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Sajjad1994
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The table represents the work performance of four machines during one 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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51% (03:53) correct 49% (03:47) wrong based on 345 sessions

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Question # 11 Date: 07-Oct-2021
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Start TimeEnd Time# Defective# of Shut Downs# Usable UnitsMachine
9:00 AM10:00 AM284434Machine B
10:00 AM11:00 AM281579Machine B
11:00 AM12:00 PM150582Machine D
12:00 PM1:00 PM220545Machine B
1:00 PM2:00 PM292394Machine A
2:00 PM3:00 PM172410Machine D
3:00 PM4:00 PM243565Machine A
4:00 PM5:00 PM170557Machine C
5:00 PM6:00 PM191375Machine D
6:00 PM7:00 PM180518Machine A
7:00 PM8:00 PM203357Machine C
8:00 PM9:00 PM284317Machine C
(Sort ↕ the table by clicking on the headers)

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.

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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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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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The table represents the work performance of four machines during one 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
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The table represents the work performance of four machines during one 07 Oct 2021, 10:28
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1. True
2. True
3. True

Explanation in pictures.

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The table represents the work performance of four machines during one 07 Oct 2021, 18:01
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A- False
B- True
C- False
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The table represents the work performance of four machines during one 07 Oct 2021, 20:01
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1. 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.
1. no. of usable units 393 for the slots 2:00 pm and 3:00 pm, now an increase of 111% will be 393*1.11 which is 436. On comparing this we will observe that when we will sort the list according to number of usable units from highest to lowest earlier it is below 434 which was 9:10 am slot, but a 10% increase will shift it above 434 , Hence true.

2. 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.
a= Most unit produced during 11-12 567
b= east unit produced 8-9 pm is 289
Ratio a:b is 1.96

Total hours 12 out of which 4 includes shutdown and 8 non-shutdown
ratio is 8/4 which is 0.5
Hence false as 1.96 > 0.5

3. The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns.
Machine with most shutdown has produced
a. 406 now 111% increase will be 451
b. 289 now 111% increase will be 321

Now machines with fewest shutdowns produced.
a. 567 slot 11-12
b. 523 slot 12-1.00
c. 540 slot 4-5
d. 500 slot 6-7
Hence false as a, b are not close to c and d,
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The table represents the work performance of four machines during one Updated on: 11 Oct 2021, 07:18
Originally posted by ktzsikka on 07 Oct 2021, 22:38.
Last edited by ktzsikka on 11 Oct 2021, 07:18, edited 2 times in total.
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Q1) 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.
A1) No. of units produced = 410. 10% increase means \(1.1*410 = 451\).
As this is more than the next in rank i.e.434 (9-10 AM). Rank will change. TRUE

Q2) The ratio(say R1) 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 (say R2) of hours with shutdowns to hours without shutdowns.
A2)\( R1 = 607/345 = 1.76\)
While \(R2 = 8/2 = 2\)
As \(R1<R2\), Statement is TRUE

Q3) The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns.
A3) Machine with fewest shutdowns (Machine D) produced = \(1367\)
And Machine with fewest shutdowns (Machine C) produced = \(731\)
As 1367 is approx. 87% more than 731, the Statement is FALSE
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The table represents the work performance of four machines during one 08 Oct 2021, 00:25
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1. 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.

- IMO True

Explanation -
Looking at the table the number of usable units between 2 to 3 are 410.
A 10% increase will be (410*1.1.0) =451, which is more than the usable units between 9 to 10.

2. 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.

IMO True

Explanation -
1.The ratio of units made during the hour producing the most units -
Find the total units produced by adding both defective and usable units. We don't really need to calculate the sum for each hour. Just by looking at the table I can see that either the line with values of 579 or 582 will be the highest.
So the highest value is that of the 10 Am to 11 AM = 607

2.Units made during the hour producing the fewest units -
Looking at the table, I can see through the lowest value, which is in between 8 to 9 = 345

Ratio1 = 607/345 < 2

3.The ratio of hours with shutdowns -
8 hours are with shutdowns.

