JAMB Mathematics Past Questions & Answers - Page 154

766.

Given that \(a*b = ab + a + b\) and that \(a ♦ b = a + b = 1\). Find an expression (not involving * or ♦) for (a*b) ♦ (a*c) if a, b, c, are real numbers and the operations on the right are ordinary addition and multiplication of numbers

A.

ac + ab + bc + b + c + 1

B.

ac + ab + a + c + 2

C.

ab + ac + a + b + 1

D.

ac + bc + ab + b + c + 2

E.

ab + ac + 2a + b + c + 1

Correct answer is E

Soln. a*b = ab + a + b,

a ♦ b = a + b + 1

a*c = ac + a + c

(a*b) ♦ (a*c) = (ab + a + b + ac + a + c + 1)

= ab + ac + 2a + b + c + 1

767.

Father reduced the quantity of food bought for the family by 10% when he found that the cost of living had increased 15%. Thus the fractional increase in the family food bill is now

A.

\(\frac{1}{12}\)

B.

\(\frac{6}{35}\)

C.

\(\frac{19}{300}\)

D.

\(\frac{7}{200}\)

E.

\(\frac{5}{100}\)

Correct answer is D

Let the cost of living = y.

The new cost of living = \(y + \frac{15y}{100} = 1.15y\)

The food bill now = \((1 - \frac{90}{100})(1.15y)\)

= \(1.035y\)

The fractional increase in food bill = \((1.035 - 1) \times 100% = 3.5%\)

= \(\frac{35}{1000} = \frac{7}{200}\)

768.

If y = 2x2 + 9x - 35. Find the range of values for which y < 0.

A.

7 < x < \(\frac{5}{2}\)

B.

-5 < 7 < x

C.

-7 < x < 5

D.

-7 < x < \(\frac{5}{2}\)

Correct answer is D

y = 2x2 + 9x - 35

2x2 + 9x = 35

x2 + \(\frac{9}{2}\) = \(\frac{35}{2}\)

x2 + \(\frac{9}{2}\) + \(\frac{81}{16}\) = \(\frac{35}{2}\) = \(\frac{81}{16}\)

(x + \(\frac{9}{4}\))2 = \(\frac{361}{16}\)

x = \(\frac{-9}{4}\) + \(\frac{\sqrt{361}}{16}\)

x = \(\frac{-9}{4}\) + \(\frac{19}{4}\)

= 2.5 or -7

-7 < x < \(\frac{5}{2}\)

769.

Five years ago, a father was 3 times as old as his son, now their combined ages amount to 110years. thus, the present age of the father is

A.

75 years

B.

60 years

C.

98 years

D.

80 years

E.

105 years

Correct answer is D

Let the present ages of father be x and son = y

five years ago, father = x - 5

son = y - 5

x - 5 = 3(y - 5)

x - 5 = 3y - 15

x - 3y = -15 + 5 = -10 ......(i)

x + y = 110......(ii)

eqn(ii) - eqn(i)

4y = 120

y = 30

sub for y = 30 in eqn(i)

x -10 = 3(30)

x -10 = 90

x = 80

770.

What is the number whose logarithm to base 10 is 2.3482?

A.

223

B.

2228

C.

2.235

D.

22.37

E.

0.2229

Correct answer is A

\(\begin{array}{c|c} Nos. & log \\ \hline 222.9 & 2.3482\end{array}\)

N : b Look for the antilog of .3482 which gives 222.9