Question

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Answer 1

Step-by-step explanation:

Given:In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course.

To find: The number of students who did not opt three courses is for any of the

(1) 102

(2) 42

(3) 1

(4) 38

Solution:

Step 1: No. of students opted Mathematics are

2,4,6..., 140

it form an A.P.

here, first term a=2

d=2

n=?

last term: 140

$a_n = a + (n - 1)d \\ \\ 140 = 2 + (n - 1)2 \\ \\ 140 - 2 = 2(n - 1) \\ \\ \frac{138}{2} = n - 1 \\ \\ n - 1 = 69 \\ \\ n = 70$

there are total 70 students who opted Mathematics.

Step 2: No. of students who opted physics

3,6,...,138

$138 = 3 + 3(n - 1) \\ \\ n = 46$

Step 3: No. of students opted chemistry

5,10,..., 140

$140 = 5 + 5(n - 1) \\ \\ n = 28$

Step4: No. of student who had not opted any subject is 1.

*Step 5: Remember that 6,12,18... 138 are even as well as divisible by 3,thus there are some students who opted both maths and physics,as well as 10,20,30... 140 are those numbers which are even and divisible by 5, these are those who opted maths and chemistry.

There are numbers which are divisible by 3 as well as 5,for eg: 15,30,45...

these are the students who opt physics as well as chemistry.

Final answer:

The number of students who did not opt three courses is 1.

Option (3) is correct.

Remark: For a shortcut, we know that 1 is neither even nor divisible by 3 and 5.

So, 1 student does not opt any of three stream.

Hope it helps you.

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Answer 2

Answer:

Step-by-step explanation:

Given:In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course.

To find: The number of students who did not opt three courses is for any of the

(1) 102

(2) 42

(3) 1

(4) 38

Solution:

Step 1: No. of students opted Mathematics are

2,4,6..., 140

it form an A.P.

here, first term a=2

d=2

n=?

last term: 140

\begin{lgathered}a_n = a + (n - 1)d \\ \\ 140 = 2 + (n - 1)2 \\ \\ 140 - 2 = 2(n - 1) \\ \\ \frac{138}{2} = n - 1 \\ \\ n - 1 = 69 \\ \\ n = 70 \\\end{lgathered}

a

n

=a+(n−1)d

140=2+(n−1)2

140−2=2(n−1)

2

138

=n−1

n−1=69

n=70

there are total 70 students who opted Mathematics.

Step 2: No. of students who opted physics

3,6,...,138

\begin{lgathered}138 = 3 + 3(n - 1) \\ \\ n = 46 \\\end{lgathered}

138=3+3(n−1)

n=46

Step 3: No. of students opted chemistry

5,10,..., 140

\begin{lgathered}140 = 5 + 5(n - 1) \\ \\ n = 28 \\\end{lgathered}

140=5+5(n−1)

n=28

Step4: No. of student who had not opted any subject is 1.

*Step 5: Remember that 6,12,18... 138 are even as well as divisible by 3,thus there are some students who opted both maths and physics,as well as 10,20,30... 140 are those numbers which are even and divisible by 5, these are those who opted maths and chemistry.

There are numbers which are divisible by 3 as well as 5,for eg: 15,30,45...

these are the students who opt physics as well as chemistry.

Final answer:

The number of students who did not opt three courses is 1.

Option (3) is correct.

Remark: For a shortcut, we know that 1 is neither even nor divisible by 3 and 5.

So, 1 student does not opt any of three stream.

Hope it helps you.

To learn more on brainly:

Answer Fast to be brainliest.

https://brainly.in/question/45608378