2020 - JAMB Mathematics Past Questions and Answers - page 2
If log\(_{10}\)2 = 0.3010 and log\(_{10}\)3 = 0.4771, evaluate log\(_{10}\)4.5 without using the logarithm tables.
0.3010
0.4771
0.6532
0.9542
Given that:
log\(_{10}\)2 = 0.3010
log\(_{10}\)3 = 04771
Then,
log\(_{10} 4.5 = log_{10}\) (\(\frac{3 \times 3}{2}\))
log\(_{10}\) 3 + log\(_{10}\) 3 - log\(_{10}\)2 = 0.4471 + 0.771 - 0.3010
= 0.6532 Ans.
Simplify \(\frac{324 - 4x^2}{2x + 18}\)
2(x - 9)
2(9 + x)
81 - x\(^2\)
-2(x - 9)
***
234 - 4x\(^2\) = 18\(^2\) - (2x)\(^2\) = (18 - 2x)(18 + 2x)
2x + 18 = 2x + 18 = (2x + 18)
18 - 2x = 2(a - x)
= -2(x - a)
In preparing rice cutlets, a cook used 75g of rice, 40g of margarine, 105g of meat and 20g of bread crumbs. Find the angle of the sector which represent meat in a pie chart?
30\(^o\)
60\(^o\)
112.5\(^o\)
157.5\(^o\)
Rice = 75g
Margarine = 40g
Meat = 105g
Bread = 20g
Total = 240
The angle of sector represented by meat:
= \(\frac{105}{240} \times \frac{360^o}{1}\)
= 157.5
The angle of a sector of a circle radius 10.5 cm is 48\(^o\). Calculate the perimeter of the sector.
25.4cm
25.4cm
25.6cm
29.8cm
Lenght of arc AB = \(\frac{Q}{360}\) 2\(\pi\)r
= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x 10.5
= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\)
= \(\frac{4 \times 22 \times 3}{30} \times \frac{88}{10}\) = 8.8cm
The perimeter = 8.8 + 2r = 8.8 + 2r
= 8.8 + 2(10.5)
= 8.8 + 21
= 29.8cm
What is the product of \(\frac{27}{5}\), \((3)^{-3}\) and (\(\frac{1}{5})^{-1}\)?
5
3
1
\(\frac{1}{25}\)
\(\frac{27}{5} \times 3^{-3} \times \frac{1^{-1}}{5}\)
= \(\frac{27}{5} \times \frac{1}{3^3} \times \frac{1}{\frac{1}{5}}\)
\(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{5}{1}\) = 1
A crate of soft drinks contains 10 bottle of Coca-Cola 8 of Fanta and 6 of Sprite. If one bottle is selected at random, what is the probability that it is Not a Coca-Cola bottle?
\(\frac{5}{12}\)
\(\frac{1}{3}\)
\(\frac{3}{4}\)
\(\frac{7}{12}\)
Coca-Cola = 10
Fanta = 8
Spirite = 6
Total = 24
P(Coca-Cola) = \(\frac{10}{24}\); P(not Coca-Cola)
1 - \(\frac{10}{24}\)
\(\frac{24 - 10}{24} = \frac{14}{24} = \frac{7}{12}\)
In this fiqure, PQ = PR = PS and SRT = 68\(^o\). Find QPS
136\(^o\)
124\(^o\)
112\(^o\)
68\(^o\)
PQRS is quadrilateral. Hence,
2y + 2x + QPS = 360\(^o\)
i.e. (y + x) + QPS = 360\(^o\)
QPS = 360\(^o\) - 2 (y + x)
But x + y + 68\(^o\) = 180\(^o\)
Therefor,
x + y = 180\(^o\) - 68\(^o\) = 112\(^o\)
QPS = 360 - 2(112\(^o\))
= 360\(^o\) - 224
= 136\(^o\)
Find the gradient of the line passing through the points (-2, 0) and (0, 4)
2
-4
-2
4
Given (-2, 0) and (0, -4),
The gradient is of a line is the difference in height (y co-ordinates) ÷ The difference in width (x co-ordinates):
= \(\frac{-4 - 0}{0 - (-2)}\)
= \(\frac{-4}{2}\)
= -2
Find the equation of the line through the points (5, 7) parallel to the line 7x + 5y = 12.
5x + 7y = 120
7x + 5y = 70
x + y = 7
15x + 17y = 90
The general equation of a straight line is y = mx + c, where m is the gradient, and y = c is the value where the line cuts the y-axis.
Equation through (5,7) parallel to the line 7x + 5y = 12
5y = 7x + 12
y = \(\frac{-7x}{5} + \frac{12}{5}\)
Gradient = \(\frac{-7}{5}\)
Required equation = \(\frac{y - 7}{x - 5} = \frac{-7}{5}\)
5y - 35 = -7x + 35
5y + 7x = 70
If N225.00 yields N27.00 in x years simple interest at the rate of 4% per annum, find x.
3
4
12
17
Principal = N225.00
Interest = N27.00
Year = x
Rate = 4%
1 = \(\frac{PRT}{100}\)
27 = \(\frac{225 \times 4 \times x}{100}\) = 2700 = 900T
T = \(\frac{2700}{900}\)
= 3 years