Ask & Answer Question

Sets Ncert Solutions Maths class 11th Maths Class 11th Sets

63. In a survey of 60 people, it was found that 25 people read newspaper H, 26 readnewspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T,8 read both T and I, 3 read all three newspapers. Find:

(i) The number of people who read at least one of the newspapers.

(ii) The number of people who read exactly one newspaper

17 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

63. Let H, T and I be of people who reads newspaper H, T and I respectively.

Then,

number of people who reads newspaper H, n (H) = 25.

number of people who T, n (T) = 26.

number of people who I, n (I) = 26

number of people who both H and T, n (H $\cap$ I) = 9

number of people who both H and T, n (H $\cap$ T) = 11

number of people who both T and I, n (T $\cap$ I) = 8

number of people who reads all newspaper, n (H $\cap$ T $\cap$ I) = 3.

Total no. of people surveyed = 60

The given sets can be represented by venn diagram

(i) The number of people who reads at least one of the newspaper.

in (H∪T $\cup$ I) = n (H) + n (T) + n (I) n (H $\cap$ T) n (H $\cap$ I) n (T $\cap$ I) + n (H $\cap$ T $\cap$ I)

= 25 + 26 + 26 11 9 8 + 3

= 80

...more

63. Let H, T and I be of people who reads newspaper H, T and I respectively.

Then,

number of people who reads newspaper H, n (H) = 25.

number of people who T, n (T) = 26.

number of people who I, n (I) = 26

number of people who both H and T, n (H $\cap$ I) = 9

number of people who both H and T, n (H $\cap$ T) = 11

number of people who both T and I, n (T $\cap$ I) = 8

number of people who reads all newspaper, n (H $\cap$ T $\cap$ I) = 3.

Total no. of people surveyed = 60

The given sets can be represented by venn diagram

(i) The number of people who reads at least one of the newspaper.

in (H∪T $\cup$ I) = n (H) + n (T) + n (I) n (H $\cap$ T) n (H $\cap$ I) n (T $\cap$ I) + n (H $\cap$ T $\cap$ I)

= 25 + 26 + 26 11 9 8 + 3

= 80 28

= 52

(ii) From venn diagram,

a = number of people who reads newspapers H and T only.

b = number of people who reads newspapers H and I only.

c = number of people who reads newspapers T and I only.

d = number of people who reads all newspaper.

So, n (H $\cap$ T) = a + d.

n (H $\cap$ I) = b + d

n (T $\cap$ I) = c + d

So, a + d + c + d + b + d = n (H $\cap$ T) + n (H $\cap$ I) + n (T $\cap$ I)

a + d + c + b + 2d = 11 + 9 + 8

a + b + c +d = 28 - 2d

= 28 2 3 [d = n (H $\cap$ T $\cap$ I) = 3]

= 28 - 6

= 22

Number of people who reads exactly one newspaper

= Total no. of people - No. of people who reads more than one newspaper

= 52 - (a + b +c + d)

= 52 - 22

= 30

less

63. Let H, T and I be of people who reads newspaper H, T and I respectively.Then, number of people who reads newspaper H, n (H) = 25.number of people who T, n (T) = 26.number of people who I, n (I) = 26number of people who both H and T, n (H<math><mrow><mo>∩</mo></mrow></math>I) = 9number of people who both H and T, n (H <math><mrow><mo>∩</mo></mrow></math>T) = 11number of people who both T and I, n (T<math><mrow><mo>∩</mo></mrow></math>I) = 8number of people who reads all newspaper, n (H<math><mrow><mo>∩</mo></mrow></math>T<math><mrow><mo>∩</mo></mrow></math>I) = 3.Total no. of people surveyed = 60The given sets can be represented by venn diagram<picture><img src="https://images.shiksha.com/mediadata/images/articles/1729492115phpO8dWgg.jpeg" alt="" width="272" height="228"></picture> (i) The number of people who reads at least one of the newspaper.in (H∪T<math><mrow><mo>∪</mo></mrow></math>I) = n (H) + n (T) + n (I) n (H<math><mrow><mo>∩</mo></mrow></math>T) n (H<math><mrow><mo>∩</mo></mrow></math>I) n (T<math><mrow><mo>∩</mo></mrow></math>I) + n (H<math><mrow><mo>∩</mo></mrow></math>T<math><mrow><mo>∩</mo></mrow></math>I)= 25 + 26 + 26 11 9 8 + 3= 80 28= 52 (ii) From venn diagram, a = number of people who reads newspapers H and T only.b = number of people who reads newspapers H and I only.c = number of people who reads newspapers T and I only.d = number of people who reads all newspaper.So, n (H<math><mrow><mo>∩</mo></mrow></math>T) = a + d.n (H<math><mrow><mo>∩</mo></mrow></math>I) = b + dn (T<math><mrow><mo>∩</mo></mrow></math>I) = c + dSo, a + d + c + d + b + d = n (H<math><mrow><mo>∩</mo></mrow></math>T) + n (H<math><mrow><mo>∩</mo></mrow></math>I) + n (T<math><mrow><mo>∩</mo></mrow></math>I)a + d + c + b + 2d = 11 + 9 + 8a + b + c +d = 28 - 2d= 28 2 3 [d = n (H<math><mrow><mo>∩</mo></mrow></math>T<math><mrow><mo>∩</mo></mrow></math>I) = 3]= 28 - 6= 22Number of people who reads exactly one newspaper= Total no. of people - No. of people who reads more than one newspaper= 52 - (a + b +c + d)= 52 - 22= 30

