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07 Oct 2021, 08:00
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A 10% increase in the number of usable units produced from 2:00PM to 3:00PM would cause a change in this hour's rank according to number of usable units produced.: True
The ratio of units made during the hour producing the most units to units made during the hour producing the fewest units is less than the ratio of hours with shutdowns to hours without shutdowns.: True
The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns.: False
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Participate and Win! GMAT Club IR Mini Marathon
Question # 11 Date: 07-Oct-2021
This question is a part of GMAT Club IR Mini Marathon. Click here for Details
Question # 11 Date: 07-Oct-2021
This question is a part of GMAT Club IR Mini Marathon. Click here for Details
| Start Time | End Time | # Defective | # of Shut Downs | # Usable Units | Machine |
|---|---|---|---|---|---|
| 9:00 AM | 10:00 AM | 28 | 4 | 434 | Machine B |
| 10:00 AM | 11:00 AM | 28 | 1 | 579 | Machine B |
| 11:00 AM | 12:00 PM | 15 | 0 | 582 | Machine D |
| 12:00 PM | 1:00 PM | 22 | 0 | 545 | Machine B |
| 1:00 PM | 2:00 PM | 29 | 2 | 394 | Machine A |
| 2:00 PM | 3:00 PM | 17 | 2 | 410 | Machine D |
| 3:00 PM | 4:00 PM | 24 | 3 | 565 | Machine A |
| 4:00 PM | 5:00 PM | 17 | 0 | 557 | Machine C |
| 5:00 PM | 6:00 PM | 19 | 1 | 375 | Machine D |
| 6:00 PM | 7:00 PM | 18 | 0 | 518 | Machine A |
| 7:00 PM | 8:00 PM | 20 | 3 | 357 | Machine C |
| 8:00 PM | 9:00 PM | 28 | 4 | 317 | Machine C |
The table represents the work performance of four machines during one working day. Each machine produces units that are either usable or defective. Performance throughout the working day is represented per hour, and only one machine is active at a time.
For each statement, indicate whether the statement is true or false, based on the information in the table.
Attachment:
1.jpg [ 91.67 KiB | Viewed 8244 times ]
| True | False | |
| A 10% increase in the number of usable units produced from 2:00PM to 3:00PM would cause a change in this hour's rank according to number of usable units produced. | ||
| The ratio of units made during the hour producing the most units to units made during the hour producing the fewest units is less than the ratio of hours with shutdowns to hours without shutdowns. | ||
| The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns. |
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Official Answer
A 10% increase in the number of usable units produced from 2:00PM to 3:00PM would cause a change in this hour's rank according to number of usable units produced.: True
The ratio of units made during the hour producing the most units to units made during the hour producing the fewest units is less than the ratio of hours with shutdowns to hours without shutdowns.: True
The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns.: False
10 Oct 2021, 09:45
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Official Explanation
A 10% increase in the number of usable units produced from 2:00PM to 3:00PM would cause a change in this hour's rank according to number of usable units produced.
Refer to the table for the information corresponding to the hour of 2:00PM to 3:00PM and sort by the number of usable units to find the rank according to number of usable units produced per hour.
From least to greatest, the hour of 2:00 PM to 3:00 PM ranks fifth lowest in number of usable units produced, with 410 units.
A 10% increase would bring this number from 410 to 451 (410 x 1.1 = 451), which would change the rank from fifth lowest to sixth lowest, bringing it past the 9:00 AM to 10:00 AM number of 434. So this statement is true.
Answer: True
The ratio of units made during the hour producing the most units to units made during the hour producing the fewest units is less than the ratio of hours with shutdowns to hours without shutdowns.
The ratio of total units during the most productive hour to total units during least productive hour can also be calculated as: \(\frac{units during most productive hours}{units during least productive hours}\). Since usable units produced per hour are much greater than defective units produced per hour, the total units produced per hour is going to be determined mostly by the usable units produced per hour, but it's still important to note that "units produced" includes both usable and defective units. Therefore, sort the table by usable units produced per hour, then add the amounts for usable units produced per hour to defective units produced per hour.
