Advanced quantitative strategy: plug in

Scoring higher requires different strategy. One should not stick with just algebraic approach, rather use different strategy, such as plug in. Today, I am going to explain how to use plug in method. Sometimes, when  a quantitative math requires many steps , thus complicated, it is difficult to solve that math using traditional method. At that time plug in method is boon. You must be prepared to use different strategies in the quant section.

What is plug in method?
Plug in method means using smart numbers in math to solve the question efficiently. For instance, If there is variable in the question, you can replace the variable by any number. Using the numbers in lieu of the variables is easier and faster to solve that math. 


When to use Plug in method:

1. You can use plug in method when there are variables in the question

2.You can use plug in method when there are variables in the answer choices

3. When the algebraic approaches requires many steps

4. You are not good in algebra or afraid of algebra

5. Algebraic method might require longer time, but considering the time limit in GRE you should use plug in method.

6. But mind it when the math is too simple, use algebraic approach.

Now lets see the math below:

1.At the end of each year, the value of a certain Champagne is Q percent more than its value one year earlier, where Q has the same value each year. If the value of the Champagne was t dollars on January 1, 2002, and p dollars on January 1, 2004, then in terms of p and t, what was the value of the Champagne, in dollars on January 1st, 2005?
1. p + (1/2)(p-t)
2. p + (1/2)((p-t)/t)p
3. (p√p)/√t
4. p² /4t
5. tp²

We have to plug in smart numbers: 10,100,100 etc.

So lets take Q=100 and t=10

 so in 2002, the price is=10
in 2003. the price is =10+100% of 10=20
in 2004, the  price is=20+100% of 20=40
 that is p
so p=40
so in 2005, the price will be 40+100% of 40=80

So now we will put p=40 and t=10, and the answer choice that will yield 80 will be our answer!!

1. p + (1/2)(p-t) =40+(1/2)(40-10)=40+15=55 :: Eliminate it as we need 80

2. p + (1/2)((p-t)/t)p=40+(1/2)((40-10)/10)40=not 80=Eliminate it!

3. (p√p)/√t=(40√40)/√10=(40√(4*10))/√10=40*2*√10/√10=80=Bingo!

4. p² /4t=40² /4*10=1600/40=40=Elimiate it!

5. tp²=10*40²=16000=Eliminate it!

So 3 is our answer!!