Average
- The average of 30 numbers is 40 and that of other 40 numbers is 30. The average of all the numbers is
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∴ Average of all numbers = 30 × 40 + 40 × 30 70 Average of all numbers = 240 7 Average of all numbers = 34 2 7
Second method to find the required average ,
Here, n1 = 30, a1 = 40
n2 = 40, a2 = 30∴ Average = n1a1 + n2a2 n1 + n2
Correct Option: A
∴ Average of all numbers = 30 × 40 + 40 × 30 70 Average of all numbers = 240 7 Average of all numbers = 34 2 7
Second method to find the required average ,
Here, n1 = 30, a1 = 40
n2 = 40, a2 = 30∴ Average = n1a1 + n2a2 n1 + n2 ∴ Average = 30 × 40 + 40 × 30 70 Average = 240 7 Average = 34 2 7
- 12 kg of rice costing 30 per kg is mixed with 8 kg of rice costing 40 per kg.The average per kg price of mixed rice is
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Total cost price of 20kg of mixed rice = (12 × 30 + 8 × 40) = ₹ 680
∴ Average per kg price = 680 = ₹ 34 20
Second method to find the required average ,
y1 = 12, A1 = 30
y2 = 8, A2 = 40∴ Average = A1y1 + A2y2 y1 + y2
Correct Option: D
Total cost price of 20kg of mixed rice = (12 × 30 + 8 × 40) = ₹ 680
∴ Average per kg price = 680 = ₹ 34 20
Second method to find the required average ,
y1 = 12, A1 = 30
y2 = 8, A2 = 40∴ Average = A1y1 + A2y2 y1 + y2 Average = 30 × 12 + 40 × 8 12 + 8 Average = 360 + 320 20 Average = 680 = ₹ 34 20
- The average of x numbers is y and average of y numbers is x. Then the average of all the numbers taken together is
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Sum of x numbers = xy
Sum of y numbers = xy∴ Required average = Sum of x numbers + Sum of y numbers = 2xy x + y x + y
Second method to find the required average ,
Here, n1 = x, a1 = y
n2 = y, a2 = x∴ Average = n1a1 + n2a2 n1 + n2
Correct Option: B
Sum of x numbers = xy
Sum of y numbers = xy∴ Required average = Sum of x numbers + Sum of y numbers = 2xy x + y x + y
Second method to find the required average ,
Here, n1 = x, a1 = y
n2 = y, a2 = x∴ Average = n1a1 + n2a2 n1 + n2 Average = xy + yx = 2xy x + y x + y
- A library has an average number of 510 visitors on Sunday and 240 on other days. The average number of visitors per day in a month of 30 days beginning with Sunday is :
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That month will have 5 sundays.
∴ Required average = 5 × 510 + 25 × 240 40 Required average = 2550 + 6000 30 Required average = 8550 = 285 30
Second method to find the required average ,
Here,
n1 = 5, a1 = 510
n2 = 25, a2 = 240∴ Average = n1a1 + n2a2 n1 + n2
Correct Option: A
That month will have 5 sundays.
∴ Required average = 5 × 510 + 25 × 240 40 Required average = 2550 + 6000 30 Required average = 8550 = 285 30
Second method to find the required average ,
Here,
n1 = 5, a1 = 510
n2 = 25, a2 = 240∴ Average = n1a1 + n2a2 n1 + n2 Average = 5 × 510 + 25 × 240 40 Average = 2550 + 6000 30 Average = 8550 = 285 30
- The average of three numbers is 40. The first number is twice the second and the second one is thrice the third number. The difference between the largest and the smallest numbers is
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Let the third number be p.
∴ Second number = 3p
First number = 6p∴ 6p + 3p + p = 40 3
⇒ 10p = 120 ⇒ p = 12
∴ Required difference = 6p – p = 5p = 5 × 12 = 60
Second method to solve this question with the help of given formulas :
Here, a = 2, b = 3, X = 40Largest Number = First Number = 3ab × X 1 + b + ab Largest Number = 3 × 2 × 2 × 40 1 + 3 + 2 × 2
Correct Option: D
Let the third number be p.
∴ Second number = 3p
First number = 6p∴ 6p + 3p + p = 40 3
⇒ 10p = 120 ⇒ p = 12
∴ Required difference = 6p – p = 5p = 5 × 12 = 60
Second method to solve this question with the help of given formulas :
Here, a = 2, b = 3, X = 40Largest Number = First Number = 3ab × X 1 + b + ab Largest Number = 3 × 2 × 2 × 40 1 + 3 + 2 × 2 Largest Number = 18 × 40 = 72 10 Smallest Number = Third Number = 3 × X 1 + b + ab Smallest Number = 3 × 40 1 + 3 + 2 × 3 Smallest Number = 3 × 40 = 12 10
Difference = 72 – 12 = 60
