There are a bunch of tricks in this one!

Let's start by properly phrasing the question. If we say that Laura sold x encyclopedias in March, then her income = 1000 + (x - n)*c, assuming that x is greater than n. (If x is NOT greater than n, then her income is just $1000.) So we need to solve for x, n, and c.

S1 is really tricky. Remember that Laura gets c dollars for each set she sells beyond the first n. Now let's think about a few scenarios:

Suppose that in March Laura sold (n+1) encyclopedias; in other words, one more than she needs to get her bonus (x = n+1). That means that that ONE extra encyclopedia gives her a $600 bonus, so c = $600.

Now suppose that in March Laura sold (n+2) encyclopedias, or x = (n+2). Now TWO extra encyclopedias get her the $600 bonus, so 2c = $600 and c = $300.

Now suppose that in March Laura sold (n+3) (or more) encyclopedias, or x = (n+3). Then three encyclopedias are worth $600 in bonus, so 3c = $600 = c = $200.

So we have three possible values for c, let alone x and n; INSUFFICIENT.

S2 gives us the equation:
Laura's income for ten sets > $4000, which (using our earlier equation) is
$1000 + (10-n)*c > $4000, or
(10-n) * c > $3000, or
10c - nc > $3000

Since n must be positive, 10c > 10c - nc. Since 10c - nc > $3000, 10c > $3000, and c > $300. On its own this is also INSUFFICIENT.

Together we know that c must be $600. This tells us that Laura sold exactly (n+1) encyclopedias in March, so she gets $1000 for the first n and $600 for the additional one, making her income $1600. (Algebraically, her income is $1000 + ((n+1)-n)*$600.)

Yeesh! That's some question - where is it from?

Hello Matt,

Thanks for the reply,but the explanation is not very clear.

S1-It states that "If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was"
that means 3 sets relate to $600 => 1 set is equal to $200.Hence,C is $200.There is only one possible solution for C.

S2- 1000 + (10 - n) C > 4000

=> (10-n) C > 3000
=> (10-n) 200 > 3000
Solving this equation further we will not get fixed value of n and hence cannot comment on the salary.

So,I will go for E.

I am sure that I may be missing some key points as my answer is not matching with the OA,but please explain where am I going wrong.

Agreed this is a good one , what is the source ?

IMO C) is the right answer.

S1 says Fewer than 3 (3,2,1) ... I was also confused on that one...

=

GaneshMalkar wrote:Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000, and a variable component of $C for each set of encyclopaedias that she sells in that month over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000

OA - C

The first n sets do not earn a bonus.

Statement 1: If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
Case 1: 1 set earned a bonus of $600, and the remaining n sets did not earn a bonus.
Case 2: 2 sets each earned a bonus of $300, and the remaining n sets did not earn a bonus.
Case 3: 3 sets each earned a bonus of $200, and the remaining n sets did not earn a bonus.
Case 4: More than 3 sets each earned a bonus of $200, and the remaining n sets did not earn a bonus.
In any of these cases, selling 3 fewer sets would decrease the total income by $600.
No way to determine the total income.
INSUFFICIENT.

Statement 2: If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000
Since the first n sets earn a total of $1000, the total bonus earned by the remaining sets is more than $3000.
The minimum value of c -- the bonus earned by each set sold after the first n sets -- occurs if n=1, with the result that 9 of the 10 sets here each earn a bonus.
If 9 sets here each earn a bonus of c dollars, we get:
9c > 3000
c > 333.33.
No way to determine the total income.
INSUFFICIENT.

Statements combined:
Since c > 333.33, only Case 1 above is viable:
1 set earned a bonus of $600, and the remaining n sets did not earn a bonus.
Thus:
Total income = base income + bonus = 1000 + 600 = 1600.
SUFFICIENT.

The correct answer is C.

Thank you very much Matt for the detailed explanation...
I got this question from the "700 Tough DS questionnaire".

I am still confused with the statement 1 explanation

Stat #1 clearly says that Laura had sold 3 fewer sets so why we should consider the case of one and two set sold?

Matt@VeritasPrep wrote:There are a bunch of tricks in this one!

Let's start by properly phrasing the question. If we say that Laura sold x encyclopedias in March, then her income = 1000 + (x - n)*c, assuming that x is greater than n. (If x is NOT greater than n, then her income is just $1000.) So we need to solve for x, n, and c.

