The volume of water in a certain tank is x percent greater

Best approach for solving this specific problem would be "Pick numbers". You can also solve it in Algebraic way, but it could be time consuming and tedious.

Given that, the water was increased by x percent -> then reduced by r percent -> to 90 percent of what it was one week ago.
Consider, the 100 units of water was increased by 20% to 120 units -> then reduced by 25% of 120 (i.e. by 30 units) -> to 90 units.
i.e. Pick numbers as x = 20% and r = 25%

Substitute x=20 and identify the answer that gives r=25
A) x+10 --20+1=30
b) 10x+1 --200+1=201
c) 100(x+10) --100(20+10)=3000
d) 100( x+10/x+100) --100(20+10/20+100) = 100(30/120)=25
e) 100 ( 10x+1/10x+10) -- 100(200+1/200+10)= 100(201/210)

Hence choice(D) is the answer.

Volume now is x% greater than volume one week ago
=> Vnow = Vweekago (1+x/100)

If r percent of the current volume is removed, the resulting volume will be 90 percent of the volume a week ago
=> Vnow (1-r/100) = 0.9*Vweekago

Using the first equation, Vnow/Vweekago = (1+x/100)
Putting this in the second equation,
(1-r/100) (1+x/100) = 0.9
=> (100 - r) (100 + x) = 9000
=> r = 100 - [9000/(100+x)]
=> r = 100* (10+x)/(100+x)

Option D

I think that in this scenario under pressure someone could not visualize 25% to obtain 90 and to go in the wrong way.

better an hybrid approach and in this case algebraic translation maybe is a bit safer
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Yes I agree, not all problems can be solved using pick numbers strategy or not always will be able to come up with ideal pick numbers under pressure. At times for complex problems, you need pen down that lengthy/heavy algebraic computations. However, giving 5 seconds to think about execution method (pick numbers, backsolving or algebraic) after reading the problem, can help saving over 40-50 seconds of 'extra' time taken by algebraic method.

In this specific problem, I tried to visualize how can I increase 100 to certain number and decrease to obtain 90?? First comes 110, but reducing it by 20 doesnt give integer percentage on 110. Next we have 120 that includes 20% increase and then 25% decrease gives 90..bingo! It took 5-7 seconds to think & 15-20 seconds to back-solve, but saved 40-50 seconds of algebraic calculations!

Picking number requires strategy of identifying optimum numbers using LCM, LCD, prime factors, multiples of 10/100s.

How do we get used to this method? Practice!!

When solving OG or practice problems, always see if you can solve it using multiple ways. Try same problem with Algebra, back-solving, pick numbers, hybrid approach (whichever methods are applicable) and even try guessing it. You may also automatically build that intuition over the time.

Another consideration. be flexible

here we have something go up for a certain % and the go down for a certain % to have some result, here is 90. Ok 90 is our target value or our landmark

but what about if we think in a more abstract way ?? even if I increase my % of 20 or 30 and then decrease the same by any % to obtain a value of less than 100........the process is the same.

The gmat questions are so consistent and coherent (two word to say the same thing, indeed) that if you use several approachs they conduct you to the same answer, MUST conduct you to the same place. For this reason you can attack a problem with different stategies

Infact, even if you have 85 and not 90 as target value the answer is still D. Of course , this kind of resilience is acquired after you repeat over and over again the concepts behind
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How did you get from here to here?
=> r = 100 - [9000/(100+x)]
=> r = 100* (10+x)/(100+x)

Thanks

Cheers!
J

\(r = 100 - \frac{9000}{100+x}\);

\(r = \frac{100*(100+x) - 9000}{100+x}\);

\(r = \frac{100*(10+x)}{100+x}\).

Hope it's clear.
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(1-r/100) (1+x/100) = 0.9
=> (100 - r) (100 + x) = 9000

I am seriously having a brain fart, how do you get the 9000. I multiplied everything by 100.

You should have multiplied by 100*100:

\((1-\frac{r}{100}) (1+\frac{x}{100}) =0.9\);

\((\frac{100-r}{100}) (\frac{100+x}{100}) = 0.9\);

\((100 - r) (100 + x) = 0.9*100*100=9000\).

Hope it's clear.
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As per question if r percent is removed then remaining is 90 % of original. Can we infer that removed water is 10% of the original. Hence i used this equation

(r/100)*[1+x/100] V= 0.1 V..Which gave me wrong answer....Help please

No, that's not correct.

The volume of water one week ago = 100 (that's what I believe you call original);

The volume of water now is x=20% greater = 120.

The volume of water after removal of r% of 120 = 90.

So, as you can see r% is not 10% of 100, it's r% of 120, which result in final volume of 90.

Does this make sense?
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We can let the volume of water in the tank one week ago = n.

Since the volume is now x percent greater than it was one week ago, the volume today is:

(1 + x/100)n

Next we are given that r percent of the current volume is removed. Thus we now have:

(1 - r/100)(1 + x/100)n

Since the resulting volume will be 90 percent of what it was a week ago, we can create the following equation and isolate r:

(1 - r/100)(1 + x/100)n = 0.9n

(1 - r/100)(1 + x/100) = 0.9

[(100 - r)/100][(100 + x)/100] = 0.9

[10,000 - rx + 100x - 100r]/10,000 = 0.9

10,000 - rx + 100x - 100r = 9,000

1,000 + 100x = rx + 100r

1,000 + 100x = r(x + 100)

100(10 + x) = r(x + 100)

100(10 + x)/(x + 100) = r

Answer: D
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I used numbers to make this easier.

Let original capacity = 100
Let X% = 1%

So this week we have capacity of 101.

Target capacity is 90.

% change needed (i.e. r %) to bring the capacity from 101 to 90 is: [101 - 90 / 101] * 100
so the target value of r = 1100/101.

Now, to test the answer choices, I know A, B and C are unlikely to be right because my value for "r" is not an integer, i.e. I need a numerator and a denominator term. So I would start by testing options D and E

Option D: 100 [1+10] / [100 + 1] = 1100 / 101

Hope this helps.

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