# Mastering the GRE: A Deep Dive into ‘Age and Digit’ Problems – Part 1

When it comes to the Graduate Record Examination (GRE), understanding concepts, formulas, and shortcuts is integral to acing the test. This series aims to shed light on one of the more intriguing concepts tested in the GRE – Age and Digit problems. By the end of this five-part series brought to you by MKSprep, you’ll be confident enough to tackle these types of questions with ease. Let’s start with the first part: understanding the concept.

## What are ‘Age and Digit’ Problems?

‘Age and Digit’ problems are a type of algebraic word problems often found in the GRE quantitative section. They involve the use of basic arithmetic operations and require a good understanding of algebra and number properties. These problems usually involve situations that ask you to find a person’s age based on certain conditions or the value of digits in a number.

For instance, a typical ‘Age and Digit’ problem might ask, “Ten years from now, Alice will be twice as old as Bob was ten years ago. If Alice is 40 now, how old is Bob?” Solving this involves forming and manipulating algebraic equations.

## The Basic Formula

In its simplest form, the formula for age problems can be written as:

A = B + C

where:

• A is the current age of a person.
• B is the age of the person at a different time.
• C is the number of years before or after the current time.

In terms of digit problems, the basic concept is understanding the place value of each digit in a number. For example, in the number 345, 3 is in the hundreds place, 4 in the tens place, and 5 in the ones place.

## Reasoning Behind ‘Age and Digit’ Problems

Reasoning through these problems requires the ability to translate words into mathematical expressions. Your task is identifying the variables, establishing a relationship between them, and solving the equation.

For instance, if the problem states, “John is twice as old as Mary was four years ago,” you would first identify John and Mary’s ages as variables, say J and M. Then, translate the sentence into an equation: J = 2*(M-4).

The reasoning behind ‘digit problems’ involves understanding how numbers are structured. You’ll often have to break down the number into its individual digits and sometimes even rearrange them.

Stay tuned for the second part of this series, where we’ll delve deeper into the art of solving ‘Age and Digit’ problems, reveal some useful shortcuts, and discuss their importance in the GRE. By the time we’re done, you’ll have a whole new appreciation for these intriguing GRE problems!

# Mastering the GRE: A Deep Dive into ‘Age and Digit’ Problems – Part 2

Welcome back to our second part in this five-part series on ‘Age and Digit’ problems in the GRE, brought to you by MKSprep. In the first part, we introduced the concept and the basic formula for these problems. Now, we’ll dive deeper into the methods of solving these problems, providing shortcuts, and discussing their importance in the GRE.

## Solving ‘Age and Digit’ Problems

### Age Problems

The key to solving age problems is setting up the correct equations based on the information given in the problem. Let’s use the problem from Part 1 as an example:

“Ten years from now, Alice will be twice as old as Bob was ten years ago. If Alice is 40 now, how old is Bob?”

Given that Alice is currently 40, in ten years, she will be 50. We can set up the equation as 50 = 2*(B-10). Solving for B, we find that Bob is currently 35 years old.

### Digit Problems

Digit problems often involve understanding the place value of digits in a number. Let’s consider a problem:

“If the digit in the tens place of a two-digit number is three times the digit in the units place, and the number itself is 4 times the sum of its digits, what is the number?”

Let’s say the ten’s digit is 3x, and the unit’s digit is x. The number is then 30x + x = 31x. If the number is also 4 times the sum of the digits, we have 31x = 4*(3x + x). Solving for x gives us x = 2. Hence the number is 62.

## Shortcuts

A good shortcut for age problems is always to set the current time as the reference point. This simplifies the problem by reducing the number of variables you have to consider.

For digit problems, a handy shortcut is to express the number in terms of its digit. This allows you to form equations that can be solved easily.

## Importance of ‘Age and Digit’ Problems in the GRE

‘Age and Digit’ problems are crucial in the GRE because they test your ability to translate verbal descriptions into mathematical expressions, a critical skill in many areas of quantitative reasoning. Moreover, they assess your understanding of basic algebra and number properties, fundamental concepts that the GRE aims to evaluate.

In the next part of this series, we will cover the practical uses of these concepts and explore more complex examples. Stay tuned as we continue to unravel the secrets of ‘Age and Digit’ problems in the GRE.

# Mastering the GRE: A Deep Dive into ‘Age and Digit’ Problems – Part 3

In this third installment of our five-part series, we’re continuing our exploration of ‘Age and Digit’ problems in the GRE. Brought to you by MKSprep, we’ve already covered the concept, basic formulas, reasoning, shortcuts, and their importance in the GRE. Now, let’s explore these problems’ practical use and provide more complex examples.

## Practical Use of ‘Age and Digit’ Problems

While the ‘Age and Digit’ problems seem like abstract mathematical exercises, they actually have quite a few real-world applications.

Age problems help us understand and predict scenarios that involve time progression, such as calculating retirement age, understanding population demographics, or even in medical fields to calculate dosages over time.

Digit problems, on the other hand, play a crucial role in computing and data science. Understanding the value of individual digits within a number is fundamental to number systems, coding, data encryption, and much more.

## More Complex Examples

Let’s delve deeper into some more complex examples.

### Age Problems

Consider this problem: “Five years ago, a mother was four times as old as her son. In four years, she’ll be twice as old. How old are they now?”

We can start by defining the current ages of the mother (M) and the son (S). The problem gives us two equations: M – 5 = 4 * (S – 5) and M + 4 = 2 * (S + 4). Solving this system of equations gives us their current ages.

### Digit Problems

Now let’s consider this digit problem: “A two-digit number is such that the product of the digits is 24. When the digits are reversed, the number increases by 18. What is the number?”

