1. **Problem statement:** There are 90 members in a sports club playing at least one of tennis, football, and volleyball. Given: - Tennis and football players: 10 - Football and volleyball players: 19 - Tennis and volleyball players: 29 - Let $n$ be the number of players playing all three games. - $2n$ people play only one game. Find the total number of football players. 2. **Formula and rules:** Use the principle of inclusion-exclusion for three sets $T$, $F$, and $V$: $$|T \cup F \cup V| = |T| + |F| + |V| - |T \cap F| - |F \cap V| - |T \cap V| + |T \cap F \cap V|$$ 3. **Define variables:** - Let $a$, $b$, and $c$ be the number of people who play only tennis, only football, and only volleyball respectively. - Given $2n = a + b + c$ (since $2n$ people play only one game). 4. **Express total players:** $$90 = a + b + c + (10 - n) + (19 - n) + (29 - n) + n$$ Here, $(10 - n)$, $(19 - n)$, and $(29 - n)$ are players playing exactly two games (subtracting those who play all three). 5. **Simplify:** $$90 = (a + b + c) + (10 + 19 + 29 - 3n) + n = (a + b + c) + 58 - 2n$$ Substitute $a + b + c = 2n$: $$90 = 2n + 58 - 2n = 58$$ This is a contradiction, so re-express carefully. 6. **Re-express total players:** Total players = only one game + exactly two games + all three games $$90 = (a + b + c) + ((10 - n) + (19 - n) + (29 - n)) + n$$ $$90 = (a + b + c) + (58 - 3n) + n = (a + b + c) + 58 - 2n$$ Given $a + b + c = 2n$, substitute: $$90 = 2n + 58 - 2n = 58$$ Again, contradiction. 7. **Check assumptions:** The problem states $2n$ people play only one game, so $a + b + c = 2n$. Rearranged: $$90 = 2n + 58 - 2n = 58$$ This suggests an error in the problem or interpretation. 8. **Assuming $2n$ people play only one game, and $n$ play all three, then the number playing exactly two games is:** $$90 - (2n + n) = 90 - 3n$$ But sum of exactly two games is: $$(10 - n) + (19 - n) + (29 - n) = 58 - 3n$$ So, $$90 - 3n = 58 - 3n$$ $$90 = 58$$ Contradiction again. 9. **Conclusion:** The problem data is inconsistent or incomplete. --- **Second problem:** Make $q$ the subject of $p = xqp - q$. 1. Rearrange: $$p = xqp - q$$ $$p + q = xqp$$ $$p + q = qxp$$ 2. Factor $q$ on the right: $$p + q = qxp$$ $$p = qxp - q = q(xp - 1)$$ 3. Solve for $q$: $$q = \frac{p}{xp - 1}$$ --- **Third problem:** Evaluate $\log 42$ given $\log 2=0.3010$, $\log 3=0.4771$, $\log 7=0.8451$. 1. Express 42 as product: $$42 = 2 \times 3 \times 7$$ 2. Use log product rule: $$\log 42 = \log 2 + \log 3 + \log 7 = 0.3010 + 0.4771 + 0.8451 = 1.6232$$ --- **Fourth problem:** Express $2 \log 3 + \log 6$ as a single logarithm. 1. Use power rule: $$2 \log 3 = \log 3^2 = \log 9$$ 2. Sum of logs: $$\log 9 + \log 6 = \log (9 \times 6) = \log 54$$ --- **Fifth problem:** Evaluate $2 + \log 20 - \log 1.6$ without tables. 1. Express $2$ as $\log 100$ (since $\log 100 = 2$): $$2 + \log 20 - \log 1.6 = \log 100 + \log 20 - \log 1.6$$ 2. Combine logs: $$= \log \left( \frac{100 \times 20}{1.6} \right) = \log \left( \frac{2000}{1.6} \right) = \log 1250$$ 3. $\log 1250 = \log (5^4) = 4 \log 5$ but since $\log 5 = \log \frac{10}{2} = 1 - 0.3010 = 0.6990$, 4. Calculate: $$4 \times 0.6990 = 2.7960$$ --- **Sixth problem:** Use tables to evaluate $2.0670^{0.0348} \times 0.538$. 1. Approximate $2.0670^{0.0348} = 10^{0.0348 \log 2.0670}$. 2. Calculate $\log 2.0670 \approx 0.3157$ (approximate from tables). 3. Multiply: $$0.0348 \times 0.3157 = 0.0110$$ 4. So, $$2.0670^{0.0348} = 10^{0.0110} \approx 1.0255$$ 5. Multiply by 0.538: $$1.0255 \times 0.538 = 0.552$$ --- **Seventh problem:** Make $x$ the subject of $mx - 5 = t$. 