backsolving
- backsolving involves starting with one of the numerical answers and working backwards to see if it makes sense.
- a recommended strategy is to begin with answer choice c, which statistically allows for the elimination of three answers at once if it's incorrect.
- an alternate strategy suggests starting with answer choice b, potentially increasing the chances of finding the correct answer on the first guess.
- practical application of backsolving is demonstrated through practice questions, showing how to efficiently narrow down answer choices.
- the content underscores the importance of understanding the concept of backsolving as a potent tool for tackling numerical answer choice questions on the gre.
faq: will the answers always be arranged in ascending order?
a: yes, numerical answer choices are usually arranged in increasing order. it’s not very likely, but it’s possible that some answers will be in decreasing order -- but the backsolving strategy still works in this case.
even if the answer choices weren't in increasing order, you could still apply the same strategy, but you would have to be careful to track ordering of the answer choices yourself.
faq: how would the first practice problem (about taxes) be solved algebraically?
a: this problem was chosen to illustrate backsolving in part because it's difficult to do algebraically! this explanation will make more sense if you first watch our lesson on mixture problems.
since we'll eventually need to solve for the money spent on food, we can begin by calling that amount f. we’ll call the amount spent on stationery 500-f.
we already know the following:
- the tax on food = 0.02f
- the tax on stationary = 0.08 (500-f) = 40 - 0.08 f
- the total tax spent on both = 0.02f + 40 - 0.08 f = 40 - 0.06f
- the total tax spent on both is given as 19, so 40 - 0.06 f = 19
now we can solve this using algebra:
40 - 0.06f = 19
subtract 19 and add 0.06f to each side.
21 = 0.06f
multiply each side by 100 to clear the decimal fraction.
2100 = 6f
divide each side by 6.
f=350.
faq: how would we solve the second example problem algebraically?
a: we can begin by finding the original amounts of hno3 and water in the solution. it's 40% hno3 and 60% water of 60 total liters:
.40 * 60 = 24 l hno3
.60 * 60 = 36 l water
we know that we want water to be half of the solution, and we won't be adding anymore water. the amount of water will stay the same and will be half of the total solution. so, 50% of the total (x) will equal 36 l, and we end up with 72 l total:
.50 * x = 36 l
x = 72 l total
half of this will be hno3, so we can divide 72 by 2 to get 36. now we have to subtract the original 24 l of nho3 to get the added amount:
36 - 24= 12 l
so, 12 l of hno3 were added to get a 50% solution.
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