I’ve looked up explanations but they are not thorough enough…
If 75 percent of m is equal to k percent of 25, where k > 0, what is the value of m/k
A) 3/16
CORRECT ANSWER B) 1/3
C) 3/4
D) 3
E 16/3
A telephone company charges x cents for the first minute of a call and charges for any additional time at the rate of y cents per minute. If a certain call costs $5.55 and lasts longer than 1 minute, which of the following expressions represents the length of the call, in minutes?
A) (555-x) / y
B) (555±y)/y
CORRECT ANSWER: C) (555-x+y)/y
D) (555-x-y)/y
E) (555) / (x + y)
3/4 m = 25 k/100
multiply both sides by 100
75 m = 25 k
Divide both sides by k
75m/k = 25
Divide both sides by 75
m/k = 25/75 = 1/3
Number 2:
x + (minutes -1) y = 555 (because that first minute is the x, so the remaining minutes are (min - 1)
distribute
x + y(minutes) - y = 555
get middle term alone:
y(minutes) = 555 - x + y
divide by y to isolate minutes
minutes = (555 - x + y)/ y
For number 2, if you don’t want to use algebra, you can make up numbers that fit. This is a classic SAT method that should be in your bag of tricks (for the new SAT as well). In this case, you could make x .55 and y .50 so that an 11 minute call comes out to $5.55. Or you could make x $5 and y .11 and that way a 6 minute call comes out to $5.55 – play around until you can make up your own numbers. Then put your x and y into the answer choice and rule out anything that does not crank out an answer that matches your number of minutes.
For number 1 – if you notice that k % of 25 is the same thing as 25% of k, you can get the answer by thinking: If 75% of one pie is the same as 25% of another pie, then the second pie has to be 3 times as big as the first pie!
This is a common trait of many SAT questions: you can do them by algebra but they are also designed in a way that lets you avoid the algebra by playful thinking.
It’s also pretty easy to make up numbers for 1. Say m = 100. 75% of 100 is 75, and this equals k% of 25. Since 300% of 25 is 75, we have k = 300. Then m/k = 100/300 = 1/3.
@kyloren2200 note that I didn’t say plugging in numbers is the best way. It might be the best way for many students, although many others might prefer the algebraic route.
Just to expand on what @MITer94 stated, there is no one strategy that fits all, “all” being both students and questions.
Some students may have trouble solving certain problems algebraically, but plugging in numbers could help them crack those problems with just a bit of brute work.
Other students have such a degree of algebraic mastery that for them solving problems would be a much easier route.
And then, there are students who can decide on the fly which approach works better/faster for a particular question.
I would probably do #2 algebraically - but once again, this is a matter of a personal preference.
After the first minute the cost of the call is (555 - x) cents;
dividing it by the going rate of “y” cents/min we are getting the number of additional (after the first one) minutes:
(555 - x) / y.
To calculate the total duration of the call we now add that first minute
(555 - x) / y + 1 =
(555 - x + y) / y.
Answering question #1, on the other hand, is easier IMHO either by reasoning (@pckeller) or by working with good numbers (@MITer94).
Combining the best of both worlds - a playful @pckeller’s thinking trick (k% of 25 = 25% of k) and @MITer94’s efficient approach - and converting percents into fractions would save even more time:
Assuming m = 1,
3/4 of 1 = 1/4 of k,
k = 3,
m / n = 1/3.
