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

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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

Hi

How did you guys determine if it was an increase or decrease?

is it because the value of c > c1?

I am using the same approach as above:

50% increase = 1.5r
20% incease = 1.2s
30% decrease = .7r

So new value = (1.5r)(1.2s)(.7r)^2=(15/10*12/10*49/100)(.03rst^2)
Old value = (.03rst^2)

Now I am approx. 49 to 50 for faster calculation -
New value = (15/10*12/10*50/100) = 180/100*1/2=(90/100)(.03rst^2)

New value is 90% of the old value
So 10% less ....Let me see any value near 10% decreases
B is the closest...So I go with B

Hope it is clear to you how it is decreasing ... Taking a larger value like - 50 even decreased,so obviously a smaller value - 49 will always decrease

The reason for approximating is sometimes we make silly mistakes in calculation.
So the best way is to pick smart numbers when needed.........Again it is up to you to decide the approach that works nest for you.....
Some people can calculate this in head in (5-10) secs....then do not approximate...This is for the other folks who spend 20-30 secs on multiplication(calculation)

Thank you so much you guys, Why I haven't thought about using Plug-in strategy.

This question is very simple after I saw your explanation. Thank you again for the effot.

tracyyahoo - Give kudos to jamifahad if you liked his approach...

That is one way of saying thanks for the person spending time on your questions.....

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.

[(1.5)*(1.2)*(0.7)*(0.7) - 1]*100% = [1.8*0.49 - 1]*100% = - 12%

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

50 % Increase = 1.5r
20 % increase = 1.2 s
30 % decrease = 0.7t --> Formula says we have t² hence 0.7²

multiply: 1.5*1.2*0.7² = 0.882 = 12 % decrease. (1-0.882)

The cost C of manufacturing a product can be estimated by the formula C= 0.03 rs t^2, where r ans s are amounts, in pounds, of two major ingredients and t is production time in hours. if r is increased by 50%, s increased by 20% and t decreased by 30%, by approximately what % will the estimated cost of manufacturing the product change.

C(old) = 0.03 * r * s * t^2

C(new) = 0.03 * [(3/2)*r] * [(6/5)*s] * [(7/10*t)^2]
C(new) = [0.03 * r * s * t^2 ][(3/2)*(6/5)*(49/100)]
C(new) = C(old) * (9/10) ......................(approximate by taking 49/100 = 1/2)
C(new) = C(old) * (0.9)

So, C(new) is 10% less than C(new). So, a 10% decrease.

Option (D) i.e. 12% decrease is the closest.

jamifahad 150*120=18000, you missed a zero, kindly edit.

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

This formula is very straightforward so we can "ignore" it and focus on multiplying percentages.

15/10 * 12/10 * 7/10 * 7/10 = 49*180 / 10 000

50*180 = 9000, so 49*180 = 8820.

8820 is 11,8 percent less than 10 000.

Given: The cost C of manufacturing a product can be estimated by the formula \(C= 0.03 rs t^2\), where r and s are amounts, in pounds, of two major ingredients and t is production time in hours.

Asked: If r is increased by 50%, s increased by 20% and t decreased by 30%, by approximately what % will the estimated cost of manufacturing the product change.

\(C' = 0.03(1.5r)(1.2s)(.7t)^2 = .03rst^2 * 1.5*1.2*.49 = 0.88 * C = (100% - 12%) C\)

IMO D