JAMB Mathematics Past Questions & Answers - Page 36

176.
177.

A surveyor walks 500m up a hill which slopes at an angle of 30\(^o\). Calculate the vertical height through which he rises

A.

252m

B.

500m

C.

250m

D.

255m

Correct answer is C

\(\frac{h}{500}\) = sin 30\(^o\) 

= 500 sin 30\(^o\) 

= 500 x \(\frac{1}{2}\) 

= 250m 

178.

The mean age group of some students is 15years. When the age of a teacher, 45 years old, is added to the ages of the students, the mean of their ages become 18 years. Find the number of students in the group. 

A.

7

B.

9

C.

15

D.

42

Correct answer is B

x \(\frac{\sum x}{N}\)

15 = \(\frac{\sum x }{N}\)

\(\sum x\) = 15N........(i)

y = \(\frac{\sum y}{Ny} = \frac{\sum x + 45}{N + 1}\)

\(\frac{18}{1} = \frac{15N + 45}{N + 1}\) 

18(N + 1) = 15N + 45

18N + 18 = 15N + 45

18N - 15N = 45 - 18 

3N = 27 

N = \(\frac{27}{3}\) 

= 9

179.

The sum of the interior angle of pentagon is 6x + 6y. Find y in terms of x.

A.

y = 6 - x

B.

y = 90 - x

C.

y = 120 - x

D.

y = 150 - x

Correct answer is B

Sum of interior angles = (2n - 4) 90\(^o\) 

For perntagon, n = 5 

Sum of interior angles = 6 x 90\(^o\) = 540\(^o\) 

6x + 6y = 540\(^o\) 

6(x + y) = 540\(^o\) 

x + y = \(\frac{540^o}{6}\) = 90\(^o\) 

y = 90\(^o\) 

y = 90 - x