# Problem-3 (Word Problem)

**According to the stock policy of a company, each employee working in only the Technical division is given 15 shares of the company stock and each employee in only the Recruitment division is given 13 shares. Employees belonging to both divisions get 20 shares each. There are 20 employees in the company, and each employee belongs to at least one division.**

- If the total number of the shares distributed is 298 and there is at least one employee working in both divisions, then which of the following could be the number of employees working only in the Technical division and the number of employees working only in the Recruitment division?

Technical-Division | Recruitment-Division | |

3 | ||

5 | ||

7 | ||

9 | ||

11 |

# Solution-3 (Word Problem)

For example, if 15 people work in Technical, and 13 work in Recruitment, then

Thus, we have that

Based on this distribution, the total number of shares can be represented as

*m*+*n*â€“ 20 = 15 + 13 â€“ 20 = 28 â€“ 20 = 8 work in both Technical and Recruitment. This means that 15 â€“ 8 = 7 work in Technical only and 13 â€“ 8 = 5 work in recruitment only.Thus, we have that

*m*â€“ (*m*+*n*â€“ 20) =*m*â€“*m*â€“*n*+ 20 = 20 â€“*n*^{ }working in technical; and, similarly, 20 â€“*m*working in recruitment.Based on this distribution, the total number of shares can be represented as

Now, we need to find the possible solutions to the last equation above such that

*m*and*n*are positive integers less than 20.So, substitute the possible values for

*n*(1 though 20) into this equation, and accept the resulting values of*m*that are integers:**Note:**the official GMAT would expect that you to just do the calculations roughly on the paper, so just check which subtraction yields numbers that divide evenly by 5. Hence, the problem is not as long as it appears. It is not suggestible to use calculator here as it might be time consuming.

From the above list of calculations, we have three possible solution pairs:

*m*= 15 and

*n*= 9

*m*= 8 and

*n*= 14

*m*= 1 and

*n*= 19

For the possible solution pair

*m*= 15 and*n*= 9, the number of people working in both departments is*m*+*n*â€“ 20 = 15 + 9 â€“ 20 = 24 â€“ 20 = 4. This number (4) is acceptable because the problem requires that there be at least 1 person working in both departments.Here, there are 20 â€“

For the possible solution pair

Here, there are 20 â€“

For the possible solution pair

Hence, the answer must consist of 11 or 6 in Technical and 5 or 12 in Recruitment.

*n*= 20 â€“ 9 = 11 people working in the Technical division only, and 20 â€“*m*= 20 â€“ 15 = 5 people working in the Recruitment division only.For the possible solution pair

*m*= 8 and*n*= 14, the number of people working in both departments is*m*+*n*â€“ 20 = 8 + 14 â€“ 20 = 22 â€“ 20 = 2. This number (2) is acceptable because the problem requires that there be at least 1 person working in both departments.Here, there are 20 â€“

*n*= 20 â€“ 14 = 6 people working in the Technical division only, and 20 â€“*m*= 20 â€“ 8 = 12 people working in the Recruitment division only.For the possible solution pair

*m*= 1 and*n*= 19, the number of people working in both departments is*m*+*n*â€“ 20 = 1 + 19 â€“ 20 = 20 â€“ 20 = 0. This number (0) is NOT acceptable because the problem requires that there be at least 1 person working in both departments.Hence, the answer must consist of 11 or 6 in Technical and 5 or 12 in Recruitment.

Technical-Division | Recruitment-Division | |

3 | ||

5 | ||

6 | ||

9 | ||

11 |