Test: Set Theory- 3 - CAT MCQ

According to the question, it is given that there was a survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of the three popular options- air-conditioning, radio, and power windows- were already installed.

15 cars had to air-condition. This doesn’t mean that it only had to air-condition. It can also have a radio, or power-windows, or both.
2 cars had air-conditioning and power windows but no radios.
12 cars had power windows. This doesn’t mean that it only had power windows. It can also have a radio, or air-conditioning, or both.
6 cars had air-conditioning and radio but no power windows.
11 cars had a radio. This doesn’t mean that it only had a radio. It can also have power-windows, or air-conditioning, or both.
4 cars had radio and power windows. It may or may not have to air-condition.
3 cars had all three options.

Now, on drawing Venn diagram, we get

When we add all the values, we get a total of 23 cars. 

So, 2 cars don’t have the air conditioning or the radio, or the power windows.

Let us first understand the meaning of the statement given in the question—

It is given that the numbers are from 1 to 100, so while counting we will include both the limits, i.e., 1 and 100. Had this been “How many numbers in between 1 to 100 are..." then we would not have included either 1 or 100.

Now to solve this question, we will first find out the number of numbers from 1 to 100 which are divisible by either 2 or 5 (since all the numbers which are not divisible by 2 will not be divisible by 4 also, so we do not need to find the numbers divisible by 4). And then we will subtract this from the total number of numbers i.e., 100. It can be seen below:

Total number of numbers = numbers which are divisible + numbers which are not divisible

So, n(2U5) = n(2) + n(5) - n (2∩5)

Now, n(2) = 50 

n(5) = 20 

n(2∩5) = 10

n(2U5) = 50 + 20 - 10 = 60

Numbers which are not divisible = total number of numbers-numbers which are divisible

=100 - 60 = 40

We know that n (A∪B ) = n(A ) + n(B) - n(A∩B) 

Assuming that n(Q) = number of students passing QA and n(R) = number of students passing RC.

Hence, n(Q∪R ) = n(Q) + n(R) - n(Q∩R)

Or, n(Q∩R) = n(Q) + n(R) - n(Q∪R) = 70% + 60% - n(Q∪R)

Now the minimum value of n(Q∩R) will occur for the maximum value of n(Q∪R).

The maximum possible value of n(Q∪R) = 100%

So, the minimum value of n(Q∩R) = 30%

This question is just the extension of the earlier question.

The minimum value of EU and (QA∩RC) = 90% + 30% - 100% = 20%

The minimum value of DI and EU and  (QA∩RC) = 20% +85% - 100% = 5%

We can do this question by using either the formula for three sets, or we can simply keep on applying the formula for two sets required a number of times.

We know that n(A∪B) = n (A ) + n(B) - n(A∩B)

Assuming that n(Q) = number of students passing QA, n(E) = number of students passing EU and n(R) = number of students passing RC

Hence, n(Q∪R) = n(Q) + n(R) - n(Q∩R).

Or, n(Q∩R) = n(Q) + n(R) - n(QuR) = 70% + 60% - n(Q∪R)

Now the minimum value of n(Q∩R) will occur for the maximum value of n(Q∪R). The maximum possible value of n(Q∪R) =100%

So, the minimum value of n(Q∩R) = 30%

Now, the minimum value of EU, QA and RC will be obtained by finding the minimum of EU and (QA∩RC).

The minimum value of EU and (QA∩RC) = 90% + 30% - 100% = 20%

From the figure, it is clear that the number of  people at  the party were 30+10+15 = 55.

we can of course solve this mathematically as below:

Let  n(A) = No. of  persons    who kissed Sherry = 40

n(B) = No. of  persons  who    shake hands with Sherry  =  25

and n(A « B) = No. of   persons who shook  hands    with Sherry and kissed him    both =10

Then using  the formula, n(A    »B) = n(A) + n(B) -n(A « B)

n(A »B) =40 + 25  -10 = 55

The total number of peoples = 100 + 220+130 = 450

So, Option D is correct

The required answer would be = 20000 - 4000 - 5000 - 8000 = 3000

In order to think of the maximum   number of  athletes who can play   none of  the games, we can think   of  the fact that since there  are 70 athletes who play tennis, fundamentally there are a maximum of 50 athletes who would be possibly in the can play none of the games’.

No other constraint in the problem necessitates a reduction of this number and hence the answer to this question is 50.

since x = 6 , the figure becomes:

The answer would be :

6. option (B) is correct

21 + 21 +21 = 63

Option (B) is correct

(21+ 21 +21 +6+4+3+2)/21 = 78/21

Option (A) is correct

If we try to consider each of the households watching Zee TV and   StarPlus as independent of  each   other, we would get a total of 1000 + 750 = 1750 households.However, wehave a totalof only1500 householdsi the locality and hence, there has to be a minimum interference of atleast 250 households whowould  be watching both Zee TV and Star Plus. Hence, the answer to this question is 250.

In this case,the number of households watching ZeeTV and NDTV can be separate from each other since there is no interference required between the households     watching ZeeTV and the households watching NDTV  as their individual sum(1000 +300) is smaller than the 1500 available householdsin the locality.Hence,the answer in this question is 0.

For this  to occur  the following situation would give us the required solution:

As you can clearly see from the figure, all the requirements of each category of viewers is fulfilled by the given allocation of 1000 households. In this situation,the maximum number of households who do not watch any of the three channels is visible as 1500 –1000 = 500.