Average
- Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, then difference of first and third number is
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Let the numbers be 2k, k and 4k respectively
∴ Average = 2k + k + 4k 3 ⇒ 7k = 56 3 ⇒ k = 3 × 56 = 24 7
∴ First number = 2k = 2 × 24 = 48
Third number = 4k = 4 × 24 = 96
∴ Required difference = Third number - First number = 96 – 48 = 48
Second method to solve this question with the help of given formulas :
Here, a = 2, b = 1/4 , X = 56First number = 3ab × X 1 + b + ab First number = 3 × 2 × 1 × 56 4 1 + 1 + 2 × 1 4 4
Correct Option: D
Let the numbers be 2k, k and 4k respectively
∴ Average = 2k + k + 4k 3 ⇒ 7k = 56 3 ⇒ k = 3 × 56 = 24 7
∴ First number = 2k = 2 × 24 = 48
Third number = 4k = 4 × 24 = 96
∴ Required difference = Third number - First number = 96 – 48 = 48
Second method to solve this question with the help of given formulas :
Here, a = 2, b = 1/4 , X = 56First number = 3ab × X 1 + b + ab First number = 3 × 2 × 1 × 56 4 1 + 1 + 2 × 1 4 4 First number = 3 × 4 × 56 = 48 2 4 + 1 + 2 Third number = 3 × X 1 + b + ab Third number = 3 × 56 1 + 1 + 2 × 1 4 4 Third number = 3 × 4 × 56 = 96 4 + 4 + 2
Required difference = 96–48 = 48
- The average of three numbers is 77. The first number is twice the second and the second number is twice the third. The first number is :
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Let the third number = p
∴ Second number = 2p
First number = 4p
The average of three numbers = 77
Now, p + 2p + 4p = 3 × 77
⇒ 7p = 3 × 77⇒ p = 3 × 77 = 33 7
∴ First number = 33 × 4 = 132
Second method to solve this question with the help of given formula :
Here, a = 2, b = 2, X = 77First number = 3ab × X 1 + b + ab
Correct Option: D
Let the third number = p
∴ Second number = 2p
First number = 4p
The average of three numbers = 77
Now, p + 2p + 4p = 3 × 77
⇒ 7p = 3 × 77⇒ p = 3 × 77 = 33 7
∴ First number = 33 × 4 = 132
Second method to solve this question with the help of given formula :
Here, a = 2, b = 2, X = 77First number = 3ab × X 1 + b + ab First number = 3 × 2 × 2 × 77 1 + 2 + 2 × 2 First number = 12 × 77 7
First number = 12 × 11 = 132
- Of the three numbers, first is twice the second and second is twice the third. The average of three numbers is 21. The smallest of the three numbers is
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Let the third number be n.
∴ The second number = 2n
and the third number = 2 × 2n = 4nAccording to the question, 4n + 2n + n = 21 3
⇒ 7n = 21 × 3⇒ n = 21 × 3 = 9 7
Second method to solve this question :
a = 2, b = 2, y = 21First number = 3ab y 1 + b + ab First number = 3 × 2 × 2 × 21 1 + 2 + 4
Correct Option: A
Let the third number be n.
∴ The second number = 2n
and the third number = 2 × 2n = 4nAccording to the question, 4n + 2n + n = 21 3
⇒ 7n = 21 × 3⇒ n = 21 × 3 = 9 7
Second method to solve this question :
a = 2, b = 2, y = 21First number = 3ab y 1 + b + ab First number = 3 × 2 × 2 × 21 1 + 2 + 4 First number = 12 × 21 = 36 7 Second number = 3b y 1 + a + ab Second number = 3 × 2 × 21 = 18 7 Third number = 3 y 1 + a + ab Third number = 3 × 21 = 9 7
- Of the three numbers, second is twice the first and also thrice the third. If the average of the three numbers is 44, the largest number is :
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Let third number be p.
∴ Second number = 3pand first number = 3 p 2 Now, p + 3p + 3p = 3 × 44 2 ⇒ 8p + 3p = 3 × 44 2
⇒ 11p = 6 × 44
Correct Option: B
Let third number be p.
∴ Second number = 3pand first number = 3 p 2 Now, p + 3p + 3p = 3 × 44 2 ⇒ 8p + 3p = 3 × 44 2
⇒ 11p = 6 × 44⇒ p = 6 × 44 = 24 11
The largest number = 3p = 3 × 24 = 72
- A man spends in 8 months as much as he earns in 6 months. He saves Rs. 6000 in a year. His average monthly income is :
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Let the average monthly income of man be Rs. p.
∴ Man’s annual income = Rs. 12p∴ Man’s annual expenses = Rs. 
6p × 12 
8
Man’s annual expenses = Rs. 9p
∴ Savings = 12p – 9p = Rs. 3p
Correct Option: B
Let the average monthly income of man be Rs. p.
∴ Man’s annual income = Rs. 12p∴ Man’s annual expenses = Rs. 6p × 12 8
Man’s annual expenses = Rs. 9p
∴ Savings = 12p – 9p = Rs. 3p
∴ 3p = 6000⇒ p = 6000 3
∴ p = Rs. 2000
