GMAT Decimal
Introduction
Decimals are a vital concept in mathematics and frequently appear in various sections of the GMAT exam. It is essential to understand decimal algebra, as it will help you solve problems involving decimals quickly and accurately. In this four-part series, we will explore the intricacies of decimal algebra, including decimal representation, arithmetic operations, decimal-fraction conversions, and real-world applications.
Part 1: Decimal Representation and Basics
1.1 What are Decimals?
A decimal is a number system representation that uses a base of 10. It utilizes digits from 0 to 9 and a decimal point to represent fractional parts. Decimals are commonly used in everyday life for measurements, currency, and expressing values with varying degrees of precision.
1.2 Decimal Representation
A decimal number can be expressed as a whole number part and a fractional part, separated by a decimal point. The whole number part comprises the digits to the left of the decimal point, while the fractional part consists of the digits to the right of the decimal point.
For example, in the decimal number 12.345, 12 is the whole number part, and .345 is the fractional part.
1.3 Decimal Places
The position of a digit to the right of the decimal point determines its place value. The first digit after the decimal point represents tenths, the second digit represents hundredths, the third digit represents thousandths, and so on.
For example, in the decimal number 0.123, 1 is in the tenths place, 2 is in the hundredths place, and 3 is in the thousandths place.
1.4 Rounding Decimals
Rounding decimals is a technique used to approximate a decimal value to a specific number of decimal places. This simplifies calculations and makes it easier to work with decimals.
To round a decimal number:
- Identify the digit in the desired rounding place.
- Check the digit to the right of the rounding place.
- If the digit to the right is 5 or greater, round up the digit in the rounding place.
- If the digit to the right is less than 5, leave the digit in the rounding place unchanged.
- Remove all digits to the right of the rounding place.
For example, to round 12.345 to two decimal places, we check the third digit (5). Since it is greater than or equal to 5, we round up the second digit (4) to 5, resulting in 12.35.
1.5 Comparing Decimals
To compare decimal numbers, follow these steps:
Align the decimal points vertically.
Compare the whole number parts; the larger whole number part indicates a larger decimal.
If the whole number parts are equal, compare the digits in each decimal place, starting from the left. The decimal with the larger digit in the first unequal decimal place is the larger decimal.
For example, to compare 0.123 and 0.132, we can see that the whole number parts are equal (0). Comparing the digits from left to right, we find that the first unequal digits are 2 (in 0.123) and 3 (in 0.132). Since 3 is greater than 2, 0.132 is the larger decimal.
In the next part, we will delve into arithmetic operations involving decimals, including addition, subtraction, multiplication, and division.
Part 2: Arithmetic Operations with Decimals
2.1 Adding Decimals
To add decimal numbers, follow these steps:
Align the decimal points vertically.
Add zeros to the right of the shorter decimal to ensure equal decimal places.
Perform column-wise addition, starting from the rightmost digit and moving left, carrying over as needed.
Place the decimal point in the sum, aligned with the decimal points of the addends.
Example: Add 12.34 and 5.678
12.34
+ 5.678
________
18.018
2.2 Subtracting Decimals
To subtract decimal numbers, follow these steps:
Align the decimal points vertically.
Add zeros to the right of the shorter decimal to ensure equal decimal places.
Perform column-wise subtraction, starting from the rightmost digit and moving left, borrowing as needed.
Place the decimal point in the difference, aligned with the decimal points of the minuend and subtrahend.
Example: Subtract 9.52 from 12.34
12.34
– 9.52
______
2.82
2.3 Multiplying Decimals
To multiply decimal numbers, follow these steps:
Ignore the decimal points and multiply the numbers as if they were whole numbers.
Count the total number of decimal places in both factors.
Place the decimal point in the product so that the total number of decimal places in the product is equal to the sum of the decimal places in the factors.
Example: Multiply 1.2 by 3.4
12
x 34
________
48
+ 36
________
408
1.2 has one decimal place, and 3.4 has one decimal place. The product should have two decimal places. Thus, the product is 4.08.
2.4 Dividing Decimals
To divide decimal numbers, follow these steps:
Remove the decimal point from the divisor by multiplying it by a power of 10. Perform the same multiplication on the dividend.
Divide the resulting whole numbers using long division or any other preferred method.
Place the decimal point in the quotient to ensure the same number of decimal places as in the original dividend.
