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Please provide solutions for these DS questions.

by snakedoc » Wed Dec 14, 2011 9:35 am
1) In the xy plane, at what point does y = (x + a)(x + b) cross the x axis?
a. a + b = -1
b. graph intersects y axis at (0, -6)

2) S is finite set of numbers. Does S contain more negative numbers than positive numbers?
a. Product of all numbers in S is -1200
b. There are 6 numbers in S


3) Each employee is either a director or a manager. What % of the employees are directors?
a. The average salary of managers is $5000 less than the average of all employees
b. The average salary of directors is $15000 greater than the average of all employees


4) The ratio of women to children on a tour is 5:2. How many men are there?
a. Ratio of children to women is 5:11
b. There are less than 30 women on the tour.


5) Sum of positive integers x and y is 72. What is xy?
a. x = y + 1
b. x and y have same tens digit
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by rijul007 » Wed Dec 14, 2011 7:29 pm
snakedoc wrote:1) In the xy plane, at what point does y = (x + a)(x + b) cross the x axis?
a. a + b = -1
b. graph intersects y axis at (0, -6)
y = (x + a)(x + b)
y = x^2 + x(a+b) + ab
x^2 + x(a+b) +ab = 0
for x- intercept
put y=0


a. a + b = -1
x^2 - x + ab = 0

value of x depends on ab

Insufficient



b. graph intersects y axis at (0, -6)

y = (x+a)(x+b)
at y intercept, x=0

y = ab = -6


Combining the two statements

x^2 +x(a+b) + ab = 0
x^2 - x - 6 = 0
(x-3)(x+2) = 0

y= (x+a)(x+b) crosses x-axis at (3,0) and (-2,0)

Option C
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by Anurag@Gurome » Wed Dec 14, 2011 7:32 pm
snakedoc wrote:
3) Each employee is either a director or a manager. What % of the employees are directors?
a. The average salary of managers is $5000 less than the average of all employees
b. The average salary of directors is $15000 greater than the average of all employees

Let the average salary of managers of the task force = S(m), the average salary of the directors on the task force = S(d), and the average salary of all the employee on the task force = S(e).
Let the no. of managers = m and no. of directors = d. We have to find d/(m + d).

(1) S(m) = S(e) - 5000, which ALONE is NOT SUFFICIENT.

(2) S(d) = S(e) + 15000, which ALONE is NOT SUFFICIENT.

Combining (1) and (2), we know that S(e) = {m * S(m) + d * S(d)}/(m + d)
So, S(e) = {m * [S(e) - 5000] + d * [S(e) + 15000]}/(m + d)
m * S(e) + d * S(e) = m * S(e) - 5000m + d * S(e) + 15000d
15000d = 5000m
3d = m
So, d/(m + d) = d/4d = 1/4, which is SUFFICIENT.

The correct answer is C.
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by rijul007 » Wed Dec 14, 2011 7:35 pm
snakedoc wrote:2) S is finite set of numbers. Does S contain more negative numbers than positive numbers?
a. Product of all numbers in S is -1200
b. There are 6 numbers in S

a. Product of all numbers in S is -1200

This statement tells us
there are odd no of negative numbers

No idea about whether -ve no are more than +ve nos

Isufficient

b. There are 6 numbers in S

This statement is clearly Insufficient


Combining the two statements
Product of all numbers in S is -1200
There are 6 numbers in S

We know that odd no of nos are -ve

Possible case : 1 no is negative and all the other are positive
Another possible case : 5 nos are negative and only one is positive

So not sufficient


Option E
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by Anurag@Gurome » Wed Dec 14, 2011 7:37 pm
snakedoc wrote:5) Sum of positive integers x and y is 72. What is xy?
a. x = y + 1
b. x and y have same tens digit
The correct question says that x + y = 77 (if the sum is 72 then either x and y are both odd, or x and y are both even. According to statement 1, if one of them is odd, the other one will be even, which is a contradiction). So, I have done the question, assuming x + y = 77.

(1) x = y + 1.
We have two equations and two variables, so we can solve them x and y, hence we can find xy; SUFFICIENT.

(2) x and y have same tens digit.
The tens digit should be 3 because if it is 2, then maximum possible sum can be 58 and if it is 4, then minimum possible sum will be 80.
The sum of x and y is 77, so the sum of unit places is 17 and it is possible only with the digits 8 and 9, as 9 + 8 = 17
So, the numbers are 38 and 39; SUFFICIENT.

