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E
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Difficulty:
35%
(medium)
Question Stats:
74% (01:49) correct
26%
(01:58)
wrong
based on 704
sessions
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Date
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If a certain company purchased computers at $2000 each and printers $300 each , how many computers did it purchase?
(1) More than three printers were purchased
(2) The total amount for the purchase of the computers and printers was $15,000
(1) More than three printers were purchased
(2) The total amount for the purchase of the computers and printers was $15,000
06 Sep 2015, 00:45
Kudos
Bookmarks
Yes E is correct. I will explain how.
Statement 1 is clearly not sufficient.
Statement 2 says That the total was 15,000.
So (c)2000 + (p)300 = 15,000
where p is for printers and c is for computers.
when p=10 then the remaining amount left for computers is 12000 which means 6 computers.
but when p=30 the then the remaining amount left is 15,000 - 30(300) = 6000 which means 3 computers.
Different values means the statement is insufficient
Combining:
St. 1 says p>3 St. 2 says (c)2000 + (p)300 = 15,000.
But we know that the number of printers has to be a minimum of 10 for the remaining amount to be a multiple of 2000. ( from ST. 2)
Combining does not help because p>10 from statement 2. and statement 1 says p>3. p>10 is the more limiting inequality so we use that but it will not get us the answer.
So the answer is E.
The easiest approach according to me:
Start subtracting multiples of 300 from 15000 till you get a multiple of 2000.
15,000-300 = 14,700( not a multiple of 2000)
15,000- 600 = 14,400(not a multiple of 2000)
till you get to 15,000 - 3000 = 12,000 (Bingo you found the multiple)
Now you have to check if any other value of p satisfies the equation.
15,000- 9,000 = 6000( multiple of 2000)
So answer is E.
Hope this helps!
Statement 1 is clearly not sufficient.
Statement 2 says That the total was 15,000.
So (c)2000 + (p)300 = 15,000
where p is for printers and c is for computers.
when p=10 then the remaining amount left for computers is 12000 which means 6 computers.
but when p=30 the then the remaining amount left is 15,000 - 30(300) = 6000 which means 3 computers.
Different values means the statement is insufficient
Combining:
St. 1 says p>3 St. 2 says (c)2000 + (p)300 = 15,000.
But we know that the number of printers has to be a minimum of 10 for the remaining amount to be a multiple of 2000. ( from ST. 2)
Combining does not help because p>10 from statement 2. and statement 1 says p>3. p>10 is the more limiting inequality so we use that but it will not get us the answer.
So the answer is E.
The easiest approach according to me:
Start subtracting multiples of 300 from 15000 till you get a multiple of 2000.
15,000-300 = 14,700( not a multiple of 2000)
15,000- 600 = 14,400(not a multiple of 2000)
till you get to 15,000 - 3000 = 12,000 (Bingo you found the multiple)
Now you have to check if any other value of p satisfies the equation.
15,000- 9,000 = 6000( multiple of 2000)
So answer is E.
Hope this helps!
06 Sep 2015, 20:42
Kudos
Bookmarks
Forget conventional ways of solving math questions. In DS, Variable approach is
the easiest and quickest way to find the answer without actually solving the
problem. Remember equal number of variables and equations ensures a solution.
If a certain company purchased computers at $2000 each and printers $300 each , how many computers did it purchase?
1) more than three printers were purchased
2) The total amount for the purchase of the computers and printers was $15,000
Transforming the original condition by variable approach method, we have 2 variables (number of computers;x, number of printers:y) and since we need to match the number of variables and equations, we need 2 equations. Since there is 1 each in 1) and 2), C is likely the answer. Using both 1) & 2) together, 2000x+300y=15000 ==> 20x+3y=150. then we have pair (x,y)=(3,30), (6,10), and since the answer is Not unique the conditions are not sufficient. Therefore the answer is E.
the easiest and quickest way to find the answer without actually solving the
problem. Remember equal number of variables and equations ensures a solution.
If a certain company purchased computers at $2000 each and printers $300 each , how many computers did it purchase?
1) more than three printers were purchased
2) The total amount for the purchase of the computers and printers was $15,000
Transforming the original condition by variable approach method, we have 2 variables (number of computers;x, number of printers:y) and since we need to match the number of variables and equations, we need 2 equations. Since there is 1 each in 1) and 2), C is likely the answer. Using both 1) & 2) together, 2000x+300y=15000 ==> 20x+3y=150. then we have pair (x,y)=(3,30), (6,10), and since the answer is Not unique the conditions are not sufficient. Therefore the answer is E.
