[x] denotes the greatest integer less than or equal to X. If X is a po

[x] denotes the greatest integer less than or equal to X. If X is a positive integer and \([\frac{x}{5}] − [\frac{x}{7}] = 1\). If the minimum value of x is a and the maximum value is b, then a + b = ?

\([\frac{14}{5}] - [\frac{7}{7}]= 2-1 =1\) --> \(a+b = 14+7 =21\)

\([\frac{19}{5}] - [\frac{14}{7}]= 3-2 =1\) --> \(a+b = 19+14 =33\)

\([\frac{24}{5}] - [\frac{21}{7}]= 4-3 =1\) --> \(a+b = 24+21 =45\)
------------------------------------------------------------------------------------------------------------
\([\frac{10}{5}] - [\frac{13}{7}]= 2-1 =1\) --> \(a+b = 10+13 =23\)

\([\frac{15}{5}] - [\frac{20}{7}]= 3-2 =1\) --> \(a+b = 15+20 =35\)

\([\frac{20}{5}] - [\frac{27}{7}]= 4-3 =1\) --> \(a+b = 20+27 =47\)

------------------------------------------------------------------------------------------------------------

I got the value of B and D.
I would go to B.
Answer(B)

x is a positive integer
if x=3, then [x]=3
if x=3.5, then [x]=3

min x for [x/5]-[x/7]=1
x=5: [1]-[5/7]=[1]-[0]=1
anything below x=5, both [x/5] [x/7] would be [0]

minimum a=5 then b=?
A. 40-5=b=35, 35/5=[7], 35/7=[5], 7-5≠1
B. 35-5=b=30, b/5=[6], b/7=[4], 6-4≠1
C. 34-5=b=29, b/5=[5], b/7=[4], 5-4=1
D. 33-5=b=28, b/5=[5], b/7=[4], 5-4=1
E. 32-5=b=27, b/5=[5], b/7=[3], 5-3≠1

C and D are valid, but C has a greater b

Ans (C)

[x] denotes the greatest integer less than or equal to X. If X is a positive integer and [\(\frac{x}{5}\)] − [\(\frac{x}{7}\)] = 1. If the minimum value of x is a and the maximum value is b, then a + b = ?

A. 40
B. 35
C. 34
D. 33
E. 32

[\(\frac{x}{5}\)] − [\(\frac{x}{7}\)] should be such that difference is always '1'.

So minimum possible value 'a' of x is 5 since
[\(\frac{5}{5}\)] = [1.0] = 1
[\(\frac{5}{7}\)] = [0.71] = 0

Thus for minimum value of x [\(\frac{x}{5}\)] − [\(\frac{x}{7}\)] = 1 - 0 = 1
a = 5

Similarly, maximum possible value 'b' of x is 29 since values beyond 29 doesn't satisfy the condition [\(\frac{x}{5}\)] − [\(\frac{x}{7}\)] = 1
[\(\frac{29}{5}\)] = [5.8] = 5
[\(\frac{29}{7}\)] = [4.1] = 4

Thus for maximum value of x [\(\frac{x}{5}\)] - [\(\frac{x}{7}\)] = 5 - 4 = 1
b = 29

a + b = 5 + 29 = 34


The catch is that
1 = [1.0] - [0.1] = [1.1] - [0.1]= [1.2] - [0.1] = ....... = [1.9] - [0.1]
OR
1 = [1.0] - [0.9]= [1.1] - [0.9]= [1.2] - [0.9] = ....... = [1.9] - [0.9]
So range of multipliers of 5 & 7 = 5 * (1.9 - 0.1) = 9

To better understand:
Since denominator is 5 and 7, range of multiple of 5 to multiple of 7 so that condition [\(\frac{x}{5}\)] - [\(\frac{x}{7}\)] = 1 is satisfied should always be less than 10.
Consider table below:
Multiple of 5 / Multiple of 7
5 7
10 14
15 21
20 28
25 35
30 42
35 49
....


See the red text the difference in multipliers is ≥10. The suitable next difference in multiplier is 8 so x should be 28 at least.
But since \([\frac{29}{5}] = [\frac{28}{5}]\) = 5
Hence x should be less than 30. So x = 29.

Hope you guys like this. Took me a while to write down in detail.

Answer C.

[x] denotes the greatest integer less than or equal to X. If X is a positive integer and [x/5] − [x/7] = 1. If the minimum value of x is a and the maximum value is b, then a + b = ?
minimum value of x = 10 = a
Since [10/5] − [10/7] = 2-1 = 1

A. 40
b = 30
[30/5] − [30/7] = 6 - 4 = 2 <>1
B. 35
b = 25
[25/5] − [25/7] = 5 - 3 = 2 <>1
C. 34
b = 24
[24/5] − [24/7] = 4 - 3 = 1
D. 33
E. 32

IMO C

x must be same in entire expression \([\frac{x}{5}] − [\frac{x}{7}] = 1\)

which it seems that your solution has assumed two different values (such as 14 and 7 both as mentioned below) of x in the same calculation
\([\frac{14}{5}] - [\frac{7}{7}]= 2-1 =1\) --> \(a+b = 14+7 =21\)

Yes, you’re right. I didn’t get the question well.
Kudos for you.

Posted from my mobile device

The minimum value of x is 5 since [5/5] = 1 and [5/7] = 0 (but [4/5] = 0 and [4/7] = 0). So a = 5.

The maximum value of x is 29 since [29/5] = 5 and [29/7] = 4 (but [30/5] = 6 and [30/7] = 4). So b = 29.

Therefore, a + b = 5 + 29 = 34.

Answer: C

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.