The correct answer is D.
Last edited by Anurag@Gurome on Wed Dec 14, 2011 7:50 pm, edited 1 time in total.
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by rijul007 » Wed Dec 14, 2011 7:39 pm
snakedoc wrote:1)
5) Sum of positive integers x and y is 72. What is xy?
a. x = y + 1
b. x and y have same tens digit
a. x = y + 1
x+y = 72
x-y = 1

you can find the value of x and y

Sufficient

b. x and y have same tens digit
Possible pairs: 36+36 , 37 + 35 , 38 + 34 , 39 + 33

Insufficient


Option A
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by rijul007 » Wed Dec 14, 2011 7:43 pm
Anurag@Gurome wrote:
snakedoc wrote:5) Sum of positive integers x and y is 72. What is xy?
a. x = y + 1
b. x and y have same tens digit
x + y = 77

(1) x = y + 1.
We have two equations and two variables, so we can solve them x and y, hence we can find xy; SUFFICIENT.

(2) x and y have same tens digit.
The tens digit should be 3 because if it is 2, then maximum possible sum can be 58 and if it is 4, then minimum possible sum will be 80.
The sum of x and y is 77, so the sum of unit places is 17 and it is possible only with the digits 8 and 9, as 9 + 8 = 17
So, the numbers are 38 and 39; SUFFICIENT.

The correct answer is D.
That should have been 72..
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by Anurag@Gurome » Wed Dec 14, 2011 7:43 pm
snakedoc wrote:(4) The ratio of women to children on a tour is 5:2. How many men are there?
a. Ratio of children to men is 5:11
b. There are less than 30 women on the tour.
Let no. of women = W, no. of men = M, and no. of children = C
Then W : C = 5 : 2, we have to find M = ?

(1) C : M = 5 : 11
So, W : C : M = 25 : 10 : 22, but we do not know how many men are there; NOT sufficient.

(2) There are less than 30 women on the tour does not give us the no. of men; NOT sufficient.

Combining (1) and (2), we know that no. of women = 25, so no. of men = 22; SUFFICIENT.

The correct answer is C.
Last edited by Anurag@Gurome on Fri Dec 16, 2011 5:15 am, edited 1 time in total.
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by rijul007 » Wed Dec 14, 2011 7:46 pm
snakedoc wrote: 4) The ratio of women to children on a tour is 5:2. How many men are there?
a. Ratio of children to women is 5:11
b. There are less than 30 women on the tour.

The information given in the question and statement 1 contradict each other...
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by Anurag@Gurome » Wed Dec 14, 2011 7:51 pm
rijul007 wrote:
Anurag@Gurome wrote:
snakedoc wrote:5) Sum of positive integers x and y is 72. What is xy?
a. x = y + 1
b. x and y have same tens digit
x + y = 77

(1) x = y + 1.
We have two equations and two variables, so we can solve them x and y, hence we can find xy; SUFFICIENT.

(2) x and y have same tens digit.
The tens digit should be 3 because if it is 2, then maximum possible sum can be 58 and if it is 4, then minimum possible sum will be 80.
The sum of x and y is 77, so the sum of unit places is 17 and it is possible only with the digits 8 and 9, as 9 + 8 = 17
So, the numbers are 38 and 39; SUFFICIENT.

The correct answer is D.
That should have been 72..
I have edited my reply above and explained why x + y = 72 is not possible. The correct question says that x + y = 77.
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by pemdas » Wed Dec 14, 2011 9:12 pm
agree about (-3;2) and x-intercepts but disagree about answer C
Need to find only one possible point at which the graph intercepts x axis ...
find x-intercept when y=0 -> (x+a)(x+b)=0, x=-a and/or x=-b
st(1) a+b=-1 Not Sufficient
st(2) plug into the equation -6=(0+a)(0+b), -6=ab Not Sufficient
Combined st(1&2): a+b=-1 and -6=ab -> 6=-ab and a=-b-1, 6=b(b+1), b^2+b-6=0 Not Sufficient
we should have only one unique value for x-intercept
>>> check, D=1-4*(-6)=25, b(1,2)=(-1+-5)/2=(-3;2); a=(2;-3)
y=(x+a)(x+b) and y=0 -> (x+2)(x-3)=0 and (x-3)(x+2)=0
e
p.s. you can right away select option e if aware of the quadratics graphics and up/down-ward direction. The question requests only one point for y=0 which is only possible if the graph is neither up nor downward but skewed left or right. The quadratics will either have > or < than 0 value and must have -ve discriminant then.
rijul007 wrote:
snakedoc wrote:1) In the xy plane, at what point does y = (x + a)(x + b) cross the x axis?
a. a + b = -1
b. graph intersects y axis at (0, -6)
y = (x + a)(x + b)
y = x^2 + x(a+b) + ab
x^2 + x(a+b) +ab = 0
for x- intercept
put y=0


a. a + b = -1
x^2 - x + ab = 0

value of x depends on ab

Insufficient



b. graph intersects y axis at (0, -6)

y = (x+a)(x+b)
at y intercept, x=0

y = ab = -6


Combining the two statements

x^2 +x(a+b) + ab = 0
x^2 - x - 6 = 0
(x-3)(x+2) = 0

y= (x+a)(x+b) crosses x-axis at (3,0) and (-2,0)

