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What is the average (arithmetic mean) of a list of 6

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by BTGmoderatorDC » Tue Sep 03, 2019 5:17 pm

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What is the average (arithmetic mean) of a list of 6 consecutive two-digit integers?

(1) The remainder when the fourth integer is divided by 5 is 3.
(2) The ratio of the largest integer to the smallest integer is 5:4.

OA B

Source: Princeton Review
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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Thu Sep 05, 2019 11:29 pm
BTGmoderatorDC wrote:What is the average (arithmetic mean) of a list of 6 consecutive two-digit integers?

(1) The remainder when the fourth integer is divided by 5 is 3.
(2) The ratio of the largest integer to the smallest integer is 5 : 4.

OA B

Source: Princeton Review
Say the 6 consecutive two-digit integers are x, (x + 2), (x + 4), (x + 6), (x + 8), and (x + 10). Since the numbers are evenly spaced, the average of the 6 consecutive two-digit integers would be the average of the two middle-most numbers = [(x + 4) + (x + 6)]/2 = x + 5

So, we have to get the value of x + 5.

Let's take each statement one by one.

(1) The remainder when the fourth integer is divided by 5 is 3.

=> x + 4 = 5q + 3. where q is a posotive integer

Thus, x + 5 = 5q + 4. Since the value of q is not known, we cannot get the unique value of x + 5. insufficient.

(2) The ratio of the largest integer to the smallest integer is 5 : 4.

=> (x + 10)/x = 5/4 => x = 40 => x + 5 = 40 + 5 = 45. Unique value. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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by deloitte247 » Sat Sep 07, 2019 1:39 pm
Find the average of 6 consecutive two-digit integers
$$i.e.\ \frac{x,\left(x+1\right),\left(x+2\right),\left(x+3\right),\left(x+4\right),\left(x+5\right)}{6}=\frac{6x+15}{6}$$

Statement 1 => The remainder when the fourth integer is divided by 5 is 3
i.e. 4th integer = ( multiple of 5 ) + 3
The consecutive integers could be either; 10, 11, 12, 13, 14, 15 or 15, 16, 17, 18, 19, 20
Information given is not enough to find the average as the 6 consecutive integers remain unknown
Hence, statement 1 is NOT SUFFICIENT

Statement 2 => The ratio of the largest integer to the smallest integer is 5:4
Largest integer = 6th integer = x + 5
Smallest integer = 1st integer = x
Largest : smallest = x + 5 : x = $$\frac{x+5}{x}$$

$$\frac{x+5}{x}=\frac{5}{4}$$ => 4x + 20 = 5x and x = 20
=> integers => 20, 21, 22, 23, 24, 25 => $$average=\frac{22+23}{2}=22.5$$
Statement 2 alone is SUFFICIENT
Answer = option B
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