Average - Study Mode

[#76] B was born when A was 4 years 7 months old and C was born when B was 3 years 4 months old. When C was 5 years 2 months old, then their average age was :
Correct Answer

(D) 8 years 11 months

Explanation

Solution: According to the question, A - B = 4y 7m . . . . . . . (i) B - C = 3y 4m . . . . . . . (ii) (+) xa0(+) xa0 (+) A - C = 7y 11m . . . . . . . (ii) Given : When,
C = 5 years 2 months ∴ A = 13 years 1 month B = 8 years 6 months ∴ Average of $$frac{A + B + C}{3}$$ $$ = frac{{{ ext{26 years 9 months}}}}{3}$$ = 8 years 11 months

[#77] The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is -
Correct Answer

(D) 174

Explanation

Solution: Max - Min = 172.....(i) Max + Min = 50 × 40 - 48 × 38 Max + Min = 2000 - 1824 Max + Min = 176.....(ii) From equation (i) and (ii) Max = 174

[#78] The average of some natural numbers is 15. If 30 is added to the first number and 5 is subtracted from the last number the average becomes 17.5 then the number of natural numbers is -
Correct Answer

(D) 10

Explanation

Solution: Let the number of natural numbers = n ∵ The average of some natural numbers = 15 ⇒ Sum of these natural number = 15 × n = 15n ∵ 30 is added and 5 is subtracted So, now addition of these number = 15n + 30 - 5 = 15n + 25 According to the question, ⇒ $$frac{15n + 25}{n}$$ = 17.5 ⇒ 15n + 25 = 17.5n ⇒ 2.5n = 25 ⇒ n = 10 Therefore, the numbers of natural numbers n = 10

[#79] A student finds the average of ten 2 digits numbers. While copying numbers, by mistake, he writes in number with its digits interchanged. As a result his answer is 1.8 less than the correct answer. The difference of digits of the number, in which he made mistake is ?
Correct Answer

(A) 2

Explanation

Solution: Let us consider by mistake he writes 10th number with its digits interchanged. $$ herefore frac{{10x + y - left( {10y + x}
ight)}}{{10}} = 18$$ (In this remaining nine numbers are same and they cancel out) ∴ 10x + y - 10y - x = 18 ⇒ 9x - 9y = 18 ⇒ x - y = 2

[#80] The average marks obtained by 22 candidate in an examination are 45. The average marks of the first 10 candidates is 55 and those of the last eleven is 40. The number of marks obtained by the eleventh candidate is ?
Correct Answer

(B) 0

Explanation

Solution: Given that average marks obtained by 22 students = 45 And average of first 10 students = 55 Average of last 11 students = 40 So using, average = $$frac{{{ ext{sum}}}}{{{ ext{number of elements}}}}$$ Sum of marks of all 22 students = 22 x 45 = 990 Sum of students of first 10 students = 55 x 10 = 550 Sum of last 11 students = 11 x 40 = 440 So marks obtained by 11 th student = 990 - (550 + 440) = 990 - 990 = 0