Example: Divide 4.2 by 1.4
First, multiply both the dividend and divisor by 10 to remove the decimal points:
Copy code
42 ÷ 14
3.0
The quotient is 3.0.
In the next part, we will explore decimal-fraction conversions and strategies for simplifying and
Part 3: Decimal-Fraction Conversions and Equations
3.1 Converting Decimals to Fractions
To convert a decimal to a fraction, follow these steps:
Determine the place value of the last digit of the decimal.
Write the decimal as a fraction with the numerator equal to the decimal number without the decimal point, and the denominator equal to the place value of the last digit.
Simplify the fraction, if possible.
Example: Convert 0.75 to a fraction
The last digit, 5, is in the hundredths place. Write the decimal as a fraction:
75/100
Simplify the fraction:
3/ 4
The fraction is 3/4.
3.2 Converting Fractions to Decimals
To convert a fraction to a decimal, follow these steps:
Divide the numerator by the denominator.
If necessary, round the result to the desired number of decimal places.
Example: Convert 3/8 to a decimal
Perform the division:
3 ÷ 8 = 0.375
The decimal is 0.375.
3.3 Simplifying Decimal Equations
When solving equations involving decimals, it’s often helpful to eliminate the decimals by multiplying both sides of the equation by a power of 10. This can make the equation easier to solve.
Example: Solve the equation 0.3x + 1.2 = 2.4
First, multiply both sides of the equation by 10 to eliminate the decimals:
10(0.3x + 1.2) = 10(2.4) 3x + 12 = 24
Now, solve the equation:
3x = 12 x = 4
3.4 Solving Decimal Equations with Fractions
When solving equations that involve both decimals and fractions, it can be helpful to convert the decimals to fractions. This allows you to work with a single number system, making the equation easier to solve.
Example: Solve the equation 0.5x – 1/4 = 1.25
First, convert the decimals to fractions:
1/2x – 1/4 = 5/4
Now, solve the equation:
1/2x = 1/4 + 5/4 1/2x = 6/4 x = (6/4) / (1/2) x = 3
In the final part, we will discuss real-world applications of decimal algebra and practice problems to reinforce your understanding of the concepts.
Part 4: Real-World Applications and Practice Problems
4.1 Real-World Applications
Decimal algebra is frequently used in everyday life and various fields, such as finance, engineering, and science. Some common real-world applications include:
Currency calculations: Decimals are used to represent money and perform calculations involving income, expenses, and investments.
Measurements: Decimals are used in various units of measurement, such as length, weight, and temperature, to represent values with varying degrees of precision.
Percentage calculations: Decimals are used to express percentages, which are often used to compare quantities, calculate discounts, or determine interest rates.
Scientific notation: Decimals are used in scientific notation to express very large or very small numbers in a more manageable form.
4.2 Practice Problems
Solve the following problems to reinforce your understanding of decimal algebra concepts:
Add the following decimals: 14.56 and 3.009
Solution: Align the decimals and add:
14.56
+ 3.009
_________
17.569
2. Subtract the following decimals: 8.1 – 2.37
Solution: Align the decimals and subtract:
8.10
– 2.37
_________
5.73
3. Multiply the following decimals: 1.5 x 2.4
Solution: Multiply as whole numbers, then add the decimal places:
15
x 24
________
60
+ 30
________
360
Both numbers have one decimal place, so the product has two decimal places: 3.60
Divide the following decimals: 2.7 ÷ 0.9
Solution: Multiply both numbers by 10 to remove the decimal points:
27 ÷ 9 = 3
The quotient is 3.
Convert the decimal 0.875 to a fraction.
Solution: The last digit is in the thousandths place:
875/ 1000
Simplify the fraction:
7 /8
The fraction is 7/8.
Remember that mastering decimal algebra requires practice. Continue working on problems involving decimals to build your confidence and proficiency in preparation for the GMAT exam.
Practice Questions
Question 1: Add the following decimals: 5.31 + 2.9
A) 8.21 B) 7.31 C) 8.12 D) 7.21 E) 8.01
Answer: A) 8.21
Question 2: Subtract the following decimals: 4.5 – 1.25
A) 3.15 B) 3.25 C) 3.35 D) 3.45 E) 3.55
Answer: B) 3.25
Question 3: Multiply the following decimals: 0.6 x 0.3
A) 0.18 B) 0.19 C) 0.20 D) 0.21 E) 0.22
Answer: A) 0.18
Question 4: Divide the following decimals: 1.2 ÷ 0.4
A) 2 B) 3 C) 4 D) 5 E) 6
Answer: B) 3
Question 5: Convert the decimal 0.2 to a fraction.
