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41. The average of five different positive numbers is 25. x is the decrease in the average when the smallest number among them is replaced by 0. What can be said about x?
We are given that the average of the five numbers is 25. Hence, we have the equation
a + b + c + d + e = 125 ----------- (1) by multiplying by 5.
The smallest number in a set is at least less than the average of the numbers in the set if at least one number is different.
For example, the average of 1, 2, and 3 is 2, and the smallest number in the set 1 is less than theaverage 2. Hence, we have the inequality
0 < e < 25
0 > -e > -25 by multiplying both sides of the inequality by -1 and flipping the directions of the inequalities.Adding this inequality to equation (1) yields
0 + 125 > (a + b + c + d + e) + (-e) > 125 - 25
125 > (a + b + c + d) > 100
125 > (a + b + c + d + 0) > 100 by adding by 0
25 > ⇒ 20 by dividing the inequality by 5
25 > The average of numbers a, b, c, d and 0 > 20
Hence, x equals
(Average of the numbers a, b, c, d and e) – (Average of the numbers a, b, c, and d)
= 25 − (A number between 20 and 25)
⇒ A number less than 5
Hence, x is less than 5
Solution:
Let a, b, c, d, and e be the five positive numbers in the decreasing order of size such that e is the smallest number.We are given that the average of the five numbers is 25. Hence, we have the equation
a + b + c + d + e = 125 ----------- (1) by multiplying by 5.
The smallest number in a set is at least less than the average of the numbers in the set if at least one number is different.
For example, the average of 1, 2, and 3 is 2, and the smallest number in the set 1 is less than theaverage 2. Hence, we have the inequality
0 < e < 25
0 > -e > -25 by multiplying both sides of the inequality by -1 and flipping the directions of the inequalities.Adding this inequality to equation (1) yields
0 + 125 > (a + b + c + d + e) + (-e) > 125 - 25
125 > (a + b + c + d) > 100
125 > (a + b + c + d + 0) > 100 by adding by 0
25 > ⇒ 20 by dividing the inequality by 5
25 > The average of numbers a, b, c, d and 0 > 20
Hence, x equals
(Average of the numbers a, b, c, d and e) – (Average of the numbers a, b, c, and d)
= 25 − (A number between 20 and 25)
⇒ A number less than 5
Hence, x is less than 5
42. In 2011, the arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800. The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800, and the arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800. What is the arithmetic mean of the incomes of the three?
Now, we are given that
The arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800.
Hence,
= 3800
⇒ a + b = 2 × 3800 = 7600
The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800.
Hence,
= 4800
⇒ b + c = 2 × 4800 = 9600
The arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800.
Hence,
= 5800
⇒ c + a = 2 × 5800 = 11,600
Adding these three equations yields:
(a + b) + (b + c) + (c + a) = 7600 + 9600 + 11,600
2a + 2b + 2c = 28,800
a + b + c = 14,400
The average of the incomes of the three equals the sum of the incomes divided by 3,
Solution:
Let a, b, and c be the annual incomes of Ramesh, Suresh, and Pratap, respectively.Now, we are given that
The arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800.
Hence,
= 3800
⇒ a + b = 2 × 3800 = 7600
The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800.
Hence,
= 4800
⇒ b + c = 2 × 4800 = 9600
The arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800.
Hence,
= 5800
⇒ c + a = 2 × 5800 = 11,600
Adding these three equations yields:
(a + b) + (b + c) + (c + a) = 7600 + 9600 + 11,600
2a + 2b + 2c = 28,800
a + b + c = 14,400
The average of the incomes of the three equals the sum of the incomes divided by 3,
43. In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
According to Arun,
65 < X < 72
According to Arun's brother,
60 < X < 70
According to Arun's mother,
X 68
The values satisfying all the above conditions are 66, 67 and 68.
Required average,
Solution:
Let Arun's weight by X kg.According to Arun,
65 < X < 72
According to Arun's brother,
60 < X < 70
According to Arun's mother,
X 68
The values satisfying all the above conditions are 66, 67 and 68.
Required average,
44. A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits a and b?
After interchanging the digits, the new number becomes (10b + a)
The question states that the average of 10 numbers has become 1.8 less than the original average. Therefore, the sum of the original 10 numbers will be (10 × 1.8 = 18 more than the sum of the 10 numbers with the digits interchanged.
10a + b = 10b + a + 18
9a - 9b = 18
a - b = 2
Solution:
Let the original number be (10a + b)After interchanging the digits, the new number becomes (10b + a)
The question states that the average of 10 numbers has become 1.8 less than the original average. Therefore, the sum of the original 10 numbers will be (10 × 1.8 = 18 more than the sum of the 10 numbers with the digits interchanged.
10a + b = 10b + a + 18
9a - 9b = 18
a - b = 2
45. The average of first five multiples of 3 is:
3, 6, 9, 12, 15.
Alternatively,
Average =
Solution:
First five multiples of three are:3, 6, 9, 12, 15.
Alternatively,
Average =
46. There are two sections A and B of a class, consisting of 36 and 44 students' respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class.
= (36 × 40 + 44 × 35) kg = 2980 kg.
Therefore average weight of the whole class = kg
Therefore average weight = 37.25 kg
Solution:
The total weight of (36 + 44) Students of A and B,= (36 × 40 + 44 × 35) kg = 2980 kg.
Therefore average weight of the whole class = kg
Therefore average weight = 37.25 kg
47. Distance between two stations A and B is 778km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km per hour. Find the average speed of train during the whole journey.
Solution:
The required average speed given by the formula,
48. The average of 50 numbers is 30. If two numbers, 35 and 40 are discarded, then the average of the remaining numbers is nearly:
Solution:
Total sum of 48 numbers,
= (50 × 30) – (35 +40)
= 1500 – 75
= 1425
Average = = 29.68
49. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, then find the average for the last four matches.
= (10 × 38.9) - (6 × 42)
= 389 - 252 = 137
Average = = 34.25
Solution:
Total sum of last 4 matches,= (10 × 38.9) - (6 × 42)
= 389 - 252 = 137
Average = = 34.25
50. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning.
Then average after 16th innings = (x - 3)
Therefore 16 × (x - 3) + 87 = 17x
Therefore x = 39
Solution:
Let the average after 17th innings = xThen average after 16th innings = (x - 3)
Therefore 16 × (x - 3) + 87 = 17x
Therefore x = 39