# Sec 2 Maths Direct and Inverse Proportion

I realise that many of my Sec 2 students find proportion question pretty easy except questions on the type of questions where by1 variable is changed and the question ask how will the other variable be affected.

Hence, I am going to list down the 3 steps method taught during my lessons.

Qns) Given that x is directly proportional to y2(square) and that x=17 for a certain value of y. When y doubles in value, what is the value of x?

Solution:

FIRST STEP: WRITE THE MAIN EQUATION

x=ky2 , k is a constant ————-> Equation 1

Input any details from the question

17= k(y square)

SECOND STEP: MAKE K THE SUBJECT

k = 17/y square ———————–> Equation 2

THIRD STEP: EXPRESS NEW Y IN TERMS OF ORIGINAL Y

y(new) = 2y ———————-> Equation 3

LASTLY SUB EQUATION 3 and 2 IN EQUATION 1

x (new) = 17 /y square times ( 2y ) square

= 17 / y square times 4 y square

= 17 times 4 ( The y square got camcelled off)

= 68

These 3 steps method can be used always for this type of questions. So the key is consistency!

I am going to do another similar question on inverse proportion via youtude video in my next post. 🙂

Im expecting my sec 2 kids to excel in this type of questions soon. 🙂