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A student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.
- a)3
- b)4
- c)5
- d)6
Correct answer is option 'B'. Can you explain this answer?
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A student was given 8 two-digit numbers to add,by a teacher.If the stu...
Let us say that the 8 numbers are a1 b1, a2 b2, a3 b3 … a8 b8 …{eg: 38}
They actually represent a1*10 + b1, a2*10 + b2 … a8*10 + b8 …{eg: 3*10 + 8}
Sum of the numbers = (a1 + a2 + a3 .. a8)*10 + (b1 + b2 + b3 .. b8)
Reversed numbers will be b1 a1, b2 a2, … b8 a8
Sum of the reversed numbers will be (b1 + b2 + b3 .. b8)*10 + (a1 + a2 + a3 .. a8)
Reverse sum – actual sum = 36 … {Given in the question}
(b1 + b2 + b3 .. b8)*9 – (a1 + a2 + a3… a8)*9 = 36
=> (b1 + b2 + b3 … b8) – (a1 + a2 + a3 … a8) = 4 {9 can be taken common and cancelled from both sides}
=> Sum of the units digits – Sum of the tens digits = 4
Most Upvoted Answer
A student was given 8 two-digit numbers to add,by a teacher.If the stu...
To solve this problem, let's break it down step by step:
Step 1: Let's assume the ten's digit of the first number is a and the unit's digit is b. So the original number is 10a + b.
Step 2: If we reverse this number, we get 10b + a.
Step 3: Now, let's add the original number and the reversed number:
(10a + b) + (10b + a) = 11a + 11b = 11(a + b)
Step 4: According to the problem, the sum of the reversed numbers is 36 more than the actual sum. So we can write the equation as:
11(a + b) = (a + b) + 36
Step 5: Simplifying the equation:
10(a + b) = 36
Step 6: Divide both sides of the equation by 10:
a + b = 3.6
Step 7: Since the ten's digit and unit's digit are integers, we can conclude that a + b must be equal to 4.
Step 8: Now, let's consider the excess of the sum of the unit's digits over that of the ten's digits. The unit's digit sum is b + b + b + b + b + b + b + b = 8b. The ten's digit sum is a + a + a + a + a + a + a + a = 8a.
Step 9: We need to find the excess of the unit's digit sum over the ten's digit sum, which is 8b - 8a.
Step 10: Since a + b = 4, we can rewrite it as b = 4 - a.
Step 11: Substitute b = 4 - a into the expression 8b - 8a:
8(4 - a) - 8a = 32 - 8a - 8a = 32 - 16a
Step 12: Since we know that a + b = 4, we can substitute the value of a into the expression:
32 - 16a = 32 - 16(4 - a) = 32 - 64 + 16a = 48 - 16a
Step 13: Simplify the expression:
48 - 16a = 48 - 16(4) = 48 - 64 = -16
Step 14: The excess of the sum of the unit's digits over that of the ten's digits is -16.
Step 15: Since the question asks for the excess as a positive value, we take the absolute value of -16, which is 16.
Therefore, the correct answer is option B) 4.
Step 1: Let's assume the ten's digit of the first number is a and the unit's digit is b. So the original number is 10a + b.
Step 2: If we reverse this number, we get 10b + a.
Step 3: Now, let's add the original number and the reversed number:
(10a + b) + (10b + a) = 11a + 11b = 11(a + b)
Step 4: According to the problem, the sum of the reversed numbers is 36 more than the actual sum. So we can write the equation as:
11(a + b) = (a + b) + 36
Step 5: Simplifying the equation:
10(a + b) = 36
Step 6: Divide both sides of the equation by 10:
a + b = 3.6
Step 7: Since the ten's digit and unit's digit are integers, we can conclude that a + b must be equal to 4.
Step 8: Now, let's consider the excess of the sum of the unit's digits over that of the ten's digits. The unit's digit sum is b + b + b + b + b + b + b + b = 8b. The ten's digit sum is a + a + a + a + a + a + a + a = 8a.
Step 9: We need to find the excess of the unit's digit sum over the ten's digit sum, which is 8b - 8a.
Step 10: Since a + b = 4, we can rewrite it as b = 4 - a.
Step 11: Substitute b = 4 - a into the expression 8b - 8a:
8(4 - a) - 8a = 32 - 8a - 8a = 32 - 16a
Step 12: Since we know that a + b = 4, we can substitute the value of a into the expression:
32 - 16a = 32 - 16(4 - a) = 32 - 64 + 16a = 48 - 16a
Step 13: Simplify the expression:
48 - 16a = 48 - 16(4) = 48 - 64 = -16
Step 14: The excess of the sum of the unit's digits over that of the ten's digits is -16.
Step 15: Since the question asks for the excess as a positive value, we take the absolute value of -16, which is 16.
Therefore, the correct answer is option B) 4.
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A student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.a)3b)4c)5d)6Correct answer is option 'B'. Can you explain this answer?
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