1.On a sightseeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sightseeing tour?
a. On the sightseeing tour, the ratio of the number of children to the number of men was 5 to 11.
b. The number of women on the sight-seeing tour was less than 30
2. If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?
a. n is not divisible by 2
b. n is not divisible by 3
3. Ix-yI > IxI - IyI
a. Y < x
b. Xy < 0
3. The points A, B, C, D are on a number line, not necessarily in that order. If the distance between A and B is 18 and the distance between C and D is 8, what is the distance between B and D?
a. The distance between C and A is the same as the distance between C and B
b. A is to the left of D on the number line
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Data Sufficiency
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Night reader
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call women=w, children=c, men=m; w/c=5/2 find m-?
st(1) c/m=5/11 Not Sufficient as we are only given ratios w:c:m=25:10:55 from (w:c=5:2 and c:m=5:11) multiplied by 5
st(2) the number of women was less than 30, the Only starting condition for all three ratios is w:c:m=25:10:55. Hence the number of men is 55
IOM C
st(1) c/m=5/11 Not Sufficient as we are only given ratios w:c:m=25:10:55 from (w:c=5:2 and c:m=5:11) multiplied by 5
st(2) the number of women was less than 30, the Only starting condition for all three ratios is w:c:m=25:10:55. Hence the number of men is 55
IOM C
Tagne wrote:1.On a sightseeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sightseeing tour?
a. On the sightseeing tour, the ratio of the number of children to the number of men was 5 to 11.
b. The number of women on the sight-seeing tour was less than 30
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Night reader
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given: n>0, (n-1)(n+1)/24= i {i is integer} +r {remainder}, find r-?
st(1) n is not divisible by 2 means n is odd. Hence (n-1) and (n+1) are even numbers. There are different even numbers {2,4,6,...1000} we don't know which numbers are (n+1) and (n-1) and cannot define remainder (r), Not Sufficient;
st(2) n is not divisible by 3 means n cannot be divided by 3 Only. Not Sufficient;
Combined st(1&2): n is odd and n is not divisible by 3, there can be various numbers --> 1,5,7 their (n+1)(n-1) will give various remainders // (1+1)(1-1)/24, (5+1)(5-1)/24, ... Not Sufficient
IOM E
st(1) n is not divisible by 2 means n is odd. Hence (n-1) and (n+1) are even numbers. There are different even numbers {2,4,6,...1000} we don't know which numbers are (n+1) and (n-1) and cannot define remainder (r), Not Sufficient;
st(2) n is not divisible by 3 means n cannot be divided by 3 Only. Not Sufficient;
Combined st(1&2): n is odd and n is not divisible by 3, there can be various numbers --> 1,5,7 their (n+1)(n-1) will give various remainders // (1+1)(1-1)/24, (5+1)(5-1)/24, ... Not Sufficient
IOM E
Tagne wrote:2. If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?
a. n is not divisible by 2
b. n is not divisible by 3
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Night reader
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|x-y| > |x| - |y| ?
st(1) y<x OR x-y>0, given this condition we can take x-y > |x| - |y|
for x>=0 and y>=0 we take x-y > x-y, we spot that the expression cannot be greater itself, Not Sufficient
st(2) xy<0 OR two possible takes x>0 y<0, x<o y>0
first take --> |x-y| > x+y, condition (x-y) >=0 And x-y > x+y, y<0
second take --> |x-y| > -x -y, condition (x-y) >=0 And x-y > -x-y, x>y Sufficient as x>y is true for x>0 ONLY from xy<0 and |x-y| is the sum (+ve x - [-ve y]) <--- effect of summing and individual |x| - |y| will be less than Left-Hand-Side (LHS)
IOM B
st(1) y<x OR x-y>0, given this condition we can take x-y > |x| - |y|
for x>=0 and y>=0 we take x-y > x-y, we spot that the expression cannot be greater itself, Not Sufficient
st(2) xy<0 OR two possible takes x>0 y<0, x<o y>0
first take --> |x-y| > x+y, condition (x-y) >=0 And x-y > x+y, y<0
second take --> |x-y| > -x -y, condition (x-y) >=0 And x-y > -x-y, x>y Sufficient as x>y is true for x>0 ONLY from xy<0 and |x-y| is the sum (+ve x - [-ve y]) <--- effect of summing and individual |x| - |y| will be less than Left-Hand-Side (LHS)
IOM B
Tagne wrote:3. Ix-yI > IxI - IyI
a. Y < x
b. Xy < 0
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Night reader
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because in case of 20 and 15 ... you cannot preserve the ratios with multiplier 5
w:c:m=25:10:55 from (w:c=5:2 and c:m=5:11) multiplied by 5
w:c:m=25:10:55 from (w:c=5:2 and c:m=5:11) multiplied by 5
Tagne wrote:In question 1 above, the condition given was "The number of women on the sight-seeing tour was less than 30". less than 30 could be 25, 20, 15, 10. I am curious why 25 was chosen?
