Mathematics questions and answers

Mathematics Questions and Answers

How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.

1,786.

Simplify \(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\)

A.

\(\frac{4ab}{a - b}\)

B.

\(\frac{-4ab}{a^2 - b^2}\)

C.

\(\frac{-4ab}{a^{-2} - b}\)

D.

\(\frac{4ab}{a^{-2} - b^{-2}}\)

Correct answer is B

\(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) = \(\frac{(a - b)^2}{(a + b)}\) - \(\frac{(a + b)^2}{(a - b0}\)

applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b)

= (a = b + a = b)(a = b - a = b)

2a(-2b) = -4ab

= \(\frac{-4ab}{(a + b)(a - b)}\)

= \(\frac{-4ab}{a^2 - b^2}\)

1,787.

If the four interior angles of a quadrilateral are (p + 10)°, (p - 30)°, (2p - 45)°, and (p + 15)°, then p is

A.

125o

B.

82o

C.

135o

D.

105o

E.

60o

Correct answer is B

Sum of interior angles of a polygon = (2n - 4) x 90

where n is the no. of sides

sum of interior angles of the quad. = ({2 x 4} -4) x 90

(8 - 4) x 90 = 4 x 90o

= 360o

(p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o

p + 10o + p - 30o + 2p - 45o + p + 15 = 360o

5p = 360o + 50o

= 82o

1,788.

If a circular paper disc is trimmed in such a way that its circumference is reduced in the ratio 2:5, In what ratio is the surface area reduced?

A.

8 : 125

B.

2 : 5

C.

8 : 25

D.

4 : 25

E.

4 : 10

Correct answer is D

surface area of formula = πr\(^2\) 

If the radius is reduced then let its radius be x.

Its area is πx\(^2\) .

x : r = 2 : 5,

so \(\frac{x}{r}\) =  \(\frac{2}{5}\)

 → 5x = 2r

Hence, x = 0.4r.

Hence the area of the new circle is π (0.4r)\(^2\)  = 0.16π r\(^2\) .

The ratio of the two areas is

0.16πr\(^2\) : πr\(^2\) 

 = 0.16 : 1 = 16 : 100

= 4 : 25.

1,789.

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

1,790.

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 = \((1.035 - 1) \times 100% = 3.5%\)

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