Social Distancing with Data Sufficiency Challenge! – Day 24

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This series is designed to help you step up your Data Sufficiency practice while we all spend a little extra time at home during the coronavirus situation; I’m going to publish one practice Data Sufficiency question each day! You can choose to do all of them, or none of them. 

Click on the tag “Social Distancing DS Challenge” at the bottom of this post to see all of the questions in this series! 

And remember to take whatever precautions you need to stay healthy over the next few weeks! 


Question #24

4.7☆◎3

If and each represent single digits in the decimal above, what digit does represent?

(1) When the decimal is rounded to the nearest tenth, 4.8 is the result.

(2) When the decimal is rounded to the nearest hundredth, 4.77 is the result.


Explanation

By statement (1), we know that the hundredths digit (represented by ) must be equal to 5, 6, 7, 8, or 9. We cannot narrow it down further than that, so statement (1) is not sufficient.

Statement (2) shows us the hundredths digit, but it has been rounded. Since we do not know the thousandths digit, we cannot be sure whether the hundredths digit is indeed 7, or if it was rounded up from 6. Because we cannot be sure of ‘s value, this is not sufficient.

Even combined, we do not have enough information to answer the question. The correct answer is (E).

Social Distancing with Data Sufficiency Challenge! – Day 23

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This series is designed to help you step up your Data Sufficiency practice while we all spend a little extra time at home during the coronavirus situation; I’m going to publish one practice Data Sufficiency question each day! You can choose to do all of them, or none of them. 

Click on the tag “Social Distancing DS Challenge” at the bottom of this post to see all of the questions in this series! 

And remember to take whatever precautions you need to stay healthy over the next few weeks! 


Question #23

What distance did Marty drive?

(1) Wendy drove 15 miles in 20 minutes.

(2) Marty drove at the same average speed as Wendy.


Explanation

We know that statement (1) is not sufficient because it doesn’t tell us anything about Marty.

Statement (2) by itself also fails to provide enough information, since statement (2) doesn’t tell us anything about Wendy’s distance.

But it looks like it might, combined with statement (1), give us enough information to answer the question. We know how far Marty would get in any 20 minute interval, but the problem is that we don’t know how long Marty drove. The correct answer is (E).

Social Distancing with Data Sufficiency Challenge! – Day 22

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This series is designed to help you step up your Data Sufficiency practice while we all spend a little extra time at home during the coronavirus situation; I’m going to publish one practice Data Sufficiency question each day! You can choose to do all of them, or none of them. 

Click on the tag “Social Distancing DS Challenge” at the bottom of this post to see all of the questions in this series! 

And remember to take whatever precautions you need to stay healthy over the next few weeks! 


Question #22

In triangle ABC, if AB = x, BC = x – 1, AC = y, which of the three angles of triangle ABC has the smallest degree measure?

(1) y = x – 2

(2) y = 5


 Explanation

In a triangle, the smallest side will be opposite the angle with the smallest degree measure. If we can identify which side is shortest, we can identify the smallest angle.

Statement (1) tells us that AC = y = x – 2. Since the other sides have measures of x and x-1, we know that AC will be the smallest side. You can test this out by making up a value for x. Since we know AC is the smallest side, we can find the angle with the smallest degree measure. Sufficient.

Statement (2) tells us the length of y. This does not help us determine which side is smallest, so it is not sufficient. The correct answer is (A).

Social Distancing with Data Sufficiency Challenge! – Day 21

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This series is designed to help you step up your Data Sufficiency practice while we all spend a little extra time at home during the coronavirus situation; I’m going to publish one practice Data Sufficiency question each day! You can choose to do all of them, or none of them. 

Click on the tag “Social Distancing DS Challenge” at the bottom of this post to see all of the questions in this series! 

And remember to take whatever precautions you need to stay healthy over the next few weeks! 


Question #21

What is the value of x?

