If S is the sum of reciprocals of a list of consecutive

I got 0.2 too (B). Please help. Here is my approach
(1/45) + (1/46) + .....+(1/54) = (45-44)/45 +(46-45)/46 +.....+(54-53)/54
= 10 -(44/45 + 45/46 + ...53/54). Each term in the bracket is > 0.977, and there are ten of them. Then their sum is roughly 9.8
10-9.8 = 0.2. But is this doable in 2 mn under stress and heat?
Brother Karamazov



Brother Karamazov

my method was 1/45 + 1/46 + ... => 10/45 ~ 2/9 ~ .2

Such questions do not test your calculations. (In fact most of GMAT questions) They test your logic and how easily can you simplify a problem.
That is why an approximate result is asked and not the absolute value.

Out of the given numbers, from 45 to 54, carrying out calculations with 50 would be easiest.

Hene instead of 1/45 + 1/46 + ... 1/50 + ... + 1/54,
We can increase some terms and decrease some terms by changing the numbers with 50

Hence we have a sequence = 1/50 + 1/50 + ... + 1/50 (Total 20 terms)
= 10/50 = 1/5 = 0.2 approximately
Option B

Hi,

aren't there 10 terms and not 20 ?

regards

We need to determine the approximate value of the sum of the reciprocals from 45 to 54 inclusive; thus, we need the approximate value of the following:

1/45 + 1/46 + 1/47 + 1/48 + 1/49 + 1/50 + 1/51 + 1/52 + 1/53 + 1/54

Rather than adding each of these numbers (which would be incredibly time-consuming), let’s strategically select one of the fractions in our list and add it to itself 10 times. That sum will give us an approximate value for S.

Scanning the list, we see the best number to add to itself 10 times is 1/50. However, instead of actually adding 1/50 ten times, we will simply multiply it by 10:

1/50 x 10 = 10/50 = 1/5 = 0.2

Thus, we see that S is approximately 0.2.

Answer: B

median=49.5
reciprocal=2/99
10*(2/99)=20/99≈.2
B

Mid point of the range 45-54 is 50. So if we take the reciprocal of 50 and add it 10 times as per the question, we get 1/5 which is 0.2

We take 50 as the average of the range and do the simplified math.

# of Terms = (54-45)/1 + 1 = 10

Method 1: Using First & Last Term
Mean = ((1/45) + (1/54))/2 = (54+45)/(45*54*2) = 99/(45*54*2)
= 33/(15*54*2) = 11/(5*54*2) = 11/540
Sum = 10 * 11/540 = 110/540 = 0.2 => B

Method 2: Using Middle Terms (Better)
Mean Number = (1/49 + 1/50)/2 = ~1/50
Sum = 10/50 = 0.2 => B

That's not something you need to do here - if you notice we're adding ten fractions roughly equal to 1/50, the sum will roughly be 10/50. Or you can notice we're adding ten things that are all less than or equal to 1/45, and also are adding ten things greater than or equal to 1/54. So

10/54 < S < 10/45

and 10/50 = 0.2 is the only value from among the answer choices that S can approximately equal.

If you did want to find the decimal equivalents of 10/45 and 10/54, in the first case, 10/45 = 2/9. You can find the decimal of 2/9 if you know the decimal of 1/3:

1/3 = 0.33333....

so dividing by 3 on both sides

1/9 = 0.1111....

and multiplying by 2 on both sides

2/9 = 0.2222....

We can do something similar for 10/54, though it's a bit trickier. Since 10/54 = 5/27, we can start from

1/9 = 0.111111....

and divide by 3 on both sides (using the fact that 111/3 = 37) to get

1/27 = 0.037037....

and then multiply by 5 to get

5/27 = 0.185185....

So for these specific fractions, you can find their decimal equivalent without using long division, because their denominators are somewhat convenient to work with. But in general, if you invent a fraction with a random denominator, most of the time you'd have no practical choice other than long division to find the decimal. So if you need the decimal equivalent of 3/29, say, you'd need to use long division (or a calculator) to find it. Fortunately you never need to do that on the GMAT -- all you might need to notice on the GMAT is that 3/29 is very slightly larger than 3/30, so is very slightly larger than 0.1.

We have the sum of 10 numbers:

(1/45) + (1/46) + …… + (1/54) =

(1st)
The sum must be less to the value if we were to assume that all 10 numbers were equal to the highest value: 1/45

1/45 = (1/9) * (1/5) = (1/9) * (.2) = .2 / 9

= .0222222…..

And

(10) (.02222….) = .222222…

sum < .222222…..

only A and B are possible


(2nd)
The sum must be greater than > (10) (1/54)

(10 / 54) = (5/27) = (5/9) * (1/3) = (.555..) * (1/3)

When we divide 3 into the recurring decimal (.555…) we get approximately .185


(.185) < sum < (.2222…)

*B* is the only possible answer

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