Dear All,

We will use this thread and fb to discuss solutions to questions of iCAT where you have a doubt

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Dear All,

We will use this thread and fb to discuss solutions to questions of iCAT where you have a doubt

We will use this thread and fb to discuss solutions to questions of iCAT where you have a doubt

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

[biCAT-3[/b]

Please start posting your doubts fo iCAT -3 here

Please start posting your doubts fo iCAT -3 here

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

iCAT 3 Doubts QA

Question Nos : 70, 79, 82, 87, 96, 98

Question Nos : 70, 79, 82, 87, 96, 98

- shubham50
**Posts:**2**Joined:**Mon Jul 10, 2017 1:03 pm

iCAT-3

Q70

One way of solving the question is provided in the solution. I would solve it this way in the exam:

By putting some values, I can quickly estimate that the function is increasing as the value of X increases and there are values like -5 or -6 for which the function is going to have a value equal to zero. So discard all positive values of x. Positive values of x are 1,2 and 4, therefore the answer must be option 3.

Q70

One way of solving the question is provided in the solution. I would solve it this way in the exam:

By putting some values, I can quickly estimate that the function is increasing as the value of X increases and there are values like -5 or -6 for which the function is going to have a value equal to zero. So discard all positive values of x. Positive values of x are 1,2 and 4, therefore the answer must be option 3.

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

iCAT-3

79

To solve such a question, one needs to understand the arrangements of terms when boys and girls alternate. When boys and girls alternate there are two possible cases:

1) When the number of boys is equal to the number of girls :Let us say n boys and n girls making it 2n students

2) When the number of boys is one more than the number of girls or the number of girls is one more than the number of boys - so one of them is n and the other is n+1

Case- 1 - If the number of students is 2n and the number of boys and girls is each equal to n, then

the number of arrangements is given by: 2 (n!)(n!)

How ??

Case a - when boys occupy all odd positions - 1,3,5,7... and so one and

Case b - when the girls occupy all even positions- 2,4,6,8 and so on....

So in such a case the number of arrangements in total will be n! x n! - where the boys can arranged on n seats in n! ways and the girls can be arranged on n seats in n! ways - making the total number of ways as n! x n!

Similarly, we have the number of arrangements when the girls sit at odd position and all boys sitting at even positions also as n! x n!

Therefor the total number of arrangements = 2 n! x n!

When one student increases, it might be a boy or a girl . But the number of student will now be odd and thus the possible arrangements of the 2n+1 students where n belong to one gender and n+1 belong to the other gender is:

n+1 ! x n!

Since the number of ways increases by 400%, the new number of ways- after the number of students increase by 1 is => (n+1 !) x( n ! )/ 2 x (n!) x (n!) = 5

=> n+1/2 = 5 => n = 9 or the total number of students = n + n+1 = 19

Case-II

Following a similar reasoning as above we have: if the initial number of students is odd - let us say the number of students is 2m+1, where m students belong to one gender and m+1 students belong to the other gender, the inital number of ways = m ! x m+1 !

When one of the student increases , then the increase can be of a girl or a boy, then the number of arrangements can be either: the increase can only be such as to ensure that the number of boys and girls, which was unequal initially, becomes equal

m+1 ! x m+1! x 2

Thus we have the ratio of the two ways as => (m+1 ! x m+1!) x 2/(m ! x m+1 ! ) = 5

2 ( m+1) = 5

In this case the value of m is not an integer. which is impossible. Thus, the 2nd case is impossible.

79

To solve such a question, one needs to understand the arrangements of terms when boys and girls alternate. When boys and girls alternate there are two possible cases:

1) When the number of boys is equal to the number of girls :Let us say n boys and n girls making it 2n students

2) When the number of boys is one more than the number of girls or the number of girls is one more than the number of boys - so one of them is n and the other is n+1

Case- 1 - If the number of students is 2n and the number of boys and girls is each equal to n, then

the number of arrangements is given by: 2 (n!)(n!)

How ??

Case a - when boys occupy all odd positions - 1,3,5,7... and so one and

Case b - when the girls occupy all even positions- 2,4,6,8 and so on....