Let’s say the tens digit is T and the ones digit is O. Our problem gives us two equations: T * O = 24 and 10O + T = 10T + O + 18. By solving these equations, we can find the original number.

By now, you should be starting to feel more comfortable with ‘Age and Digit’ problems. In the next part of this series, we’ll cover strategies to approach these problems and avoid common mistakes. Stay tuned for more from MKSprep!

# Mastering the GRE: A Deep Dive into ‘Age and Digit’ Problems – Part 4

Welcome back to part four of our series on ‘Age and Digit’ problems, brought to you by MKSprep. So far, we’ve covered the concept, formulas, reasoning, shortcuts, importance, and practical use of these problems in the GRE. Now, it’s time to provide you with strategies to approach these problems and tips on avoiding common mistakes.

## Strategies to Approach ‘Age and Digit’ Problems

### Age Problems

1. Identify the Variables: Start by identifying the individuals involved and assign variables to their current ages.
2. Translate Words into Equations: Translate the given sentences into algebraic equations. Look for keywords such as “is,” “was,” “will be, “”years ago,” and “in x years” to guide you.
3. Solve the Equations: Use algebraic methods to solve for the variables. In most cases, you will have to solve simultaneous equations.

### Digit Problems

1. Understand the Number Structure: Break down the number into its digits. Understand the value of each digit based on its position.
2. Formulate the Equations: Translate the problem into equations. The digit in the tens place, the digit in the ones place, and the number itself can all be expressed in terms of these digits.
3. Solve the Equations: Solve the resulting equations to find the value of the digits.

## Common Mistakes to Avoid

When solving ‘Age and Digit’ problems, there are a few common pitfalls to avoid:

1. Misinterpreting the Problem: Be careful to translate the words into mathematical equations correctly. A common mistake is misinterpreting the timeframe or the relationship between the ages.
2. Overcomplicating the Problem: Try to keep your equations as simple as possible. Use the current time as the reference point to simplify age problems and express the number in terms of its digits to simplify digit problems.
3. Arithmetic Errors: Be careful with your calculations. Even a small mistake can lead to an incorrect answer.

Mastering ‘Age and Digit’ problems requires understanding the concept, carefully translating words into equations, and meticulous calculations. In the final part of this series, we will look at some GRE-level examples and provide their detailed solutions. Stay tuned for more from MKSprep!

# Mastering the GRE: A Deep Dive into ‘Age and Digit’ Problems – Part 5

As we reach the conclusion of our five-part series on ‘Age and Digit’ problems in the GRE, brought to you by MKSprep, it’s time to put all the theory into practice. We have covered the concept, formulas, reasoning, shortcuts, importance, practical use, strategies, and common mistakes. In this final part, we will take you through a couple of GRE-level examples and explain their solutions.

## GRE-Level ‘Age and Digit’ Problems

### Age Problem

Here’s a GRE-level age problem:

“Three years from now, the combined age of Alice and Bob will be twice as much as their combined age was five years ago. If Bob is three years older than Alice, how old is Alice?”

To solve this, let’s denote Alice’s current age as ‘A’ and Bob’s current age as ‘A + 3’ (since Bob is three years older than Alice). According to the problem, we have:

(A + 3) + (A + 3 + 3) = 2 * ((A – 5) + (A + 3 – 5))

Solving this equation will give us Alice’s age.

### Digit Problem

Here’s a GRE-level digit problem:

“The difference between a two-digit number and the number obtained by reversing the digits is 36. The number is 4 times the sum of its digits. What is the number?”

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. The number then is 10T + O, and the number with reversed digits is 10O + T. We have two equations:

10T + O – (10O + T) = 36 and 10T + O = 4 * (T + O)

Solving this system of equations will give us the original number.

## Conclusion

Mastering ‘Age and Digit’ problems for the GRE can seem daunting at first, but with practice, these problems become much more approachable. By understanding the underlying concepts, developing a strategy, and avoiding common mistakes, you’ll be well on your way to acing these problems on the GRE.

Thank you for joining us in this series. We at MKSprep are dedicated to providing you with the tools and knowledge to succeed in your GRE. Stay tuned for more helpful resources and guides to help you in your GRE preparation journey!

## GRE Age and Digit Practice Question

Question 1:

Alice is now twice as old as Bob was four years ago. In 6 years, Alice will be 60 years old. How old is Bob now?

A) 22 B) 24 C) 26 D) 28 E) 30

Solution:

Given that Alice will be 60 years old in 6 years, this means that Alice is currently 60 – 6 = 54 years old.

Also, it is given that Alice is now twice as old as Bob was four years ago. Therefore, Bob was 54/2 = 27 years old four years ago.

This means Bob is currently 27 + 4 = 31 years old.

So, the answer is 31 years old. This option is not available in the multiple-choice answers, so it would be the write-in answer.

Question 2:

The sum of the digits of a two-digit number is 11. If the digits are reversed, the number decreases by 45. What is the original number?

A) 29 B) 38 C) 65 D) 74 E) 92

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. From the problem, we have two equations:

1. T + O = 11 (The sum of the digits is 11)
2. 10T + O – (10O + T) = 45 (The number decreases by 45 when the digits are reversed)

Solving these equations gives us T = 7 and O = 4. So, the original number is 74.

So, the correct answer is D) 74.

Question 3:

Three years from now, a father will be thrice as old as his son. Six years ago, the father was seven times as old as the son. How old is the son currently?

A) 9 B) 12 C) 15 D) 18 E) 21

Solution:

Let’s denote the current ages of the father as ‘F’ and the son as ‘S.’ From the problem, we can form two equations:

1. F + 3 = 3 * (S + 3) (Three years from now, the father will be thrice as old as the son)
2. F – 6 = 7 * (S – 6) (Six years ago, the father was seven times as old as the son)

Solving these equations gives us F = 39 and S = 12.

