Linear Functions and Graphs

Chapter 6 Study Notes · O-Level 4052

Section 8 of 9
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6.7Real-World Applications of Linear Functions

The gradient is usually the rate of change; the yy-intercept is usually the starting value.

n (kWh used)B (dollars)(0, 8)5101520510152025303540O
The bill against units used. The intercept is the fixed charge; the gradient is the rate per kWh.
  • Fixed charge + rate × quantity: phone, electricity and hire-charge bills all take this shape.
  • Distance-time: distance == speed ×\times time; the gradient is the speed.

Electricity bill

A household pays a fixed charge of $8\displaystyle \$8 per month plus $0.25\displaystyle \$0.25 per kWh of electricity used. Write an equation for the monthly bill B\displaystyle B in terms of units used u\displaystyle u. Find the bill when 240 kWh are used.

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B=8+0.25ufixed charge + rate × unitsB = 8 + 0.25u \quad \scriptsize\textit{fixed charge + rate }\times\textit{ units}
When u=240u = 240:
B\displaystyle B
=8+0.25(240)\displaystyle = 8 + 0.25(240)
=8+60\displaystyle = 8 + 60
=$68\displaystyle = \$68
The gradient (0.250.25) is the cost per kWh; the yy-intercept (88) is the fixed monthly charge.

Now you try

A phone plan costs $15\$15 per month plus $3\$3 per GB of data used. Find Daniel's total monthly cost for 6 GB.

Show solution
Total cost\displaystyle \text{Total cost}
=15+3×6\displaystyle = 15 + 3 \times 6
=15+18\displaystyle = 15 + 18
=$33\displaystyle = \$33

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Finding the linear model from two data points

A catering company charges a delivery fee plus a fixed amount per guest. For 30 guests the total cost is $410\displaystyle \$410, and for 50 guests it is $610\displaystyle \$610. Find the cost for 80 guests.

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The rate per guest is the gradient of the cost line.
r\displaystyle r
=6104105030\displaystyle = \dfrac{610 - 410}{50 - 30}
=20020\displaystyle = \dfrac{200}{20}
=10= 10, so each guest adds $10\$10.
Delivery fee\displaystyle \text{Delivery fee}
=41030×10\displaystyle = 410 - 30 \times 10
=$110\displaystyle = \$110
Cost for 8080 guests =110+10(80)= 110 + 10(80)
=$910= \$910

Now you try

Mei Ling sells cookies at $1.50\$1.50 each. The fixed costs are $55\$55. How many cookies must be sold to earn at least $75\$75 in profit?

Show solution
Let the number of cookies be nn.
Revenue\displaystyle \text{Revenue}
=1.50n\displaystyle = 1.50n
Profit\displaystyle \text{Profit}
=1.50n55\displaystyle = 1.50n - 55
1.50n55751.50n - 55 \geq 75
1.50n1301.50n \geq 130
n1301.50n \geq \frac{130}{1.50}
n86.7n \geq 86.7
n=87n = 87 (round up to nearest whole number)

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