A) 1/5 B) 1/4 C) 1/3 D) 1/2 E) 1/1
Answer: A) 1/5
Question 6: Add the following decimals: 32.156 + 17.39
A) 49.456 B) 49.546 C) 49.556 D) 49.666 E) 49.546
Answer: B) 49.546
Question 7: Subtract the following decimals: 15.8 – 3.129
A) 12.671 B) 12.761 C) 12.851 D) 12.941 E) 13.031
Answer: A) 12.671
Question 8: Multiply the following decimals: 1.4 x 3.5
A) 4.90 B) 5.10 C) 5.20 D) 5.30 E) 5.40
Answer: A) 4.90
Question 9: Divide the following decimals: 3.6 ÷ 1.2
A) 2 B) 3 C) 4 D) 5 E) 6
Answer: B) 3
Question 10: Convert the decimal 0.625 to a fraction.
A) 1/2 B) 3/5 C) 2/3 D) 5/8 E) ¾
Answer: D) 5/8
Question 11:
The value of p is derived by summing d, e, and f and then rounding the result to the tenths place. The value of q is derived by first rounding d, e, and f to the tenths place and then summing the resulting values. If d = 6.35, e = 4.78, and f = 2.46, what is q – p?
A.-0.1 B. 0 C 0.0 D 0.1 E. 0 .2
Solution:
For p, you sum d, e, and f and then round to the tenths place:
p = round(6.35 + 4.78 + 2.46, 1) = round(13.59, 1) = 13.6
For q, you round d, e, and f to the tenths place first and then sum the results:
q = round(6.35, 1) + round(4.78, 1) + round(2.46, 1) = 6.4 + 4.8 + 2.5 = 13.7
So, q – p = 13.7 – 13.6 = 0.1
The answer is D. 0.1
Question 12:
The value of m is derived by summing g, h, and i and then rounding the result to the tenths place. The value of n is derived by first rounding g, h, and i to the tenths place and then summing the resulting values. If g = 3.84, h = 7.56, and i = 2.39, what is n – m?
A.-0.2 B. -0.1 C 0.0 D 0.1 E. 0.2
Solution:
For m, you sum g, h, and i and then round to the tenths place:
m = round(3.84 + 7.56 + 2.39, 1) = round(13.79, 1) = 13.8
For n, you round g, h, and i to the tenths place first and then sum the results:
n = round(3.84, 1) + round(7.56, 1) + round(2.39, 1) = 3.8 + 7.6 + 2.4 = 13.8
So, n – m = 13.8 – 13.8 = 0
The answer is C. 0.0
Question 13:
The value of p is derived by summing j, k, and l and then rounding the result to the tenths place. The value of q is derived by first rounding j, k, and l to the tenths place and then summing the resulting values. If j = 4.67, k = 5.89, and l = 3.21, what is q – p?
A.-0.2 B. -0.1 C 0.0 D 0.1 E. 0.2
Solution:
For p, you sum j, k, and l and then round to the tenths place:
p = round(4.67 + 5.89 + 3.21, 1) = round(13.77, 1) = 13.8
For q, you round j, k, and l to the tenths place first and then sum the results:
q = round(4.67, 1) + round(5.89, 1) + round(3.21, 1) = 4.7 + 5.9 + 3.2 = 13.8
So, q – p = 13.8 – 13.8 = 0
The answer is C. 0.0
Question 14:
If the fraction 1/7 equals the repeating decimal 0.142857142857 . . . , what is the 68th digit after the decimal point of the repeating decimal?
(A) 1(B) 2(C) 4(D) 5(E) 7
Solution:
The decimal representation of 1/7 repeats every 6 digits (142857). So, we can find the 68th digit by finding the remainder when 68 is divided by 6. The remainder is 2, so the 68th digit is the 2nd digit in the repeating block, which is 4.
The answer is (C) 4.
Question 15:
If the fraction 1/13 equals the repeating decimal 0.076923076923 . . . , what is the 45th digit after the decimal point of the repeating decimal?
(A) 0(B) 1(C) 6(D) 7(E) 9
Solution:
The decimal representation of 1/13 repeats every 6 digits (076923). So, we can find the 45th digit by finding the remainder when 45 is divided by 6. The remainder is 3, so the 45th digit is the 3rd digit in the repeating block, which is 6.
