1
In five seconds, decide whether the choices can be tested.
Three conditions, and all three have to hold.
| Condition |
The opening problem |
| the choices are plain numbers | yes — −2, 0, 4, 5 |
|---|
| the question asks for one specific value | yes — the solution x |
|---|
| testing one value is quick | yes — a square root and a subtraction |
|---|
When all three hold you never have to build the equation. You only have to
recognise the right answer when it appears.
Radical and rational equations are where this pays best. Squaring both sides
can manufacture a root that the original equation rejects; a tested choice
cannot be extraneous, because you tested it in the equation as written.
2
Numeric choices ascend, so the middle splits the list.
On the Digital SAT, four numeric choices are printed smallest to largest. One test in the middle can kill half of them.
A
−2
B
0
C
4
D
5
With four choices the middle sits between B and C, so start at either one —
whichever is easier to compute. Here B = 0 is the easier arithmetic.
try B: x = 0
the easy half of the middle
left = √0 + 1 = 1
a root is never negative
right = 0 − 1 = −1
left is too big
raising x raises the right side faster
so A and B are both dead
try C: √9 = 3 and 4 − 1 = 3
two tests, finished
C. x = 4
Direction only works when the relationship is steady
"Too small, go up" assumes a bigger input gives a bigger output. With x², with
|x − 3|, or with a fraction that has x on the bottom, that is false — −4 and 4
give the same square, and the value of 1x falls as x rises. In those cases test
the choices in order and stop at the hit.
3
A context item is often faster tested than translated.
Student tickets to a school fair cost 4 dollars each and adult tickets cost 7 dollars each. A total of 200 tickets were sold for 1,010 dollars. How many adult tickets were sold?
A60
B70
C130
D140
try C: 130 adult, so 70 student
the two counts add to 200
7(130) + 4(70) = 910 + 280 = 1,190
too big — C and D are dead
try B: 70 adult, so 130 student
go down one
7(70) + 4(130) = 490 + 520 = 1,010
exactly the total
B. 70 adult tickets
130 is on the list for a reason
130 is the number of student tickets — the right work attached to the
wrong quantity. Set up the system properly and you still have to remember which
letter the question asked about. Back-solving cannot make that slip, because the
number you tested was the number the question named.
4
When the choices are expressions, pick a number instead.
Which expression is equivalent to (3x + 2)(x − 4) + 5x?
let x = 3
not 0, not 1, and not a number in the problem
(9 + 2)(3 − 4) + 15
substitute everywhere
= 11(−1) + 15 = 4
write 4 down — it is the target
| Choice | At x = 3 | Is it 4? |
|---|
| A. 3x² − 15x − 8 | 27 − 45 − 8 = −26 | no |
|---|
| B. 3x² − 5x − 8 | 27 − 15 − 8 = 4 | yes |
|---|
| C. 3x² + 5x − 8 | 27 + 15 − 8 = 34 | no |
|---|
| D. 3x² − 5x + 8 | 27 − 15 + 8 = 20 | no |
|---|
B. 3x² − 5x − 8
Never 0, never 1, never a number already in the problem
0 and 1 behave too well — several wrong choices collapse onto the right value and
you cannot tell them apart. Reusing 2 or 4 here would let a sign error hide inside
a coincidence. If two choices survive, do not guess: run a second number, and the
survivors separate almost every time.
5
For a percent with no total given, the friendly number is 100.
A store raises the price of a jacket by 30%, then takes 10% off the increased price. The final price is what percent of the original price?
let the original price be 100 dollars
no total is stated, so you may choose one
after the raise: 100 + 30 = 130
30% of 100
10% of 130 = 13
of the new price, not the old one
130 − 13 = 117
117 out of 100 → 117%
117% of the original price
what the numbers say
1.30 × 0.90 = 1.17
the multipliers multiply
what the trap says
30 − 10 = 20, so 120%
percents taken on different bases never add
Picking 100 turns the whole item into mental arithmetic and exposes the trap in one
line. The same trick handles "in terms of n" items: let n = 3, get a number, and
measure every choice against it.
6
Know when testing is the slow route.
Back-solving is a tool, not a religion. Four tests at 20 seconds each is 80 seconds — most of a question's whole budget.
| The item | Back-solve? | Why |
|---|
| radical or rational equation | yes | no extraneous root to catch |
|---|
| context item you cannot translate | yes | testing skips the set-up |
|---|
| choices are expressions | plug in a number | nothing numeric to test |
|---|
| 5(x + 3) = 40 | no | x + 3 = 8 is one line |
|---|
| "what is the value of 6x + 5" | no | solve for 6x and stop |
|---|
| "which must be true" | no | one example can only eliminate |
|---|
| two equations, two unknowns | graph it | Desmos is faster than four tests |
|---|
Desmos is open the whole time — let it do the testing
The Digital SAT gives you a graphing calculator in both modules. For the opening
problem, type y = √2x + 1 on one line and y = x − 1 on the next, then click the
intersection: it reads (4, 3), and the false root at x = 0 never appears, because
the graphs simply do not meet there. For a system, enter both equations and read
the crossing point. Reach for Desmos when an item is graphable and for the
choices when it is not.
✓
Check: the two routes agree, and one of them is shorter.
The algebra on the opening problem, for comparison.
square both sides
2x + 1 = x² − 2x + 1
watch the middle term
collect
x² − 4x = 0, so x = 0 or x = 4
two candidates
reject
x = 0 gives 1 = −1
extraneous ✗
back-solved
two tests, no squaring, no rejection step
x = 4 ✓
plugged in
x = 3 gave the target 4
one choice matched ✓
Remember the flow
Numbers in the choices? · Start at B or C · Move by direction · Letters instead? Pick a number
And before either one: check that solving is not simply shorter.
