When the choices are full of letters, turn the whole question into arithmetic.
The problem
If a = 2b and b = 3c, which expression equals a + b + c?
Let c = 2. Then b = 6 and a = 12, so a + b + c = 20 — and only 10c produces 20. The algebra agrees: a = 2(3c) = 6c, so 6c + 3c + c = 10c. 9c is the trap: it adds a + b = 6c + 3c and forgets the c itself, and 6c stops at a alone. Picking a number turns the whole chain into arithmetic.
Do not pick at random — pick so the arithmetic stays whole.
let c = 2
the letter everything depends on
b = 3c = 6
follow the chain
a = 2b = 12
now everything is a number
They behave too well. With c = 1 several wrong choices give the same value as the right one, and you cannot tell them apart.
a + b + c
what was asked
= 12 + 6 + 2
substitute
= 20
the target number
Write 20 down and circle it. Every choice now gets measured against it.
| Choice | With c = 2 | Is it 20? |
|---|---|---|
| A. 10c | 20 | yes |
| B. 11c | 22 | no |
| C. 6c | 12 | no |
| D. 9c | 18 | no |
Testing all four costs seconds and rules out a coincidence. Stopping at the first match is how two-answer situations get missed.
A second number nearly always separates them.
retry with c = 3
a fresh value
b = 9, a = 18
rebuild the chain
a + b + c = 30
the new target
10c = 30 ✓
10c survives again
Avoid values that already appear in the problem, and avoid making two different letters equal. Both create false matches.
Two places where the number you pick is forced on you by the question.
If −1 < x < 0, which is the smallest: x, 2x, x², −x or 1x?
let x = −12
a value strictly inside the range
x = −0.5, 2x = −1, x² = 0.25, −x = 0.5, 1x = −2
every choice as a number
smallest: 1x = −2
dividing by a small fraction makes it big
Boys to girls is 2 : 3. If 40% of the boys and 10% of the girls joined a club, what percent of the class joined?
let there be 20 boys and 30 girls
2 : 3, scaled so both percents are whole
40% of 20 = 8, 10% of 30 = 3
members from each group
11 out of 50 = 22%
out of the whole class
(40% + 10%) ÷ 2 = 25% is the classic wrong answer: it treats the groups as equal in size. The ratio says the girls outnumber the boys, so the answer leans toward 10%. Likewise, never pick 0 or −1 for a range like −1 < x < 0 — the ends are not inside.
“Must be true” needs different kinds of number: if x > y, which must be true — x² > y², 1x < 1y, x − y > 0, or xy > 0?
x = 3, y = 2: all four hold
positives alone prove nothing
x = 1, y = −2: x² = 1, y² = 4 ✗
now try a negative
1x = 1, 1y = −12 ✗ and xy = −2 ✗
three die at once
x − y > 0
the only survivor — and it is just x > y rewritten
What is different: one friendly number keeps every choice alive. For “must”, pick a second number of a different kind — a negative, a fraction or zero — to kill the rest.
A unit change hidden in the letters: a car travels d km in t hours. At the same speed, how many km does it travel in m minutes?
let d = 120, t = 2, m = 30
60 km/h, and half an hour
target: 30 km
half of 60
dm60t = 3600120 = 30 ✓
60dmt = 108,000 and dm/t = 1,800 fail
What is different: hours and minutes are mixed. Choosing m = 30 makes the conversion visible, and the 60 must sit underneath.
A figure that grows: the radius of a circle increases by 50%. By what percent does the area increase?
let r = 2: area = 4π
an easy radius
new r = 3: area = 9π
2 up by 50%
increase = 5π ÷ 4π = 125%
compare with the original
What is different: the area depends on r², so 50% on the radius is not 50% on the area. The multiplier is 1.5² = 2.25.
| Question looks like | Numbers to pick |
|---|---|
| choices are expressions in x | x = 2 or 3 |
| percentage of an unstated total | total = 100 |
| “in terms of n” | n = 2, then n = 5 |
| even, odd or sign questions | test one of each kind |
| a fraction of a fraction | a number both denominators divide |
For “what percent” questions with no total given, 100 turns every step into mental arithmetic.
a price p rises 30%, then falls 10%: let p = 100 → 130 → 130 − 13 = 117
final price 1.17p — each percent acts on the price before it
a jar is 25 full and 14 of what is in it is used: let the jar hold 20 → 8 in it, 2 used, 6 left = 310
20 is divisible by both 5 and 4
the average of p and q is k; p goes up by 8: let p = 2, q = 4, k = 3 → new numbers 10 and 4, average 7
7 = k + 4: the choice k + 4 survives
One warning: this finds the choice that matches, not a proof. For “which must be true”, a single example can only eliminate — never confirm.
Look for choices written in letters: “in terms of S”, successive percent changes, averages that shift, ratios with percents. Pick 100 for prices, a multiple of the ratio for groups, a value strictly inside any range — then run all four choices against your one target number.
Plug in numbers for this one
A shirt costs p pesos. The price rises by 20%, then the new price is cut by 20%. The final price is
Try p = 100. Up 20% gives 120; down 20% of 120 — not of 100 — is 120 − 24 = 96. So the answer is 0.96p. A is the trap: the two 20% moves are taken on different bases, so they never cancel.
Hint: let p = 100 — percentage questions become trivial. Work the rise, then take 20% off the new price, and compare with 100.
