Discussion

Question:

In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is

Explanation:

Let the average marks of a boy and girl be b and g respectively.

Given, (4g + 6b)/10 = 24
⇒ 4g + 6b = 240
⇒ 2g + 3b = 120 ...(1)

Also, b ≤ g ≤ 2b
⇒ 2b ≤ 2g ≤ 4b
⇒ 5b ≤ 2g + 3b ≤ 7b ...(2)

From (1) & (2), we get
5b ≤ 120 ≤ 7b
⇒ b ≤ 24 and b ≤ 17(1/7) ...(3)

Now we need to find integral values of 2g + 6b
= 2g + 3b + 3b
= 120 + 3b
120 + 3 × 120/7 ≤ 120 + 3b ≤ 120 + 3 × 24 ...from (3)
⇒ 171.42 ≤ 120 + 3b ≤ 192

∴ Integral possible values of 2g + 6b are from 172 till 192 i.e., 21 possible integral values.

Hence, option (c).

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Doubts

Ashiti said (2024-06-11 05:43:43)

I didn't understand the condition if the marks of any girl is most double the marks of any boy ....

Reply from Admin:

All the boys have same marks, let's say b.

All the girls have same marks, let's say g.

If the marks of any girl is at most double the marks of any boy. It means g ≤ 2b

but not less than the marks of any boy, It means b ≤ g

Combining both we get, b ≤ g ≤ 2b.

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