Question
The average score in Mathematics of 90 students of
section A and B of class IX was 63. The number of students in A were 10 more than those in B. The average score of students in A was 30% more than that of students in B. The average score of students in B is:Solution
Let the students in section B be ‘x’ Students in section A = x + 10 ⇒ (Sum of the score of students)/number of the students = 63 ⇒ Sum of the score of students = 90 × 63 ⇒ Sum of the score of students = 5670 Number of the students = 90 ⇒ x + x + 10 = 90 ⇒ x = 40 Students in Section B = 40 Students in section A = 40 + 10 = 50 Let the average be ‘p’. Average of students in section A = 130% of students in section B ⇒ (Average of students in section A/Average of students in section B) = 13p/10p ⇒ 13p × 50 + 10p × 40 = 5670 ⇒ 650p + 400p = 5670 ⇒ 1050p = 5670 ⇒ p = 5.4 Average of students in section B = 10p ⇒ 5.4 × 10 ⇒ 54
1219.98 ÷ 30.48 × 15.12 = ? × 2.16
What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)...
`1804/898-:99/699xx749/751=?`
(√845 ×19.932+ √4230 ×14.385)/(√1765 ×4.877 ) = ?
Direction: Please solve the following expression and choose the closest option
`(13.022)^(2)+ (42.93)^(2)-(53.125)^(2)+(192.33xx14.88)=?- (88.44)^(2)- (42.03 xx 23.12)`
√49 + 6.66% of 1725 + 22² = ?² - √361Â
What approximate value will replace the question mark (?) in the following?
? =...
? = 49.97% of 38.09% of 1998.95
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)