Average - Study Mode
[#311] If average of 20 observations x 1 , x 2 , . . . . . x 20 is y, then the average of x 1 - 101, x 2 - 101, x 3 - 101, . . . . . x 20 - 101 is :
Correct Answer
(B) y - 101
Explanation
Solution: According to the question, $$eqalign{
& Rightarrow frac{{{x_1} + {x_2} + {x_3} + {x_4} + .... + {x_{20}}}}{{20}} = y cr
& Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + .... + {x_{20}} = 20y cr} $$ $$ = frac{{{x_1} - 101 + {x_2} - 101 + {x_3} - 101 + {x_4} - 101 + .... + {x_{20}} - 101}}{{20}}$$ $$ = frac{{left( {{x_1} + {x_2} + {x_3} + {x_4} + .... + {x_{20}}}
ight) - 20 imes 101}}{{20}}$$ $$eqalign{
& = frac{{20y - 20 imes 101}}{{20}} cr
& = y - 101 cr} $$
[#312] The average of 5 consecutive integers starting with 'm' is n.
What is the average of 6 consecutive integers starting with (m + 2)?
Correct Answer
(A) $$frac{(2n + 5)}{2}$$
Explanation
Solution: According to the question, Let M = 1 ∴ 5 consecutive integers are = 1, 2, 3, 4, 5 ∴ $$frac{1 + 2 + 3 + 4 + 5}{5}$$ xa0 = n n = $$frac{15}{5}$$ = 3 ∴ 6 consecutive integers starting with (m + 2) are = 3, 4, 5, 6, 7, 8 ∴ $$frac{3 + 4 + 5 + 6 + 7 + 8 }{6}$$ xa0 xa0= $$frac{33}{6}$$ xa0= $$frac{11}{2}$$ Now check from option to put n = 3 Option : (A) $$frac{(2n + 5)}{2}$$ = $$frac{2 × 3 + 5}{2}$$ xa0 = $$frac{11}{2}$$ (satisfied)
[#313] The average marks of 14 students was 71. It was later found that the marks of one of the student has been wrongly entered as 42 instead of 56 and another as 74 instead of 32. What is the correct average ?
Correct Answer
(D) 69
Explanation
Solution: According to the question, Wrong marks = 42 + 74 = 116 Correct marks = 56 + 32 = 88 Difference = 116 - 88 = 28 marks ∵ This difference effect the 14 students = $$frac{28}{14}$$ = 2 And, incorrect average = 71 ∴ Correct average = 71 - 2 = 69
[#314] In a 20 over match, the required run rate to win is 7.2. If the run rate is 6 at the end of the 15th over, the required run rate to win the match is :
Correct Answer
(C) 10.8
Explanation
Solution: According to the question, 20 over match required run rate = 7.2 Total runs are = 7.2 × 20 = 144 runs If the run rate is 6 at the end of the 15th over ∴ Required runs = 144 - 90 = 54 runs Required run rate = $$frac{54}{5}$$ = 10.8
[#315] The total number of students in section A and B of a class is 110. The number of students in section A is 10 more than that of section B. The average score of the students in B, in a test, is 20% more than that of students in A. If the average score of all the students in the class is 72, then what is the average score of the students in A?
Correct Answer
(A) 66
Explanation
Solution: A + B = 110 x + 10 + x = 110 x = 50 A = 60 B = 50 $$frac{{6{ ext{x}} - 72}}{{72 - 5{ ext{x}}}} = frac{6}{5}$$ (x - 12) × 5 = 72 - 5x 5x - 60 = 72 - 5x 10x = 132 A = 5x = 66