AMC 10 Step by Step

Ratios, Percents & Averages

Ratio, proportion, percent change, means/averages as arithmetic word problems

79
primary-topic problems (6.1% of all)
58
more as a secondary topic
Where it appears
63
P1-10
12
P11-15
3
P16-20
1
P21-25

What you need to know

  • A ratio means "the same number of parts." If a:b:c=p:q:ra:b:c=p:q:r, then a=pka=pk, b=qkb=qk, and c=rkc=rk for one number kk (the size of one part). Work with 3k3k and 5k5k instead of fractions. Example: boys to girls is 3:53:5 in a class of 3232. Then 3k+5k=8k=323k+5k=8k=32, so k=4k=4: 1212 boys and 2020 girls.
  • Percent changes multiply. A p%p\% increase multiplies by 1+p1001+\frac{p}{100}, and a p%p\% decrease multiplies by 1−p1001-\frac{p}{100}. Several changes in a row multiply together. Example: up 20%20\% and then down 20%20\% is 1.2×0.8=0.961.2\times0.8=0.96, so you end 4%4\% lower.
  • A percent is always a percent "of" something, and that something is the base. Example: 5050 is 25%25\% more than 4040 (base 4040), but 4040 is 20%20\% less than 5050 (base 5050).
  • Mean ×\times count == sum. Example: 44 tests averaging 8585 add up to 340340.
  • A combined average is a weighted average. Add the sums and divide by the total count. The result is closer to the mean of the larger group. Example: 1010 students average 8080 and 3030 students average 6060. All 4040 average 800+180040=65\frac{800+1800}{40}=65, which is closer to 6060.
  • Chaining ratios. If a:b=2:3a:b=2:3 and b:c=4:5b:c=4:5, scale both so bb matches. Multiply the first by 44 and the second by 33: a:b=8:12a:b=8:12 and b:c=12:15b:c=12:15, so a:b:c=8:12:15a:b:c=8:12:15.

How AMC 10 tests it

  • Problems 1–6: one percent or ratio computation (a markup followed by a discount, or the ratio of areas after scaling a side).
  • Problems 4–10: "the average of nn numbers is mm; after one number is removed or added, the average becomes ..." Change the averages into sums.
  • Problems 6–12: chained ratios among three or more groups, leading to a small linear equation.
  • Problems 8–14: percents where the base is unknown, such as "what percent increase brings the price back to the original?"

Standard approaches

  1. Pick a handy total: 100100 for percents, the LCM of the ratio parts, or parts pkpk, qkqk, … with kk unknown.
  2. Change every average into a sum right away.
  3. Write each percent change as a multiplier and multiply them. Change back to a percent only at the end.
  4. For "what percent" answers, decide carefully what the base is: the original amount, the new amount, or the other quantity. Example: after a 20%20\% discount, a 25%25\% increase brings the price back, since 0.8×1.25=10.8\times1.25=1.
  5. Check that the answer goes the right way: a discount lowers the price, and a combined average lies between the group averages.

Worked example

In a club the ratio of freshmen to sophomores is 3:53:5. After 1212 freshmen join and 88 sophomores leave, the ratio becomes 5:65:6. How many members did the club have originally?

(A) 9696 (B) 112112 (C) 128128 (D) 144144 (E) 160160

Say there were 3k3k freshmen and 5k5k sophomores at first. Afterward there are 3k+123k+12 freshmen and 5k−85k-8 sophomores, and their ratio is 5:65:6:
3k+125k−8=56. \frac{3k+12}{5k-8}=\frac56.
Cross-multiply: 6(3k+12)=5(5k−8)6(3k+12)=5(5k-8), so 18k+72=25k−4018k+72=25k-40.

Move the kk terms to one side: 72+40=25k−18k72+40=25k-18k, so 7k=1127k=112 and k=16k=16.

The question asks for the whole club, which is 3k+5k=8k=1283k+5k=8k=128 members.

Check: 4848 freshmen and 8080 sophomores become 6060 and 7272, and 60:72=5:660:72=5:6. The answer is (C) 128\boxed{\textbf{(C)}\ 128}.

Pitfalls

  • Taking a percent of the wrong base. A 20%20\% discount and then a 20%20\% increase does not bring you back: $100\$100 becomes $80\$80 and then $96\$96.
  • Adding percents that have different bases. Up 10%10\% and then up 10%10\% is 1.1×1.1=1.211.1\times1.1=1.21, a 21%21\% increase, not 20%20\%.
  • Averaging group averages without weighting by group size. In the example above, 80+602=70\frac{80+60}{2}=70 is wrong; the true average is 6565.
  • Giving one part when the whole is asked, or the whole when one part is asked. A ratio of 3:53:5 means 88 parts in all: in the worked example, 4848 is only the freshmen.

Traps that recur

  • Approximating each cost against the whole allowance (20% + 5% = 25%) instead of against the stated differences. (2017 AMC 10A #14)
  • Taking 40% of the 30 swimmers instead of 40% of the 60 soccer players, or dividing the 36 by 100 rather than by the 70 non-swimmers. (2009 AMC 10A #18)
  • Finding n = 4 coins and 80 cents but then guessing a coin mix (like 25 + 25 + 10 + 10 + 10) without checking it uses exactly four coins. (2004 AMC 10A #14)
  • Pouring half of the second cup's original 4 ounces (2 oz) back instead of half of its new 6 ounces (3 oz, in a 1:2 coffee-to-cream ratio). (2002 AMC 10A #17)

Problems, easiest first

Open any of them to try it here; the worked solution is on the problem's own page.