## Mathematics (2018) Study Mode

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1.

Find the values of x for which

$$\frac {x+2}{4}$$ - $$\frac{2x - 3}{3}$$ < 4

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

2.

Find the values of x for which

$$\frac {x+2}{4}$$ - $$\frac{2x - 3}{3}$$ < 4

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

A.
65
B.
23
C.
17
D.
91

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

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A.
$$\frac{5}{8}$$
B.
$$\frac{5}{16}$$
C.
$$\frac{1}{2}$$
D.
$$\frac{3}{8}$$
Explanation. Give a better Explanation

Each multiplication of elements gives 16 results out of which the ones that are not greater than 6 are asterisked (*) and they are 8 in numbers

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?

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A.
1000
B.
2000
C.
3000
D.
4000

6.

Evaluate 5$$^{-3}$$ log$$^2$$ x 2 $$^{2 \log 3}$$

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

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?

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A.
7
B.
6
C.
5
D.
4
Explanation. Give a better Explanation

Profit (P) = 10$$_x$$ − $$_x$$2

Maximum profit can be achieved when the differential of profit with respect to number of bags(x) is 0

i.e. $$\frac{dp}{dx}$$ = 0

$$\frac{dp}{dx}$$ = 10 - 2x = 0

10 = 2x

Then x = $$\frac{10}{2}$$ = 5

8.

Find the value of x in the diagram

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A.
10°
B.
28°
C.
36°
D.
40°
Explanation. Give a better Explanation

The diagram shows angles at a point, the total angle at a point is 360

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

10x - 40 = 360

10x = 360 + 40

10x = 400

x = $$\frac{400}{10}$$

x = 40

9.

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

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A.
$$\frac{9}{13}$$
B.
$$\frac{7}{13}$$
C.
5
D.
9$$\frac{3}{5}$$
Explanation. Give a better Explanation

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

Multiply through by the LCM of 4 and 3 which 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 $$\frac{3}{5}$$

10.

A school girl spends $$\frac{1}{4}$$ of her pocket money on books and $$\frac{1}{3}$$ on dress. What fraction remains?

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A.
$$\frac{5}{6}$$
B.
$$\frac{7}{12}$$
C.
$$\frac{5}{12}$$
D.
$$\frac{1}{6}$$