4.hours without shutdowns -
4 hours (No of 0's) are without shutdowns

Ratio2 = 8/4 =2

Ratio 1 < Ratio2 Hence True


3. The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns.

IMO FALSE

Explanation-
Total no. of shutdowns for each machine -
B has (4+1+0) =5
D has (0+2+1) =3 [Fewest]
A has (2+3+0) =5
C has (0+3+4) =7 [Most]

Total usable units by D = (582+410+375) = 1367
Total usable units by C = (557+357+317) =1231
Definitely the % increase is not 111% more.
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The table represents the work performance of four machines during one 08 Oct 2021, 00:41
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ANSWERS : TRUE TRUE FALSE

I've attached a table I used for the Statements 1 and 2, Statement 3 can be done from the original table.

1. 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

10% increase in usable units produced in hour 2:00pm - 3:00pm would change 410 units to 451 units. The current rank of hour 2:00 pm - 3:00 pm is 8, and this increase would change it to rank 7 which is currently held by hour 9:00 am - 10:00 am (434 usable units)

2. 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

REMEMBER that Total Units = Defective Units + Usable Units

The hour that produced the most units (USABLE +DEFECTIVE) is HOUR 10:00 am - 11:00 am ==> 607 units
The hour the produced the least units is HOUR 8:00 pm - 9:00 pm ==> 345 units
Total hours with shutdowns ==> 8
Total hours without shutdowns ==> 4

607 / 345 < 8/2
607 / 345 < 2 YES (345 is more than half of 607 so the ratio is certainly lesser than 2)

3. The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns. FALSE

Machine with the fewest shutdowns —> MACHINE D (3 SHUTDOWNS) USABLE UNITS PRODUCED ( 1367)

Machine with the most shutdowns —> MACHINE C (7 SHUTDOWNS) USABLE UNITS PRODUCED (1237)

1367 is 211% of 1237? UNTRUE. ( 211% should be more than DOUBLE the original amount, this can be solved without a calculator )
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The table represents the work performance of four machines during one 08 Oct 2021, 08:30
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OA of this question is:

True, True and False

Only 800gal and akash2703 got it right. I will post the OE.
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The table represents the work performance of four machines during one 14 Mar 2024, 05:57
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Sajjad1994
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Question # 11 Date: 07-Oct-2021
This question is a part of GMAT Club IR Mini Marathon. Click here for Details

Start TimeEnd Time# Defective# of Shut Downs# Usable UnitsMachine
9:00 AM10:00 AM284434Machine B
10:00 AM11:00 AM281579Machine B
11:00 AM12:00 PM150582Machine D
12:00 PM1:00 PM220545Machine B
1:00 PM2:00 PM292394Machine A
2:00 PM3:00 PM172410Machine D
3:00 PM4:00 PM243565Machine A
4:00 PM5:00 PM170557Machine C
5:00 PM6:00 PM191375Machine D
6:00 PM7:00 PM180518Machine A
7:00 PM8:00 PM203357Machine C
8:00 PM9:00 PM284317Machine C
(Sort ↕ the table by clicking on the headers)

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.

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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.
Currently usable units by D=410, after 10% increase, 410+41=451. It will take D to 7th rank from 8th rank. True.

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.
582/317=1.8 and 8/4=2. 1.8<2. True.

The machine that had the fewest shutdowns produced approximately 111% more usable units than did the machine with the most shutdowns.
Shutdown ranks are
D (3 times), A (5 times), B (5 times), C (7 times).
D total usable units 410+375+582=1367 and C total usable units 317+357+557=1231.
Therefore, 1367-1231 / 1231 ~ 11%, not 111%. False.
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Sajjad1994
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

Start TimeEnd Time# Defective# of Shut Downs# Usable UnitsMachine
9:00 AM10:00 AM284434Machine B
10:00 AM11:00 AM281579Machine B
11:00 AM12:00 PM150582Machine D
12:00 PM1:00 PM220545Machine B
1:00 PM2:00 PM292394Machine A
2:00 PM3:00 PM172410Machine D
3:00 PM4:00 PM243565Machine A
4:00 PM5:00 PM170557Machine C
5:00 PM6:00 PM191375Machine D
6:00 PM7:00 PM180518Machine A
7:00 PM8:00 PM203357Machine C
8:00 PM9:00 PM284317Machine C
(Sort ↕ the table by clicking on the headers)

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.

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­Hi Sajjad, is it really GMAT like question because I dont think it is solvable under 2 mins
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