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Re: A certain toy store sold 20 toys yesterday, each of which was either a
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04 Feb 2017, 06:02
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SajjadAhmad wrote:
A certain toy store sold 20 toys yesterday, each of which was either a $40 toy or a $20 toy. How many $20 toys did the toy store sell?
1) The average price of the toys sold yesterday was $35. (2) The total price of the 20 toys sold yesterday was between $650 and $750.
So we know total toys are 20, say a of $20 and b of $40..
Let's see the statements.. (1) The average price of the toys sold yesterday was $35. Using weighted average method, we can say that the ratio of b/a=(35-20)/(40-35)=15/5=3.. B=3a.... a+b=20....3a+a=20....4a=20....A=5.. B=15 Sufficient
(2) The total price of the 20 toys sold yesterday was between $650 and $750.[/quote] 20a+40b>=660..... a+2b>=33.... a+2(20-a)=>33...a+40-2a>=33.... a<=7 20a+40b<=740..... a+2b<=37.... a+2(20-a)=<37...a+40-2a<=37.... a>=3... So various possibilities 3,4,5,6 or 7.. Insufficient
A certain toy store sold 20 toys yesterday, each of which was either a
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04 Feb 2017, 09:21
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SajjadAhmad wrote:
A certain toy store sold 20 toys yesterday, each of which was either a $40 toy or a $20 toy. How many $20 toys did the toy store sell?
1) The average price of the toys sold yesterday was $35. (2) The total price of the 20 toys sold yesterday was between $650 and $750.
Given: A certain toy store sold 20 toys yesterday, each of which was either a $40 toy or a $20 toy. Let C = NUMBER of $20 toys sold (C for cheap) Let E = NUMBER of $40 toys sold (E for expensive) Since 20 toys were sold, we can write: C + E = 20
Target question:How many $20 toys did the toy store sell? In other words, our goal is to determine the value of C
Statement 1: The average price of the toys sold yesterday was $35 In other words, (total revenue from the toys)/20 = $35 How do we determine total revenue? Well, revenue from the $20 toys = 20C, and revenue from the $40 toys = 40E
So, we can write: (20C + 40E)/20 = 35 Multiply both sides by 20 to get: 20C + 40E = 700
We now have a system of two equations: C + E = 20 20C + 40E = 70
Since we COULD solve this system for C, we COULD determine the number of $20 toys sold. Since we COULD answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: The total price of the 20 toys sold yesterday was between $650 and $750 This statement does not feel sufficient, so let's test some values. Case a: the store sold 5 $20 toys and 15 $40 toys. This yields a total revenue of $700. In this case, the store sold 5 $20 toys Case b: the store sold 6 $20 toys and 14 $40 toys. This yields a total revenue of $680. In this case, the store sold 6 $20 toys Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Re: A certain toy store sold 20 toys yesterday, each of which was either a
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26 Sep 2018, 07:46
chetan2u wrote:
SajjadAhmad wrote:
A certain toy store sold 20 toys yesterday, each of which was either a $40 toy or a $20 toy. How many $20 toys did the toy store sell?
1) The average price of the toys sold yesterday was $35. (2) The total price of the 20 toys sold yesterday was between $650 and $750.
So we know total toys are 20, say a of $20 and b of $40..
Let's see the statements.. (1) The average price of the toys sold yesterday was $35. Using weighted average method, we can say that the ratio of b/a=(35-20)/(40-35)=15/5=3.. B=3a.... a+b=20....3a+a=20....4a=20....A=5.. B=15 Sufficient
(2) The total price of the 20 toys sold yesterday was between $650 and $750.
20a+40b>=660..... a+2b>=33.... a+2(20-a)=>33...a+40-2a>=33.... a<=7 20a+40b<=740..... a+2b<=37.... a+2(20-a)=<37...a+40-2a<=37.... a>=3... So various possibilities 3,4,5,6 or 7.. Insufficient
A
Another, maybe quicker way to look at S(2) with inequalities is:
\(650<40b+20a<750\) From the question stem we know that a+b=20 so we can substitute b to find a.
\(650<40*(20-a)+20a<750\)
\(650<-800-40a+20a<750\)
\(-150<-20a<-50\)
\(150>20a>50\)
\(7.5>a>2.5\) Since a need to be an integer --> \(7=>a=>3\)
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