Monday, June 19, 2017

Solution to a tricky NECO SSCE 2017 question on Variation

Suppose y varies as the sum of two quantities, one of which varies
directly as x and the other inversely with x. When x = 2, y=4 and when
x = 5 , y = 7. Find the relationship between x and y.
















Solution.

Let the two quantities be a and b.
Then,
y= k (a + b ) ...........................................(1).
a= mx .......................................................(2)
b = n/x ......................................................(3).
K, m and n are the constants of variation. Using the same letter for
the 3 constants lead to a deadlock.
Substitute Equations 2 and 3 into Equation 1.

y = kmx + (kn /x) ..................................(4).

x =2 when y= 4 turns equation 4 into:

4 = 2mk + (kn/2).
8 = 4mk + kn ...................................(5).
x=5 when y =7 turns equation 4 into :
7 = 5mk + (kn /5).
35 = 25mk + kn .............................(6).
Solve equations 5 and 6 simultaneously.

km = 9/7 and kn = 20/7.
Hence, equation 4 becomes :
y = (9/7)x + [ (20/7) ÷ x ]
= (9x/7) + (20/7x)
= (9x² + 20)/(7x).
This is the required relationship.

Confirm that y =4 when x =2 and y =7 when x =5 by substituting 2 for x
and later 5 for x in the derived relationship.