Option C
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by rijul007 » Wed Dec 14, 2011 10:12 pm
pemdas wrote:agree about (-3;2) and x-intercepts but disagree about answer C
Need to find only one possible point at which the graph intercepts x axis ...
My take

You would be right if instead of asking for the intercept the ques would have just said
what is the value of x?
here, x is either 3 or -2
Insufficient


But in this question,
you are asked for the intercept

if you draw the graph of the given equation x^ -x -6 = y
you would ALWAYS have two x-intercept points (3,0) and (-2,0)

its not like it would be either this or that
Both the points MUST be the intercepts


So according to me the answer choice should be C


Experts please comment!
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by pemdas » Wed Dec 14, 2011 10:15 pm
@Anurag, you're assuming the sum of two positive integers in your solution?

question stem -> ax+by=72 implies x+x+...+x+y+...y=72 and a,b must be whole numbers, as we are suggested about the (number) of integers; it would be not correct mathematically to state 2.5 or 2.4 integers. Hence a,b are integers.
st(1) x=y+1 implies a(y+1)+by=72
when y is odd, x is even and vice verse, x and y can be any values depending on a,b Not Sufficient
st(2) x,y=T(U,u). The smallest ten's digit possible for the two-digit numbers is 10. Many variations are present with st(2), bifurcated -> x=10, y=13, a=3,b=4 OR x=17,y=19, a=2,b=2 Not Sufficient
Combined st(1&2): a(y+1)+by=72 considered with the condition x,y=T(U,u)
x=11, y=10
a=2, b=5 -> 2(10+1)+5*10=72

x=15, y=14
a=2, b=3 -> 2(14+1)+3*14=72

xy=15*14 and xy=11*10 Not Sufficient

e
Anurag@Gurome wrote:
snakedoc wrote:5) Sum of positive integers x and y is 72. What is xy?
a. x = y + 1
b. x and y have same tens digit
The correct question says that x + y = 77 (if the sum is 72 then either x and y are both odd, or x and y are both even. According to statement 1, if one of them is odd, the other one will be even, which is a contradiction). So, I have done the question, assuming x + y = 77.

(1) x = y + 1.
We have two equations and two variables, so we can solve them x and y, hence we can find xy; SUFFICIENT.

(2) x and y have same tens digit.
The tens digit should be 3 because if it is 2, then maximum possible sum can be 58 and if it is 4, then minimum possible sum will be 80.
The sum of x and y is 77, so the sum of unit places is 17 and it is possible only with the digits 8 and 9, as 9 + 8 = 17
So, the numbers are 38 and 39; SUFFICIENT.

The correct answer is D.
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by rijul007 » Wed Dec 14, 2011 10:21 pm
pemdas wrote:@Anurag, you're assuming the sum of two positive integers in your solution?

question stem -> ax+by=72 implies x+x+...+x+y+...y=72 and a,b must be whole numbers, as we are suggested about the (number) of integers; it would be not correct mathematically to state 2.5 or 2.4 integers. Hence a,b are integers.
st(1) x=y+1 implies a(y+1)+by=72
when y is odd, x is even and vice verse, x and y can be any values depending on a,b Not Sufficient
st(2) x,y=T(U,u). The smallest ten's digit possible for the two-digit numbers is 10. Many variations are present with st(2), bifurcated -> x=10, y=13, a=3,b=4 OR x=17,y=19, a=2,b=2 Not Sufficient
Combined st(1&2): a(y+1)+by=72 considered with the condition x,y=T(U,u)
x=11, y=10
a=2, b=5 -> 2(10+1)+5*10=72

x=15, y=14
a=2, b=3 -> 2(14+1)+3*14=72

xy=15*14 and xy=11*10 Not Sufficient

e
Anurag@Gurome wrote:
snakedoc wrote:5) Sum of positive integers x and y is 72. What is xy?
a. x = y + 1
b. x and y have same tens digit
The correct question says that x + y = 77 (if the sum is 72 then either x and y are both odd, or x and y are both even. According to statement 1, if one of them is odd, the other one will be even, which is a contradiction). So, I have done the question, assuming x + y = 77.

(1) x = y + 1.
We have two equations and two variables, so we can solve them x and y, hence we can find xy; SUFFICIENT.

(2) x and y have same tens digit.
The tens digit should be 3 because if it is 2, then maximum possible sum can be 58 and if it is 4, then minimum possible sum will be 80.
The sum of x and y is 77, so the sum of unit places is 17 and it is possible only with the digits 8 and 9, as 9 + 8 = 17
So, the numbers are 38 and 39; SUFFICIENT.

The correct answer is D.
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by pemdas » Wed Dec 14, 2011 10:22 pm
recalled the post, I think this question asks about the point from 0 to the left and right, on x-axis (like mod). So right with ans C, unless some very specific function is considered here, when y=!0 and there will be no x-intercept at all, but we tested this functions and saw it's y=0 at the given found points. Hence C
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