Question 118·200 Super-Hard SAT Math Questions·Algebra

0 / 200 Questions completed

In a community poll, the number of people who initially responded yes was 4 times the number of people who initially responded no. Later, $\frac{1}{3}$ of the people who had responded yes changed their response to no, and then 150 people changed their response from no to yes. After these changes, the number of yes responses was twice the number of no responses.

Which choice is the number of people who initially responded no?

A one-variable equation works here because the starting no count determines the starting yes count. Track each switch in both groups, then use the final “twice as many” comparison. A Desmos regression can find the starting count that makes the two sides equal. The tempting slip is to add switchers to one group without removing them from the other.

Hints

Hint 1
Let a variable count the people who initially said no. The starting yes count is 4 times that count, so you can describe both groups using one letter. What expression counts the starting yes responses?
Hint 2
A switch from yes to no is a transfer: subtract those people from yes and add them to no. Take one-third of the original yes group, not one-third of the starting no group.
Hint 3
The later transfer goes the other way: add $150$ to yes and subtract $150$ from no. The final ratio says there are twice as many yes responses as no responses. How do you write that comparison?

Step-by-step

Approach 1: Track both response counts
Step 1
Name the starting no count
Let $x$ be the number of people who initially responded no. The problem says 4 times that many initially responded yes, so $\text{initial yes}=4x$ .
Step 2
Track the first switch
One-third of the original yes group moves to no, so two-thirds of that group stays yes. Add the people who moved to the no count: $\begin{aligned}\text{yes after the first switch}&=\frac{2}{3}(4x),\text{no after the first switch}&=x+\frac{1}{3}(4x).\end{aligned}$
Step 3
Track the switch back to yes
The next 150 people move from no to yes. Add $150$ to yes and subtract $150$ from no; the same people can't remain in both counts: $\begin{aligned}\text{final yes}&=\frac{2}{3}(4x)+150,\text{final no}&=x+\frac{1}{3}(4x)-150.\end{aligned}$
Step 4
Write the final comparison
The final yes count is twice the final no count. Use the two counts you built, not the starting counts: $\frac{2}{3}(4x)+150=2\left(x+\frac{1}{3}(4x)-150\right)$ .
Step 5
Find the starting no count
Type $\frac{2}{3}(4x_1)+150\sim2\left(x_1+\frac{1}{3}(4x_1)-150\right)$ in Desmos. Use $x_1$ for the unknown and $\sim$ to ask Desmos to find the value that balances both sides. Under PARAMETERS, Desmos shows $x_1=225$ . Since $x$ counts the people who initially said no, 225 people initially responded no. Choice C.

Approach 2: Set aside the group already in balance
Step 1
Look at the original yes group alone
No Desmos needed. The first switch already puts this group in the final ratio. Of the people who initially said yes, $\frac{1}{3}$ now say no and $\frac{2}{3}$ still say yes. So this group already has two yes responses for every one no response. A group already in the required ratio can be set aside; the rest must meet that ratio too.
Step 2
Find who stayed with no
Of the people who initially said no, $150$ switched to yes. For this group to have two yes responses for every one no response, half as many must have stayed with no: $\frac{150}{2}=75$ .
Step 3
Rebuild the original no group
The original no group includes the $150$ who switched and the $75$ who stayed. Add both parts: $150+75=225$ . So 225 people initially responded no. Choice C.

Question 118·200 Super-Hard SAT Math Questions·Algebra

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this