Problem Solving

BTGmoderatorDC wrote:The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student's weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students' weights changed. What is the value of x?

A. 85
B. 86
C. 88
D. 90
E. 92
Source: Veritas Prep

(All weights are in pounds.)
$$? = x$$
Excellent opportunity to use the homogeneity nature of the average:
$$\sum\nolimits_n { = \,\,nx\,\,\,\,\,\left( {n\,\,{\rm{students}}} \right)} $$
$$\left\{ \matrix{
80 + \sum\nolimits_n {\, = \sum\nolimits_{n + 1} {\, = \,\,\left( {n + 1} \right)\left( {x - 1} \right)} } \hfill \cr
110 + \sum\nolimits_n {\, = \sum\nolimits_{n + 1} {\, = \,\,\left( {n + 1} \right)\left( {x + 4} \right)} } \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,110 - 80 = \left( {n + 1} \right)\left[ {\left( {x + 4} \right) - \left( {x - 1} \right)} \right]$$
$$30 = 5\left( {n + 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 5$$
$$80 + 5x = \left( {5 + 1} \right)\left( {x - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = x = 86$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

BTGmoderatorDC wrote:The average weight of a class is x pounds. When a new student weighing 80 pounds joins the class, the average decreases by 1 pound. In a few months the student's weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students' weights changed. What is the value of x?

A. 85
B. 86
C. 88
D. 90
E. 92

We can PLUG IN THE ANSWERS and use alligation.
To learn about alligation, check my posts here:
https://www.beatthegmat.com/statistics- ... 88903.html
https://www.beatthegmat.com/alligation- ... 93459.html
https://www.beatthegmat.com/mean-t266852.html

Let O = the original number of students and N = the new student.
Whether the new student weighs 80 pounds or 110 pounds, alligation must yield the SAME VALUE for O/N -- the ratio of old students to new student -- in each case.

D: x = 90
When the new student weighs 80 pounds, the average for all the students = x-1 = 90-1 = 89.
The following alligation is yielded:
O 90-----1-----89-----9-----80 N
Here, O/N = 9/1.
When the new student weighs 110 pounds, the average for all the students = x+4 = 90+4 = 94.
The following alligation is yielded:
O 90-----4-----94-----16-----110 N
Here, O/N = 16/4 = 4/1.
Since each alligation yields a different value for O/N, eliminate D.

B: x = 86
When the new student weighs 80 pounds, the average for all the students = x-1 = 86-1 = 85.
The following alligation is yielded:
O 86-----1-----85-----5-----80 N
Here, O/N = 5/1.
When the new student weighs 110 pounds, the average for all the students = x+4 = 86+4 = 90.
The following alligation is yielded:
O 86-----4-----90-----20-----110 N
Here, O/N = 20/4 = 5/1.
Success!
Each alligation yields the same value for O/N.

The correct answer is B.

Algebra:

Let n = the number of students.

Before the new student:
Sum of the weights = (number of students)(average weight) = nx.

After the new student:
(new number of students)(new average) = old sum + new student's weight
(n+1)(x-1) = nx + 80
nx + x - n - 1 = nx + 80
x - n = 81.

After new student gains weight:
(new number of students)(new average) = old sum + new student's heavier weight
(n+1)(x+4) = nx + 110
nx + x + 4n + 4 = nx + 110
x + 4n = 106.

Subtracting the first equation from the second, we get:
(x + 4n) - (x - n) = 106-81
5n = 25
n = 5.

Substituting n=5 into x - n = 81, we get:
x - 5 = 81
x = 86.

The correct answer is B.