The answer is (C) 6.
Question 16:
If the fraction 2/11 equals the repeating decimal 0.181818181818 . . . , what is the 72nd digit after the decimal point of the repeating decimal?
(A) 1(B) 2(C) 8(D) 6(E) 7
Solution:
The decimal representation of 2/11 repeats every 2 digits (18). So, we can find the 72nd digit by finding the remainder when 72 is divided by 2. The remainder is 0, so the 72nd digit is the last digit in the repeating block, which is 8.
The answer is (C) 8.
Question 17:
The decimal 0.456456456… repeats every three digits. If we were to write out the first 100 digits after the decimal point, how many times would the digit 6 appear?
(A) 33 (B) 34 (C) 35 (D) 36 (E) 37
Solution:
The decimal repeats every three digits, so in every cycle of three digits, the digit 6 appears once. Hence, every three digits, there’s one 6. To find out how many 6’s there would be in the first 100 digits, we divide 100 by 3. This gives approximately 33.33. Since we can’t have a partial cycle, and we know that the 34th cycle would start after the 100th digit, there are 33 complete cycles within the first 100 digits. Therefore, the digit 6 would appear 33 times.
The answer is (A) 33.
Question 18:
The decimal 0.789789789… repeats every three digits. If we were to write out the first 200 digits after the decimal point, how many times would the sequence ’89’ appear?
(A) 66 (B) 67 (C) 68 (D) 69 (E) 70
Solution:
The decimal repeats every three digits (789), and the sequence ’89’ appears once in each cycle. Therefore, every three digits, there’s one ’89’. To find out how many ’89’s there would be in the first 200 digits, we divide 200 by 3. This gives approximately 66.67. Since we can’t have a partial cycle, and we know that the 67th cycle would start after the 200th digit, there are 66 complete cycles within the first 200 digits. Therefore, the sequence ’89’ would appear 66 times.
The answer is (A) 66.
Question 19:
The decimal 0.12341234… repeats every four digits. What is the 157th digit after the decimal point?
(A) 1 (B) 2 (C) 3 (D) 4 (E) None of the above
Solution:
The decimal repeats every four digits (1234). We can find the 157th digit by finding the remainder when 157 is divided by 4. This gives a remainder of 1. So, the 157th digit would be the first digit in the repeating block, which is 1.
The answer is (A) 1.
Question 20:
The decimal 0.56785678… repeats every four digits. If we were to write out the first 150 digits after the decimal point, how many times would the digit 5 appear?
(A) 37 (B) 38 (C) 39 (D) 40 (E) 41
Solution:
The decimal repeats every four digits (5678), and the digit 5 appears once in each cycle. Therefore, every four digits, there’s one 5. To find out how many 5’s there would be in the first 150 digits, we divide 150 by 4. This gives 37.5. Since we can’t have a partial cycle, and we know that the 38th cycle would start after the 150th digit, there are 37 complete cycles within the first 150 digits. Therefore, the digit 5 would appear 37 times.
The answer is (A) 37.
Question 21:
The decimal 0.57685768… repeats every five digits. If we were to write out the first 500 digits after the decimal point, how many times would the sequence ‘5768’ appear?
(A) 99 (B) 100 (C) 101 (D) 102 (E) 103
Solution:
The decimal repeats every five digits (57685), and the sequence ‘5768’ appears once in each cycle. However, because the sequence ‘5768’ also overlaps between two cycles (as in …57685768…), we have to account for those as well.
First, let’s find out how many cycles fit into 500 digits. To do this, we divide 500 by 5 to get 100 cycles. Each cycle has one ‘5768’, so that’s 100 occurrences.
Next, let’s account for the overlapping sequences. Because the sequence ‘5768’ overlaps between two cycles, there will be an extra ‘5768’ for every consecutive pair of cycles. Since we have 100 cycles, we have 99 pairs of consecutive cycles, so that’s an additional 99 ‘5768’s.
Therefore, the total number of ‘5768’s is 100 (from individual cycles) + 99 (from overlaps) = 199.
Unfortunately, this number doesn’t match any of the options given. This is a common occurrence in real-world problems and a good reminder to always check your work. If the options had included 199, that would have been the correct choice.
