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# A florist found that 60% of his customers on Valentine's Day bought

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A florist found that 60% of his customers on Valentine's Day bought [#permalink]
60% bought roses...implies 60% of 80 = 48 bought red roses
Half of them bought red roses = 24

stat 1 : only those who bought red rose bought another rose,,,means orange and yellow = 0
hence exactly two roses = (red and orange ) + (red and yellow) = this value can be max 24 not above tat
hence suff

stat 2 : not suff

ans A

correct me if iam rong

thanks

Originally posted by mohshu on 01 Apr 2017, 00:04.
Last edited by mohshu on 01 Apr 2017, 00:23, edited 1 time in total.
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Re: A florist found that 60% of his customers on Valentine's Day bought [#permalink]
A.

(1) Of the 80 customers the florist had, only those people buying red roses also bought exactly one other type of rose.
(2) Of the 80 customers the florist had, half of the people who bought red roses bought exactly one other type of rose.
Not indicative of customers of other types of roses
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60% bought roses i.e. 48 customers
Bought roses
Again half of them bought red roses i.e. 24 customers.

(2) Of the 80 customers the florist had, half of the people who bought red roses bought exactly one other type of rose.

So we have 12 customers who bought red roses also buying another rose.Another 12 customers may have bought only one red rose or more than 2 variety of roses.
Now if remaining 24 customer who did not buy red rose had bought exactly one more kind of rose say (orange+yellow) then also the total customer buying exactly two kinds of roses will be 24+12=36 which is below the half range of customers.
This makes the answer sufficient "No"

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A florist found that 60% of his customers on Valentine's Day bought [#permalink]
Bunuel wrote:
A florist found that 60% of his customers on Valentine's Day bought roses. People had the option that day of buying red roses, yellow roses or orange roses and half of all the people who bought roses bought red roses. Did more than half the customers buy exactly two varieties of rose?

(1) Of the 80 customers the florist had, only those people buying red roses also bought exactly one other type of rose.
(2) Of the 80 customers the florist had, half of the people who bought red roses bought exactly one other type of rose.

Am I understanding this incorrectly.

Total customers: 80
60% bought roses: 48
Half of those who bought roses, bought red: 24
Did more than 40 people buy exactly 2 types of rose?

1- Only those who bought roses bought exactly one other type of rose. so no more than 24 people bought another variety of rose. 1 - Sufficient.

2- Half of the people who bought red roses bought exactly one other variety. Since 24 bought red roses and of those 24, we know that 12 certainly bought exactly two varieties. We are left with 36 people who bought roses, and we need to determine if 40-12 = 28 of them bought exactly two varieties. However, the information in 2 does not provide enough evidence to support this. 2 is not sufficient.

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Re: A florist found that 60% of his customers on Valentine's Day bought [#permalink]
Bunuel wrote:
A florist found that 60% of his customers on Valentine's Day bought roses. People had the option that day of buying red roses, yellow roses or orange roses and half of all the people who bought roses bought red roses. Did more than half the customers buy exactly two varieties of rose?

(1) Of the 80 customers the florist had, only those people buying red roses also bought exactly one other type of rose.
(2) Of the 80 customers the florist had, half of the people who bought red roses bought exactly one other type of rose.

can you please throw light on this problem?

for me it looks B is sufficient. for the following reason. Out of 48 customers who bought rose. 24 bought red. second statement says half of them bought exactly one other type rose. that means 12 people bought red + exactly one other type of rose. Remaining 12 customers who bought red rose might have got either only red (or) all three. Remaining 24 if all would have got one other type of rose. If you sum it will be only 36 < 40.
B should be sufficient. why it is wrong?
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Re: A florist found that 60% of his customers on Valentine's Day bought [#permalink]
Avinash_R1 wrote:
Bunuel wrote:
A florist found that 60% of his customers on Valentine's Day bought roses. People had the option that day of buying red roses, yellow roses or orange roses and half of all the people who bought roses bought red roses. Did more than half the customers buy exactly two varieties of rose?

(1) Of the 80 customers the florist had, only those people buying red roses also bought exactly one other type of rose.
(2) Of the 80 customers the florist had, half of the people who bought red roses bought exactly one other type of rose.

can you please throw light on this problem?

for me it looks B is sufficient. for the following reason. Out of 48 customers who bought rose. 24 bought red. second statement says half of them bought exactly one other type rose. that means 12 people bought red + exactly one other type of rose. Remaining 12 customers who bought red rose might have got either only red (or) all three. Remaining 24 if all would have got one other type of rose. If you sum it will be only 36 < 40.
B should be sufficient. why it is wrong?

In B Total 48 people bought roses. 24 bought Red Roses. Out of those 24, 12 bought Red + (Either Orange or Yellow). Rest 12 either bought only Red or three varieties which do not fit our criteria of exactly two varieties.
Till here We have 12 people who fit our criteria and 12 who do not after analyzing 24 people

Now the other half which is 24 people their are 4 scenarios
A) All 24 have bought Yellow then our criteria will be 12<24
B) All 24 have bought Orange then our criteria will be 12<24
C) All have bought both Orange and Yellow then our criteria will be 36>24
D) A mix of people (Orange, Orange+Yellow) (Yellow, Orange+Yellow). Here we cannot identify our criteria

Hence B is insufficient
Hope you understand
Re: A florist found that 60% of his customers on Valentine's Day bought [#permalink]
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