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Bunuel
For the first quarter of 1980, the profits of a certain food-market chain amounted to $49,350,000. If this figure is 5 percent higher than the profits for the first quarter of 1979, what was the amount of the increase?

A. $350,000
B. $2,350,000
C. $2,467,500
D. $4,688,250
E. $4,700,000


PS20396

1980 = 4935*10^4 which is 1.05 of 1979
4935*10^4 = x *1.05
x= 47*10^6
so ∆ = 4935*10^4-47*10^6 ; 2350,000
OPTION B
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B.

Let the profit in 1979 = X
Profit in 1980 = 49,350,000 = 5% more than Profit from 1979 = X + 5% of X
OR 49,350,000 = (105X)/100

Solving for X; X = 47,00,000

Increase in Profit: 49,350,00 - 47,000,000 = 2,350,000
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ABH06
Does anyone have any tips for completing this question in under 2 minutes? The maths is easily done on a calculator, but I'm struggling to do it in the allowed time with just pen and paper.

You can use the written/long division method once you figured that X=98700000/21
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Does anyone have any tips for completing this question in under 2 minutes? The maths is easily done on a calculator, but I'm struggling to do it in the allowed time with just pen and paper.


Let the amount in 1979 be X
Then,

49350000= X(105/100)

X= (49350000 x 20) / 21

Instead of solving this keep going,

Difference between the two amounts=

(49350000 ) - ((49350000 x 20) / 21)

= ((49350000 x 21) / 21)) - ((49350000 x 20) / 21)

Take 4935000 in common which gives you

= (49350000 (21-20)) / 21)

= (49350000 x 1) / 21

As you can see the division is quite simple.

Ans: 2350000 (B)
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Why cant 49350000 be multiplied by 0.95 then that product is the amount the increase occurred from and can't you just subtract those two to get the difference? You can use the multiplier for percent increase or decrease so you add to 1 or subtract from one accordingly? When I do that I get C. Or am I thinking of this wrong?
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GMATNinja, Bunuel, can you please advice on the most efficient way to divide 49,350,000 by 1.05?
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Batman117
1979 FQ profit is X
1980 FQ profit is 1.05X
Now 1.05X = 49350000
So X =47000000
And amount increase is =2350000

Posted from my mobile device


Your approach is the exact approach if you use your calculator, not the one that shows step by step the long addition and subtraction procces.

If you donot mind, please provide the step by step solution.
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Here is how i did it.

Let X be the profit of Quarter-1 1979 and we know from the question that the profit of Quarter 1 1980 is $49350000, which is 5% more than the profit of Quarter 1 in 1979, meaning the profit in 1980 is 105% of the profit in 1979.

Since X represents the profit in 1979, then the profit of Quarter 1 in 1980 versus the profit of Quarter 1 in 1979 looks like this algebraically:

$49350000=1.05X.

To slove the question we need X without using the calculator, which is very tempting to use for this kind of questions, coz that way you will have the solution for X within seconds. But that is of not what we gonna do, considering we are not allowed to use calculator on Exam day.

For simplicity i will temporarily remove the 4 zero's behind the profit of Quarter 1 in 1980, which is $49350000, making it just 4935.

4935=1.05X

Step 1/

We need the fraction equivalent of 1.05 and that is 21/20.

If you wonder how you should know that, here is how: 20/20=1 and 1/20=0.05, thus 20/20 + 1/20= 21/20=1.05

Step 2/

Now that we know 1.05=21/20, then:

4935=1.05X becomes 4935=21/20X.

Now we isolate X to solve for X.

Divide both sides of (4935=21/20X) by 21/20)), notice when you divide 4935 by 21/20, it becomes 4935 * 20/21(you take the reciprocal of the fraction when you divide fractions by fractions or numbers by fractions)

We get X= 20/21 *4935 here is the challenging part of the question, where you need creativity and shapness.

For an integer to be divisible by 3, the sum of the integers digits must be divisible by 3 and for an integer to be divisible by 7 dubble the units digit of that integer and subtract that from the rest of the integer excluding the units digit.

Example: 21 is divisible by 3, coz the sum of the integers digit is 2+1=3 which is divisible by 3 or 21 is divisible by 7, coz if we dubble the units digit of 21, which is 1, it becomes 2, then we subtract 2 from the rest of the integer excluding the units digit, which is 2 then we get 2-2=zero(0) which is divisible by 7. That same goes for the 4935.

Now that we know 21 is divisible by 3 and 7 and 4935 is divisible 3 and 7, both are divisible by 21.

Divide 21 by 21 and 4935 by 21.

X= 20/21 *4935 21/21=1 and 4935/21= 235

X= 20 * 235 = 4700 here is where we add back the 4 zeros we removed at beginning of the question, which then becomes:

X=47000000, which is the Profit of Quarter 1 1979.

Now we subtract the profit of Quarter 1 in 1979 (X= $47000000) from the profit of Quarter 1 in 1980 $49350000, we get:


$49350000(profit 1980) - $47000000(Profit 1979)= $ 2350000 is the answer.

Answer choice B

I hope it is clear.
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tkorzhan1995
GMATNinja, Bunuel, can you please advice on the most efficient way to divide 49,350,000 by 1.05?
Have you tried reviewing this post? Better to hold off on calculating anything until you absolutely have to.

And when you get to the final step (49,350,000/21), you can always break that up into (49,350,000/[7*3]) and divide by 7 first to make the numbers even more manageable: 49,350,000/7 = 7,050,000, so we're left with 7,050,000/3.

And if you REALLY don't want to do long division here, you could recognize that 3 divides very cleanly into 6,000,000, and then you're left with something like this: [6,000,000 + 1,050,000]/3 = 6,000,000/3 + 1,050,000/3 = 2,000,000 + 350,000 = 2,350,000.

I hope that helps!
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tkorzhan1995 Most of the explanations given here are a little time consuming and calculation intensive , so thought of suggesting a shortcut approach to solve this question in less than a minute.

For the first quarter of 1980, the profits of a certain food-market chain amounted to $49,350,000. If this figure is 5 percent higher than the profits for the first quarter of 1979, what was the amount of the increase?

A. $350,000
B. $2,350,000
C. $2,467,500
D. $4,688,250
E. $4,700,000


This Question is clearly based on percentage increase concept. In this case, an initial value is increased by 5 % to get a final value.
Let's assume that the initial value be 100 and it is increased by 5% , i.e 100 is increased by 5 and we will get 105 as the final value.

So in order to get back to 100 from 105, by what percentage of final value should we decrease ?

If you get the answer to this question, then 90 % of the task is done .

Is it 5 % ? No, clearly not .

But we know that 105 has to be decreased by 5 to get 100.
So the percentage decrease should be 5 /105 * 100 or the amount to be decreased = \(\frac{5}{105} \)* final value = \(\frac{5}{105}\) *105 = 5

So lets summarize what we have discussed.

If an initial value is increased by 5 % to get a final value, then the amount increase here is 5 % of initial value or \(\frac{5}{105}\) of final value i.e \(\frac{1}{21 }\) of final value.

Lets apply the above discussed concept here in this question.

The amount increase = \(\frac{1}{21}\) * Final value = \(\frac{1}{21}\) * $49,350,000 = 23...

You don't need to do the entire division, first 2 digit of the quotient is enough to figure the correct answer as only one answer option starts with 23...

Option B is the right answer.

If you get any different value as percentage increase instead of 5 , then assume initial value as 100 and find the corresponding percentage decrease needed to bring the final value back to 100.

Thanks,
Clifin J Francis,
GMAT Quant SME
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Ok but the whole point is doing it without a calculator:

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