0 Upvotes 0 Downvotes

Similar Questions for you

The number of elements in the set {x ∈ R : (|x| - 3)|x + 4| = 6} is equal to :

Raj Pandey

(|x| - 3)|x + 4| = 6

Case (i) x < -4
(-x - 3) (- (x + 4) = 6
(x + 3) (x + 4) = 6 ⇒ x² + 7x + 12 = 6 ⇒ x² + 7x + 6 = 0
(x + 1) (x + 6) = 0 ⇒ x = -6 (since x < -4)

Case (ii) -4 ≤ x < 0
(-x - 3) (x + 4) = 6
⇒ -x² - 7x - 12 = 6
⇒ x² + 7x + 18 = 0
The discriminant is D = 7² - 4 (1) (18) = 49 - 72 < 0, so no real solution.

Case (iii) x ≥ 0
(x - 3) (x + 4) = 6
⇒ x² + x - 12 = 6
⇒ x² + x - 18 = 0
x = [-1 ± √ (1² - 4 (1) (-18)] / 2 = [-1 ± √73] / 2
Since x ≥ 0 ⇒ x = (√73 - 1) / 2
Only two solutions.

A natural number has prime factorization given by n = $2^{x} 3^{y} 5^{z},$ where y and z are such that y + z = 5 and $y^{- 1} + z^{- 1} = \frac{5}{6}, y > z .$ Then the number of odd divisors of n, including 1, is:

Vishal Baghel

Given n = 2^x. 3^y. 5^z . (i)

$y + z = 5 & \frac{1}{y} + \frac{1}{z} = \frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

Let x denote the total number of one-one function from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions form the set A to the set A × B. Then

Payal Gupta

Let A = {a, b, c}, B = {1, 2, 3, 4, 5} n (A × B) = 15

x = number of one-one functions from A to B.

$=^{5} C_{3} . 3! = 60$

y = number of one-one functions for A to (A × B)

$=^{15} C_{3} . 3! = 15 \times 14 \times 13 = 2730$

$\frac{Y}{x} = \frac{2730}{60} \Rightarrow 2 y = 91 x$

The number of elements in the set S = ${x \in R : 2 c o s (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} i s :$

Payal Gupta

2cos $(\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$

$- 2 \leq L H S \leq 2 L H S = 2 & R H S = 2 \Rightarrow x = 0 o n l y \to t h e n L H S = 2 a l s o$

RHS $\geq$ 2

66. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + ...

Payal Gupta

66. Given series is 1× 2× 3 + 2× 3 ×4 + 3× 4 ×5 + … to n term

a_n = (n^th term of A. P. 1, 2, 3, …) ´× (n^th terms of A. P. 2, 3, 4) ×

i e, a = 1, d = 2- 1 = 1i e, a = 2, d = 3- 2 = 1

(nth term of A. P. 3, 4, 5)

i e, a = 3, d = 3 -4 = 1.

= [1 + (n -1) 1] ×[2 + (n -1):1]× [3 + (n- 1) 1]

= (1 + n -1)×(2 + n -1)×(3 + n -1)

= n (n + 1)(n + 2)

= n(n² + 2n + n + 2)

=n³ + 2n² + 2n.

S_n = ∑n³ + 3 ∑n² + 2 ∑n

$= {[\frac{m (n + 1)}{2}]}^{2} + \frac{3 n (n + 1) (2 n + 1)}{6} + \frac{2 n (n + 1)}{2}$

= $\frac{n (n + 1)}{2} [\frac{n (n + 1)}{2} + \frac{3}{3} (2 n + 1) + 2]$

= $\frac{n (n + 1)}{2} [\frac{n^{2} + n}{2} + 2 n + 1 + 2]$

$= \frac{n (n + 1)}{2} [\frac{n^{2} + n + 2 \times 2 n + 2 \times 3}{2}]$

$= \frac{n (n + 1)}{2} [\frac{x^{2} + n + 4 x + 6}{2}]$

$= \frac{n (n + 1)}{2} \times \frac{[n^{2} + 5 n + 6]}{2}$

$= \frac{n (n + 1)}{4} [n^{2} + 2 n + 3 n + 6]$

$= \frac{n (n + 1)}{4} [n (n + 2) + 3 (n + 2)]$

$= \frac{n (n + 1) (n + 2) (n + 3)}{4}$

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question