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# A student finds the average of 10 positive integers. Each integer cont

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A student finds the average of 10 positive integers. Each integer cont  [#permalink]

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21 Feb 2019, 00:30
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Difficulty:

45% (medium)

Question Stats:

69% (02:14) correct 31% (02:19) wrong based on 75 sessions

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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?

A. 8
B. 6
C. 4
D. 2
E. 1

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Re: A student finds the average of 10 positive integers. Each integer cont  [#permalink]

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21 Feb 2019, 10:11
2
1
Bunuel wrote:
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?

A. 8
B. 6
C. 4
D. 2
E. 1

given
100a+10b=10x
or say
10a+b=x

now
error done makes
it
90a+9b+10b+a=(x-1.8)*10
or say
91a+19b= 10x-18
so
91a+19b=100a+10b-18
or say
9a-9b=18
a-b=2
IMO D
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A student finds the average of 10 positive integers. Each integer cont  [#permalink]

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22 May 2019, 20:56
Bunuel wrote:
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?

A. 8
B. 6
C. 4
D. 2
E. 1

I think my method is correct. Please correct me if I'm wrong.

$$\frac{Sum of 10 Positive Integers}{10} = X$$ <------ Average #1

Sum of 10 Positive Integers = $$X(10)$$ <------ Sum #1

$$\frac{Sum of 10 Positive Integers}{10} = (X - 1.8)$$ <------ Average #2 with reversed digits.

Sum of 10 Positive Integers = $$10(X - 1.8)$$ <------ Sum #2
Sum of 10 Positive Integers = $$10X - 18$$

Sum differences:
= [Sum #2] - [Sum #1]
= (10X - 18) - 10X
= 10X - 18- 10X
= -18

Now start taking the lowest double digit numbers and subtracting from their reversed counterparts.
10: 10 - 01 = 9 <---- Nope
11: 11 - 11 = 0 <---- Nope
12: 12 - 21 = -9 <---- Nope
13: 13 - 31 = -18 <---- YASS!!!
The difference between digits is (3 - 1) = 2
24: 24 - 42 = -18 <---- YASS!!!
The difference between digits is (4 - 2) = 2
35: 35 - 53 = -18 <---- YASS!!!
The difference between digits is (5 - 3) = 2
46: 46 - 64 = -18 <---- YASS!!!
The difference between digits is (6 - 4) = 2

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Re: A student finds the average of 10 positive integers. Each integer cont  [#permalink]

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09 Jul 2019, 00:06
1
Bunuel wrote:
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?

A. 8
B. 6
C. 4
D. 2
E. 1

The correct number ab can be written as 10a+b

The wrongly taken number be ba which can be written as 10b+a

the average of 10 numbers is brought down by 1.8

so the difference of these two numbers (10a +b) - (10b+a) = 18

9a - 9b = 18

a - b = 2
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Re: A student finds the average of 10 positive integers. Each integer cont  [#permalink]

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10 Jul 2019, 17:28
Bunuel wrote:
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?

A. 8
B. 6
C. 4
D. 2
E. 1

Just as the number 36 can be expressed as 3 x 10 + 6, we can express the two-digit number ab as 10a + b. Similarly, the number ba can be expressed as 10b + a.

With the above in mind, we can let the sum of the other 9 (correct) numbers be x and create the equation:

(x + 10a + b)/10 = (x + 10b + a)/10 - 1.8

Multiplying both sides by 10, we have:

x + 10a + b = x + 10b + a - 18

10a + b = 10b + a - 18

9a - 9b = -18

9(a - b) = -18

a - b = -2

b - a = 2

So the difference between the two digits is 2.

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Re: A student finds the average of 10 positive integers. Each integer cont   [#permalink] 10 Jul 2019, 17:28
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