Question

Average of a, b and c is 11; average of c, d and e is 17; average of e and f is 22 and average of e and c is 17.What is the average of a, b, c, d, e and f?

• A. 15[2/3]
• B. 18[1/2]
• C. 16[1/2]
• D. 16[2/3]

Answer 1

The Correct Option is A

 To find the average of a, b, c, d, e, and f, we will first set up equations based on the given averages: 

1. \(\frac{a + b + c}{3} = 11\). Therefore, \(a + b + c = 33\).
2. \(\frac{c + d + e}{3} = 17\). Therefore, \(c + d + e = 51\).
3. \(\frac{e + f}{2} = 22\). Therefore, \(e + f = 44\).
4. \(\frac{e + c}{2} = 17\). Therefore, \(e + c = 34\).

Let's solve these equations step-by-step to find \(a + b + c + d + e + f\):

1. From equations 2 and 4, we have:
   • \(c + d + e = 51\)
   • \(e + c = 34\)
2. Now, consider equations 3 and 4:
   • \(e + f = 44\)
   • \(e + c = 34\)
3. To find all variables, use the equation from step 1, repeat for additional insights, and then aggregate:
   • From step 1, \(a + b + c = 33\).
   • Using the equations \(c + d + e = 51\) and \(e + c = 34\), we already found \(d = 17\) and \(c = f + 10\).
   • From \(f = c - 10\) and \(e + f = 44\), solve for \(e\ge 12\).
   • Sum of all: \(a + b + c + d + e + f = [c - 14] - 0 + [c - 4] = 94 + 10 = 96.\)

So, \(a+b+c+d+e+f = 94\). Hence, the average is:

\(\frac{a+b+c+d+e+f}{6} = \frac{94}{6} = 15 \frac{2}{3}\).

The correct option is 15[2/3].