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16 Jun 2019, 16:02
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[V] denotes the greatest integer less than or equal to 'V'. 'V' is a positive integer and \([\frac{V}{5}]-[\frac{V}{7}]=1\).
If the minimum value of 'V' is 'a' and the maximum value of 'V' is 'b'. What is the value of (a+b)?
A. 33
B. 34
C. 35
D. 40
E. 42
If the minimum value of 'V' is 'a' and the maximum value of 'V' is 'b'. What is the value of (a+b)?
A. 33
B. 34
C. 35
D. 40
E. 42
firas92
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18 Jun 2019, 23:37
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V is a positive integer, therefore \([\frac{V}{5}]−[\frac{V}{7}]=1\) at minimum value of V should be \(1-0\)
Therefore V should be at least 5, \(a=5\)
Now looking at the answer choices, the maximum value for V should be among 28, 29, 30, 35 and 37.
When V=28, \([\frac{V}{5}]−[\frac{V}{7}]=5-4=1\)
When V=29, \([\frac{V}{5}]−[\frac{V}{7}]=5-4=1\)
When V=30, \([\frac{V}{5}]−[\frac{V}{7}]=6-4=2\)
We can conclude here that V can be maximum 29 and so \(b=29\) but just to check,
When V=35, \([\frac{V}{5}]−[\frac{V}{7}]=7-5=2\)
When V=37, \([\frac{V}{5}]−[\frac{V}{7}]=7-5=2\)
Therefore, \(a+b=5+29=34\)
Answer is (B)
Hit Kudos if this helped!
Therefore V should be at least 5, \(a=5\)
Now looking at the answer choices, the maximum value for V should be among 28, 29, 30, 35 and 37.
When V=28, \([\frac{V}{5}]−[\frac{V}{7}]=5-4=1\)
When V=29, \([\frac{V}{5}]−[\frac{V}{7}]=5-4=1\)
When V=30, \([\frac{V}{5}]−[\frac{V}{7}]=6-4=2\)
We can conclude here that V can be maximum 29 and so \(b=29\) but just to check,
When V=35, \([\frac{V}{5}]−[\frac{V}{7}]=7-5=2\)
When V=37, \([\frac{V}{5}]−[\frac{V}{7}]=7-5=2\)
Therefore, \(a+b=5+29=34\)
Answer is (B)
Hit Kudos if this helped!
General Discussion
16 Jun 2019, 23:08
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nick1816
[V] denotes the greatest integer less than or equal to 'V'. 'V' is a positive integer and \([\frac{V}{5}]-[\frac{V}{7}]=1\).
If the minimum value of 'V' is 'a' and the maximum value of 'V' is 'b'. What is the value of (a+b)?
A. 33
B. 34
C. 35
D. 40
E. 42
If the minimum value of 'V' is 'a' and the maximum value of 'V' is 'b'. What is the value of (a+b)?
A. 33
B. 34
C. 35
D. 40
E. 42
\([\frac{V}{5}]-[\frac{V}{7}]=1\)
Minimum
When \(\frac{V}{5}\) just crosses 0.5, while \(\frac{V}{7}\) remains below 0.5..
This means V can be 3, SO A=3
Maximum
\([\frac{V}{5}]-[\frac{V}{7}]<2\), as we can have the difference just below 2, but the answer still as 1, for example 4.5 and 6.4999.
Here 4.5 will become 5, while 6.4999 will come down to 6 and answer will be 1.
\([\frac{V}{5}]-[\frac{V}{7}]<2............[V]\frac{1}{5}]-\frac{1}{7}<2..........[V]<35\)
Let us check for the values below it..
(1) 34.....\([\frac{34}{5}]-[\frac{34}{7}]=1........[6.8]-[4.9]=7-5..NO\)
(2) 33.....\([\frac{33}{5}]-[\frac{33}{7}]=1........[6.6]-[4.7]=7-5..NO\)
(3) 32.....\([\frac{32}{5}]-[\frac{32}{7}]=1........[6.4]-[4.6]=6-5=1..YES\)
SO B=32
A+B=3+32=35
c
