Question

In a survey of 25 student, it was found that 15 had taken mathematics, 12 had taken physics and 11 had taken chemistry, 5 had taken mathematics and chemistry and 9 had taken mathematics and physics, 4 had taken physics and chemistry and 3 had taken all three subjects. Find the no. of student that had taken (1) only chemistry

Answer 1

Solution:-

Let M: Set of students who have taken Maths

P: Set of students who have taken Physics

C: Set of students who have taken Chemistry

Given,

Total students n(U) = 25

=> n (M) = 15 n(P) = 12, n(C) = 11

=> n(M ∩ C) = 5, n(P ∩ C) = four, n(M ∩ P) = 9

=> n(M ∩ P ∩ C) = 3

1. range of scholars taking solely Chemistry = n(C - (M ∪ P))

= n(C) - n(C ∩ (M ∪ P))

= n(C) - [n(C ∩ M) + n(C ∩ P) - n((C ∩ M) ∩ (C ∩ P)) ]

= n(C) - n(C ∩ M) - n(C ∩ P) + n(C ∩ M ∩ P)

= 11 - 5 - 4 + 3

= 14 - 9

= 5

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@GauravSaxena01

Answer 2

Answer:

Question:-

the length of a rectangle is 8m more than its breadth if its perimeter is 128m, find its length , breadth and Area

Answer:-

The length of Rectangle is 36 m

The breadth of rectangle is 28 m

The area of Given rectangle is 1008 m².

To find:-

Length and breadth of rectangle

Area of rectangle

Solution:-

Let the breadth be x

Length = 8 + x

Perimeter = 128 m

$\boxed{ \large{ \mathfrak{perimeter = 2(l + b)}}}$

According to question,

$\large{ \tt: \implies \: \: \: \: \: 2(8 + x + x) = 128}$

$\begin{gathered} \large{ \tt: \implies \: \: \: \: \: 8 + 2x = \frac{128}{2} } \\ \end{gathered}:$

$\large{ \tt: \implies \: \: \: \: \: 8 + 2x = 64}$

$\large{ \tt: \implies \: \: \: \: \: 2x = 64 - 8}$

$\large{ \tt: \implies \: \: \: \: \: 2x = 56}$

$\large{ \tt: \implies \: \: \: \: \: x = 28}$

The breadth of rectangle is 28 m

Length = 8 + x = 28 + 8 = 36 m

$\large{ \boxed{ \mathfrak{area = l \times b}}}$

$\large{ \tt: \implies \: \: \: \: \: area = 28\times 36}$

$\large{ \tt: \implies \: \: \: \: \: area = 1008 \: {m}^{2} }$

The area of Given rectangle is 1008 m².