Here’s and example of a **SMART MATH** problem for **RATIO PROPORTION.**

**Problem**

**Problem**

A can give B a 30 meters start and C an 80 meters start in a kilometer race. What start can B give to C in the same race?

(1 Kilometer = 1000 meters)

- 50 meters
- 61 meters
- 52 meters
- 45 meters
- 40 meters

**The Usual Method**

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When A travels 1000 meters,

B travels 970 meters and

C travels 920 meters.

Hence, when B has to travel 1000 meters, C will travel meters.

This comes from cross multiplication as shown below:

A B C

1000 970 920

1000 x

= or 948.45 meters

Therefore B can give a start of 1000 – 948.45 = 51.55 meters 52 meters

**(Ans: 3)**

*Estimated Time to arrive at the answer = 60 seconds*

**Using Technique**

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From the data in the question, it is clear that the descending order of speed is A>B>C, i.e. A being the fastest, followed by B and C is the slowest. Also in a race of 970 meters, B can give a start of 50 meters (970 – 920) to C, Hence, it is obvious that in a 1000 meters race, the start for C has to be proportionately greater than 50 as 1000 is greater than 970. Hence, the answer has to be little over the difference of 970 and 920 meters or 50 meters; plus something. Looking at the options, the only option matching this condition is option ‘3’ or 52 meters

**(Ans: 3)**

*Estimated Time to arrive at the answer = 10 seconds*

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