The cost C of manufacturing a product can be estimated by the formula

From the formula we deduce that C directly proportional to r, s and t^2
hence an increase or decrease in these values will directly change the value of C by that factor

There new r => 1.5r
new s => 1.2s
new t^2 = (0.7t)^2 = .49t^2
let the constant 0.03 be termed as A

Putting these new values: C1 = A*1.5r*1.2s*.49t^2
C1 = 0.882A*r*s*t^2 = 0.882(C)

There the net value decrease by 12% approx

Hence D

C = 0.03 r s t^2
New C = C1 = \(0.03(150/100)r(120/100)s((70/100)^2)t^2\)

=(882/1000)[0.03rst^2]
as C>C1 => its a decrease

% decrease = ((original - new) /original)*100

= (118/1000) *100
~= 12%

Answer is D.

lets initial values be as follows
r=10, s=10, t=10

Initial cost = \(\frac{3}{100}\) X 10 X 10 X 100

Initial Cost = 300

Revised values will be

r=15 -----------------> increase of 50%

s=12 -----------------> increase of 20%

t=7 -------------------> decrease of 30%

Revised cost = \(\frac{3}{100}\) X 15 X 12 X 49

Revised cost = 265 approximatly

Reduction of 35 approx

reduction of 12% approx

Solution:

If we let r = s = t = 10, the original cost would be C = 0.03 x 10 x 10 x 10^2 = $300. Since r is increased by 50%, s increased by 20% and t decreased by 30%, the new cost would be C = 0.03 x 15 x 12 x 7^2 = $264.60. We see this is approximately $36 less than the original $300, or $12 less per every $100, i.e., a 12% decrease.

Answer: D

Video solution from Quant Reasoning starts at 9:08
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It would sure be nice if we knew what r, s, and t were to begin so we don't have to work with the variables. Let's just make them up! Here's an interesting caveat on technique for plugging in numbers. If there are variables in the question that are repeated in the answer choices, we want to be careful to pick numbers that won't lead to ambiguities among the answer choices; that is, numbers that are likely to cause more than one answer choice to be correct, which would just cause us to have to try different numbers and cost us time. On questions where there is no variable in the answer choices, we do not need to worry about that, and we can just pick whatever numbers make solving easiest.

Let's use r=1, s=1, and t=1.

So, we start with \(C= 0.03(1)(1)(1^2)=0.03\).

And we go to \(C= 0.03(1.5)(1.2)(0.7^2)=0.03(1.8)(0.49)\). Let's just use 0.5 instead of 0.49. \(0.03(1.8)(0.5)=0.03(0.9)=0.027\).

We started at 0.03 and went to 0.027. That's a 10% decrease.

Answer choice D.


ThatDudeKnowsPluggingIn

ThatDudeKnows GMATinsight MartyTargetTestPrep how do we do these calculations fast?

Elite097

I don't think the calculations should take all that long on this one if you use an approach that minimizes them. Here's an annotated version of my answer from above.
That took almost no time at all.


Getting from \((0.7^2)\) to 0.49 to 0.5 takes a few seconds. 1.5*1.2 is 1.2 plus half of 1.2, so 1.2 + 0.6. That's 1.8. Now we need to multiply by 0.5. So 0.9. We can stop at this point, since we are comparing 0.9*0.3 to 0.3 and 0.9 is 90%. That's a 10% decrease from 0.3.

The only calculations we needed to do were 1.5*1.2 and then multiplying that by 0.5.

By plugging in, we've minimized the number of manipulations we need to make, and by picking easy numbers, we've made it even more manageable.

­\(C= 0.03 rs t^2\)

Use percent multipliers to get the answer quickly.
r is increased by 50%, s increased by 20% and t decreased by 30% - So r becomes 3r/2, s becomes 6s/5 and t becomes 7t/10

So the overall multiplying factor for C is \(\frac{3}{2} * \frac{6}{5} * \frac{7}{10} * \frac{7}{10} = \frac{18 * 49}{50 * 10}\)

Since we are looking for an approximate value, we can take 49/50 as 1 to get a decrease of 2/20 which is 10% decrease. But actual decrease will be slightly more (because of the approximation) hence 12% decrease is correct.

Answer (D)

Video on Percentages: https://youtu.be/HxnsYI1Rws8­