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JAMB Mathematics Past Questions & Answers - Page 228

1,136.

If x is the addition of the prime numbers between 1 and 6; and y the H.C.F. of 6, 9, 15. Find the product of x and y

Correct answer is B

Prime numbers between 1 and 6 are 2, 3 and 5 x = 2 + 3 = 5 = 10 H.C.F. of 6, 9, 15 = 3 ∴ y = 3 X x y = 10 x 3 = 30

1,137.

$\begin{array}{c|c} Scores(x) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency(f) & 7 & 11 & 6 & 7 & 7 & 5 & 3\end{array}$
In the distribution above, the mode and median respectively are

1,3

1, 2

3, 3

0, 2

View answer & explanation

Correct answer is B

From the distribution, Mode = 1 and

Median = $\frac{2 + 2}{2}$

= 2

= 1, 2

1,138.

In a class of 150 students, the sector in a pie chart representing the students offering Physics has angle 12^o. How many students are offering Physics?

View answer & explanation

Correct answer is D

No of students offering Physics are $\frac{12}{360}$ x 150

= 5

1,139.

Find correct to one decimal place, 0.24633 $\div$ 0.0306

0.8

1.8

8.0

8.1

View answer & explanation

Correct answer is D

$\frac{0.24633}{0.03060}$ multiplying throughout by 100,000

= $\frac{24633}{3060}$

= 8.05

= 8.1

1,140.

If 7 and 189 are the first and fourth terms of geometric progression respectively, find the sum of the first three terms of the progression

182

View answer & explanation

Correct answer is B

$T_{n} = ar^{n - 1}$ (nth term of a G.P)

$T_{4} = ar^{3} = 189$

$7 \times r^{3} = 189 \implies r^{3} = 27$

$r = \sqrt[3]{27} = 3$

$S_{n} = \frac{a(r^{n} - 1)}{r - 1}$

$S_{3} = \frac{7(3^{3} - 1)}{3 - 1}$

= $\frac{7 \times 26}{2} = 91$

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