When a problem gives a limit, such as a budget, a capacity, a pass mark or a minimum amount, the condition can be written as an inequality. Solving it gives every allowed value; the context then decides which answer is wanted. **The question asks for** | **Typical words** | **Rounding a fractional boundary** a greatest number | at most, cannot exceed, can afford, has enough for | round down to the integer below a least number | at least, needs, must reach, to carry all | round up to the integer above Why the rounding differs: if n 12.5, then 13 is too many, so the greatest whole number is 12. If n 12.5, then 12 is too few, so the least whole number is 13. Rounding to the nearest whole number would give a wrong answer in one of the two cases. Method - Let a letter stand for the unknown number and say what it counts. If two kinds share a known total, write the second kind as the total minus the first. - Write an expression for the quantity that is limited, using the same unit throughout (for example, all amounts in cents). - Use the words to choose the symbol, then solve. - Interpret the solution: decide whether the answer must be a whole number and round in the direction the question needs. Check the answer and the next whole number in the original condition. Example A minibus can carry at most 19 passengers. Find the least number of minibuses needed to take 150 students on an outing. Let b be the number of minibuses. They must carry all the students, so 19b 150. 19b 150 b >= 15019=7 1719 The number of minibuses is a whole number, so b=8. Check: 7 minibuses carry 133 students, which is too few; 8 carry 152, which is enough. A quantity that falls: a water tank holds 500 litres and drains at 35 litres per minute. At least 80 litres must remain. After t minutes, 500-35t 80, so -35t >= -420 and t 12 (dividing by -35 reverses the sign). The tank can drain for at most 12 minutes. Two kinds sharing a total: a jar holds 20 coins, some 50-cent and the rest 20-cent, worth more than $7.30. If x coins are 50-cent, then 20-x are 20-cent. In cents, 50x+20(20-x)>730, so 30x+400>730, 30x>330 and x>11. There are at least 12 fifty-cent coins. Check: in a “greatest” question, n 12.5 gives 12, never 13; in a “least” question, b 7 1719 gives 8, never 7. Always test the answer in the words of the question.

Forming and solving inequalities from a context

Secondary 2

When a problem gives a limit, such as a budget, a capacity, a pass mark or a minimum amount, the condition can be written as an inequality. Solving it gives every allowed value; the context then decides which answer is wanted.

The question asks for	Typical words	Rounding a fractional boundary
a greatest number	at most, cannot exceed, can afford, has enough for	round down to the integer below
a least number	at least, needs, must reach, to carry all	round up to the integer above

Why the rounding differs: if $n\le12.5$, then $13$ is too many, so the greatest whole number is $12$. If $n\ge12.5$, then $12$ is too few, so the least whole number is $13$. Rounding to the nearest whole number would give a wrong answer in one of the two cases.

Method

Let a letter stand for the unknown number and say what it counts. If two kinds share a known total, write the second kind as the total minus the first.
Write an expression for the quantity that is limited, using the same unit throughout (for example, all amounts in cents).
Use the words to choose the symbol, then solve.
Interpret the solution: decide whether the answer must be a whole number and round in the direction the question needs. Check the answer and the next whole number in the original condition.

Example

A minibus can carry at most $19$ passengers. Find the least number of minibuses needed to take $150$ students on an outing.

Let $b$ be the number of minibuses. They must carry all the students, so $19b\ge150$.

$\begin{aligned} 19b&\ge150 \\ b&\ge\dfrac{150}{19}=7\tfrac{17}{19} \end{aligned}$

The number of minibuses is a whole number, so $b=8$. Check: $7$ minibuses carry $133$ students, which is too few; $8$ carry $152$, which is enough.

A quantity that falls: a water tank holds $500$ litres and drains at $35$ litres per minute. At least $80$ litres must remain. After $t$ minutes, $500-35t\ge80$, so $-35t\ge-420$ and $t\le12$ (dividing by $-35$ reverses the sign). The tank can drain for at most $12$ minutes.

Two kinds sharing a total: a jar holds $20$ coins, some $50$-cent and the rest $20$-cent, worth more than $\$7.30$. If $x$ coins are $50$-cent, then $20-x$ are $20$-cent. In cents, $50x+20(20-x)>730$, so $30x+400>730$, $30x>330$ and $x>11$. There are at least $12$ fifty-cent coins.

Check: in a “greatest” question, $n\le12.5$ gives $12$, never $13$; in a “least” question, $b\ge7\tfrac{17}{19}$ gives $8$, never $7$. Always test the answer in the words of the question.

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