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The table above shows the numbers of hours of television programs that Jane recorded last week and the numbers of hours she spent viewing these recorded programs. No recorded program was viewed more than once. If h is the number of hours of recorded programs she had not yet viewed by the end of Friday, which of the following intervals represents all of the possible values of h?
A. \(0 \leq h \leq 1\)
B. \(1 \leq h \leq 2\)
C. \(2 \leq h \leq 3\)
D. \(0 \leq h \leq 2\)
E. \(1 \leq h \leq 3\)
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sujoykrdatta
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gmatt1476
Total hours recorded = 4 + 2 = 6 hours
Minimum hours watched = 1 (Wednesday) + 2 (Friday) = 3 hours
=> Hours left to be watched = 6 - 3 = 3 hours (maximum hours left)
Maximum hours watched = 2 (Wednesday) + 3 (Friday) = 5 hours
=> Hours left to be watched = 6 - 5 = 1 hours (minimum hours left)
Thus, the number of hours left to watch lies between 1 hour and 3 hours (inclusive)
Answer E
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General Discussion
the numbers of hours of television programs that Jane recorded last week = 4+0+2+0= 6hrs
Minimum numbers of hours she spent viewing these recorded programs = 0+1+0+2 = 3hrs
Maximum numbers of hours she spent viewing these recorded programs = 0+2+0+3= 5hrs
(6-5) ≤ h ≤ (6-3)
1 ≤ h ≤ 3
Minimum numbers of hours she spent viewing these recorded programs = 0+1+0+2 = 3hrs
Maximum numbers of hours she spent viewing these recorded programs = 0+2+0+3= 5hrs
(6-5) ≤ h ≤ (6-3)
1 ≤ h ≤ 3
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max viewing hours ; 2 to 3 hours ; 5 hours
and min viewing hours ; 1 to 2 hours ; 1 hour
total recording time ; 6 hours
hence net viewing hours
(6-5)<=h<=(6-3)
\(1 \leq h \leq 3\)
OPTION E
and min viewing hours ; 1 to 2 hours ; 1 hour
total recording time ; 6 hours
hence net viewing hours
(6-5)<=h<=(6-3)
\(1 \leq h \leq 3\)
OPTION E
gmatt1476
gmatt1476
Refer snapshot.
Attachment:
Viewing time left h.jpg [ 101.15 KiB | Viewed 20494 times ]
Tuesday: Min = 4; Max = 4
Wednesday: Min = 4 - 2 = 2; Max = 4 - 1 =3
Thursday: Min = 2 + 2 = 4; Max = 3 + 2 = 5
Friday: Min = 4 - 3 = 1; Max = 5 - 2 = 3
Hence \(1 \leq h \leq 3\)
Answer E.
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Solution:
The total recording time is 4 + 2 = 6 hours. The total viewing time is between 1 + 2 = 3 hours and 2 + 3 = 5 hours. Therefore, the number of hours of recorded programs Jane had not yet viewed by the end of Friday is between 6 - 5 = 1 hour and 6 - 3 = 3 hours. That is, 1 ≤ h ≤ 3.
Answer: E
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gmatt1476
Wednesday Balance: \(2≤h≤3\) [Deducting maximum viewing 2 and minimum 1 from recording 4]
Thursday Balance : \(4≤h≤5\) [Adding new recording of 2 hours with wednesday]
Friday Balance :\(1≤h≤3\) [Deducting maximum viewing time 3 and minimum viewing time from Thrusday]
The answer is \(E.\)
Posts: 4,849
SOLUTION:
A NEW OG 2021 QUESTION
According to the table,
Total RECORDING TIME = 4+2 (Tue+ Thur) = 6 hrs
The time of viewing is 1+2=3 hrs and a 3+2=5 hrs and
hence the time NOT invested in viewing is
6-3 = 3hrs (Maximum)&
6-5 = 1 hour (Minimum)
Hence 1<=h<=3 is the range (OPTION E)
Hope this helps
Devmitra Sen(Math)
A NEW OG 2021 QUESTION
According to the table,
Total RECORDING TIME = 4+2 (Tue+ Thur) = 6 hrs
The time of viewing is 1+2=3 hrs and a 3+2=5 hrs and
hence the time NOT invested in viewing is
6-3 = 3hrs (Maximum)&
6-5 = 1 hour (Minimum)
Hence 1<=h<=3 is the range (OPTION E)
Hope this helps
Devmitra Sen(Math)
gmatt1476
Answer: Option E
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