Digital SAT practice

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Full-length tests

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98 questions across four modules, in the digital SAT’s order and on its timing, with a highlighter, a question map and a timer you can see.

A Peakscor full-length test in progress: Reading and Writing, Test A, Module 1 of 4, question 7 of 27, with the passage, four answer choices, a highlighter, a 24:18 timer and Flag, Back and Next buttons.

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    Play the Daily.

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  2. 2

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    Every question is tagged by domain and skill, and every answer comes with an explanation.

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  4. 4

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    Come back tomorrow and climb. Progress shows your accuracy and your strongest skills.

Built around the real SAT blueprint.

Every question is tagged by section, domain, skill and difficulty, so you can practice exactly what you need. Full tests run like test day: two Reading & Writing modules, a break, then two Math modules.

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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.

Test day

What test day looks like.

Every Peakscor full test follows this order. In the standard tests, the second module of each section adapts to how you did on the first.

  1. Reading & Writing · Module 1

    27 questions32 min

  2. Reading & Writing · Module 2

    27 questions32 min

    Adapts to how you did on Module 1.
  3. Break

    10 min

  4. Math · Module 1

    22 questions35 min

  5. Math · Module 2

    22 questions35 min

    Adapts to how you did on Module 1.

98 questions · 2 hr 14 min of testing, plus the break

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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.

A quick run with three hearts. Each right answer builds your streak: the questions get harder as it grows, and your points multiply (×2 from three in a row, ×3 from six). A miss costs a heart and drops you back to easy. Best is your longest streak.

Every day at midnight. You get a fresh 12 easy, 12 medium and 12 hard.

No. Peakscor is an independent SAT practice tool, not affiliated with the College Board. Scores shown are estimates.

Your next point starts today.

Free, no card.

Linear equations in one variable · Lesson 1

No solution or infinitely many solutions

Most linear equations have exactly one solution. Two special kinds don't, and the SAT asks about them often.

Simplify each side completely first: clear fractions, distribute and combine like terms. You want each side in the form (number)xx + (number), such as 6x−10=6x+76x - 10 = 6x + 7. Then compare the two sides.

**Different xx-coefficients: exactly one solution. Same xx-coefficient, different constants**: no solution. Subtracting 6x6x from both sides of 6x−10=6x+76x - 10 = 6x + 7 leaves −10=7-10 = 7, which is false no matter what xx is. **Same xx-coefficient and same constant**: infinitely many solutions. The two sides are the same expression, so every value of xx works, and subtracting leaves a statement that is always true, such as 7=77 = 7.

When the equation contains a constant such as kk, work backward. For no solution, set the xx-coefficients equal to each other and make sure the constants are different. For infinitely many solutions, set the xx-coefficients equal and the constants equal.

How many solutions, after simplifying both sides
After simplifyingExampleNumber of solutions
Different xx-coefficients4x+1=2x+74x + 1 = 2x + 7Exactly one (x=3x = 3)
Same xx-coefficient, different constants4x+1=4x+74x + 1 = 4x + 7None
Same xx-coefficient, same constant4x+1=4x+14x + 1 = 4x + 1Infinitely many

Worked example Easy

2(3x−5)=6x+72(3x - 5) = 6x + 7

How many solutions does the given equation have?

  1. ZeroAnswer
  2. Exactly one
  3. Exactly two
  4. Infinitely many

How to solve it

  1. Distribute on the left: 6x−10=6x+76x - 10 = 6x + 7.
  2. Both sides have the same xx-coefficient, 6, but different constants, −10-10 and 7.
  3. Subtracting 6x6x from both sides leaves −10=7-10 = 7, which is never true. The equation has zero solutions.

Why each choice is right or wrong

  • A. Correct. After distributing, 6x−10=6x+76x - 10 = 6x + 7. The xx-terms cancel and leave −10=7-10 = 7, which is false for every xx.
  • B. Incorrect. An equation has exactly one solution when the xx-coefficients differ after simplifying. Here both are 6.
  • C. Incorrect. A linear equation can't have exactly two solutions. It has zero, one or infinitely many.
  • D. Incorrect. Infinitely many solutions would need the constants to match as well, but −10≠7-10 \ne 7.

Worked example Medium

5(x−2)+kx=3x+85(x - 2) + kx = 3x + 8

In the given equation, kk is a constant. The equation has no solution. What is the value of kk?

  1. −2-2Answer
  2. 22
  3. 33
  4. 88

How to solve it

  1. Distribute: 5x−10+kx=3x+85x - 10 + kx = 3x + 8. Group the xx-terms: (5+k)x−10=3x+8(5 + k)x - 10 = 3x + 8.
  2. No solution needs equal xx-coefficients: 5+k=35 + k = 3, so k=−2k = -2.
  3. Check the constants: −10-10 and 8 are different. With k=−2k = -2 the equation is 3x−10=3x+83x - 10 = 3x + 8, which has no solution.

Why each choice is right or wrong

  • A. Correct. The left side's xx-coefficient is 5+k5 + k. Setting 5+k=35 + k = 3 gives k=−2k = -2, and the constants −10-10 and 8 differ, so there is no solution.
  • B. Incorrect. This is a sign slip in 5+k=35 + k = 3. With k=2k = 2, the equation is 7x−10=3x+87x - 10 = 3x + 8, which has exactly one solution, x=4.5x = 4.5.
  • C. Incorrect. This sets kk equal to the right side's coefficient and forgets the 5x5x already on the left. With k=3k = 3, the equation is 8x−10=3x+88x - 10 = 3x + 8, which has exactly one solution.
  • D. Incorrect. This matches kk to a constant. Because kk multiplies xx, it changes the xx-coefficient, and that coefficient must equal 3. With k=8k = 8, the equation 13x−10=3x+813x - 10 = 3x + 8 has exactly one solution.

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Each test mirrors the digital SAT: 98 questions across four modules, 2 hr 14 min on official timing.

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Same 98-question format, but every question is drawn from the toughest SAT material: Challenging-tier math throughout and the hardest reading.

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