So, the correct answer is B) 12.

Question 4:

A two-digit number is four times the sum of its digits. The digit in the units place is thrice the digit in the tens place. What is the number?

A) 12 B) 24 C) 36 D) 42 E) 48

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. From the problem, we have two equations:

1. 10T + O = 4 * (T + O) (The number is four times the sum of its digits)
2. O = 3 * T (The digit in the units place is thrice the digit in the tens place)

Solving these equations gives us T = 1 and O = 3. So, the original number is 13.

However, 13 is not available in the options, which indicates there might be an error in the problem or the solution. The problem statement might be ambiguous or might contain a mistake.

Question 5:

In five years, Anna will be twice as old as Ben. Five years ago, Anna was four times as old as Ben. How old is Ben now?

A) 10 B) 15 C) 20 D) 25 E) 30

Solution:

Let’s denote the current ages of Anna as ‘A’ and Ben as ‘B.’ We can form two equations from the problem:

1. A + 5 = 2 * (B + 5) (In five years, Anna will be twice as old as Ben)
2. A – 5 = 4 * (B – 5) (Five years ago, Anna was four times as old as Ben)

Solving these equations gives us A = 35 and B = 20.

So, the correct answer is C) 20.

Question 6:

A two-digit number is five times the sum of its digits. The digit in the units place is twice the digit in the tens place. What is the number?

A) 15 B) 25 C) 35 D) 45 E) 55

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. From the problem, we have two equations:

1. 10T + O = 5 * (T + O) (The number is five times the sum of its digits)
2. O = 2 * T (The digit in the units place is twice the digit in the tens place)

Solving these equations gives us T = 2 and O = 4. So, the original number is 24.

However, 24 is not available in the options, which indicates there might be an error in the problem or the solution. The problem statement might be ambiguous or might contain a mistake.

Question 7:

Six years from now, Mike will be thrice as old as he was nine years ago. How old is Mike currently?

A) 18 B) 21 C) 24 D) 27 E) 30

Solution:

Let’s denote Mike’s current age as ‘M.’ From the problem, we can form the equation:

M + 6 = 3 * (M – 9) (Six years from now, Mike will be thrice as old as he was nine years ago)

Solving this equation gives us M = 24.

So, the correct answer is C) 24.

Question 8:

A two-digit number is seven times the sum of its digits. The digit in the units place is four times the digit in the tens place. What is the number?

A) 14 B) 28 C) 37 D) 56 E) 84

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. From the problem, we have two equations:

1. 10T + O = 7 * (T + O) (The number is seven times the sum of its digits)
2. O = 4 * T (The digit in the units place is four times the digit in the tens place)

Solving these equations gives us T = 2 and O = 8. So, the original number is 28.

So, the correct answer is B) 28.

Question 9:

Three years ago, the sum of Paul and Mary’s ages was 41. In 7 years, Paul will be twice as old as Mary. If Paul is older than Mary, how old is Paul now?

A) 22 B) 24 C) 26 D) 28 E) 30

Solution:

Let’s denote Paul’s current age as ‘P’ and Mary’s current age as ‘M.’ From the problem, we can form two equations:

1. P – 3 + M – 3 = 41 (Three years ago, the sum of Paul and Mary’s ages was 41)
2. P + 7 = 2 * (M + 7) (In 7 years, Paul will be twice as old as Mary)

Solving these equations gives us P = 24 and M = 23.

So, the correct answer is B) 24.

Question 10:

A two-digit number is six times the sum of its digits. The digit in the units place is five times the digit in the tens place. What is the number?

A) 15 B) 30 C) 45 D) 60 E) 75

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. From the problem, we have two equations:

1. 10T + O = 6 * (T + O) (The number is six times the sum of its digits)
2. O = 5 * T (The digit in the units place is five times the digit in the tens place)

Solving these equations gives us T = 2 and O = 10. However, a digit can’t be 10, which indicates there might be an error in the problem or the solution. The problem statement might be ambiguous or might contain a mistake.

Question 11:

Four years ago, the age of the mother was three times the age of her daughter. In eight years, the mother will be twice as old as her daughter. What is the current age of the daughter?

A) 12 B) 16 C) 20 D) 24 E) 28

Question 12:

Let’s denote the current ages of the mother as ‘M’ and the daughter as ‘D.’ From the problem, we can form two equations:

1. M – 4 = 3 * (D – 4) (Four years ago, the mother was three times the age of the daughter)
2. M + 8 = 2 * (D + 8) (In eight years, the mother will be twice as old as the daughter)

Solving these equations gives us M = 40 and D = 16.

So, the correct answer is B) 16.

Question 13:

A two-digit number is four times the sum of its digits. The digit in the units place is three times the digit in the tens place. What is the number?

A) 12 B) 24 C) 36 D) 48 E) 60

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. From the problem, we have two equations:

1. 10T + O = 4 * (T + O) (The number is four times the sum of its digits)
2. O = 3 * T (The digit in the units place is three times the digit in the tens place)

Solving these equations gives us T = 3 and O = 9. So, the original number is 39.

However, 39 is not available in the options, which indicates there might be an error in the problem or the solution. The problem statement might be ambiguous or might contain a mistake.

Question 14:

In 4 years, Henry will be twice as old as he was 8 years ago. How old is Henry now?

A) 16 B) 20 C) 24 D) 28 E) 32

Solution:

Let’s denote Henry’s current age as ‘H.’ From the problem, we can form the equation:

H + 4 = 2 * (H – 8) (In 4 years, Henry will be twice as old as he was 8 years ago)

Solving this equation gives us H = 20.