So in such a case the number of arrangements in total will be n! x n! - where the boys can arranged on n seats in n! ways and the girls can be arranged on n seats in n! ways - making the total number of ways as n! x n!

Similarly, we have the number of arrangements when the girls sit at odd position and all boys sitting at even positions also as n! x n!

Therefor the total number of arrangements = 2 n! x n!

When one student increases, it might be a boy or a girl . But the number of student will now be odd and thus the possible arrangements of the 2n+1 students where n belong to one gender and n+1 belong to the other gender is:

n+1 ! x n!

Since the number of ways increases by 400%, the new number of ways- after the number of students increase by 1 is => (n+1 !) x( n ! )/ 2 x (n!) x (n!) = 5

=> n+1/2 = 5 => n = 9 or the total number of students = n + n+1 = 19

Case-II

Following a similar reasoning as above we have: if the initial number of students is odd - let us say the number of students is 2m+1, where m students belong to one gender and m+1 students belong to the other gender, the inital number of ways = m ! x m+1 !

When one of the student increases , then the increase can be of a girl or a boy, then the number of arrangements can be either: the increase can only be such as to ensure that the number of boys and girls, which was unequal initially, becomes equal

m+1 ! x m+1! x 2

Thus we have the ratio of the two ways as => (m+1 ! x m+1!) x 2/(m ! x m+1 ! ) = 5

2 ( m+1) = 5

In this case the value of m is not an integer. which is impossible. Thus, the 2nd case is impossible.

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

iCAT-3

Q 82

The milk man in question has a mixture that contains three things - Milk, Liquid X and Water

In the first container - let us assume he has 50 l of the mixture, which contains- x units of liquid 40% liquid X mixture and the rest of it-i.e. 50-x units of milk.

In the second container too- there must be 50 l of the mixture, which contains y units of 30% liquid X mixture and the rest of it i.e. 50-y units of milk.

So the contents of the two containers are as follows:

Milk Liquid X Water

Container-1 - 50-x 0.4x 0.6x

Container -2 50-y 0.3y 0.7 y

The questions suggests that on mixing the two contents- we get the total percentage of X and water to b 8% and 16%. Therefore the percentage of milk in the final mixture must be - 100 -8-16 = 76 %

The total quantity of mixture of the two containers= 100 units ( 50 units from each)

76% OF THE same is milk - therefore, the final amount of milk in the mixture must be 76 units

(50-x) + (50-y) = 76 units => x + y = 24 ..........(i)

Similarly the quantity of liquid X must be 8 units of the total 100 units of the mixture.

0.4x + 0.3 y = 8 ...............(ii)

On solving i and ii we get x = 8 units and y =16 units

The quantity of milk in the first mixture = 50 -x = 50 -8 = 42

Required percentage = 42/50 x 100 = 84%

Q 82

The milk man in question has a mixture that contains three things - Milk, Liquid X and Water

In the first container - let us assume he has 50 l of the mixture, which contains- x units of liquid 40% liquid X mixture and the rest of it-i.e. 50-x units of milk.

In the second container too- there must be 50 l of the mixture, which contains y units of 30% liquid X mixture and the rest of it i.e. 50-y units of milk.

So the contents of the two containers are as follows:

Milk Liquid X Water

Container-1 - 50-x 0.4x 0.6x

Container -2 50-y 0.3y 0.7 y

The questions suggests that on mixing the two contents- we get the total percentage of X and water to b 8% and 16%. Therefore the percentage of milk in the final mixture must be - 100 -8-16 = 76 %

The total quantity of mixture of the two containers= 100 units ( 50 units from each)

76% OF THE same is milk - therefore, the final amount of milk in the mixture must be 76 units

(50-x) + (50-y) = 76 units => x + y = 24 ..........(i)

Similarly the quantity of liquid X must be 8 units of the total 100 units of the mixture.

0.4x + 0.3 y = 8 ...............(ii)

On solving i and ii we get x = 8 units and y =16 units

The quantity of milk in the first mixture = 50 -x = 50 -8 = 42

Required percentage = 42/50 x 100 = 84%

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

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