What this lesson covers
- Turning "has $60" into ≤ 60, fixed fee included
- Solving down to a decimal bound instead of an answer
- Why a maximum steps down and a minimum steps up
- Checking the neighbour value, because the neighbour is the distractor
Questions worked in the video
- 0:14$60 to spend, paperbacks cost $7 each, plus a one-time $12 entry fee. What is the greatest number of paperbacks she can buy?
Worked examples
The questions the video works, written out: the setup, each step, the answer and the trap.
0:14Example 1
A student has $60 to spend at a book fair. Each paperback costs $7, and she must also pay a one-time $12 entry fee. What is the greatest number of paperbacks she can buy? A) 4 B) 6 C) 7 D) 8
- Let p be the number of paperbacks. The total cost is 7p + 12, and has $60 means the cost can be at most 60: .
- Subtract the fee from both sides: .
- Divide by 7: . That is a bound, not an answer.
- Books come in whole numbers, and the allowed side is below the bound, so step down to 6. For a maximum you step down, never to the nearest.
- Check the neighbours: 6 books cost 7(6) + 12 = 42 + 12 = 54, within the 60. 7 books cost 7(7) + 12 = 49 + 12 = 61, one dollar over.
Answer: B) 6
Since 6 books fit the budget and 7 do not, 6 is the greatest she can buy. Choice C rounds 6.857 to the nearest, which puts her a dollar over. Choice D ignores the fee and takes 60 divided by 7, rounded down to 8; eight books plus the fee is $68. Choice A divides the leftover $48 by the $12 fee instead of the $7 price; 4 books are affordable, but 4 is not the greatest.
1:05Example 2
Forty people need rides, and each van seats 9. What is the least number of vans needed?
- Let v be the number of vans. The seats must cover everyone: , so .
- Vans come in whole numbers, and the allowed side is above the bound, so a minimum steps up: v = 5.
- Check the neighbours: 4 vans seat 36 and leave 4 people standing, so 4 fails. 5 vans seat 45, enough for 40.
Answer: 5 vans
This is the mirror of the book-fair problem: a maximum steps down to the whole number below the bound, a minimum steps up to the whole number above it. The trap answer 4 comes from rounding 4.44 down by habit, and it leaves four people without a seat.
Chapters
- 0:006.857 books
- 0:14The setup
- 0:29Solve to the bound, then step down
- 1:05Three ways this goes wrong
- 1:28At most means step down
Lesson transcript
The narration of the video, word for word, under its chapter headings.
0:14The setup
Sixty dollars at a book fair. Paperbacks cost seven dollars each, and there is a one time twelve dollar entry fee. What is the greatest number of paperbacks she can buy?
0:29Solve to the bound, then step down
Let p be the number of paperbacks. The cost is seven p plus twelve, and she has sixty dollars, so seven p plus twelve is at most sixty. Subtract the fee from both sides. Seven p is at most forty eight. Divide by seven. p is at most six point eight five seven. Books come in whole numbers, and the allowed side is below that bound, so step down to six. Not to the nearest. Down. Six books cost fifty four dollars. Seven books cost sixty one, and she only has sixty.
1:05Three ways this goes wrong
Three ways this goes wrong. One, rounding six point eight five seven up to seven, which puts her a dollar over. Two, forgetting the fixed fee and just dividing sixty by seven. Three, rounding a minimum down. Forty people and nine seats to a van is four point four, and four vans leave four people standing.
1:28At most means step down
Solve it down to the bound, then step to the side you are allowed to be on.
Read the written version: the Linear inequalities strategy guide, then try 25 hard Linear inequalities questions with full explanations. Both are free, no account needed.