Maths

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

0 Follower 17 Views

Contributor-Level 10

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 2

...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

0 0

New answer posted

6 months ago

22. $t a n^{- 1} (\frac{c o s x - s i n x}{c o s x + s i n x}), 0 \frac{- π}{4} < x < \frac{3 π}{4}$

0 Follower 2 Views

Contributor-Level 10

Given tan ^-1 $\frac{c o s x - s i n x}{c o s x + s i n x}$ , $\frac{- π}{4},$ x< $\frac{3 x}{4}$

(M) Dividing numerator & denominator cos x we get,

tan ^-1 $\frac{\frac{c o s x - s i n x}{c o s x}}{\frac{c o s x + s i n x}{c o s x}} = t a n^{- 1} \frac{\frac{c o s x}{c o s x} - \frac{s i n x}{c o s x}}{\frac{c o s x}{c o s x} + \frac{s i n x}{c o s x}} = t a n^{- 1} \frac{1 - t a n x}{1 + t a n x .}$

We know that $t a n \frac{π}{4} = 1$ = 1 so,

$= t a n^{- 1} \frac{t a n \frac{π}{4} - t a n x}{1 + t a n \frac{π}{4} t a n x} = t a n^{- 1} [t a n (\frac{π}{4} - x)]$ ${? t a n (x - y) = \frac{t a n x - t a n y}{1 + t a n x \cdot t a n y}}$

$= \frac{π}{4} - x$ .

0 0

New answer posted

6 months ago

62. In a group of students, 100 students know Hindi, 50 know English and 25 knowboth. Each of the students knows either Hindi or English. How many studentsare there in the group?

0 Follower 3 Views

Contributor-Level 10

62. Let H and E be set of students who known Hindi and English respectively.

Then, number of students who know Hindi = n (H) = 100

Number of students who know English = n (E) = 50

Number of students who know both English & Hindi = 25 = n (H $\cap$ E)

As each of students knows either Hindi or English,

Total number of students in the group,

n (H $\cup$ E) = n (H) + n (E) - n (H $\cap$ E)

= 100 + 25

= 125,

0 0

New answer posted

6 months ago

18. $2 t a n^{- 1} \frac{1}{2} + t a n^{- 1} \frac{1}{7} = t a n^{- 1} \frac{31}{17}$

0 Follower 2 Views

Contributor-Level 10

L.H.S= 2 tan ^-1 $\frac{1}{2}$ + tan ^-1 $\frac{1}{7}$

(E) Using2 tan ^-1x= tan ^-1 $\frac{2 a}{1 - x^{2}}$ we can write.

L.H.S = tan ^-1 $\frac{2 * \frac{1}{2}}{1 - {(\frac{1}{2})}^{2}}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{1}{1 \frac{- 1}{4}}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{1}{4 \frac{- 1}{4}}$ + tan ^-1 $\frac{1}{7}$ = tan ^-1 $\frac{4}{3}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} * \frac{1}{7}}$ { Ø tan -1 x + tan -1 y = tan - 1 $\frac{x + y}{1 - x y}$ }

= tan ^-1 $\frac{\frac{7 * 4 + 1 * 3}{3 * 7}}{\frac{3 * 7 - 4 * 1}{3 * 7}}$

= tan -1 $\frac{28 + 3}{21 - 4}$ = tan ^-1 $\frac{31}{17}$ = R H S

0 0

New answer posted

6 months ago

61. In a survey of 600 students in a school, 150 students were found to be taking teaand 225 taking coffee, 100 were taking both tea and coffee. Find how manystudents were taking neither tea nor coffee?

0 Follower 5 Views

Contributor-Level 10

61. Let T and C be sets of students taking tea and coffee.

Then, n (T) = 150, number of students taking tea

n (C) = 225, number of students taking coffee

n (T $\cap$ C) = 100, number of students taking both tea and coffee.

