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

**Problem**

**Problem**

It takes 60 turns for one vessel to empty the tank. How much turns would it take totally to empty the tank if the vessel is used alternately with another one 1/3^{rd} its capacity?

- 30
- 90
- 100
- 60
- 45

**The Usual Method**

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Let the vessels be A and B.

Vessel A takes 60 turns to empty the tank so vessel B will take 60 x 3 = 180 turns to empty the tank as it is only 1/3^{rd} the size of the vessel A.

If both vessels work together of the tank will be emptied by vessel A and of it will be emptied by vessel B.

Hence, + of the tank will be emptied by one turn of each vessel A and B.

, 45 turns of vessel A and 45 turns of vessel B will empty the tank completely. Hence, total number of turns = 45 + 45 = 90.

**(Ans: 2)**

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

**Using Technique**

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Note that capacities of vessel A to that of B = 3 : 1. Hence in one turn of each 3*x* + 1*x* = 4*x* tank will be emptied (where ‘*x*’ the constant of proportionality). The tank in turn is large.

Hence, turns of each vessel will be needed. Hence total 90 turns are needed.

**(Ans: 2)**

*Estimated Time to arrive at the answer = 15 seconds.*

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