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Ajay attempted to add ten two-digit numbers. One of -them, a, was the reverse of one of the others. If a was replaced by another two-digit number, b and the reverse of a was replaced by the reverse of b and the average was found, it would be 2.2 more. The sum of the digits in b exceeds the sum of the digits in a by?
Most Upvoted Answer
Ajay attempted to add ten two-digit numbers. One of -them, a, was the ...
Solution:
Let's begin by assuming that the ten two-digit numbers are: a, b, c, d, e, f, g, h, i, j.
1. Identifying the Reverse Pairs
Given that one of the numbers is the reverse of another, we can assume that a is the reverse of b (or vice versa). Therefore, we can write the following pairs of numbers:
- a, b
- c, d
- e, f
- g, h
- i, j
2. Finding the Value of a
Since a is the reverse of b, we know that a = 10x + y and b = 10y + x, where x and y are digits. Therefore, we can write:
a + b + c + d + e + f + g + h + i + j = (10x + y) + (10y + x) + c + d + e + f + g + h + i + j
Simplifying the above equation, we get:
2a + 2b + c + d + e + f + g + h + i + j = 110x + 110y
a + b + c + d + e + f + g + h + i + j = 55x + 55y
Since a is a two-digit number, x and y must be distinct and non-zero digits. The only possible values of x and y in this case are 1 and 2 (or 2 and 1). Therefore, a = 12 (or 21) and b = 21 (or 12).
3. Finding the Value of b
Let's assume that a = 12 and b = 21. We are given that if a is replaced by another two-digit number, b and the reverse of a is replaced by the reverse of b, then the average of the ten numbers is increased by 2.2. Therefore, we can write:
(a - 12) + (b - 21) + c + d + e + f + g + h + i + j = 2.2
Simplifying the above equation, we get:
a + b + c + d + e + f + g + h + i + j = 123
Let's replace a with b and the reverse of a with the reverse of b:
b + a + c + d + e + f + g + h + i + j = 123
12 + 21 + c + d + e + f + g + h + i + j = 123
c + d + e + f + g + h + i + j = 90
The sum of the digits in b is equal to 2 + 1 = 3, and the sum of the digits in a is equal to 1 + 2 = 3. Therefore, the sum of the digits in b exceeds the sum of the digits in a by 0.
4. Verifying the Solution
We can verify our solution by checking the sum of the ten two-digit numbers:
a + b + c + d + e + f + g + h + i + j = 12 + 21 + c + d + e + f + g + h + i + j
Since c, d, e, f, g, h, i,
Let's begin by assuming that the ten two-digit numbers are: a, b, c, d, e, f, g, h, i, j.
1. Identifying the Reverse Pairs
Given that one of the numbers is the reverse of another, we can assume that a is the reverse of b (or vice versa). Therefore, we can write the following pairs of numbers:
- a, b
- c, d
- e, f
- g, h
- i, j
2. Finding the Value of a
Since a is the reverse of b, we know that a = 10x + y and b = 10y + x, where x and y are digits. Therefore, we can write:
a + b + c + d + e + f + g + h + i + j = (10x + y) + (10y + x) + c + d + e + f + g + h + i + j
Simplifying the above equation, we get:
2a + 2b + c + d + e + f + g + h + i + j = 110x + 110y
a + b + c + d + e + f + g + h + i + j = 55x + 55y
Since a is a two-digit number, x and y must be distinct and non-zero digits. The only possible values of x and y in this case are 1 and 2 (or 2 and 1). Therefore, a = 12 (or 21) and b = 21 (or 12).