Question 22:
The decimal 0.89738973… repeats every four digits. If we were to write out the first 200 digits after the decimal point, how many times would the sequence ‘8973’ appear?
(A) 49 (B) 50 (C) 51 (D) 52 (E) 53
Solution:
The decimal repeats every four digits (8973), and the sequence ‘8973’ appears once in each cycle. To find out how many ‘8973’s there would be in the first 200 digits, we divide 200 by 4 to get 50 cycles. Each cycle has one ‘8973’, so that’s 50 occurrences.
The answer is (B) 50.
Data Sufficiency Questions
In Data Sufficiency questions, you are given a question and two statements. You must determine whether the information given in the statements is sufficient to answer the question. Choose from the following answer choices:
A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.
Question 23:
p = 0.25q8
If q denotes the thousandths digit in the decimal representation of p above, what digit is q?
(1) If p were rounded to the nearest hundredth, the result would be 0.25.
(2) If p were rounded to the nearest thousandth, the result would be 0.252.
Solution:
Statement (1) tells us that when rounded to the nearest hundredth, p = 0.25. This means that the thousandths place could be 5 or greater. However, we don’t know the exact value of q, so this statement is insufficient.
Statement (2) tells us that when rounded to the nearest thousandth, p = 0.252. This gives us the exact value of q, which is 2. Therefore, this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 24:
n = 0.15r9
If r denotes the ten-thousandths digit in the decimal representation of n above, what digit is r?
(1) If n were rounded to the nearest thousandth, the result would be 0.159. (2) If n were rounded to the nearest ten-thousandth, the result would be 0.1598.
Solution:
Statement (1) tells us that when rounded to the nearest thousandth, n = 0.159. This means that the ten-thousandths place could be 5 or greater. However, we don’t know the exact value of r, so this statement is insufficient.
Statement (2) tells us that when rounded to the nearest ten-thousandth, n = 0.1598. This gives us the exact value of r, which is 8. Therefore, this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 25:
m = 0.27s6
If s denotes the hundred-thousandths digit in the decimal representation of m above, what digit is s?
(1) If m were rounded to the nearest ten-thousandth, the result would be 0.276.
(2) If m were rounded to the nearest hundred-thousandth, the result would be 0.2769.
Solution:
Statement (1) tells us that when rounded to the nearest ten-thousandth, m = 0.276. This means that the hundred-thousandths place could be 5 or greater. However, we don’t know the exact value of s, so this statement is insufficient.
Statement (2) tells us that when rounded to the nearest hundred-thousandth, m = 0.2769. This gives us the exact value of s, which is 9. Therefore, this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 26:
z = 0.123×456
If x denotes the ten-thousandths digit in the decimal representation of z above, what digit is x?
(1) If z were multiplied by 10, then rounded to the nearest thousandth, the result would be 1.234.
(2) If z were divided by 10, then rounded to the nearest ten-thousandth, the result would be 0.01235.
Solution:
Statement (1) tells us that when z is multiplied by 10 and then rounded to the nearest thousandth, the result is 1.234. This means that the original number before rounding was between 1.2335 and 1.2345. Dividing these bounds by 10 gives us a range for z of 0.12335 to 0.12345. This means that x could be either 3 or 4, so this statement is insufficient.
Statement (2) tells us that when z is divided by 10 and then rounded to the nearest ten-thousandth, the result is 0.01235. This means that the original number before rounding was between 0.012345 and 0.012355. Multiplying these bounds by 10 gives us a range for z of 0.12345 to 0.12355. This means that x must be 4, so this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 27:
y = 0.789w123
If w denotes the hundred-thousandths digit in the decimal representation of y above, what digit is w?
(1) If y were multiplied by 100, then rounded to the nearest tenth, the result would be 78.9.
(2) If y were divided by 100, then rounded to the nearest millionth, the result would be 0.000007892.
Solution:
Statement (1) tells us that when y is multiplied by 100 and then rounded to the nearest tenth, the result is 78.9. This means that the original number before rounding was between 78.85 and 78.95. Dividing these bounds by 100 gives us a range for y of 0.7885 to 0.7895. This means that w could be either 8 or 9, so this statement is insufficient.
Statement (2) tells us that when y is divided by 100 and then rounded to the nearest millionth, the result is 0.000007892. This means that the original number before rounding was between 0.0000078915 and 0.0000078925. Multiplying these bounds by 100 gives us a range for y of 0.00078915 to 0.00078925. This means that w must be 9, so this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 28:
Let a = 0.1234b56
If b denotes the ten-thousandths digit in the decimal representation of a above, what digit is b?
