2018 JAMB Mathematics Past Questions & Answers - page 1

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1

In a class of 40 students, 32 offer Mathematics, 24 offer Physics and 4 offer neither Mathematics nor Physics. How many offer both Mathematics and Physics?

A 4
B 8
C 16
D 20


Correct Option: D



2

Find the values of x for which

\(\frac {x+2}{4}\) - \(\frac{2x - 3}{3}\) < 4

A x < 8
B x > -6
C x < 4
D x > -3


Correct Option: B



3

Find \N\

A 65
B 23
C 17
D 91


Correct Option: C



4

If X, Y can take values from the set (1, 2, 3 ,4), find the probability that the product of X and Y is not greater than 6

 

A \(\frac{5}{8}\)
B \(\frac{5}{16}\)
C \(\frac{1}{2}\)
D \(\frac{3}{8}\)


Correct Option: D


Solution

  Pr(product of x and y NOT > 6 ) = \(\frac{8}{16}\) = \(\frac{1}{2}\)


5

The pie chart shows the monthly expenditure of a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is N7200?

A 1000
B 2000
C 3000
D 4000


Correct Option: B



6

Evaluate 5\(^{-3}\) log\(^2\) x 2 \(^{2 \log 3}\)

A 8
B A. \(\frac{1}{8}\)
C B. 1\(\frac{1}{8}\)
D C. \(\frac{2}{5}\)


Correct Option: B



7

A trader realises 10x - x\(^2\) Naira profit from the sale of x bags of corn. How many bags will give him the maximum profit?

A 7
B 6
C 5
D 4


Correct Option: C


Solution

Profit (P) = 10\(_x\) − \(_x\)2

At max profit, the differential of profit with respect to number of bags(x) = 0

Hence,

  \(\frac{dp}{dx}\) = 0

 \(\frac{dp}{dx}\) = 10 - 2x = 0

  10 = 2x

  Then x = \(\frac{10}{2}\) = 5


8

If y = 23\(_{five}\) 101\(_{three}\) , find y, leaving your answer in base two

A 1110
B 10111
C 11101
D 111100


Correct Option: B



9

Find the value of x in the diagram

A 10°
B 28°
C 36°
D 40°


Correct Option: D


Solution

Total angle at a point = 360

  x - 10 + 4x - 50 + 2x + 3x + 20 = 360

  10x - 40 = 360

  10x = 360 + 40

  10x = 400

  x = \(\frac{400}{10}\)

  x = 40


10

Solve for t in the equation \(\frac{3}{4}\)t + \(\frac{1}{3}\)(21 - t) = 11

A \(\frac{9}{13}\)
B \(\frac{7}{13}\)
C 5
D 9\(\frac{3}{5}\)


Correct Option: D


Solution

\(\frac{3}{4}\) t + \(\frac{1}{3}\) (21 - t) = 11

  Multiply through by the LCM of 4 and 3 which, that is 12

  12 x(\(\frac{3}{4}\) t) + 12 x (\(\frac{1}{3}\) (21 - t)) = (11 x 12)

  9t + 4(21 - t) = 132

  9t + 84 - 4t = 132

  5t + 84 = 132

  5t = 132 - 84 = 48

  t = \(\frac{48}{5}\)

  t = 9 \(\frac{3}{5}\)


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