Answer

Verified

423k+ views

**Hint:**Here using the information we draw a Venn diagram and assume the number of people living in the town as a variable and calculate each percentage in terms of fraction. We denote each value in terms of probability of set and using the formula for set union find the values.

* Set theory formula of number of elements in the union of 2 sets : \[n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)\]

**Complete step-by-step answer:**

We draw a Venn diagram of the question

Step 1: Solving for option (c)

Let x be the number of families living in the town.

P denotes families owning a phone.

C denotes families owning a car.

So families owning both a phone and car will be denoted by \[P \cap C\].

Now we convert percentage into fraction form.

Number of families owning a phone is \[25\% \] of \[x\].

\[

\Rightarrow n(P) = \dfrac{{25}}{{100}}x \\

\Rightarrow n(P) = \dfrac{x}{4} \\

\]

Number of families owning a car is \[15\% \] of \[x\].

$

\Rightarrow n(C) = \dfrac{{15}}{{100}}x \\

\Rightarrow n(C) = \dfrac{{3x}}{{20}} \\

$

We are given a total number of families owning both a car and phone is 2000.

$ \Rightarrow n(P \cap C) = 2000$

So, number of families owning neither car nor phone will be total number of families owning car or phone subtracted from total number of families

Thus \[n\](Neither Phone nor Car)$ = x - n(P \cup C)$

Number of families owning neither a car nor a phone is \[65\% \]of \[x\].

$ \Rightarrow \dfrac{{65}}{{100}}x = x - n(P \cup C)$

Shifting all the values except $n(P \cup C)$ to one side of the equation.

$ \Rightarrow n(P \cup C) = x - \dfrac{{65x}}{{100}}$

Take LCM on the right hand side of the equation.

$

\Rightarrow n(P \cup C) = \dfrac{{100x - 65x}}{{100}} \\

\Rightarrow n(P \cup C) = \dfrac{{35x}}{{100}} \\

$

Cancelling out factors we get.

$ \Rightarrow n(P \cup C) = \dfrac{{7x}}{{20}}$ … (i)

Now using the set formula we find the value of x.

Substitute the values \[n(P) = \dfrac{x}{4},n(C) = \dfrac{{3x}}{{20}}, n(P \cap C) = 2000, n(P \cup C) = \dfrac{{7x}}{{20}}\] in the formula \[n(P \cup C) = n(P) + n(C) - n(P \cap C)\] .

$ \Rightarrow \dfrac{{7x}}{{20}} = \dfrac{x}{4} + \dfrac{{3x}}{{20}} - 2000$

Taking LCM on the right hand side of the equation, only values along with the variable x.

$

\Rightarrow \dfrac{{7x}}{{20}} = \dfrac{{5x + 3x}}{{20}} - 2000 \\

\Rightarrow \dfrac{{7x}}{{20}} = \dfrac{{8x}}{{20}} - 2000 \\

$

Shift all values along with variable x to one side.

$ \Rightarrow 2000 = \dfrac{{8x}}{{20}} - \dfrac{{7x}}{{20}}$

Take LCM on the right hand side of the equation.

$

\Rightarrow 2000 = \dfrac{{8x - 7x}}{{20}} \\

\Rightarrow 2000 = \dfrac{x}{{20}} \\

$

Multiply both sides of the equations by 20

$

\Rightarrow 2000 \times 20 = \dfrac{x}{{20}} \times 20 \\

\Rightarrow 40000 = x \\

$

Thus \[x = 40000\]. Total number of families living in town is 40000

So, option (c) is correct.

Step 2: Checking for option (a)

Now we calculate \[5\% \] of the total number of families which gives us the number of families owning both a car and a phone.

\[5\% x = \dfrac{5}{{100}} \times 40000 = 2000\]

$n(P \cap C) = 2000$ which is \[5\% \] of \[40000\].

So, option (a) is correct.

Step 3: Checking for option (b)

Now we calculate \[35\% \] of the total number of families which gives us the number of families owning either a car or a phone.

\[35\% x = \dfrac{{35}}{{100}} \times 40000 = 14000\]

$n(P \cup C) = 14,000$ which is \[35\% \]of \[40000\]

So, option (b) is correct.

Therefore, All (a), (b) and (c) are correct.

**So, the correct answer is “Option A”.**

**Note:**Students many times make the mistake of assuming the neither nor value as the value for union but they should keep in mind that neither a car nor a phone means families which have none of the items, so we subtract families having either car or phone from the total number of families. That is represented by the empty space outside the circles in the Venn diagram.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE