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# A certain toy store sold 20 toys yesterday, each of which was either a

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A certain toy store sold 20 toys yesterday, each of which was either a  [#permalink]

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04 Feb 2017, 04:23
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55% (hard)

Question Stats:

69% (02:23) correct 31% (01:59) wrong based on 151 sessions

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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.

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Re: A certain toy store sold 20 toys yesterday, each of which was either a  [#permalink]

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04 Feb 2017, 06:02
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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 _________________ CEO Joined: 12 Sep 2015 Posts: 3853 Location: Canada A certain toy store sold 20 toys yesterday, each of which was either a [#permalink] ### Show Tags 04 Feb 2017, 09:21 1 Top Contributor 1 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 Answer: A RELATED VIDEO FROM OUR COURSE _________________ Test confidently with gmatprepnow.com Manager Joined: 07 Aug 2018 Posts: 111 Location: United States (MA) GMAT 1: 560 Q39 V28 GMAT 2: 670 Q48 V34 Re: A certain toy store sold 20 toys yesterday, each of which was either a [#permalink] ### Show Tags 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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Re: A certain toy store sold 20 toys yesterday, each of which was either a   [#permalink] 26 Sep 2018, 07:46
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