(1) If a were multiplied by 10 and then divided by 3, then rounded to the nearest thousandth, the result would be 0.411.
(2) If a were divided by 5, then rounded to the nearest hundred-thousandth, the result would be 0.02468.
Solution:
Statement (1) tells us that when a is multiplied by 10 and then divided by 3, then rounded to the nearest thousandth, the result is 0.411. This means that the original number before rounding was between 0.4105 and 0.4115. Multiplying these bounds by 3 and then dividing by 10 gives us a range for a of 0.12315 to 0.12345. This means that b could be either 1 or 4, so this statement is insufficient.
Statement (2) tells us that when a is divided by 5, then rounded to the nearest hundred-thousandth, the result is 0.02468. This means that the original number before rounding was between 0.024675 and 0.024685. Multiplying these bounds by 5 gives us a range for a of 0.123375 to 0.123425. This means that b must be 4, so this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 29:
Let c = 0.5678d901
If d denotes the millionths digit in the decimal representation of c above, what digit is d?
(1) If c were multiplied by 100 and then divided by 7, then rounded to the nearest tenth, the result would be 8.1.
(2) If c were divided by 9, then rounded to the nearest ten-millionth, the result would be 0.000063101.
Solution:
Statement (1) tells us that when c is multiplied by 100 and then divided by 7, then rounded to the nearest tenth, the result is 8.1. This means that the original number before rounding was between 8.05 and 8.15. Multiplying these bounds by 7 and then dividing by 100 gives us a range for c of 0.5635 to 0.5705. This means that d could be any digit from 3 to 5, so this statement is insufficient.
Statement (2) tells us that when c is divided by 9, then rounded to the nearest ten-millionth, the result is 0.000063101. This means that the original number before rounding was between 0.0000631005 and 0.0000631015. Multiplying these bounds by 9 gives us a range for c of 0.5679045 to 0.5679145. This means that d must be 9, so this statement is sufficient.
The answer is (B), statement (2) alone is sufficient.
Question 30:
What is the hundredths digit in the decimal representation of a certain number?
(1) The number is less than 1/6.
(2) The number is greater than 1/8.
Solution
Statement (1) tells us that the number is less than 1/6, which is approximately 0.1666… This suggests that the hundredths digit could be 0, 1, 2, 3, 4, 5, or 6. But we can’t determine the exact digit, so this statement is insufficient.
Statement (2) tells us that the number is greater than 1/8, which is 0.125. This suggests that the hundredths digit could be 2 or any digit greater than 2. But again, we can’t determine the exact digit, so this statement is also insufficient.
Taking both statements together, we know that the number is between 0.125 and 0.1666… This suggests that the hundredths digit could be 2, 3, 4, 5, or 6. Since we can’t determine the exact digit, both statements together are still insufficient.
So, the answer is (E), both statements together are still not sufficient.
Question 31:
If s is represented by the decimal 0.p7, what is the digit p?
(1) s < 1/4
(2) s < 1/15
Solution:
Statement (1) tells us that s is less than 1/4, which is 0.25. This suggests that the digit p could be 0, 1, or 2. But we can’t determine the exact digit, so this statement is insufficient.
Statement (2) tells us that s is less than 1/15, which is approximately 0.0666… This suggests that the digit p must be 0 because the decimal representation of s, 0.p7, must be less than 0.1. So, this statement is sufficient.
Therefore, the answer is (B), statement (2) alone is sufficient.
Question 32:
If e denotes a decimal, is e ≥ 0.6?
(1) When e is rounded to the nearest tenth, the result is 0.6. (2) When e is rounded to the nearest integer, the result is 1.
Solution:
Statement (1) tells us that when e is rounded to the nearest tenth, the result is 0.6. This means that e could be anywhere from 0.55 (which rounds up to 0.6) to 0.6499999… (which rounds down to 0.6). This suggests that e could be less than 0.6, so this statement is insufficient.
Statement (2) tells us that when e is rounded to the nearest integer, the result is 1. This means that e could be anywhere from 0.5 (which rounds up to 1) to 1.4999999… (which rounds down to 1). This suggests that e could be less than 0.6, so this statement is also insufficient.
Therefore, the answer is (E), both statements together are still not sufficient.