Question: A restaurant records the number of
customers at breakfast, lunch and
dinner during one day.
The ratio of customers at breakfast
to customers at lunch was 3 : 2
The ratio of customers at breakfast
to customers at dinner was 11 : 8
The number of customers at lunch
was 10 less than the number of
customers at dinner.
How many customers were at
lunch?
The members of a martial arts club
are all ranked as beginner,
intermediate or expert.
The ratio of beginners to
intermediates is 9 : 2 and the ratio
of intermediates to experts is 3 : 5
The head of the club takes a sample
of 86 members.
The proportions of beginners,
intermediates and experts in the
sample are the same as the
proportions in the whole club.
How many intermediates are in the
sample?
The ratio a : c is 10 : 7
The ratio b : c is 3 : 2
The median of a, b and c is 80
Work out the values of a, b and c.
1. Problem 1: Find the number of customers at lunch given ratios and difference.
2. Let the number of customers at breakfast, lunch, and dinner be $B$, $L$, and $D$ respectively.
3. Given ratios:
- $\frac{B}{L} = \frac{3}{2}$ so $B = \frac{3}{2}L$
- $\frac{B}{D} = \frac{11}{8}$ so $B = \frac{11}{8}D$
4. From these, equate $B$:
$$\frac{3}{2}L = \frac{11}{8}D$$
5. Given $L = D - 10$, substitute into the equation:
$$\frac{3}{2}(D - 10) = \frac{11}{8}D$$
6. Multiply both sides by 8 to clear denominators:
$$8 \times \frac{3}{2}(D - 10) = 8 \times \frac{11}{8}D$$
$$12(D - 10) = 11D$$
7. Expand left side:
$$12D - 120 = 11D$$
8. Subtract $11D$ from both sides:
$$12D - 11D - 120 = 0$$
$$D - 120 = 0$$
9. Solve for $D$:
$$D = 120$$
10. Find $L$:
$$L = D - 10 = 120 - 10 = 110$$
11. Answer for problem 1: Number of customers at lunch is $\boxed{110}$.
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12. Problem 2: Find the number of intermediates in a sample of 86 members given ratios.
13. Let beginners, intermediates, and experts be $B$, $I$, and $E$.
14. Given ratios:
- $\frac{B}{I} = \frac{9}{2}$ so $B = \frac{9}{2}I$
- $\frac{I}{E} = \frac{3}{5}$ so $E = \frac{5}{3}I$
15. Total members in club proportional to:
$$B + I + E = \frac{9}{2}I + I + \frac{5}{3}I$$
16. Find common denominator 6:
$$= \frac{27}{6}I + \frac{6}{6}I + \frac{10}{6}I = \frac{43}{6}I$$
17. The sample size is 86, so:
$$\frac{43}{6}I = 86$$
18. Multiply both sides by 6:
$$43I = 516$$
19. Solve for $I$:
$$I = \frac{516}{43} = 12$$
20. Answer for problem 2: Number of intermediates in the sample is $\boxed{12}$.
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21. Problem 3: Find values of $a$, $b$, and $c$ given ratios and median.
22. Given:
- $\frac{a}{c} = \frac{10}{7}$ so $a = \frac{10}{7}c$
- $\frac{b}{c} = \frac{3}{2}$ so $b = \frac{3}{2}c$
- Median of $a$, $b$, $c$ is 80
23. To find median, order $a$, $b$, $c$:
- Calculate approximate values in terms of $c$:
$$a = 1.4286c, \quad b = 1.5c, \quad c = c$$
24. Since $c$ is $1c$, the order depends on $c$:
- If $c < 0$, order changes, but assume positive.
- For positive $c$, order is $a < c < b$ or $c < a < b$ depending on $c$.
25. Check which is median:
- If $c < a < b$, median is $a$.
- If $a < c < b$, median is $c$.
26. Test median = 80 for each:
- If median is $a$, then $a = 80 = \frac{10}{7}c \Rightarrow c = \frac{7}{10} \times 80 = 56$
- Then $b = \frac{3}{2} \times 56 = 84$, $c = 56$, $a = 80$
- Order: $c=56$, $a=80$, $b=84$ median is $a=80$ correct.
27. Values:
$$a = 80, \quad b = 84, \quad c = 56$$
28. Answer for problem 3: $a=80$, $b=84$, $c=56$.