JAMB Mathematics Past Questions & Answers - Page 427

Hint: Use BODMAS and algebra to arrive at the values of P = 5/2, q = -25/4 and r = -9/2.
Then substitute the values of p and q into p+2q to get -10.

Hint: apply basic mathematics rules beginning from BODMAS to algebra, and follow solution carefully to arrive at

p = \(\frac{5}{2}\), q = -\(\frac{25}{4}\) and r = -\(\frac{17}{4}\)

NUMERATOR:   a\(^{\frac{2}{1}}*{\frac{3}{4}}\)          b\(^{\frac{-3}{1}}*{\frac{3}{4}}\)          c\(^{\frac{1}{1}}*{\frac{3}{4}}\)

                           a\(\frac{3}{2}\)      b\(\frac{-9}{4}\)    c\(\frac{3}{4}\)

Using index method for numerator & denominator :  a\(^{\frac{3}{2}}-{\frac{-1}{1}}\)           b\(^{\frac{-9}{4}}-{\frac{4}{1}}\)           c\(^{\frac{3}{4}}-{\frac{5}{1}}\)

:  a\(\frac{5}{2}\)      b\(\frac{-25}{4}\)    c\(\frac{-17}{4}\)

Then p+2q will give you \(\frac{5}{2}+2\left(\frac{-25}{4}\right)= -10\)

Start your solution by cross-multiplying,

\(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\)

[x - √2] √2  = x + √2

where √2×√2=2 

x√2 - √2 * √2 = x + √2

then collect like terms

x√2 - x = √2 + 2

and factorize accordingly to get the unknown.

x(√2 - 1) = √2 + 2

x = \(\frac{√2 + 2}{√2 - 1}\)

rationalize 

x = \(\frac{√2 + 2}{√2 - 1}\) * \(\frac{√2 + 1}{√2 + 1}\)

x = \(\frac{√4 + √2 + 2√2 + 2}{√4  + √2 - √2 - 1}\)

x = \(\frac{2 +  3√2 + 2}{2 - 1}\)

x =  \(\frac{3√2 + 4}{1}\)

x = 3√2 + 4

\(log_810\) = X = \(log_8{2 x 5}\)

\(log_82\) + \(log_85\) = X

Base 8 can be written as \(2^3\)

\(log_82 = y\)

therefore \(2 = 8^y\)

\(y = \frac{1}{3}\)

\(\frac{1}{3} = log_82\)

taking \(\frac{1}{3}\) to the other side of the original equation

\(log_85 = X-\frac{1}{3}\)

Let the three items be M, Y and P.
n{M ∩ Y} only = 4-3 = 1
n{M ∩ P) only = 5-3 = 2
n{ Y ∩ P} only = 2
n{M} only = 12-(1+3+2) = 6
n{Y} only = 10-(1+2+3) = 4
n{P} only = 14-(2+3+2) = 7
n{M∩P∩Y} = 3
Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women.