1. Add 5 to both sides: $$mx = t + 5$$ 2. Divide both sides by $m$: $$x = \frac{t + 5}{m}$$ --- **Eighth problem:** Simplify $32x^4 y^{216} 2p^6 z^8$. 1. Combine constants: $$32 \times 2 = 64$$ 2. Final expression: $$64 x^4 y^{216} p^6 z^8$$ --- **Ninth problem:** Solve for $m$ in $5^{2m - 1} = 1625$. 1. Express 1625 as power of 5: $$1625 = 5^4 \times 1.3$$ Not exact power, so take logs: 2. Take log base 10: $$\log 5^{2m - 1} = \log 1625$$ $$(2m - 1) \log 5 = \log 1625$$ 3. Calculate logs: $$\log 5 = 0.6990$$ $$\log 1625 = \log (1.625 \times 10^3) = \log 1.625 + 3 = 0.2100 + 3 = 3.2100$$ 4. Solve for $m$: $$(2m - 1) = \frac{3.2100}{0.6990} = 4.59$$ $$2m = 5.59$$ $$m = 2.795$$ --- **Tenth problem:** Solve for $x$ and $y$ in: $$9(1 - x) = 27y$$ $$x - y = -112$$ 1. Simplify first equation: $$9 - 9x = 27y$$ 2. Express $y$: $$y = \frac{9 - 9x}{27} = \frac{1 - x}{3}$$ 3. Substitute into second equation: $$x - \frac{1 - x}{3} = -112$$ 4. Multiply both sides by 3: $$3x - (1 - x) = -336$$ $$3x - 1 + x = -336$$ $$4x - 1 = -336$$ 5. Solve for $x$: $$4x = -335$$ $$x = -\frac{335}{4} = -83.75$$ 6. Find $y$: $$y = \frac{1 - (-83.75)}{3} = \frac{84.75}{3} = 28.25$$ --- **Eleventh problem:** Cost of dinner partly constant and partly varies directly with number of students. Given: - Cost $= 74$ when students $= 20$ - Cost $= 96$ when students $= 30$ Find cost when students $= 15$. 1. Model: $$C = k + mn$$ where $k$ is constant cost, $m$ is cost per student, $n$ is number of students. 2. Set up equations: $$74 = k + 20m$$ $$96 = k + 30m$$ 3. Subtract first from second: $$96 - 74 = (k + 30m) - (k + 20m)$$ $$22 = 10m$$ $$m = 2.2$$ 4. Find $k$: $$74 = k + 20(2.2) = k + 44$$ $$k = 74 - 44 = 30$$ 5. Find cost for 15 students: $$C = 30 + 2.2 \times 15 = 30 + 33 = 63$$ --- **Twelfth problem:** Make $x$ the subject of $a = p 2^{2x - 1}$. 1. Divide both sides by $p$: $$\frac{a}{p} = 2^{2x - 1}$$ 2. Take log base 2: $$\log_2 \frac{a}{p} = 2x - 1$$ 3. Solve for $x$: $$2x = \log_2 \frac{a}{p} + 1$$ $$x = \frac{1}{2} \left( \log_2 \frac{a}{p} + 1 \right)$$ --- **Thirteenth problem:** Use tables to evaluate $338.32 \times 2.9648 \times 8.6637 \times 6.2852$. 1. Multiply stepwise: $$338.32 \times 2.9648 \approx 1003.5$$ $$1003.5 \times 8.6637 \approx 8693.5$$ $$8693.5 \times 6.2852 \approx 54600$$ --- **Fourteenth problem:** Make $x$ the subject of $mx - nx = 5$. 1. Factor $x$: $$x(m - n) = 5$$ 2. Solve for $x$: $$x = \frac{5}{m - n}$$ --- **Fifteenth problem:** Survey of 290 readers: - Daily Times (D): 181 - Guardian (G): 142 - Punch (P): 117 - All read at least one paper - $|D \cap G| = 75$, $|D \cap P| = 60$, $|G \cap P| = 54$ Find: - Number reading all three papers $n$ - Number reading exactly two papers - Number reading exactly one paper - Number reading Guardian alone 1. Use inclusion-exclusion: $$|D \cup G \cup P| = |D| + |G| + |P| - |D \cap G| - |D \cap P| - |G \cap P| + |D \cap G \cap P|$$ 2. Substitute values: $$290 = 181 + 142 + 117 - 75 - 60 - 54 + n$$ $$290 = 440 - 189 + n$$ $$290 = 251 + n$$ 3. Solve for $n$: $$n = 290 - 251 = 39$$ 4. Number reading exactly two papers: $$(|D \cap G| - n) + (|D \cap P| - n) + (|G \cap P| - n) = (75 - 39) + (60 - 39) + (54 - 39) = 36 + 21 + 15 = 72$$ 5. Number reading exactly one paper: $$|D| + |G| + |P| - 2 \times \text{(exactly two)} - 3 \times n$$ $$= 181 + 142 + 117 - 2 \times 72 - 3 \times 39 = 440 - 144 - 117 = 179$$ 6. Number reading Guardian alone: $$|G| - (|D \cap G| - n) - (|G \cap P| - n) - n = 142 - 36 - 15 - 39 = 52$$