When the positive number a is rounded to the nearest tenth, the result is the number b. What is the tenths digit of a?
(1) When a is rounded to the nearest integer, the result is less than a.
(2) When b is rounded to the nearest integer, the result is greater than b.
OA C
Source: Manhattan Prep
When the positive number a is rounded to the nearest tenth,
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Say,BTGmoderatorDC wrote:When the positive number a is rounded to the nearest tenth, the result is the number b. What is the tenths digit of a?
(1) When a is rounded to the nearest integer, the result is less than a.
(2) When b is rounded to the nearest integer, the result is greater than b.
OA C
Source: Manhattan Prep
a = x.yz, where x is units' digit, y is tenth digit, and z is hundredth digit and
b = x.p, where x is units' digit, and p is tenth digit (p is either y or y + 1)
We have to get the value of y.
Let's take each statement one by one.
(1) When a is rounded to the nearest integer, the result is less than a.
Say the result is a' such that a' < a
Or, a' = x, we see that x < x.yz
Here, y can be any digit 0, 1, 2, 3, or 4. No unique value of y. Insufficient.
(2) When b is rounded to the nearest integer, the result is greater than b.
Say the result is b' such that b' > b
Or, b' = x + 1, we see that (x + 1) > x.p
Here, p can be any digit 5, 6, 7, 8, or 9. The value of p will depend on the values of y and z.
Case 1: Say a = x.46, then b = x.5 and b' = x + 1. Here, y = 4
Case 2: Say a = x.56, then b = x.6 and b' = x + 1. Here, y = 5
No unique value of y. Insufficient.
(1) and (2) together
Considering Statement 1, we cannot have Case 2 as a valid case, else for a = x.56, we will have a' = x + 1 > a (= x.56); this is invalid since as per Statement 1, we must have a' < a.
This implies that a must be x.4z, where z ≥ 5
This way, we have
- a = x.4z
a' = x < a (= x.4z)
b = x.5, since z ≥ 5 and
b' = x + 1 > b (= x.5)
The correct answer: C
Hope this helps!
-Jay
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