(1) 4x + 4 = 0

(2) (x + 2)2 = x2


Explanation

Statement (1) is simple enough to solve with some algebra:
4x + 4 = 0
4x = -4
x = -1: Sufficient.

Statement (2) is trickier. We should multiply it out:
(x + 2)2 = x2
x2 + 2x + 2x + 4 = x2
x
2 + 4x + 4 = x2
4x + 4 = 0
No need to go on, we should recognize this from above. Sufficient!

The correct answer is (D).

Social Distancing with Data Sufficiency Challenge! – Day 20

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This series is designed to help you step up your Data Sufficiency practice while we all spend a little extra time at home during the coronavirus situation; I’m going to publish one practice Data Sufficiency question each day! You can choose to do all of them, or none of them. 

Click on the tag “Social Distancing DS Challenge” at the bottom of this post to see all of the questions in this series! 

And remember to take whatever precautions you need to stay healthy over the next few weeks! 


Question #20

If x – y = z, what is the value of y?

(1) x = 7

(2) z = x + 4


 Explanation

Statement (1) is insufficient because, if we plug in x = 7 we are left with: 7 – y = z … and we still are unable to solve for y‘s value.

Let’s try plugging in statement (2): x – y = z … x – y = (x + 4) … -y = 4 … y = -4:  Statement (2) alone is sufficient! The correct answer is (B).

Social Distancing with Data Sufficiency Challenge! – Day 19

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This series is designed to help you step up your Data Sufficiency practice while we all spend a little extra time at home during the coronavirus situation; I’m going to publish one practice Data Sufficiency question each day! You can choose to do all of them, or none of them. 

Click on the tag “Social Distancing DS Challenge” at the bottom of this post to see all of the questions in this series! 

And remember to take whatever precautions you need to stay healthy over the next few weeks! 


Question #19

If j and k are integers, is j + k an odd integer?

(1) j > 12

(2) j = k – 1


Explanation

This is a pretty straightforward number properties question drilling us on our odds and evens rules. Remember that when adding, the only way to end up with an odd sum is to have an odd number of odd values in the list of numbers to be added together. If this is unclear, pick a few different values– odds and evens– and test it out! 🙂

Statement (1) doesn’t give us much to work with. It’s certainly not sufficient.
Statement (2) is enough to know for certain that the sum of j and k would be odd. Make up a few numbers to check it out. You’ll find that the sum of consecutive integers is always odd. Sufficient. The correct answer is (B).

Social Distancing with Data Sufficiency Challenge! – Day 18

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This series is designed to help you step up your Data Sufficiency practice while we all spend a little extra time at home during the coronavirus situation; I’m going to publish one practice Data Sufficiency question each day! You can choose to do all of them, or none of them. 

Click on the tag “Social Distancing DS Challenge” at the bottom of this post to see all of the questions in this series! 

And remember to take whatever precautions you need to stay healthy over the next few weeks! 


Question #18

Is the number x between .5 and .9 ?

(1) 800x < 620

(2) 1200x > 620


Explanation

Let’s solve for x in each inequality statement.

Statement (1): 800x < 620 … x < .775  … That alone is not sufficient to say that x is in the presented range, since we can’t be sure x isn’t less than .5
Note: The easier way to think about this without using a calculator is to think about fractions and decimals you know well. Reduce this fraction first by ten (80x < 62), then cut it in half (40x < 31), and finally isolate the x: x < 31/40.  What’s really close to 31/40 that you know well? Let’s use 3/4, which equals .75.

Statement (2) = 1200x > 620 … x > .517 … Statement (2) is also not sufficient alone.
Suggestion: Handle this calculation the same way as you did statement (1). 1200x > 620 … reduce by ten: 120x > 62 … cut in half: 60x > 31 … isolate x: x > 31/60 which is just barely more than .5 … no lengthy calculation needed!

But combined with statement (1), we know that x is in the range: .517 < x < .775, which is within the range presented of .5 < x < .9

The correct answer is (C).