The most productive hour was 10:00 AM to 11:00 AM, with 579 usable units and 28 defective units, for a total of 607 units produced. The least productive hour was 8:00 PM to 9:00 PM, with 317 usable units and 28 defective units for a total of 345 units. The ratio of usable units for the most productive hour to usable units for the least productive hour would be 607:345 or \(\frac{607}{345}\) = 1.76
Now sort the table by number of shutdowns. The number of hours with shutdowns is 8, while the number of hours with no shutdowns is 4. Thus the ratio of hours with shutdowns to hours with no shutdowns is \(\frac{8}{4} =2\) The ratio for total units is smaller than the ratio for shutdowns, so this statement is true.
Answer: True
The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns.
Sorting the table by the Machine column lets you compare the number of shutdowns per machine and the usable units produced. Machine C had 7 shutdowns, the most of all the machines, and Machine D had 3 shutdowns, the fewest. So now compare the total usable units produced: Machine D produced a total of 1367 usable units, while Machine C produced 1231 usable units. To figure the percent more usable units that Machine D produced, use the formula for percent change = \(\frac{difference}{original}\). In this case, the original amount is 1231, since you start with a smaller value in order to increase to a larger value. So change= \(\frac{difference}{original}\) \( = \frac{136}{1231}\) = 0.11 or 11%, and thus this statement is false. Machine D did not produce 111% more units than did Machine C. Machine D produced 11% more units than did Machine C.
Answer: False
A 10% increase in the number of usable units produced from 2:00PM to 3:00PM would cause a change in this hour's rank according to number of usable units produced.
Refer to the table for the information corresponding to the hour of 2:00PM to 3:00PM and sort by the number of usable units to find the rank according to number of usable units produced per hour.
From least to greatest, the hour of 2:00 PM to 3:00 PM ranks fifth lowest in number of usable units produced, with 410 units.
A 10% increase would bring this number from 410 to 451 (410 x 1.1 = 451), which would change the rank from fifth lowest to sixth lowest, bringing it past the 9:00 AM to 10:00 AM number of 434. So this statement is true.
Answer: True
The ratio of units made during the hour producing the most units to units made during the hour producing the fewest units is less than the ratio of hours with shutdowns to hours without shutdowns.
The ratio of total units during the most productive hour to total units during least productive hour can also be calculated as: \(\frac{units during most productive hours}{units during least productive hours}\). Since usable units produced per hour are much greater than defective units produced per hour, the total units produced per hour is going to be determined mostly by the usable units produced per hour, but it's still important to note that "units produced" includes both usable and defective units. Therefore, sort the table by usable units produced per hour, then add the amounts for usable units produced per hour to defective units produced per hour.
The most productive hour was 10:00 AM to 11:00 AM, with 579 usable units and 28 defective units, for a total of 607 units produced. The least productive hour was 8:00 PM to 9:00 PM, with 317 usable units and 28 defective units for a total of 345 units. The ratio of usable units for the most productive hour to usable units for the least productive hour would be 607:345 or \(\frac{607}{345}\) = 1.76
Now sort the table by number of shutdowns. The number of hours with shutdowns is 8, while the number of hours with no shutdowns is 4. Thus the ratio of hours with shutdowns to hours with no shutdowns is \(\frac{8}{4} =2\) The ratio for total units is smaller than the ratio for shutdowns, so this statement is true.
Answer: True
The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns.
Sorting the table by the Machine column lets you compare the number of shutdowns per machine and the usable units produced. Machine C had 7 shutdowns, the most of all the machines, and Machine D had 3 shutdowns, the fewest. So now compare the total usable units produced: Machine D produced a total of 1367 usable units, while Machine C produced 1231 usable units. To figure the percent more usable units that Machine D produced, use the formula for percent change = \(\frac{difference}{original}\). In this case, the original amount is 1231, since you start with a smaller value in order to increase to a larger value. So change= \(\frac{difference}{original}\) \( = \frac{136}{1231}\) = 0.11 or 11%, and thus this statement is false. Machine D did not produce 111% more units than did Machine C. Machine D produced 11% more units than did Machine C.
Answer: False
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07 Oct 2021, 10:28
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1. True
2. True
3. True
Explanation in pictures.
Posted from my mobile device
2. True
3. True
Explanation in pictures.
Posted from my mobile device
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