Mathematics Questions & Answers - Free Online Practice

2,031.

In a restaurant, the cost of providing a particular type of food is partly constant and partially inversely proportional to the number of people. If cost per head for 100 people is 30k and the cost for 40 people is 60k, Find the cost for 50 people?

15k

20k

50k

40k

45k

View answer & explanation

Correct answer is C

C = a + k

\(\frac{1}{N}\) = c

= \(\frac{aN + k}{N}\)

CN = aN + K

30(100) = a(100) + k

3000 = 100a + k.......(i)

60(40) = a(40) + k

2400 = 40a + k.......(ii)

eqn (i) - eqn (ii)

600 = 60a

a = 10

subt. for a in eqn (i) 3000 = 100(10) + K

3000 - 1000 = k

k = 2000

CN = 10N + 2000. when N = 50,

50C = 10(50) + 2000

50C = 500 + 2000

C = \(\frac{2500}{50}\)

= 50k

2,032.

\((\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2})(\sqrt{3} + \sqrt{2})\) is equal to

(\(\sqrt{2} + 4\sqrt{2}\))

(6\(\sqrt{2}\)

View answer & explanation

Correct answer is A

\((\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2})(\sqrt{3} + \sqrt{2})\)

\((\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2}) = \sqrt{3} - \sqrt[4]{6} + \sqrt[4]{6} - \sqrt{2}\)

= \(\sqrt{3} - \sqrt{2}\)

\((\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2}) = 3 + \sqrt{6} - \sqrt{6} - 2\)

= \(3 - 2 = 1\)

2,033.

If \(e^{x} = 1 + x + \frac{x^{2}}{1.2} + \frac{x^{3}}{1.2.3} + ... \), find \(\frac{1}{e^{\frac{1}{2}}}\)

1 - \(\frac{x}{2}\) + \(\frac{x^2}{1.2^3}\) + \(\frac{x^3}{2^4.3}\) + .........

1 - \(\frac{x}{2}\) + \(\frac{x^2}{1.2^3}\) + \(\frac{x^4}{2.4.3}\) + ..

1 + \(\frac{x}{2}\) + \(\frac{x^2}{1.2}\) + \(\frac{x^3}{1.2.3}\) + \(\frac{x^4}{1.23.4}\) + .........

1 - x + \(\frac{x^2}{1.2^3}\) + \(\frac{x^3}{2^4.3}\) + .........

1 + \(\frac{x}{2}\) + \(\frac{x^2}{1.2^3}\) + \(\frac{x^4}{1.2.6}\) + .........

View answer & explanation

Correct answer is C

\(e^{x} = 1 + x + \frac{x^{2}}{1.2} + \frac{x^{3}}{1.2.3} + ...\)

\(\frac{1}{e^{\frac{1}{2}}} = e^{-\frac{1}{2}}\)

\(e^{-\frac{1}{2}} = 1 - \frac{x}{2} + \frac{x^{2}}{1.2^{3}} - \frac{x^{3}}{1.2^{4}.3} + ... \)

2,034.

Find the values of p for which the equation x² - (p - 2)x + 2p + 1 = 0

(21, 0)

(0, 12)

(1, 2)

(3, 4)

(4, 5)

View answer & explanation

Correct answer is B

Equal roots implies b² - 4ac = 0

a = 1b = - (p - 2), c = 2p + 1

[-(p - 2)]² - 4 x 1 x (2p + 1) = 0

p² - 4p + 4 - 4(2p + 1) = 0

p² - 4p = 4 - 8p - 4 = 0

p² - 12p = 0

p(p - 12) = 0

p = 0 or 12

2,035.

22\(\frac{1}{2}\)% of the Nigerian Naira is equal to 17\(\frac{1}{10}\)% of a foreign currency M. What is the conversion rate of the M to the Naira?

1M = 1\(\frac{15}{57}\)N

1M = 38\(\frac{1}{4}\)N

1M = 1\(\frac{18}{57}\)N

1M = 384\(\frac{3}{4}\)N

View answer & explanation

Correct answer is C

N = 22\(\frac{1}{2}\)%, M = 17\(\frac{1}{10}\)%

M = \(\frac{171}{10}\)%, N = \(\frac{45}{2}\)

\(\frac{45}{2}\) x \(\frac{10}{171}\)

= \(\frac{225}{171}\)

= 1 \(\frac{54}{171}\)

= 1 \(\frac{18}{57}\)