How many odd integers are greater than the integer x and less than the integer y ?
(1) There are 12 even integers greater than x and less than y.
(2) There are 24 integers greater than x and less than y.
answer：B
It didn't tell us these integers are consecutive nummbers or not and the first and last numbers are odd or even. How do I work it out?
GMAT Forum
4 postsPage 1 of 1
- BG
- Saurav
(1) There are 12 even integers greater than x and less than y.
The series has to be consecutive, since then only can there be 12 such even integers.
Let is consider the following possibilities where X can be ODD and X can be EVEN.
X (odd)
i1 (even 1), i2, i3(even 2), i4, i5(even 3), i6, i7(even 4), i8, i9(even 5), i10, i11(even 6), i12
i13(even 7), i14, i15(even 8), i16, i17(even 9), i18, i19(even10), i20, i21(even11), i22, i23(even12)
Y
Total Odd integers - 11
X (even)
i01, i02(even 1), i03, i04(even 2), i05, i06(even 3), i07, i08(even 4), i09, i10(even 5), i11, i12(even 6)
i13, i14(even 7), i15, i16(even 8), i17, i18(even 9), i19, i20(even10), i21, i22(even11),i23, i24(even12)
Y
Total Odd integers - 12
Hence not Sufficient
(2) There are 24 integers greater than x and less than y.
Again, this is only possible for a consecutive set of integers. Out of 24 integers there will always be 12 odd and 12 even.
Lets verify this-
X (odd)
i01(even 1), i02, i03(even 2), i04, i05(even 3), i06, i07(even 4), i08, i09(even 5), i10, i11(even 6), i12
i13(even 7), i14, i15(even 8), i16, i17(even 9), i18, i19(even10), i20, i21(even11),i22, i23(even12), i24
Y
Total Odd integers - 12
X (even)
i01, i02(even 1), i03, i04(even 2), i05, i06(even 3), i07, i08(even 4), i09, i10(even 5), i11, i12(even 6)
i13, i14(even 7), i15, i16(even 8), i17, i18(even 9), i19, i20(even10), i21, i22(even11),i23, i24(even12)
Y
Total Odd integers - 12
Sufficient
The series has to be consecutive, since then only can there be 12 such even integers.
Let is consider the following possibilities where X can be ODD and X can be EVEN.
X (odd)
i1 (even 1), i2, i3(even 2), i4, i5(even 3), i6, i7(even 4), i8, i9(even 5), i10, i11(even 6), i12
i13(even 7), i14, i15(even 8), i16, i17(even 9), i18, i19(even10), i20, i21(even11), i22, i23(even12)
Y
Total Odd integers - 11
X (even)
i01, i02(even 1), i03, i04(even 2), i05, i06(even 3), i07, i08(even 4), i09, i10(even 5), i11, i12(even 6)
i13, i14(even 7), i15, i16(even 8), i17, i18(even 9), i19, i20(even10), i21, i22(even11),i23, i24(even12)
Y
Total Odd integers - 12
Hence not Sufficient
(2) There are 24 integers greater than x and less than y.
Again, this is only possible for a consecutive set of integers. Out of 24 integers there will always be 12 odd and 12 even.
Lets verify this-
X (odd)
i01(even 1), i02, i03(even 2), i04, i05(even 3), i06, i07(even 4), i08, i09(even 5), i10, i11(even 6), i12
i13(even 7), i14, i15(even 8), i16, i17(even 9), i18, i19(even10), i20, i21(even11),i22, i23(even12), i24
Y
Total Odd integers - 12
X (even)
i01, i02(even 1), i03, i04(even 2), i05, i06(even 3), i07, i08(even 4), i09, i10(even 5), i11, i12(even 6)
i13, i14(even 7), i15, i16(even 8), i17, i18(even 9), i19, i20(even10), i21, i22(even11),i23, i24(even12)
Y
Total Odd integers - 12
Sufficient
- BG
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- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
saurav, well played.
for statement (1), we can express the possibilities in a more compact form; i wouldn't really want to see anyone actually expand out the sequences like that on the real exam. yikes.
break it down by whether x and y themselves are even or odd, since that's the sole determinant of what will happen between them.
here are all 4 possibilities:
(1) both x and y are odd.
then you have
x, (even #), 11 pairs of (odd # even #), y.
therefore, 11 odd numbers.
(2) x is even, y is odd.
then you have
x, 12 pairs of (odd # even #), y.
therefore, 12 odd numbers.
(3) x is odd, y is even.
then you have
x, 12 pairs of (even # odd #), y.
therefore, 12 odd numbers.
(4) x and y are both even.
then you have
x, (odd #), 12 pairs of (even # odd #), y.
therefore, 13 odd numbers.
--
for statement (2), you can just realize that the 24 numbers will pair off into 12 pairs, each of which includes one even and one odd. the pairs could be either (even # odd #) or (odd # even #), but the order is immaterial; either way, you'll get 12 of each.
for statement (1), we can express the possibilities in a more compact form; i wouldn't really want to see anyone actually expand out the sequences like that on the real exam. yikes.
break it down by whether x and y themselves are even or odd, since that's the sole determinant of what will happen between them.
here are all 4 possibilities:
(1) both x and y are odd.
then you have
x, (even #), 11 pairs of (odd # even #), y.
therefore, 11 odd numbers.
(2) x is even, y is odd.
then you have
x, 12 pairs of (odd # even #), y.
therefore, 12 odd numbers.
(3) x is odd, y is even.
then you have
x, 12 pairs of (even # odd #), y.
therefore, 12 odd numbers.
(4) x and y are both even.
then you have
x, (odd #), 12 pairs of (even # odd #), y.
therefore, 13 odd numbers.
--
for statement (2), you can just realize that the 24 numbers will pair off into 12 pairs, each of which includes one even and one odd. the pairs could be either (even # odd #) or (odd # even #), but the order is immaterial; either way, you'll get 12 of each.
4 posts Page 1 of 1