So, the correct answer is B) 20.

Question 15:

The sum of the ages of a father and his son is 70 years. Five years ago, the father was four times as old as the son. How old is the son?

A) 15 B) 20 C) 25 D) 30 E) 35

Solution:

Let’s denote the father’s current age as ‘F’ and the son’s current age as ‘S.’ We can form two equations from the problem:

1. F + S = 70 (The sum of the ages of the father and his son is 70 years)
2. F – 5 = 4 * (S – 5) (Five years ago, the father was four times as old as the son)

Solving these equations gives us F = 50 and S = 20.

So, the correct answer is B) 20.

Question 16:

A two-digit number is nine times the sum of its digits. The digit in the units place is three times the digit in the tens place. What is the number?

A) 27 B) 36 C) 45 D) 54 E) 63

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. From the problem, we have two equations:

1. 10T + O = 9 * (T + O) (The number is nine times the sum of its digits)
2. O = 3 * T (The digit in the units place is three times the digit in the tens place)

Solving these equations gives us T = 3 and O = 9. So, the original number is 39.

However, 39 is not available in the options, which indicates there might be an error in the problem or the solution. The problem statement might be ambiguous or might contain a mistake.

Question 17:

The sum of the ages of a mother and her daughter is 80 years. Eight years ago, the mother was five times as old as the daughter. How old is the daughter?

A) 20 B) 24 C) 28 D) 32 E) 36

Solution:

Let’s denote the mother’s current age as ‘M’ and the daughter’s current age as ‘D.’ From the problem, we can form two equations:

1. M + D = 80 (The sum of the ages of the mother and her daughter is 80 years)
2. M – 8 = 5 * (D – 8) (Eight years ago, the mother was five times as old as the daughter)

Solving these equations gives us M = 56 and D = 24.

So, the correct answer is B) 24.

Question 18:

A two-digit number is eight times the sum of its digits. The digit in the units place is twice the digit in the tens place. What is the number?

A) 16 B) 24 C) 32 D) 40 E) 48

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. From the problem, we have two equations:

1. 10T + O = 8 * (T + O) (The number is eight times the sum of its digits)
2. O = 2 * T (The digit in the units place is twice the digit in the tens place)

Solving these equations gives us T = 2 and O = 4. So, the original number is 24.

So, the correct answer is B) 24.

Question 19:

The sum of the ages of a mother and her son is 66 years. Six years ago, the mother was six times as old as the son. How old is the son?

A) 12 B) 18 C) 24 D) 30 E) 36

Solution:

Let’s denote the mother’s current age as ‘M’ and the son’s current age as ‘S.’ We can form two equations from the problem:

1. M + S = 66 (The sum of the ages of the mother and her son is 66 years)
2. M – 6 = 6 * (S – 6) (Six years ago, the mother was six times as old as the son)

Solving these equations gives us M = 54 and S = 12.

So, the correct answer is A) 12.

Question 20:

The sum of the ages of a father and his son is 85 years. Ten years ago, the father was seven times as old as the son. How old is the son?

A) 15 B) 20 C) 25 D) 30 E) 35

Solution:

Let’s denote the father’s current age as ‘F’ and the son’s current age as ‘S.’ We can form two equations from the problem:

1. F + S = 85 (The sum of the ages of the father and his son is 85 years)
2. F – 10 = 7 * (S – 10) (Ten years ago, the father was seven times as old as the son)

Solving these equations is challenging, but it gives us F = 70 and S = 15.

So, the correct answer is A) 15.

Question 21:

A two-digit number is ten times the sum of its digits. The digit in the units place is three times the digit in the tens place. What is the number?

A) 20 B) 30 C) 40 D) 50 E) 60

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. From the problem, we have two equations:

1. 10T + O = 10 * (T + O) (The number is ten times the sum of its digits)
2. O = 3 * T (The digit in the units place is three times the digit in the tens place)

Solving these equations, especially the first equation is challenging, but it gives us T = 2 and O = 6. So, the original number is 26.

However, 26 is not available in the options, which indicates there might be an error in the problem or the solution. The problem statement might be ambiguous or might contain a mistake.

Question 22:

The sum of the ages of a mother and her daughter is 90 years. Twelve years ago, the mother was eight times as old as the daughter. How old is the daughter?

A) 16 B) 22 C) 28 D) 34 E) 40

Solution:

Let’s denote the mother’s current age as ‘M’ and the daughter’s current age as ‘D.’ We can form two equations from the problem:

1. M + D = 90 (The sum of the ages of the mother and her daughter is 90 years)
2. M – 12 = 8 * (D – 12) (Twelve years ago, the mother was eight times as old as the daughter)

Solving these equations is challenging, but it gives us M = 70 and D = 20.

However, 20 is not available in the options, which indicates there might be an error in the problem or the solution. The problem statement might be ambiguous or might contain a mistake.

Question 23:

A mother is three times as old as her son. In how many years will the mother be twice as old as her son? Select all that apply.

A) 5 B) 10 C) 15 D) 20 E) 25 F) 30

Solution :

Let’s denote the mother’s current age as ‘M’ and the son’s current age as ‘S.’ We have the equation from the problem:

M = 3S (The mother is three times as old as the son)

We are looking for the number of years ‘Y’ when the mother will be twice as old as the son. So, we have another equation:

M + Y = 2 * (S + Y)

Substituting M = 3S into the second equation, we get:

3S + Y = 2S + 2Y S = Y

This means that the number of years ‘Y’ is equal to the son’s current age ‘S.’ Since the mother is currently three times as old as the son, the son’s age (and therefore ‘Y’) could be any multiple of 3.

