Pricing

PP1 (Shorter) Quant Section 2 (Medium) Q9

Solving the Problem Method 1 (Algebra) We can start out by assigning some variables: <ul> <li>$A$ = The price an an apple</li> <li>$O$ = The price of an orange</li> </ul> <blockquote> the price of an apple is twice the price of an orange. </blockquote> From this, we can also set up the following equation: $A=2O$ <blockquote> For which of the following combinations of apples and oranges is the total price equal to the total price of 20 oranges? </blockquote> So, we're looking for anything that is equivalent to $20O$ or $10A$. We can now go through each answer choice, write them out in terms of only Oranges, and check which equals $\textbf{20O}$.  <ul> <li>A: 2 apples and 16 oranges -> $2A+16O = 4O+16O = 20O$</li> <li>B: 3 apples and 14 oranges -> $3A+14O = 6O+14O = 20O$</li> <li>C: 4 apples and IO oranges -> $4A+10O = 8O+10O = 18O$</li> <li>D: 6 apples and 8 oranges -> $6A+80 = 12O+8O = 20O$</li> <li>E: IO apples and 5 oranges -> $10A+5O = 20O+5O = 25O$</li> <li>F: 12 apples and 4 oranges -> $12A+4O = 24O+4O = 28O$</li> </ul> From, this we can see that only A, B, and D work. Method 2 (Choosing Numbers) Let's choose the following prices: <ul> <li>Price of an Apple = $\$2$</li> <li>Price of an Orange = $\$1$</li> </ul> Now, we can look for anything that equals the price of 20 Oranges, which would be $20: <ul> <li>A: 2 apples and 16 oranges -> $2(2) + 16(1)$ = $20</li> <li>B: 3 apples and 14 oranges -> $3(2) + 14(1)$ = $20</li> <li>C: 4 apples and IO oranges -> $4(2) + 10(1)$ = $18</li> <li>D: 6 apples and 8 oranges -> $6(2) + 8(1)$ = $20</li> <li>E: IO apples and 5 oranges -> $10(2) + 5(1)$ = $25</li> <li>F: 12 apples and 4 oranges -> $12(2) + 4(1)$ = $28</li> </ul> From, this we can see that only A, B, and D work. So, A, B, and D are all valid answers.