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.
Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30
x = 3 or 7, y = 12 or 8
x = 6 or 1, y = 11 or 5
x = 2 or 5, y = 15 or 6
x = 1 or 5, y = 10 or 7
Correct answer is C
3x + y = 21 ... (i);
xy = 30 ... (ii)
From (ii), \(y = \frac{30}{x}\). Putting the value of y in (i), we have
3x + \(\frac{30}{x}\) = 21
\(\implies\) 3x\(^2\) + 30 = 21x
3x\(^2\) - 21x + 30 = 0
3x\(^2\) - 15x - 6x + 30 = 0
3x(x - 5) - 6(x - 5) = 0
(3x - 6)(x - 5) = 0
3x - 6 = 0 \(\implies\) x = 2.
x - 5 = 0 \(\implies\) x = 5.
If x = 2, y = \(\frac{30}{2}\) = 15;
If x = 5, y = \(\frac{30}{5}\) = 6.
Simplify \(\frac{0.0839 \times 6.381}{5.44}\) to 2 significant figures.
0.2809
2.51
3.5
0.098
Correct answer is D
No explanation has been provided for this answer.
If 2\(^{x + y}\) = 16 and 4\(^{x - y} = \frac{1}{32}\), find the values of x and y.
x = \(\frac{3}{4}\), y = \(\frac{11}{4}\)
x = \(\frac{3}{4}\), y = \(\frac{13}{4}\)
x = \(\frac{2}{3}\), y = \(\frac{4}{5}\)
x = \(\frac{2}{3}\), y = \(\frac{13}{4}\)
Correct answer is B
2\(^{x + y}\) = 16 ; 4\(^{x - y}\) = \(\frac{1}{32}\).
\(\implies 2^{x + y} = 2^4\)
\(x + y = 4 ... (1)\)
\(2^{2(x - y)} = 2^{-5} \)
\(2^{2x - 2y} = 2^{-5}\)
\(\implies 2x - 2y = -5 ... (2)\)
Solving the equations (1) and (2) simultaneously, we have
x = \(\frac{3}{4}\) and y = \(\frac{13}{4}\)
If 2\(^{x + y}\) = 16 and 4\(^{x - y} = \frac{1}{32}\), find the values of x and y
x = \(\frac{3}{4}\), y = \(\frac{11}{4}\)
x = \(\frac{3}{4}\), y = \(\frac{13}{4}\)
x = \(\frac{2}{3}\), y = \(\frac{4}{5}\)
x = \(\frac{2}{3}\), y = \(\frac{13}{4}\)
Correct answer is B
2\(^{x + y}\) = 16 ; 4\(^{x - y}\) = \(\frac{1}{32}\).
\(\implies 2^{x + y} = 2^4\)
\(x + y = 4 ... (1)\)
\(2^{2(x - y)} = 2^{-5} \)
\(2^{2x - 2y} = 2^{-5}\)
\(\implies 2x - 2y = -5 ... (2)\)
Solving the equations (1) and (2) simultaneously, we have
x = \(\frac{3}{4}\) and y = \(\frac{13}{4}\)
Evaluate \((\frac{6}{0.32} \div \frac{2}{0.084})^{-1}\) correct to 1 decimal place.
1.3
2.5
4.6
3.2
Correct answer is A
\((\frac{6}{0.32} \div \frac{2}{0.084})^{-1}\)
= \((\frac{600}{32} \div \frac{2000}{84})^{-1}\)
= \((\frac{600}{32} \times \frac{84}{2000})^{-1}\)
= \((\frac{63}{80})^{-1}\)
= \(\frac{80}{63}\)
= 1.3 (to 1 decimal place)