Among the options, only B) 10 and E) 25 are multiples of 3.

So, the correct answers are B) 10 and E) 25.

Question 24:

A two-digit number is five times the sum of its digits. Which of the following could be the number? Select all that apply.

A) 10 B) 15 C) 20 D) 30 E) 35 F) 40

Solution:

A two-digit number can be expressed as 10T + O, where T is the tens digit, and O is the one’s digit. From the problem, we have the equation:

10T + O = 5 * (T + O)

Solving this equation gives us 5T = 4O. This implies that the one’s digit (O) must be a multiple of 5, and the tens digit (T) must be a multiple of 4.

Among the options, only B) 15 and E) 35 meet this requirement.

So, the correct answers are B) 15 and E) 35.

Question 25:

The sum of the ages of a father and his son is 50 years. In which of the following number of years will the father be twice as old as the son? Select all that apply.

A) 5 B) 10 C) 15 D) 20 E) 25 F) 30

Solution:

Let’s denote the father’s current age as ‘F’ and the son’s current age as ‘S.’ We have the equation from the problem:

F + S = 50

We are looking for the number of years ‘Y’ when the father will be twice as old as the son. So, we have another equation:

F + Y = 2 * (S + Y)

Substituting F = 50 – S into the second equation, we get:

50 – S + Y = 2S + 2Y 50 + Y = 3S + 2Y Y = 3S – 50

This implies that the number of years ‘Y’ when the father will be twice as old as the son is equal to three times the son’s current age minus 50.

Among the options, only B) 10, C) 15 and D) 20 could be obtained for some values of S (For B) 10, S = 20; for C) 15, S = 22; for D) 20, S = 23).

So, the correct answers are B) 10, C) 15, and D) 20.

Question 26:

A two-digit number is four times the sum of its digits. Which of the following could be the number? Select all that apply.

A) 12 B) 24 C) 36 D) 48 E) 60 F) 72

Solution:

A two-digit number can be expressed as 10T + O, where T is the tens digit, and O is the one’s digit. From the problem, we have the equation:

10T + O = 4 * (T + O)

Solving this equation gives us 6T = 3O. This implies that the one digit (O) must be twice the tens digit (T).

Among the options, only B) 24, D) 48, and F) 72 meet this requirement.

So, the correct answers are B) 24, D) 48, and F) 72.

Question 27:

The sum of the ages of a mother and her daughter is 60 years. In which of the following number of years will the mother be twice as old as the daughter? Select all that apply.

A) 10 B) 15 C) 20 D) 25 E) 30 F) 35

Solution:

Let’s denote the mother’s current age as ‘M’ and the daughter’s current age as ‘D.’ We have the equation from the problem:

M + D = 60

We are looking for the number of years ‘Y’ when the mother will be twice as old as the daughter. So, we have another equation:

M + Y = 2 * (D + Y)

Substituting M = 60 – D into the second equation, we get:

60 – D + Y = 2D + 2Y 60 + Y = 3D + 2Y Y = 3D – 60

This implies that the number of years ‘Y’ when the mother will be twice as old as the daughter is equal to three times the daughter’s current age minus 60.

Among the options, only B) 15, C) 20 and D) 25 could be obtained for some values of D (For B) 15, D = 25; for C) 20, D = 27; for D) 25, D = 28).

So, the correct answers are B) 15, C) 20, and D) 25.

Question 28:

A two-digit number is six times the sum of its digits. Which of the following could be the number? Select all that apply.

A) 24 B) 36 C) 48 D) 60 E) 72 F) 84

Solution:

A two-digit number can be expressed as 10T + O, where T is the tens digit, and O is the one’s digit. From the problem, we have the equation:

10T + O = 6 * (T + O)

Solving this equation gives us 4T = 5O. This implies that the one’s digit (O) must be 0.8 times the tens digit (T).

Among the options, only D) 60 and F) 84 meet this requirement (For D) 60, T = 6 and O = 0, which is 0.8 * 6; for F) 84, T = 8 and O = 4, which is 0.8 * 8).

So, the correct answers are D) 60 and F) 84.

Question 29:

A two-digit number is seven times the sum of its digits. Which of the following could be the number? Select all that apply.

A) 14 B) 28 C) 42 D) 56 E) 70 F) 84

Solution:

A two-digit number can be expressed as 10T + O, where T is the tens digit, and O is the one’s digit. From the problem, we have the equation:

10T + O = 7 * (T + O)

Solving this equation gives us 3T = 6O. This implies that the one’s digit (O) must be 0.5 times the tens digit (T).

Among the options, only A) 14 and E) 70 meet this requirement (For A) 14, T = 1 and O = 4, which is 0.5 * 1; for E) 70, T = 7 and O = 0, which is 0.5 * 7).

So, the correct answers are A) 14 and E) 70.

Question 30:

A mother is four times as old as her son. In how many years will the mother be twice as old as her son? Select all that apply.

A) 5 B) 10 C) 15 D) 20 E) 25 F) 30

Solution:

Let’s denote the mother’s current age as ‘M’ and the son’s current age as ‘S.’ We have the equation from the problem:

M = 4S (The mother is four times as old as the son)

We are looking for the number of years ‘Y’ when the mother will be twice as old as the son. So, we have another equation:

M + Y = 2 * (S + Y)

Substituting M = 4S into the second equation, we get:

4S + Y = 2S + 2Y 2S = Y

This means that the number of years ‘Y’ is equal to twice the son’s current age ‘S.’ Since the mother is currently four times as old as the son, the son’s age (and therefore ‘Y’) could be any multiple of 2.

Among the options, all are multiples of 2.

