Let one letter stand for one unknown, then say everything else in terms of it.
Find the ages
Maria is 3 times as old as her brother. In 6 years she will be twice as old as he is then. How old is her brother now?
Let the brother be b, so Maria is 3b. In 6 years they are b + 6 and 3b + 6, and 3b + 6 = 2(b + 6). Expanding: 3b + 6 = 2b + 12, so b = 6. Check: they are 6 and 18 now; in 6 years, 12 and 24, and 24 is twice 12 ✓ D is Maria's age, not the brother's — read the last line of the question. A adds 6 to only one person, forgetting that both age.
| Words | Symbols |
|---|---|
| the sum of a number and 7 | x + 7 |
| 5 less than a number | x − 5, not 5 − x |
| twice a number, increased by 3 | 2x + 3 |
| 3 more than half a number | x2 + 3 |
| the product of a number and 4, decreased by 9 | 4x − 9 |
"5 less than x" means you start at x and take 5 away: x − 5. Writing 5 − x is the most frequent translation error in the whole topic.
Then everything else is described in terms of it.
Two numbers add to 45
one is 7 more than the other
let the smaller = x
so the larger = x + 7
x + (x + 7) = 45
the equation
2x = 38, x = 19
so the numbers are 19 and 26
Check: 19 + 26 = 45 ✓ and 26 is 7 more than 19 ✓ — both conditions, not just one.
three consecutive integers sum to 72
find them
n + (n + 1) + (n + 2) = 72
one letter, three numbers
3n + 3 = 72
collect
n = 23, so 23, 24, 25
and they do sum to 72 ✓
Even and odd consecutive numbers use the same gaps of 2 — the difference is only which value n starts at. There is no separate formula to remember.
| Brother | Maria | |
|---|---|---|
| Now | b | 3b |
| In 6 years | b + 6 | 3b + 6 |
| 6 years ago | b − 6 | 3b − 6 |
A table like this makes age problems almost mechanical. The condition in the question always compares two cells in the same row.
Writing 3b + 6 = 2b is the classic mistake — it ages Maria but not her brother. Six years pass for both of them.
A mother is 5 times as old as her daughter. In 9 years she will be 3 times as old as her daughter. How old is the daughter now?
now: d and 5d
the younger one gets the letter
in 9 years: d + 9 and 5d + 9
nine years for both
5d + 9 = 3(d + 9)
the condition uses the "in 9 years" row
5d + 9 = 3d + 27, so 2d = 18 and d = 9
distribute, gather, divide
Check: now 9 and 45; in 9 years 18 and 54, and 54 = 3 × 18 ✓
Writing 5d + 9 = 3d + 9 multiplies only the daughter's age by 3 and not her extra 9 years, giving 2d = 0 and an age of 0. "Three times as old" means three times all of d + 9.
Take off the fixed part first; what is left is shared by the rate.
An electric bill is a fixed ₱150 plus ₱11 for every kilowatt-hour used. A household's bill is ₱1,250. How many kilowatt-hours did it use?
150 + 11k = 1,250
k kilowatt-hours at ₱11 each
11k = 1,100
remove the fixed charge
k = 100 kWh
check: 150 + 11(100) = 1,250 ✓
1,250 ÷ 11 = 113.6…, which pays the ₱150 charge as if it were more electricity. An untidy answer like that is a signal the fixed amount was never taken off.
When the fixed charge already covers the first few units, only the extra units pay the rate: parking at ₱40 for the first 2 hours and ₱20 for each additional hour costs 40 + 20(6 − 2) = ₱120 for 6 hours, not 40 + 20(6).
Two jeepneys are 100 km apart and drive toward each other at 35 km/h and 45 km/h. How long until they meet?
35t + 45t = 100
together they cover the whole gap
80t = 100, t = 1.25 hours
the gap closes at 80 km/h
= 1 hour 15 minutes
0.25 hour = 15 minutes
Check: 35 × 1.25 = 43.75 km and 45 × 1.25 = 56.25 km, which add to 100 km ✓. Toward each other, the speeds add; subtracting them would answer a chase, not a meeting.
The same four moves; each of these hides one extra step.
One number is 23 of another, and together they make 40. Find both numbers.
x + 2x3 = 40
the larger is x
3x + 2x = 120, so x = 24
multiply every term by 3
24 and 16
check: 16 is 23 of 24, and 24 + 16 = 40 ✓
Letting the larger number be x keeps the fraction on the smaller one, where it belongs.
A father is now 45. Five years ago his son was half his age. How old is the son now?
five years ago: s − 5 and 40
take 5 from both
s − 5 = 40 ÷ 2 = 20
the condition is in the "ago" row
s = 25
add the 5 years back
Halving 45 to get 22.5 uses today's age in a condition about the past.
A rectangle's length is twice its width and its perimeter is 54 cm. What is its area?
2(w + 2w) = 54, so 6w = 54
width w, length 2w
w = 9 cm, length 18 cm
not the answer yet
area = 9 × 18 = 162 cm²
the question asked for area
Solving for w is only the middle of the problem; reread the last sentence before answering.
A rectangle is 4 m longer than it is wide
perimeter 52 m
width w, length w + 4
one letter
2w + 2(w + 4) = 52
P = 2l + 2w
4w + 8 = 52, w = 11
11 m by 15 m
Word problems on tests are a short list of types: number, consecutive integers, ages, perimeter, fixed-plus-rate money, and meeting or chasing. The fastest route is to name the smallest unknown and write one equation from the condition; the 10-second check is to put your answer back into the words, testing every condition, not just one.
Find the number
Three consecutive even integers add up to 96. What is the largest one?
Let them be n, n + 2 and n + 4. Then 3n + 6 = 96, so 3n = 90 and n = 30. The three numbers are 30, 32 and 34, so the largest is 34. A is the smallest and B the middle one — the question asked for the largest. C uses gaps of 1 and produces an odd number, which cannot be one of three consecutive even integers.
Hint: consecutive even numbers are 2 apart. Solve for n, then read the question again before choosing.
