CAT 2008: Quantitative Questions a Day 101-133 - The discussions

[Aarav]

Hello All,

Please carry the discussions for QQAD 101-133 here.

We will switch to 2 problems a day in mid-September, so this is the penultimate QQAD thread for 2008.

Hope you all have a great time over here

[gripened]

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Quantitative Question # 101
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A gathering of a certain number of families consists of people belonging to two generations only. It is known that the number of families is less than the number of girls, the number of girls is less than the number of boys and that the
number of boys is less than the number of parents. If the minimum number of single parent families is two, then what is the minimum number of families, given that no family has more than 3 children?

(1) 3 (2) 4 (3) 5 (4) 7 (5) none of these

[gripened]

Originally Posted by gripened

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Quantitative Question # 101
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A gathering of a certain number of families consists of people belonging to two generations only. It is known that the number of families is less than the number of girls, the number of girls is less than the number of boys and that the
number of boys is less than the number of parents. If the minimum number of single parent families is two, then what is the minimum number of families, given that no family has more than 3 children?

(1) 3 (2) 4 (3) 5 (4) 7 (5) none of these

We have
F : No. of families
G : No. of girls
B : No. of boys
P : No. of Parents

And given , F < G < B < P
Now, Min(F) = 2 , as minimum number of single parents = 2...
Min. number of single parents = 2.... --(I)
Hence we can have at best case :

___F___G___B___P___
___2____3____4___5___ ---(II) ==> P >= F + 3

But, 1 F ==> Max(P) = 2, and Min(P) = 1... -- (III)
From (I),(II) and (III), we have

No. of Parents = 2 + (No. of Families - 2)*2
==> P = 2 + (F-2)*2...
But best case we have, P = F + 3, hence
F + 3 = 2F - 2
==> F = 5.. && P = 8


Answer Option (3) 5 Families minimum

[genius_inside]

given: p>b>g>n
also, we have the min value for n=2.
inorder to minimize n, we'l maximize d rest. ie, we'l assume 3 kids n both parents at most of the places.

for the eqn to hold true, we need atleast a difference of 3 between p and n. currently the difference is 0 (as p=n=2).
the closest difference is when p=8(1+1+2+2+2) and n=5(1+1+1+1+1)
also, among children can be split across boys=7 and girls=6, total=13, (which is less than max of 3*5=15)
Hence the minimum ans is 5 families

Ans: (3) 5

[Varun Khullar]

Originally Posted by gripened

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Quantitative Question # 101
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A gathering of a certain number of families consists of people belonging to two generations only. It is known that the number of families is less than the number of girls, the number of girls is less than the number of boys and that the
number of boys is less than the number of parents. If the minimum number of single parent families is two, then what is the minimum number of families, given that no family has more than 3 children?

(1) 3 (2) 4 (3) 5 (4) 7 (5) none of these

min no of families = 5...
no of parents are to be even.. so 5 is min we can do..

[rahulformba]

no of families=5
went with the options.. 5 seemed to satisfy the conditions

[selebratinglife]

Wrong post

[rohith_s_theone]

Reasonably simple question.
Let no of families be f
No of girls be g
No of boys be b
and no of parents be p

f<g<b<p

It is given that there are atleast 2 single parent families
That means f=2 p =2 initially
In order to minimize the number of families and ensure that all the conditions are met we have to add 2 parent families to our set

Let number of families be n
In order to minimize the number of families
No of girls = n+1
No of boys =n+2
No of parents = n+3

[The difference between number of parents and number of families must be at least 3. We originally had 2 parents and 2 families. We added (n-2) families of 2 parents each.]
So we get
2+(n-2)*2- n = 3
which gives n = 5

ANS (3) 5

[Catalyst2008]

condition given - f<g<b<p
since the min no. of single parent is 2, p-2 i.e. parents who are not single (which has to be an even no.). So p is also even.
When we take f as 4(min of all the options), min p comes out to be 7...so not possible.
If we take f as 5, it satisfies and p comes out to be 8.
So ans option (3)-5

[prakharc]

Ans)Option 3--- 5 families

If there are 3 families with 2 single parent families we get 3<g<b<4 ...not possible....for 4 we get 4<g<b<5...not possible

For 5 families we get 5<g<b<8
So g=6,b=7...And in 5 families we can have 5*3=15 children>13 ...
SO minimum number of families =5