So, the correct answers are A) 5, B) 10, C) 15, D) 20, E) 25, and F) 30.

Question 31:

A father is five times as old as his son. In how many years will the father be twice as old as the son? Select all that apply.

A) 10 B) 15 C) 20 D) 25 E) 30 F) 35

Solution:

Let’s denote the father’s current age as ‘F’ and the son’s current age as ‘S.’ We have the equation from the problem:

F = 5S (The father is five times as old as the son)

We are looking for the number of years ‘Y’ when the father will be twice as old as the son. So, we have another equation:

F + Y = 2 * (S + Y)

Substituting F = 5S into the second equation, we get:

5S + Y = 2S + 2Y 3S = Y

This means that the number of years ‘Y’ is equal to three times the son’s current age ‘S.’ Since the father is currently five times as old as the son, the son’s age (and therefore ‘Y’) could be any multiple of 3.

Among the options, only B) 15, and E) 30 meet this requirement.

So, the correct answers are B) 15 and E) 30.

Question 32:

A two-digit number is eight times the sum of its digits. Which of the following could be the number? Select all that apply.

A) 16 B) 32 C) 48 D) 64 E) 80 F) 96

Solution:

A two-digit number can be expressed as 10T + O, where T is the tens digit, and O is the one’s digit. From the problem, we have the equation:

10T + O = 8 * (T + O)

Solving this equation gives us 2T = 7O. This equation can’t be fulfilled with integer values for T and O since 7 can’t evenly divide any even number.

Therefore, there are no correct options among the given answer choices.

For Each Quantity Comparison

A) The quantity in Column A is greater

B) The quantity in Column B is greater

C) The two quantities are equal

D) The relationship cannot be determined from the information given

Question 33:

Column A: The sum of the digits of a two-digit number, where the number is six times the sum of its digits.

Column B: The tens digit of the same number.

Solution:

If the number is six times the sum of its digits, the two-digit number can be expressed as 10T + O, where T is the tens digit, and O is the one’s digit. From this, we have:

10T + O = 6 * (T + O) 10T + O = 6T + 6O 4T = 5O

This implies that the one’s digit (O) is 0.8 times the tens digit (T). It shows that the tens digit is always greater than the ones digit. Therefore, the sum of the digits (which includes the ones digit) will always be greater than just the tens digit.

So, the correct answer is A) The quantity in Column A is greater.

Question 34:

Column A: The number of years from now when a mother will be twice as old as her daughter, given that the mother is currently four times as old as the daughter.

Column B: The daughter’s current age.

Solution:

Let’s denote the mother’s current age as ‘M’ and the daughter’s current age as ‘D.’ We have the equation:

M = 4D (The mother is currently four times as old as the daughter)

We are looking for the number of years ‘Y’ when the mother will be twice as old as the daughter. So, we have another equation:

M + Y = 2 * (D + Y)

Substituting M = 4D into the second equation, we get:

4D + Y = 2D + 2Y 2D = Y

This means that the number of years ‘Y’ when the mother will be twice as old as the daughter is equal to twice the daughter’s current age ‘D.’ Hence, the two quantities are equal.

So, the correct answer is C) The two quantities are equal.

Question 35:

Column A: The sum of the digits of a two-digit number, where the number is five times the sum of its digits.

Column B: The tens digit of the same number.

Solution:

If the number is five times the sum of its digits, the two-digit number can be expressed as 10T + O, where T is the tens digit, and O is the one’s digit. From this, we have:

10T + O = 5 * (T + O) 10T + O = 5T + 5O 5T = 4O

This implies that the one’s digit (O) is 1.25 times the tens digit (T). Since the one digit can only be a whole number, T has to be 4, which makes O = 5.

So, the sum of the digits is T + O = 4 + 5 = 9, and the tens digit is 4.

Therefore, the correct answer is A) The quantity in Column A is greater.

Question 36:

Column A: The number of years from now when a father will be twice as old as his son, given that the father is currently five times as old as the son.

Column B: The son’s current age.

Solution:

Let’s denote the father’s current age as ‘F’ and the son’s current age as ‘S.’ We have the equation:

F = 5S (The father is currently five times as old as the son)

We are looking for the number of years ‘Y’ when the father will be twice as old as the son. So, we have another equation:

F + Y = 2 * (S + Y)

Substituting F = 5S into the second equation, we get:

5S + Y = 2S + 2Y 3S = Y

This means that the number of years ‘Y’ when the father will be twice as old as the son is equal to three times the son’s current age ‘S.’ Hence, the quantity in Column A is greater.

So, the correct answer is A) The quantity in Column A is greater.

Question 37:

Column A: The sum of the digits of a two-digit number, where the number is seven times the sum of its digits.

Column B: The tens digit of the same number.

Solution:

If the number is seven times the sum of its digits, the two-digit number can be expressed as 10T + O, where T is the tens digit, and O is the one’s digit. From this, we have:

10T + O = 7 * (T + O) 10T + O = 7T + 7O 3T = 6O

This implies that the one’s digit (O) is 0.5 times the tens digit (T). Since the one digit can only be a whole number, T has to be 2, which makes O = 1.

So, the sum of the digits is T + O = 2 + 1 = 3, and the tens digit is 2.

Therefore, the correct answer is A) The quantity in Column A is greater.

Question 38:

Column A: The number of years from now when a mother will be thrice as old as her daughter, given that the mother is currently six times as old as the daughter.

Column B: The daughter’s current age.

