Band 4 – Page 89

Kim and Alex start jobs at the beginning of the same year. Kim's annual salary in the first year is $30,000 and increases by 5% at the beginning of each subsequent year. Alex's annual salary in the first year is $33,000, and increases by $1,500 at the beginning of each subsequent year.

Show that in the 10th year, Kim's annual salary is higher than Alex's annual salary. (2 marks)

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In the first 10 years how much, in total, does Kim earn? (2 marks)

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Every year, Alex saves `1/3` of her annual salary. How many years does it take her to save $87,500? (3 marks)

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Show Answers Only

`text{Proof (See Worked Solutions)}`
`$377\ 336.78`
`text(7 years)`

Show Worked Solutions

i. `text(Let)\ \ K_n=text(Kim’s salary in Year)\ n`

`{:{:(K_1=a=30\ 000),(K_2=ar^1=30\ 000(1.05^1)):}}{:(\ =>\ GP),(\ \ \ \ \ \ a=30\ 000),(\ \ \ \ \ \ r=1.05):}`

`vdots`

`:.K_10=ar^9=30\ 000(1.05)^9=$46\ 539.85`

`text(Let)\ \ A_n=text(Alex’s salary in Year)\ n`

`{:{:(A_1=a=33\ 000),(A_2=33\ 000+1500=34\ 500):}}{:(\ =>\ AP),(\ \ \ \ \ \ a=33\ 000),(\ \ \ \ \ \ d=1500):}`

`vdots`

`A_10=a+9d=33\ 000+1500(9)=$46\ 500`

`=>K_10>A_10`

`:.\ text(Kim earns more than Alex in the 10th year)`

ii. `text(In the first 10 years, Kim earns)`

`K_1+K_2+\ ….+ K_10`

`S_10`	`=a((r^n-1)/(r-1))`
	`=30\ 000((1.05^10-1)/(1.05-1))`
	`=377\ 336.78`

`:.\ text(In the first 10 years, Kim earns $377 336.78)`

iii. `text(Let)\ T_n=text(Alex’s savings in Year)\ n`

`{:{:(T_1=a=1/3(33\ 000)=11\ 000),(T_2=a+d=1/3(34\ 500)=11\ 500),(T_3=a+2d=1/3(36\ 000)=12\ 000):}}{:(\ =>\ AP),(\ \ \ \ a=11\ 000),(\ \ \ \ d=500):}`

`text(Find)\ n\ text(such that)\ S_n=87\ 500`

♦ Mean mark 45%.
IMPORTANT: Using the AP sum formula to create and then solve a quadratic in `n` is challenging and often examined. Students need to solve and interpret the solutions.

`S_n`	`=n/2[2a+(n-1)d]`
`87\ 500`	`=n/2[22\ 000+(n-1)500]`
`87\ 500`	`=n/2[21\ 500+500n]`

`250n^2+10\ 750n-87\ 500`	`=0`
`n^2+43n-350`	`=0`
`(n-7)(n+50)`	`=0`

`:.n=7,\ \ \ \ n>0`

`:.\ text(Alex’s savings will be $87,500 after 7 years).`