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04 Jan 2024, 04:18
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Difficulty:
85%
(hard)
Question Stats:
58% (02:11) correct
42%
(02:26)
wrong
based on 1096
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If m is the tens digit of the sum of the two-digit positive integers kr and ms, where k,m, r and s are digits of the integers, which of the following must be true?
I. k = 9
II. m < 9
III. r + s >9
A. II only
B. III only
C. I and II only
D. I and III only
E. I, II, and III
I. k = 9
II. m < 9
III. r + s >9
A. II only
B. III only
C. I and II only
D. I and III only
E. I, II, and III
04 Jan 2024, 15:59
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If m is the tens digit of the sum of the two-digit positive integers kr and ms, where k,m, r and s are digits of the integers, which of the following must be true?
Reading the question stem, we see that m is the tens digit of two positive integers. The first one is the integer ms. The second is the integer that is the sum of kr and ms.
So, the digit represented by m remains the same throughout the process of kr being added to ms.
The only way for the tens digit of the sum of multiple numbers to be the same as the tens digit of one of those numbers is for the number added to the tens digit that remains the same to be a multiple of 10.
For example, in 24 + 97 = 121, the tens digits of 24 and 121 are the same because the number added to 2 when the two numbers are added is 10, which is a multiple of 10. In other words, 7 + 4 = 11,, and we carry the 1 and add it to the 9 and 2 when we add the tens digits. So, we basically add (9 + 1) + 2 or 10 + 2 when we add the tens digits.
I. k = 9
This statement must be true. After all, the maximum values of the ones digits r and s are 9 and 9. Thus, the maximum sum of r and s is 18. So, the maximum we can carry over when we add the tens digits is 1.
So, the only way for us to add a multiple of 10 to m when we add the tens digits is for k to be 9. If K = 9, when we add the tens digits we have the carried over 1 + k + m = (1 + 9) + m = 10 + m
II. m < 9
This statement does not have to be true. Even if m is 9, we can still add 10 to m and have m be the tens digit of the sum of kr and ms.
III. r + s > 9
This statement must be true. After all, to have a multiple of 10 to add to m, we need a 1 to carry over to add to k + m. So, r + s must equal at least 10.
A. II only
B. III only
C. I and II only
D. I and III only
E. I, II, and III
Correct Answer
Reading the question stem, we see that m is the tens digit of two positive integers. The first one is the integer ms. The second is the integer that is the sum of kr and ms.
So, the digit represented by m remains the same throughout the process of kr being added to ms.
The only way for the tens digit of the sum of multiple numbers to be the same as the tens digit of one of those numbers is for the number added to the tens digit that remains the same to be a multiple of 10.
For example, in 24 + 97 = 121, the tens digits of 24 and 121 are the same because the number added to 2 when the two numbers are added is 10, which is a multiple of 10. In other words, 7 + 4 = 11,, and we carry the 1 and add it to the 9 and 2 when we add the tens digits. So, we basically add (9 + 1) + 2 or 10 + 2 when we add the tens digits.
I. k = 9
This statement must be true. After all, the maximum values of the ones digits r and s are 9 and 9. Thus, the maximum sum of r and s is 18. So, the maximum we can carry over when we add the tens digits is 1.
So, the only way for us to add a multiple of 10 to m when we add the tens digits is for k to be 9. If K = 9, when we add the tens digits we have the carried over 1 + k + m = (1 + 9) + m = 10 + m
II. m < 9
This statement does not have to be true. Even if m is 9, we can still add 10 to m and have m be the tens digit of the sum of kr and ms.
III. r + s > 9
This statement must be true. After all, to have a multiple of 10 to add to m, we need a 1 to carry over to add to k + m. So, r + s must equal at least 10.
A. II only
B. III only
C. I and II only
D. I and III only
E. I, II, and III
Correct Answer
04 Jan 2024, 14:07
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Churej
\(\quad kr\)
+
\(\quad ms\)
--------------------
\(\quad xmy\)
Given: tens digit of the sum = \(m\)
As \(m\) is a single-digit integer, the unit digit of the other number must be zero for the unit digit of the sum of both digits to be equal to \(m\).
Case 1: \(k\) can be = 0 ⇒ This is not possible as \(kr\) is a two digit number.
Case 2: The sum of \(r\) and \(s\) resulted in a value greater than 9 and 1 was carried over to the tens place. Therefore the sum \(k + 1\) resulted in 10 10 ⇒ If this was the case, 10 + m will have a unit digit of m.
For this to happen \(k + 1 = 10, k = 9\) and we should have a carry from the sum of \(r\) and \(s\). Therefore \(r + s > 9\).
Option D
