# Value Based Questions (VBQs) - Polynomials Class 9 Notes | EduRev

## Class 9 : Value Based Questions (VBQs) - Polynomials Class 9 Notes | EduRev

The document Value Based Questions (VBQs) - Polynomials Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by VP Classes.
All you need of Class 9 at this link: Class 9

Question 1. In a school function, students having 100% attendance are to be honoured. Class teacher of IX A gives the number of eligible students as whereas the class teacher of IX B gives the number of such students as If both the above numbers are equal then
(i) Find the number of prize-winners of 100% attandance in each section of class IX.
(ii) Which mathematical concept is used in the above problem?
(iii) By honouring 100% attendance holders, which value is depicted by the school administration?
Solution.
(i) Number of eligible students of class IX A

= Number of eligible students of class IX B Since, the number of eligible students in both sections are equal.  Thus, in each section number of eligible students = 8
(ii) Number system.
(iii) Regularity.

Question 2. Ravita donated to a blind school. Her friends wanted to know the amount donated by her. She did not disclose the amount but gave a hint that (i) Find the amount donated by Ravita to blind school.
(ii) Which mathematical concept is involved in this problem?
(iii) By donating an amount to blind school, which value is depicted by Ravita?

Solution. (i) ∵   Thus, the amount donated by Ravita is 322.
(ii) Polynomials.
(iii) Charity.

Question 3. A group of (a + b) teachers, (a+ b2) girls and (a3 + b3) boys set out for an ‘Adult Education Mission’. If in the group, there are 10 teachers and 58 girls then:
(i) Find the number of boys.
(ii) Which mathematical concept in used in the above problem? (iii) B
y working for ‘Adult Education’, which value is dopicted by the teachers and students?
Solution.
(i) ∵ (a + b)2 = a2 + b+ 2ab
∴ 102 = 58 + 2ab                   [∵ Number of teachers = (a + b) = 10 and Number of girls = (a2 + b2) = 58]
⇒ 100 = 58 + 2ab
⇒ 2ab = 100 – 58 = 42
⇒ ab = 42/2
= 21
Now, (a + b)3 = a3 + b3 + 3ab (a + b) ∴
(10)= a3 + b3 + 3 * 21 * 10
⇒ 1000 = a+ b+ 630 a+ b3 = 1000 – 630 = 370
∴ Number of boys = 370.
(ii) Polynomials.
(iii) Social upliftment.

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