3. Finding the Value of b
Let's assume that a = 12 and b = 21. We are given that if a is replaced by another two-digit number, b and the reverse of a is replaced by the reverse of b, then the average of the ten numbers is increased by 2.2. Therefore, we can write:
(a - 12) + (b - 21) + c + d + e + f + g + h + i + j = 2.2
Simplifying the above equation, we get:
a + b + c + d + e + f + g + h + i + j = 123
Let's replace a with b and the reverse of a with the reverse of b:
b + a + c + d + e + f + g + h + i + j = 123
12 + 21 + c + d + e + f + g + h + i + j = 123
c + d + e + f + g + h + i + j = 90
The sum of the digits in b is equal to 2 + 1 = 3, and the sum of the digits in a is equal to 1 + 2 = 3. Therefore, the sum of the digits in b exceeds the sum of the digits in a by 0.
4. Verifying the Solution
We can verify our solution by checking the sum of the ten two-digit numbers:
a + b + c + d + e + f + g + h + i + j = 12 + 21 + c + d + e + f + g + h + i + j
Since c, d, e, f, g, h, i,
Community Answer
Ajay attempted to add ten two-digit numbers. One of -them, a, was the ...
Let avg of initial 10 digital be P
Avg of 10 digits after replacement be Q
let initial number and it's reverse be A= 10x+y and 10y+x
Replaced Number and it's reverse be B= 10a+b and 10b+a
The difference of their avg is 2.2
so the difference in total sum is 2.2*10 i.e 22
considering all other numbers to be unchanged.
the difference Q-P = 22
therefore, 10a+b + 10b+a - (10x+y + 10y+x) = 22
or, 11(a+b) - 11(x+y) = 22
or, (a+b) - (x+y) = 2 , i.e. the sum of the digits in B that exceeds the sum of digits in A.
Avg of 10 digits after replacement be Q
let initial number and it's reverse be A= 10x+y and 10y+x
Replaced Number and it's reverse be B= 10a+b and 10b+a
The difference of their avg is 2.2
so the difference in total sum is 2.2*10 i.e 22
considering all other numbers to be unchanged.
the difference Q-P = 22
therefore, 10a+b + 10b+a - (10x+y + 10y+x) = 22
or, 11(a+b) - 11(x+y) = 22
or, (a+b) - (x+y) = 2 , i.e. the sum of the digits in B that exceeds the sum of digits in A.
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Ajay attempted to add ten two-digit numbers. One of -them, a, was the reverse of one of the others. If a was replaced by another two-digit number, b and the reverse of a was replaced by the reverse of b and the average was found, it would be 2.2 more. The sum of the digits in b exceeds the sum of the digits in a by?
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Ajay attempted to add ten two-digit numbers. One of -them, a, was the reverse of one of the others. If a was replaced by another two-digit number, b and the reverse of a was replaced by the reverse of b and the average was found, it would be 2.2 more. The sum of the digits in b exceeds the sum of the digits in a by? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Ajay attempted to add ten two-digit numbers. One of -them, a, was the reverse of one of the others. If a was replaced by another two-digit number, b and the reverse of a was replaced by the reverse of b and the average was found, it would be 2.2 more. The sum of the digits in b exceeds the sum of the digits in a by? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ajay attempted to add ten two-digit numbers. One of -them, a, was the reverse of one of the others. If a was replaced by another two-digit number, b and the reverse of a was replaced by the reverse of b and the average was found, it would be 2.2 more. The sum of the digits in b exceeds the sum of the digits in a by?.
Ajay attempted to add ten two-digit numbers. One of -them, a, was the reverse of one of the others. If a was replaced by another two-digit number, b and the reverse of a was replaced by the reverse of b and the average was found, it would be 2.2 more. The sum of the digits in b exceeds the sum of the digits in a by? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Ajay attempted to add ten two-digit numbers. One of -them, a, was the reverse of one of the others. If a was replaced by another two-digit number, b and the reverse of a was replaced by the reverse of b and the average was found, it would be 2.2 more. The sum of the digits in b exceeds the sum of the digits in a by? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ajay attempted to add ten two-digit numbers. One of -them, a, was the reverse of one of the others. If a was replaced by another two-digit number, b and the reverse of a was replaced by the reverse of b and the average was found, it would be 2.2 more. The sum of the digits in b exceeds the sum of the digits in a by?.
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