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MathProblem-Solving and Data AnalysisHard

A store raises the price of a jacket by 20%. Later, it lowers the new price by 20%. The final price is what percent of the original price?

A80%
B96%Correct
C100%The trap
D104%

Check with $100: $100then$120then$96

  1. Turn each change into a multiplier.

    Up 20% means × 1.20. Down 20% means × 0.80.

  2. Apply them in order.

    1.20 × 0.80 = 0.96

  3. Read the answer.

    The final price is 96% of the original. That’s B.

  4. Why not 100%?

    The 20% cut is taken from the higher price, so it removes more than the 20% increase added.

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The Daily, plus 36 practice questions every day: 12 easy, 12 medium and 12 hard, each with a full explanation. Full-length Test A, Challenge Test 1 and all 1,023 vocabulary words are free too. No card needed.

Unlimited practice with no daily limit, plus every full-length test and Challenge test. Pro is $20 a month or $180 a year. Checkout isn’t open yet, so nothing can be bought today.

98 questions across four modules, in the digital SAT’s order: Reading & Writing Module 1 and Module 2 (27 questions each), a 10-minute break, then Math Module 1 and Module 2 (22 questions each). On official timing that’s 2 hr 14 min of testing. You get estimated section scores and a total at the end.

Yes. In the standard full-length tests, Module 2 of each section is the harder version if you get 60% or more of Module 1 right, and the easier version if you don’t. Challenge tests use the hardest material throughout.

Official timing, extended 1.5× time, or untimed, on every test, including the free ones. Extended time allows 48 minutes for each Reading & Writing module and about 52 for each Math module. Untimed removes the clock so you can focus on accuracy.

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Systems of linear equations · Lesson 2

Systems in word problems

Many word problems have two unknowns and two facts, so they turn into a system of two equations. The SAT either asks which system represents the situation or asks you to solve it for one number.

Most of these problems follow one pattern. One equation counts things: the number of student tickets plus the number of adult tickets equals the total number of tickets, s+a=120s + a = 120. The other equation adds up a value: price times number for each kind, added together, equals the total money, 5s+9a=7605s + 9a = 760.

Coins work the same way. The number of dimes plus the number of quarters is the number of coins, d+q=36d + q = 36, and 0.10d+0.25q0.10d + 0.25q is their total value in dollars. Mixtures do too: the liters of each solution add up to the total liters, and each solution's concentration times its liters gives the liters of the ingredient being tracked, such as alcohol.

Read comparison words carefully. There are 3 times as many pens as notebooks means p=3np = 3n: the larger group, pens, equals 3 times the smaller group. Test it with small numbers: 1 notebook and 3 pens should work, and p=3np = 3n gives 3=3(1)3 = 3(1).

To test a which system represents choice, try one simple case from the story. If all 120 tickets were student tickets, the money would be 5⋅120=6005 \cdot 120 = 600 dollars, so the money equation's left side should give 600 when s=120s = 120 and a=0a = 0. A choice with the prices swapped or combined gives something else.

When the question asks for a number, solve the system by substitution or elimination, then check which unknown the question wants. Both unknowns, and often their difference, are usually among the choices.

Worked example Easy

A theater sold 120 tickets for one show. Student tickets cost 5 dollars each, adult tickets cost 9 dollars each, and the theater collected 760 dollars from these ticket sales. The theater sold ss student tickets and aa adult tickets.

Which system of equations represents this situation?

  1. s+a=760s + a = 760 and 5s+9a=1205s + 9a = 120
  2. s+a=120s + a = 120 and 9s+5a=7609s + 5a = 760
  3. s+a=120s + a = 120 and 5s+9a=7605s + 9a = 760Answer
  4. s+a=120s + a = 120 and 14(s+a)=76014(s + a) = 760

How to solve it

  1. Count the tickets: student tickets plus adult tickets is 120, so s+a=120s + a = 120.
  2. Add up the money: ss student tickets at 5 dollars each bring in 5s5s dollars, and aa adult tickets at 9 dollars each bring in 9a9a dollars. Together they make 760 dollars: 5s+9a=7605s + 9a = 760.
  3. Check with one simple case: if s=120s = 120 and a=0a = 0, the money should be 5⋅120=6005 \cdot 120 = 600 dollars, and 5(120)+9(0)=6005(120) + 9(0) = 600. (Solving the system gives s=80s = 80 and a=40a = 40.)

Why each choice is right or wrong

  • A. Incorrect. This swaps the two totals. The number of tickets is 120 and the money is 760 dollars, so s+as + a must equal 120.
  • B. Incorrect. This swaps the prices, charging 9 dollars for each student ticket and 5 dollars for each adult ticket. With s=120s = 120 and a=0a = 0 it gives 1,080 dollars instead of 600.
  • C. Correct. The first equation counts the tickets and the second adds the money from each kind of ticket: 5 dollars times ss plus 9 dollars times aa.
  • D. Incorrect. 14(s+a)14(s + a) charges both prices, 14 dollars in all, for every ticket. Each ticket costs either 5 or 9 dollars, and with s+a=120s + a = 120 this equation would need 1,680 dollars.

Worked example Medium

A jar contains only dimes and quarters. There are 36 coins in the jar, and their total value is 6 dollars. A dime is worth 0.10 dollars and a quarter is worth 0.25 dollars.

How many quarters are in the jar?

  1. 1010
  2. 1616Answer
  3. 2020
  4. 2424

How to solve it

  1. Let dd be the number of dimes and qq the number of quarters. Counting coins: d+q=36d + q = 36. Adding value: 0.10d+0.25q=60.10d + 0.25q = 6.
  2. Work in cents to clear the decimals: multiply the value equation by 100 to get 10d+25q=60010d + 25q = 600.
  3. Substitute d=36−qd = 36 - q: 10(36−q)+25q=60010(36 - q) + 25q = 600, so 360−10q+25q=600360 - 10q + 25q = 600, 15q=24015q = 240 and q=16q = 16.
  4. Then d=36−16=20d = 36 - 16 = 20. Check: 20(0.10)+16(0.25)=2+4=620(0.10) + 16(0.25) = 2 + 4 = 6 dollars. The question asks for quarters, which is 16.

Why each choice is right or wrong

  • A. Incorrect. This comes from writing 10(36−q)10(36 - q) as 360−q360 - q instead of 360−10q360 - 10q. That gives 360+24q=600360 + 24q = 600, so q=10q = 10. The 10 multiplies both terms in 36−q36 - q.
  • B. Correct. With d=36−qd = 36 - q, the value equation in cents is 10(36−q)+25q=60010(36 - q) + 25q = 600, so 15q=24015q = 240 and q=16q = 16. Then 20 dimes and 16 quarters are worth 2+4=62 + 4 = 6 dollars.
  • C. Incorrect. 20 is the number of dimes. The question asks for quarters: 36−20=1636 - 20 = 16.
  • D. Incorrect. This divides 6 dollars by 0.25 dollars, which is how many quarters there would be if every coin were a quarter. It ignores the dimes and the 36-coin total.

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