JAMB Mathematics Past Questions & Answers - Page 168

836.

Solve the given equation \((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\)

A.

27

B.

9

C.

\(\frac{1}{27}\)

D.

18

E.

81

Correct answer is A

\((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\)

Let \(\log_{3} x = a\).

\(a^{2} - 6a + 9 = 0\)

\(a^{2} - 3a - 3a + 9 = 0\)

\(a(a - 3) - 3(a - 3) = 0\)

\((a - 3)(a - 3) = 0\)

\(\implies a = 3 (twice)\)

\(\log_{3} x = 3 \implies x = 3^{3} = 27\)

837.

The sum of the root of a quadratic equation is \(\frac{5}{2}\) and the product of its root is 4. The quadratic equation is

A.

x + 8x2 - 5 = 0

B.

2x2 - 5x + 8 = 0

C.

2x2 - 8x + 5 = 0

D.

x2 + 8x - 5 = 0

Correct answer is B

x2 - (sum of roots) x + (product of roots) = 0

x2 - \(\frac{5x}{2}\) + 4 = 0

2x2 - 5x + 8 = 0

838.

What is the size of an exterior angle of a regular pentagon?

A.

36o

B.

60o

C.

72o

D.

120o

E.

360o

Correct answer is C

Pentagon is a polygon with 5 sides. Sum of interior angles of polygon (regular) = (2n - 4) x 90o

= 5

= (10 - 4) x 90o

= 540o

each interior angle = \(\frac{540}{5}\)

= 108o

= 180o let x represent ext. angle

x + 108o = 180v

x = 180o - 108o

= 72o

839.

The difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\) greater by

A.

\(\frac{23}{28}\)

B.

\(\frac{24}{28}\)

C.

\(\frac{50}{56}\)

D.

\(\frac{27}{28}\)

E.

\(\frac{48}{56}\)

Correct answer is C

Difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\)

\(\frac{33}{7}\) - \(\frac{9}{4}\) = \(\frac{69}{24}\)

The sum of \(\frac{1}{14}\) and 1\(\frac{1}{14}\) + \(\frac{3}{2}\)

= \(\frac{11}{7}\)

\(\frac{69}{28}\) - \(\frac{11}{7}\) = \(\frac{25}{28}\)

= \(\frac{50}{56}\)

840.

A baking recipe calls for 2.5kg of sugar and 4.5kg of flour. With this recipe some cakes were baked using 24.5kg of a mixture of sugar and flour. How much sugar was used?

A.

12.25kg

B.

6.75kg

C.

8.75kg

D.

15.75kg

E.

8.25kg

Correct answer is C

Sugar : flour = 2.5 : 4.5

Total = 7

sugar used = \(\frac{2.5}{7}\) x 24.5

= \(\frac{61.25}{7}\)

= 8.75