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Worked solution

Count a shared middle pearl once

A string has 31 pearls, with the largest and most expensive in the middle. From one end each pearl is worth $100 more than its predecessor up to the middle; from the other end each is worth $150 more. The whole string costs $63,000. Find the middle pearl’s value.

[2007/SAJC/I/9(a)]

CJC · Discussion · Question 5

Answer

The middle pearl costs $3000.

Express the two inclusive half-string sums

Let the middle pearl cost xx dollars. Each end-to-middle half contains 1616 pearls and 1515 price increases. The endpoints are x−15(150)x-15(150) and x−15(100)x-15(100). The two sums are 8(2x−2250)8(2x-2250) and 8(2x−1500)8(2x-1500).

How did this part go?

Remove the duplicated middle pearl

The combined inclusive halves count the middle pearl twice, so 8(2x−2250)+8(2x−1500)−x=630008(2x-2250)+8(2x-1500)-x=63000. This gives 31x−30000=6300031x-30000=63000, hence x=3000x=3000.

How did this part go?

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A string with a shared largest middle pearl of value x. Each end-to-middle inclusive group has 16 pearls: one end costs x minus 15 times 150 dollars, the other x minus 15 times 100 dollars. The middle belongs to both groups.
A string with a shared largest middle pearl of value x. Each end-to-middle inclusive group has 16 pearls: one end costs x minus 15 times 150 dollars, the other x minus 15 times 100 dollars. The middle belongs to both groups.
A string with a shared largest middle pearl of value x. Each end-to-middle inclusive group has 16 pearls: one end costs x minus 15 times 150 dollars, the other x minus 15 times 100 dollars. The middle belongs to both groups.

Common mistakes

Subtract the shared middle value once when adding the two halves.

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