Solution:

Let’s denote the mother’s current age as ‘M’ and the daughter’s current age as ‘D.’ We have the equation:

M = 6D (The mother is currently six times as old as the daughter)

We are looking for the number of years ‘Y’ when the mother will be thrice as old as the daughter. So, we have another equation:

M + Y = 3 * (D + Y)

Substituting M = 6D into the second equation, we get:

6D + Y = 3D + 3Y 3D = 2Y

This means that the number of years ‘Y’ when the mother will be thrice as old as the daughter is 1.5 times the daughter’s current age ‘D.’ Hence, the quantity in Column A is greater.

So, the correct answer is A) The quantity in Column A is greater.

Question 39:

Column A: The sum of the digits of a two-digit number, where the number is four times the sum of its digits.

Column B: The tens digit of the same number.

Solution:

If the number is four times the sum of its digits, the two-digit number can be expressed as 10T + O, where T is the tens digit, and O is the one’s digit. From this, we have:

10T + O = 4 * (T + O) 10T + O = 4T + 4O 6T = 3O

This implies that the one’s digit (O) is 2 times the tens digit (T). Since the one digit can only be a whole number, T has to be 2, which makes O = 4.

So, the sum of the digits is T + O = 2 + 4 = 6, and the tens digit is 2.

Therefore, the correct answer is A) The quantity in Column A is greater.

Question 40:

Column A: The number of years from now when a father will be thrice as old as his son, given that the father is currently seven times as old as the son.

Column B: The son’s current age.

Solution:

Let’s denote the father’s current age as ‘F’ and the son’s current age as ‘S.’ We have the equation:

F = 7S (The father is currently seven times as old as the son)

We are looking for the number of years ‘Y’ when the father will be thrice as old as the son. So, we have another equation:

F + Y = 3 * (S + Y)

Substituting F = 7S into the second equation, we get:

7S + Y = 3S + 3Y 4S = 2Y

This means that the number of years ‘Y’ when the father will be thrice as old as the son is 2 times the son’s current age ‘S.’ Hence, the quantity in Column A is greater.

So, the correct answer is A) The quantity in Column A is greater.

Question 41:

Column A: The sum of the digits of a two-digit number, where the number is nine times the sum of its digits.

Column B: The tens digit of the same number.

Solution:

If the number is nine times the sum of its digits, the two-digit number can be expressed as 10T + O, where T is the tens digit and O is the ones digit. From this, we have:

10T + O = 9 * (T + O) 10T + O = 9T + 9O T = 😯

This implies that the one’s digit (O) is 0.125 times the tens digit (T). This is not possible since the one digit can only be a whole number.

Therefore, the relationship cannot be determined from the information given.

So, the correct answer is D). The relationship cannot be determined from the information given.

Question 42:

Column A: The number of years from now when a man will be twice as old as his son, given that the man is currently ten times as old as his son.

Column B: The son’s current age.

Solution:

Let’s denote the man’s current age as ‘M’ and the son’s current age as ‘S.’ We have the equation:

M = 10S (The man is currently ten times as old as the son)

We are looking for the number of years ‘Y’ when the man will be twice as old as the son. So, we have another equation:

M + Y = 2 * (S + Y)

Substituting M = 10S into the second equation, we get:

10S + Y = 2S + 2Y 8S = Y

This means that the number of years ‘Y’ when the man will be twice as old as the son is 8 times the son’s current age ‘S.’ Hence, the quantity in Column A is greater.

So, the correct answer is A) The quantity in Column A is greater.

Question 43:

Column A: The sum of the digits of a two-digit number, where the number is eleven times the sum of its digits.

Column B: The tens digit of the same number.

Solution:

If the number is eleven times the sum of its digits, the two-digit number can be expressed as 10T + O, where T is the tens digit and O is the ones digit. From this, we have:

10T + O = 11 * (T + O) 10T + O = 11T + 11O -T = 10O

This implies that the tens digit (T) is -10 times the ones digit (O). This is not possible since the tens digit cannot be a negative number.

Therefore, the relationship cannot be determined from the information given.

So, the correct answer is D). The relationship cannot be determined from the information given.

Question 44:

Column A: The number of years from now when a mother will be twice as old as her daughter, given that the mother is currently twelve times as old as the daughter.

Column B: The daughter’s current age.

Solution:

Let’s denote the mother’s current age as ‘M’ and the daughter’s current age as ‘D.’ We have the equation:

M = 12D (The mother is currently twelve times as old as the daughter)

We are looking for the number of years ‘Y’ when the mother will be twice as old as the daughter. So, we have another equation:

M + Y = 2 * (D + Y)

Substituting M = 12D into the second equation, we get:

12D + Y = 2D + 2Y 10D = Y

This means that the number of years ‘Y’ when the mother will be twice as old as the daughter is 10 times the daughter’s current age ‘D.’ Hence, the quantity in Column A is greater.

So, the correct answer is A) The quantity in Column A is greater.

Question 45:

Five years ago, a father was four times as old as his son. In four years from now, the father will be twice as old as his son. How old is the son currently?

Solution:

Let’s denote the father’s current age as ‘F’ and the son’s current age as ‘S.’ We have two equations:

F – 5 = 4 * (S – 5) (Five years ago, the father was four times as old as his son)

F + 4 = 2 * (S + 4) (In four years from now, the father will be twice as old as his son)

Solving these two equations simultaneously, we get:

F = 4S – 15 and F = 2S + 4

Setting these two equal to each other:

4S – 15 = 2S + 4 2S = 19

S = 19 / 2 = 9.5 years

The son is currently 9.5 years old.

Question 46:

A two-digit number is such that the number is seven times the sum of its digits. What is the number?

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. Then, the two-digit number can be represented as 10T + O. We have the equation:

10T + O = 7 * (T + O)

Solving this equation, we get:

10T + O = 7T + 7O 3T = 6O

This implies that the one’s digit (O) is 0.5 times the tens digit (T). Since the one digit can only be a whole number, T has to be 2, which makes O = 1.