S1 is really tricky. Remember that Laura gets c dollars for each set she sells beyond the first n. Now let's think about a few scenarios:

Suppose that in March Laura sold (n+1) encyclopedias; in other words, one more than she needs to get her bonus (x = n+1). That means that that ONE extra encyclopedia gives her a $600 bonus, so c = $600.

Now suppose that in March Laura sold (n+2) encyclopedias, or x = (n+2). Now TWO extra encyclopedias get her the $600 bonus, so 2c = $600 and c = $300.

Now suppose that in March Laura sold (n+3) (or more) encyclopedias, or x = (n+3). Then three encyclopedias are worth $600 in bonus, so 3c = $600 = c = $200.

So we have three possible values for c, let alone x and n; INSUFFICIENT.

S2 gives us the equation:
Laura's income for ten sets > $4000, which (using our earlier equation) is
$1000 + (10-n)*c > $4000, or
(10-n) * c > $3000, or
10c - nc > $3000

Since n must be positive, 10c > 10c - nc. Since 10c - nc > $3000, 10c > $3000, and c > $300. On its own this is also INSUFFICIENT.

Together we know that c must be $600. This tells us that Laura sold exactly (n+1) encyclopedias in March, so she gets $1000 for the first n and $600 for the additional one, making her income $1600. (Algebraically, her income is $1000 + ((n+1)-n)*$600.)

Yeesh! That's some question - where is it from?

Hi Ganesh!

Sorry for such a late reply - I realized that I didn't have follow up alerts turned on, so I had to fix that.

We're considering the two cases you mentioned (1 fewer and 2 fewer) because we're trying to figure out how many sets Laura has sold above the first n required for her bonus. A less abstract illustration might help. Let's say that I'm an encyclopedia salesman who is paid a flat fee of $1000 every month, with a bonus $50 for each encyclopedia I sell beyond the first 20.

If I sell 20 encyclopedias in June and 17 in July, I am paid the same amount each month - the three fewer encyclopedias in July don't cause me to lose any money, so in both months I only made my $1000 flat fee.

If I sell 21 encyclopedias in June and 18 in July, I lose $50, because I got $50 for the extra encyclopedia in June, but I didn't even make it to to 20 in July.

If I sell 22 encyclopedias in June and 19 in July, I lose $100, as above.

Finally, if I sell 23 encyclopedias in June and 20 in July, I lose $150.

So you can see that selling three fewer encyclopedias could cause me to lose $0, $50, $100, or $150, respectively, which is why we have to consider each case separately.

GaneshMalkar wrote:Thank you very much Matt for the detailed explanation...
I got this question from the "700 Tough DS questionnaire".

I am still confused with the statement 1 explanation

Stat #1 clearly says that Laura had sold 3 fewer sets so why we should consider the case of one and two set sold?

Matt@VeritasPrep wrote:There are a bunch of tricks in this one!

Let's start by properly phrasing the question. If we say that Laura sold x encyclopedias in March, then her income = 1000 + (x - n)*c, assuming that x is greater than n. (If x is NOT greater than n, then her income is just $1000.) So we need to solve for x, n, and c.

S1 is really tricky. Remember that Laura gets c dollars for each set she sells beyond the first n. Now let's think about a few scenarios:

Suppose that in March Laura sold (n+1) encyclopedias; in other words, one more than she needs to get her bonus (x = n+1). That means that that ONE extra encyclopedia gives her a $600 bonus, so c = $600.

Now suppose that in March Laura sold (n+2) encyclopedias, or x = (n+2). Now TWO extra encyclopedias get her the $600 bonus, so 2c = $600 and c = $300.

Now suppose that in March Laura sold (n+3) (or more) encyclopedias, or x = (n+3). Then three encyclopedias are worth $600 in bonus, so 3c = $600 = c = $200.

So we have three possible values for c, let alone x and n; INSUFFICIENT.

S2 gives us the equation:
Laura's income for ten sets > $4000, which (using our earlier equation) is
$1000 + (10-n)*c > $4000, or
(10-n) * c > $3000, or
10c - nc > $3000

Since n must be positive, 10c > 10c - nc. Since 10c - nc > $3000, 10c > $3000, and c > $300. On its own this is also INSUFFICIENT.

Together we know that c must be $600. This tells us that Laura sold exactly (n+1) encyclopedias in March, so she gets $1000 for the first n and $600 for the additional one, making her income $1600. (Algebraically, her income is $1000 + ((n+1)-n)*$600.)

Yeesh! That's some question - where is it from?