Decimals & Fractions: Level 3
1.
In the repeating decimal \(0.123454545…\), what is the 101st digit to the right of the decimal point?
Solution (A): The digits “123” (3 digits) do not repeat.
The pattern “45” (2 digits) begins *after* the 3rd digit.
We want the 101st digit. This is the \((101 – 3) = 98\text{th}\) digit of the repeating pattern.
Now we divide 98 by the pattern length (2): \(98 \div 2 = 49\) with a remainder of 0.
A remainder of 0 means it is the *last* (2nd) digit of the pattern, which is 5.
*Correction:* Let’s re-check.
101st digit. 3 non-repeating. \(101-3 = 98\).
Pattern length 2. \(98 \div 2\) has rem 0. Last digit is 5.
Let me check my explanation logic.
4th digit: (4-3) = 1st. 1%2=1. (4)
5th digit: (5-3) = 2nd. 2%2=0. (5)
6th digit: (6-3) = 3rd. 3%2=1. (4)
7th digit: (7-3) = 4th. 4%2=0. (5)
So the 98th digit of the pattern (which is an even number) will be 5.
Let’s check 101st. \(101-3 = 98\). 98 is even, so it’s the 2nd digit (5).
My answer A is 4. Why? Let’s re-do 97.
If the question was 100th digit: \(100-3 = 97\). \(97 \div 2\) rem 1. 1st digit is 4.
Ah, I must have calculated for 100th. Let’s make the question 100th.
1.
In the repeating decimal \(0.123454545…\), what is the 100th digit to the right of the decimal point?
Solution (A): The digits “123” (3 digits) do not repeat.
The pattern “45” (2 digits) begins *after* the 3rd digit.
We want the 100th digit. This is the \((100 – 3) = 97\text{th}\) digit of the repeating pattern.
Now we divide 97 by the pattern length (2): \(97 \div 2 = 48\) with a remainder of 1.
A remainder of 1 means it is the 1st digit of the pattern, which is 4.
2.
A number \(x\) is rounded to the nearest hundredth as 10.45. Which of the following *cannot* be the original number?
Solution (D): To round to the nearest hundredth (10.45), the original number must be in the range [10.445, 10.455).
The number must be greater than or equal to 10.445.
The number must be *less than* 10.455.
The number 10.455 would be rounded up to 10.46, so it cannot be the original number.
3.
If \(a = 1.49\), \(b = 2.49\), and \(c = 3.49\), let \(x\) be the sum \((a+b+c)\) rounded to the nearest integer. Let \(y\) be the sum of \(a\) (rounded to the nearest integer), \(b\) (rounded to the nearest integer) and \(c\) (rounded to the nearest integer). What is \(y-x\)?
Solution (A):
1. Find \(x\): Sum first. \(1.49 + 2.49 + 3.49 = 7.47\).
Round 7.47 to the nearest integer: \(x = 7\).
2. Find \(y\): Round first.
\(a = 1.49 \to 1\)
\(b = 2.49 \to 2\)
\(c = 3.49 \to 3\)
Sum the rounded values: \(y = 1 + 2 + 3 = 6\).
3. Find \(y – x\): \(6 – 7 = -1\).
4.
In the repeating decimal \(0.123123…\), what is the sum of the 50th, 51st, and 52nd digits?
Solution (B): The repeating pattern is “123”, which has a length of 3.
1. 50th digit: \(50 \div 3 = 16\) rem 2. The 2nd digit is 2.
2. 51st digit: \(51 \div 3 = 17\) rem 0. The 3rd (last) digit is 3.
3. 52nd digit: \(52 \div 3 = 17\) rem 1. The 1st digit is 1.
The three digits are 2, 3, and 1. Their sum is \(2 + 3 + 1 = 6\).
5.
What is the sum of the first 100 digits to the right of the decimal point for the fraction \( \frac{1}{6} \)?
Solution (C):
1. Convert \( \frac{1}{6} \) to a decimal: \(1 \div 6 = 0.16666…\).
2. The first digit is 1 (non-repeating).
3. The next 99 digits are all 6.
4. Sum = (1st digit) + (Sum of the 99 digits)
Sum = \(1 + (99 \times 6)\)
Sum = \(1 + 594 = 595\).
6.
The fraction \( \frac{1}{7} \) equals the repeating decimal \(0.142857142857…\). What is the 1000th digit to the right of the decimal point?
Solution (D): The repeating pattern is “142857”, which has a length of 6 digits.
We need to find the remainder of \(1000 \div 6\).
\(1000 \div 6\). We know \(996\) is a multiple of 6 (since \(996/6 = 166\)).
\(1000 – 996 = 4\). The remainder is 4.
A remainder of 4 means it is the 4th digit in the pattern.
The 4th digit is 8.
7.
A value \(V\) is 4.8. This value was obtained by rounding an original number \(x\) to the nearest tenth. What is the *smallest* possible integer \(y\) such that \(y \times x > 100\)?
Solution (C):
1. If \(x\) rounds to 4.8, its true value is in the range [4.75, 4.85).
2. We need an integer \(y\) such that \(y \times x > 100\). This must be true for *all* possible values of \(x\).
3. The “worst-case” scenario is when \(x\) is at its smallest possible value, \(x = 4.75\). We must ensure \(y\) works even for this value.
4. \(y \times 4.75 > 100 \implies y > \frac{100}{4.75}\).
5. \( \frac{100}{4.75} \approx 21.05… \)
6. Since \(y\) must be greater than 21.05…, the smallest possible *integer* for \(y\) is 22.
8.
What is the repeating decimal \(0.121212…\) as a fraction in simplest form?
Solution (E):
1. Let \(x = 0.121212…\)
2. Multiply by 100 to shift the decimal two places: \(100x = 12.121212…\)
3. Subtract the original \(x\) from \(100x\):
\(100x – x = 12.1212… – 0.1212…\)
\(99x = 12\)
4. Solve for \(x\): \(x = \frac{12}{99}\).
5. Simplify by dividing the numerator and denominator by 3: \(x = \frac{4}{33}\).
9.
Which of the following values is the smallest?
Solution (D): Convert all values to decimals to compare.
(A) \( \frac{1}{9} = 0.1111… \)
(B) \( 0.11 = 0.1100 \)
(C) \( \frac{1}{10} = 0.1000 \)
(D) \( (0.1)^2 = 0.0100 \)
(E) \( 0.1 = 0.1000 \)
Comparing the decimals, 0.0100 is the smallest.
10.
A number \(N = 1.25\) is rounded to the nearest tenth to get \(N’\). What is the value of \( \frac{N^2 – (N’)^2}{N-N’} \)?
Solution (B):
1. First, find \(N’\). \(N = 1.25\) rounded to the nearest tenth is \(N’ = 1.3\).
2. The expression is a difference of squares: \( \frac{N^2 – (N’)^2}{N-N’} = \frac{(N – N’)(N + N’)}{N – N’} \).
3. Cancel the \((N – N’)\) terms. The expression simplifies to \(N + N’\).
4. \(N + N’ = 1.25 + 1.3 = 2.55\).
Score: 0 / 10