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# [Smart Math] Algebra Problem 26

Here’s and example of a SMART MATH problem for ALGEBRA.

### Problem

How much water must be added to 100 cc of a 60% solution of sulphuric acid to reduce it to a 40% solution?

1. 20 cc
2. 30 cc
3. 40 cc
4. 50 cc
5. 60 cc

### The Usual Method

Let ‘x’ cc be the amount of water added to 100 cc of sulphuric acid. Concentration of acid in water = 0%

Hence,

$\frac{x\times 0+100\times 0.60}{100+x}=0.40$

$=\frac{0+60}{100+x}=0.40$

$\therefore 60=40+0.40x$

$\therefore 20=0.40x$

$\therefore x=\frac{20}{0.40}=50$cc

(Ans: 4)

Estimated Time to arrive at the answer = 45 seconds.

### Using Technique

In order to reduce concentration from 60 to 40 (i.e. by 33.33%) we need to increase the quantity of the solution by 50% (See concept explained in section of Percentages in Stuff to Remember.)

Thus if solution initially was 100 cc, it should now be 150 cc. Thus additional 50 cc is to be added.

(Ans: 4)

Estimated Time to arrive at the answer = 10 seconds.
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## 5 replies on “[Smart Math] Algebra Problem 26”

karthiksays:

how come 33.33% ?

LazyMaths.comsays:

When you reduce from 60 to 40, you are removing 20 out of 60 which is same as 1 out of 3 or 33.33%

PhantomMathematiciansays:

We are reducing 60 to 40, thus reducing 20.
We know that percentage reduction= (Amount reduced/ Initial amount)* 100
= (20/60)*100 %
= (1/3)*100 %
= (0.3333333…)100 %
= 33.333… %
Plain, simple, but smartly solved.

mukeshsays: