JAMB Mathematics Past Questions & Answers - Page 244

1,216.

Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?

A.

\(\frac{2}{3}\)

B.

\(\frac{2}{15}\)

C.

\(\frac{1}{2}\)

D.

\(\frac{1}{4}\)

Correct answer is C

\(\begin{array}{c|c} 1 & 2 & 3 & 4\\\hline 1(1, 1) & (1, 2) & (1, 3) & (1, 4)\\ \hline 2(2, 1) & (2 , 2) & (2, 3) & (2, 4) \\ \hline 3(3, 1) & (3, 2) & (3, 3) & (3, 4)\\ \hline 4(4, 1) & (4, 2) & (4, 3) & (4, 4)\end{array}\)

sample space = 16

sum of nos. removed are (2), 3, (4), 5

3, (4), 5, (6)

(4), 5, (6), 7

(5), 6, 7, (8)

Even nos. = 8 of them

Pr(even sum) = \(\frac{8}{16}\)

= \(\frac{1}{2}\)

1,217.

Below are the scores of a group of students in a test
\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}\)
If the average scores is 3.5, find the value of x

A.

1

B.

2

C.

3

D.

4

Correct answer is B

\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}\)
Average = 3.5

3.5 = \(\frac{(1 \times 1) + (2 \times 4) + (3 \times 5) + (4 \times 6) + 5x + (6 \times 2)}{1 + 4 + 5 + 6 + x + 2}\)

\(\frac{3.5}{1}\) = \(\frac{1 + 8 + 15 + 24 + 5x + 12}{18 + x}\)

\(\frac{3.5}{1}\) = \(\frac{60 + 5x}{18 + x}\)

60 + 5x = 3.5(18 \(\div\) x)

60 + 5x = = 63 + 1.5x

5x - 1.5x = 63 - 60

1.5x = 3

x = \(\frac{3}{1.5}\)

\(\frac{30}{15}\) = 2

1,218.

The mean of ten positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number

A.

3

B.

16

C.

18

D.

32

Correct answer is D

Mean of 10 numbers = 16

The total sum of numbers = 16 x 10 = 160

Mean of 11 numbers = 18

Total sum of numbers = 11 x 18

= 198

The 11th no. = 198 - 160

= 38

1,219.

The mean of ten positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number

A.

3

B.

16

C.

18

D.

32

Correct answer is D

Mean of 10 numbers = 16

The total sum of numbers = 16 x 10 = 160

Mean of 11 numbers = 18

Total sum of numbers = 11 x 18

= 198

The 11th no. = 198 - 160

= 38

1,220.

4, 16, 30, 20, 10, 14 and 26 are represented on a pie chart. Find the sum of the angles of the bisectors representing all numbers equals to or greater than 16

A.

48o

B.

84o

C.

92o

D.

276o

Correct answer is D

Given that 4, 16, 30, 20, 10, 14 and 26

Adding up = 120

nos \(\geq\) 16 are 16 + 30 + 20 + 26 = 92

The requires sum of angles = \(\frac{92}{120}\) x \(\frac{360^o}{1}\)

= 276o