Henry Ernest Dudeney/Modern Puzzles/200 - Cricket Scores/Solution
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Modern Puzzles by Henry Ernest Dudeney: $200$
- Cricket Scores
- In a country match Great Muddleton, who went in first, made a score of which they were proud.
- Then Little Wurzleford had their innings and scored a quarter less.
- The Muddletonians in their next attempt made a quarter less than their opponents,
- who, curiously enough, were only rewarded on their second attempt by a quarter less than their last score.
- Thus, every innings was a quarter less fruitful in runs than the one that preceded it.
- Yet the Muddletonians won the match by $50$ runs.
- Can you give the exact score for every one of the four innings?
Solution
| \(\ds \text {Great Muddleton's $1$st innings}\) | \(:\) | \(\ds 128\) | ||||||||||||
| \(\ds \text {Little Wurzleford's $1$st innings}\) | \(:\) | \(\ds 96\) | ||||||||||||
| \(\ds \text {Great Muddleton's $2$nd innings}\) | \(:\) | \(\ds 72\) | ||||||||||||
| \(\ds \text {Little Wurzleford's $2$nd innings}\) | \(:\) | \(\ds 54\) |
Proof
Let $a$, $b$, $c$ and $d$ be Great Muddleton's first innings' score, Little Wurzleford's first innings' score, Great Muddleton's second innings' score and Little Wurzleford's second innings' score respectively.
Then:
| \(\text {(1)}: \quad\) | \(\ds b\) | \(=\) | \(\ds a - \dfrac a 4\) | \(\ds = \dfrac {3 a} 4\) | Then Little Wurzleford had their innings and scored a quarter less. | |||||||||
| \(\text {(2)}: \quad\) | \(\ds c\) | \(=\) | \(\ds b - \dfrac b 4\) | \(\ds = \dfrac {9 a} {16}\) | The Muddletonians in their next attempt made a quarter less than their opponents, | |||||||||
| \(\text {(3)}: \quad\) | \(\ds d\) | \(=\) | \(\ds c - \dfrac c 4\) | \(\ds = \dfrac {27 a} {64}\) | who, curiously enough, were only rewarded on their second attempt by a quarter less than their last score. | |||||||||
| \(\text {(4)}: \quad\) | \(\ds a + c\) | \(=\) | \(\ds 50 + b + d\) | Yet the Muddletonians won the match by $50$ runs. | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a + \dfrac {9 a} {16}\) | \(=\) | \(\ds 50 + \dfrac {3 a} 4 + \dfrac {27 a} {64}\) | substituting for $b$, $c$ and $d$ from $(1)$, $(2)$ and $(3)$ in $(4)$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {64 a} {64} + \dfrac {36 a} {64}\) | \(=\) | \(\ds 50 + \dfrac {48 a} {64} + \dfrac {27 a} {64}\) | putting everything over a common denominator | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 64 a + 36 a\) | \(=\) | \(\ds 3200 + 48 a + 27 a\) | multiplying both sides by $64$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds 128\) | simplifying |
The rest of the scores fall to simple arithmetic.
$\blacksquare$
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $200$. -- Cricket Scores