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Night reader
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Let's elaborate on st(1) which implies that on the number line C is in the middle of AB --> AC=BC. Now depending where D is placed to the left of C or to the right of C, D will be distanced from from B differently. If D is close to B then the distance is 1 (18/2 -8), if far the distance is 17 (18/2 +8) Not Sufficient
st(2) says A is to the left of D. Alone is Not Sufficient
Combined st(1&2): we know that A is to the left of D, hence D is close to B, because B can be only to the right of D! Still Not Sufficient, as the distance can be 1 and 17
answer E
st(2) says A is to the left of D. Alone is Not Sufficient
Combined st(1&2): we know that A is to the left of D, hence D is close to B, because B can be only to the right of D! Still Not Sufficient, as the distance can be 1 and 17
answer E
Tagne wrote: 3. The points A, B, C, D are on a number line, not necessarily in that order. If the distance between A and B is 18 and the distance between C and D is 8, what is the distance between B and D?
a. The distance between C and A is the same as the distance between C and B
b. A is to the left of D on the number line
Last edited by Night reader on Tue Mar 15, 2011 5:52 pm, edited 1 time in total.
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srcc25anu
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Q4. stat 1: B----9------C-----8----D--1--A
B--1--D-----8-----C----9-----A
A--1--D-----8----C------9-----B
A-------9-------C---8-----D--1--B
distance CA = dist CB NOT SUFFICINET AS all 4 cases satisfy the condition for different sets of distance BD. case 1: dist BD = 17 and case 2: dist BD = 1
Stat 2: A is to left of D; again 2 cases satisfy the condition. condition 3 and 4
Together still not sufficient. we have 2 cases: case 3 and 4
hence E
B--1--D-----8-----C----9-----A
A--1--D-----8----C------9-----B
A-------9-------C---8-----D--1--B
distance CA = dist CB NOT SUFFICINET AS all 4 cases satisfy the condition for different sets of distance BD. case 1: dist BD = 17 and case 2: dist BD = 1
Stat 2: A is to left of D; again 2 cases satisfy the condition. condition 3 and 4
Together still not sufficient. we have 2 cases: case 3 and 4
hence E
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Night reader
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that's correct
thanks
thanks
srcc25anu wrote:Q4. stat 1: B----9------C-----8----D--1--A
B--1--D-----8-----C----9-----A
A--1--D-----8----C------9-----B
A-------9-------C---8-----D--1--B
distance CA = dist CB NOT SUFFICINET AS all 4 cases satisfy the condition for different sets of distance BD. case 1: dist BD = 17 and case 2: dist BD = 1
Stat 2: A is to left of D; again 2 cases satisfy the condition. condition 3 and 4
Together still not sufficient. we have 2 cases: case 3 and 4
hence E
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Night reader
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Tagne wrote:The solution to question 3 is a bit convoluted. Please can anyone throw more light?
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Night reader
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what's not clear in it? very careful reasoning is needed here
Tagne wrote:I am sorry, when I said question 3, I meant this question;
3. Ix-yI > IxI - IyI
a. Y < x
b. Xy < 0
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anshumishra
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You may use PIN, if the algebraic method looks tedious to you :Tagne wrote:I am sorry, when I said question 3, I meant this question;
3. Ix-yI > IxI - IyI
a. Y < x
b. Xy < 0
Is |x-y| > |x| - |y| ?
Statement 1:
y < x
Let say, y = 1, x= 2
|x-y| = |2-1| = 1
|x|-|y| = |2|-|1| = 1 ; So |x-y| = |x| - |y| ---- (i)
Choose; y=-1, x = 2
|x-y| = |2-(-1)| = 3
|x|-|y| = |2| - |-1| = 2-1 = 1
So, |x-y| > |x|-|y| -----(ii)
Hence Insufficient
Statement 2:
xy < 0
Choose, x = 2, y=-1 (as shown above in [ii])
|x-y| > |x|-|y|
Choose, x = -1, y = 2
|x-y| = |-1-2| = 3
|x|-|y| = |-1| - |2| = 1-2 = -1
Hence, again |x-y| > |x|-|y| --- Sufficient B
Thanks
Anshu
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
