In a certain store, the price of five pens is equal to the price of a notebook. What is the cost of a ruler in this store?
- The cost of four pens and two notebooks is 10 dollars more than the cost of six rulers.
- The cost of seven notebooks is 25 dollars more than the cost of 15 rulers.
Have your answer? Great! Let’s go ahead and walk through it.
My first step is to assign variables and translate the prompt into math – I let p represent the number of pens, n represent the number of notebooks, and r represent the number of rulers. From the prompt, then, I can get the equation 5p = n, and I want to find the value of r.
Translating statement 1, I can say that 4p + 2n = 10 + 6r. Three variables in one equation doesn’t help me, but I can substitute 5p for n to get 4p + 2(5p) = 10 + 6r, or 14p = 10 + 6r. Uh-oh – I have two variables, but only one linear equation! I can’t solve for r, so statement 1 must be insufficient.
Moving on to statement 2, I can translate to get the equation 7n = 25 + 15r. I’m already suspicious, because again I have two variables and only one linear equation, but I’ll try substituting 5p from the prompt anyway. This gives me 7(5p) = 25 + 15r, or 35p = 25 + 15r. Well, that didn’t really help me – I still have two variables and only one equation, so I can’t solve for r. Statement 2 must be insufficient.
Now I consider the statements together. I can see from statement 1 that I have a linear equation with the variables r and p. In statement 2, I have another equation with the same two variables. Two variables and two linear equations? Great! I’m a good data sufficiency strategist, so I know that I shouldn’t waste time to actually do the math here – I know that I can solve for both variables in the system, so I’ll be able to find the value of r. The answer here is C.