags
Q increases production by 5%
i.e. increase = \(\frac{5}{100}\) x \(\frac{15000}{1}\)
= 750
Effective loss by the two = 1000 - 750
= 250 bags