So, the number is 10 * T + O = 10 * 2 + 1 = 21.

Question 47:

The product of the digits of a two-digit number is 12, and the difference between the number and the one obtained by reversing its digits is 18. What is the original number?

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. Then, the two-digit number can be represented as 10T + O and the reversed number as 10O + T. We have two equations:

T * O = 12 (The product of the digits is 12)

10T + O – (10O + T) = 18 (The difference between the number and the one obtained by reversing its digits is 18)

From the first equation, we find that the possible pairs of (T,O) are (3,4), (4,3), (2,6), (6,2), (1,12), (12,1). Considering only the pairs which consist of single digits, we get four possibilities: (3,4), (4,3), (2,6), (6,2).

Substituting these into the second equation, we find that the pair (4,3) satisfies the equation since 104 + 3 – (103 + 4) = 18.

So, the original number is 10 * T + O = 10 * 4 + 3 = 43.

Question 48:

A man is currently five times as old as his son. In 7 years, the man will be three times as old as his son. How old is the man currently?

Solution:

Let’s denote the man’s current age as ‘M’ and the son’s current age as ‘S.’ We have two equations:

M = 5S (The man is currently five times as old as his son)

M + 7 = 3 * (S + 7) (In seven years from now, the man will be three times as old as his son)

Solving these two equations simultaneously, we get:

M = 5S and M = 3S + 14

Setting these two equal to each other:

5S = 3S + 14 2S = 14

S = 14 / 2 = 7 years

So, the man is currently M = 5 * S = 5 * 7 = 35 years old.

Question 49:

In a two-digit number, the digit in the unit’s place is twice that in the tens place. If 18 is subtracted from the number, the digits are reversed. What is the original number?

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. Then, the two-digit number can be represented as 10T + O. We have two equations:

O = 2T (The digit in the unit’s place is twice that in the tens place)

10T + O – 18 = 10O + T (If 18 is subtracted from the number, the digits are reversed)

Substituting the first equation into the second one, we get:

10T + 2T – 18 = 20T + T 8T = 18 T = 18 / 8 = 2.25

Since T must be a whole number, no solution exists for this problem in the set of whole numbers. This could be a trick question to test your understanding of the number system.

Question 50:

A father is 24 years older than his son. In two years, the father’s age will be twice the age of the son. How old is the son currently?

Solution:

Let’s denote the father’s current age as ‘F’ and the son’s current age as ‘S.’ We have two equations:

F = S + 24 (The father is 24 years older than his son)

F + 2 = 2 * (S + 2) (In two years, the father’s age will be twice the age of the son)

Solving these two equations simultaneously, we get:

F = S + 24 and F = 2S + 2 – 2

Setting these two equal to each other:

S + 24 = 2S S = 24 years

So, the son is currently 24 years old.

Question 51:

A mother is 21 years older than her daughter. In 6 years, the mother’s age will be 3 times the age of the daughter. How old is the daughter currently?

Solution:

Let’s denote the mother’s current age as ‘M’ and the daughter’s current age as ‘D.’ We have two equations:

M = D + 21 (The mother is 21 years older than her daughter)

M + 6 = 3 * (D + 6) (In 6 years, the mother’s age will be three times the age of the daughter)

Solving these two equations simultaneously, we get:

M = D + 21 and M = 3D + 12

Setting these two equal to each other:

D + 21 = 3D + 12 2D = 9 D = 9 / 2 = 4.5 years

So, the daughter is currently 4.5 years old.

Question 52:

A two-digit number is such that the number is five times the sum of its digits. If the number is increased by 9, the result is the same number with its digits reversed. What is the original number?

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. Then, the two-digit number can be represented as 10T + O. We have two equations:

10T + O = 5 * (T + O) (The number is five times the sum of its digits)

10T + O + 9 = 10O + T (If the number is increased by 9, the result is the same number with its digits reversed)

Solving these two equations simultaneously, we get:

10T + O = 5T + 5O and 10T + O + 9 = 10O + T

Solving for T and O, we get T = 3 and O = 2.

So, the original number is 10 * T + O = 10 * 3 + 2 = 32.

Question 53:

A mother’s age is three times the sum of her twin children’s ages. In 5 years, she will be twice as old as them combined. What is the mother’s current age?

Solution:

Let’s denote the mother’s current age as ‘M’ and each twin’s current age as ‘T.’ We have two equations:

M = 3 * (T + T) (The mother’s age is three times the sum of her twin children’s ages)

M + 5 = 2 * ((T + 5) + (T + 5)) (In 5 years, she will be twice as old as them combined)

Solving these two equations simultaneously, we get:

M = 6T and M = 4T + 20

Setting these two equal to each other:

6T = 4T + 20 2T = 20 T = 10 years

So, the mother’s current age is M = 6 * T = 6 * 10 = 60 years.

Question 54:

A two-digit number is such that the number is seven times the sum of its digits. If 36 is added to the number, the result is the same number with its digits reversed. What is the original number?

Solution:

Let’s denote the tens digit as ‘T’ and the ones digit as ‘O’. Then, the two-digit number can be represented as 10T + O. We have two equations:

10T + O = 7 * (T + O) (The number is seven times the sum of its digits)

10T + O + 36 = 10O + T (If 36 is added to the number, the result is the same number with its digits reversed)

Solving these two equations simultaneously, we get:

10T + O = 7T + 7O and 10T + O + 36 = 10O + T

Solving for T and O, we get T = 4 and O = 2.

So, the original number is 10 * T + O = 10 * 4 + 2 = 42.