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If the places of last two-digits of a three digit number are interchanged, a new number greater than the original number by 36 is obtained. What is the difference between the last two digits of that number?
- a)2
- b)3
- c)4
- d)7
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If the places of last two-digits of a three digit number are interchan...
let the number be 100a + 10b + c
(100a + 10b +c) – (100a + 10c +b) = 36
b – c = 4
(100a + 10b +c) – (100a + 10c +b) = 36
b – c = 4
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Solution:
Let the number be XYZ.
According to the question, we have to interchange the last two digits of the number. So, the new number will be XZY.
It is given that the new number is greater than the original number by 36. So, we can write:
XZY = XYZ + 36
Now, let's consider the place value of the digits:
Z × 1 + Y × 10 + X × 100 = X × 1 + Z × 10 + Y × 100 + 36
Simplifying the above equation, we get:
99(Y - Z) = 36
Y - Z = 36/99
Y - Z = 4/11
Since Y and Z are both digits, the only possible values of Y - Z are 1/11, 2/11, 3/11, 4/11, 5/11, 6/11, 7/11, 8/11, 9/11.
Out of these, the only value that is an integer is 4/11. Therefore, the difference between the last two digits of the number is 4.
Hence, the correct option is (c) 4.
Let the number be XYZ.
According to the question, we have to interchange the last two digits of the number. So, the new number will be XZY.
It is given that the new number is greater than the original number by 36. So, we can write:
XZY = XYZ + 36
Now, let's consider the place value of the digits:
Z × 1 + Y × 10 + X × 100 = X × 1 + Z × 10 + Y × 100 + 36
Simplifying the above equation, we get:
99(Y - Z) = 36
Y - Z = 36/99
Y - Z = 4/11
Since Y and Z are both digits, the only possible values of Y - Z are 1/11, 2/11, 3/11, 4/11, 5/11, 6/11, 7/11, 8/11, 9/11.
Out of these, the only value that is an integer is 4/11. Therefore, the difference between the last two digits of the number is 4.
Hence, the correct option is (c) 4.
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