So, Number of students taking either tea or coffee is.

n (T $\cup$ C) = n (T) + n (C) n (T $\cap$ C)

= 150 + 225 100

= 275

Number of students taking neither tea coffee

= Total number of students No of students taking either tea or coffee

= 600 275

= 325.

0 0

New answer posted

6 months ago

17. $t a n^{- 1} \frac{2}{11} + t a n^{- 1} \frac{7}{24} = t a n^{- 1} \frac{1}{2}$

0 Follower 2 Views

Contributor-Level 10

L.H.S = tan ^-1 $\frac{2}{11}$ + tan ^-1 $\frac{7}{24}$

Using tan ^-1x+ tan ^-1y= tan ^-1 $\frac{x + y}{1 - x y}$ , xy<1

L.H.S =tan ^-1 $\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} * \frac{7}{24}}$ = tan ^-1 $\frac{\frac{2 * 24 + 7 * 2}{11 * 24}}{\frac{11 * 24 - 7 * 2}{11 * 24}}$

tan ^-1 $\frac{48 + 14}{264 - 14}$ = tan ^-1 $\frac{125}{250}$ = tan ^-1 $\frac{1}{2}$ = R.H.S

0 0

New answer posted

6 months ago

60. Find sets A, B and C such that A $\cap$ B, B $\cap$ C and A $\cap$ C are non-empty sets and A $\cap$ B $\cap$ C = $?$

0 Follower 2 Views

Contributor-Level 10

60. Let A = {x, y}

B = {y, z}

C = {x, z}

So, A $\cap$ B = {x, y} $\cap$ {y, z} = {y}≠ $?$

B $\cap$ C = {y, z} $\cap$ {x, z} = {z}≠ $?$

A $\cap$ C = {x, y} $\cap$ {x, z} = {x}≠ $?$

But A $\cap$ B $\cap$ C = (A $\cap$ B) $\cap$ C

= {y} $\cap$ (x, z}

= $?$

0 0

New answer posted

6 months ago

16. $3 c o s^{- 1} x = c o s^{- 1} (4 x^{3} - 3 x), x \in [\frac{1}{2}, 1]$

0 Follower 2 Views

Contributor-Level 10

We know that,

cos3θ= 4cos³θ - 3cosθ

Letx = cosθ Then θ = cos^-1x. We have,

Cos3 (cos^-1x) = 4x³-3x

3cos^-1x = cos^-1 (4x³- 3x)

Hence Proved

0 0

New answer posted

6 months ago

59. Let A and B be sets. If A $\cap$ X = B $\cap$ X = $?$ and A $\cup$ X = B $\cup$ X for some set X, show that A = B. (Hints A = A $\cap$ (A $\cup$ X) , B = B $\cap$ (B $\cup$ X ) and use Distributive law )

0 Follower 3 Views

Contributor-Level 10

59. Let A, B and x be sets such that,

A $\cap$ x = B $\cap$ x = $?$ and A $\cup$ x = B $\cup$ x.

We know that,

A = A $\cap$ (A $\cup$ x)

= A $\cap$ (B $\cup$ x) [? A $\cup$ x = $\cup$ Bx]

= (A $\cap$ B) $\cap$ (A $\cap$ x) [by distributive law]

= (A $\cap$ B) ∪? [? A∩x =? ]

=> A = A∩ B [? A ∪? = A]

And B = $\cap$ (B $\cup$ x)

= B $\cap$ (A $\cup$ x) [? B $\cup$ x = A $\cup$ x]

= (B $\cap$ A) $\cup$ (B $\cap$ x) [By distributive law]

= (B $\cap$ A) ∪ $?$ [? B $\cap$ x =? ]

B = B $\cap$ A [? A $\cup$ $?$ = A]

So, A = B = A $\cap$ B.

0 0

New answer posted

6 months ago

15. $t a n^{- 1} (\frac{x}{y}) - t a n^{- 1} \frac{x - y}{x = y} i s e q u a l t o$

$(A) \frac{π}{2} (B) \frac{π}{3} (C) \frac{π}{4} (D) \frac{3 π}{4}$

0 Follower 4 Views