CE Board Exam Randomizer

⬅ Back to Subject Topics

Number Problems

Number problems often involve finding unknown values that satisfy several conditions simultaneously. These conditions may include:

The key steps are:

  1. Assign variables to represent each unknown number.
  2. Translate the conditions into equations.
    For example:
    • “The sum of the three numbers is 51” becomes:
      $x + y + z = 51$
    • “The first number divided by the second gives a quotient of 2 and remainder 5” becomes:
      $x = 2y + 5$
    • “The second divided by the third gives a quotient of 3 and remainder 2” becomes:
      $y = 3z + 2$
  3. Solve the system of equations step-by-step, often starting from the simpler equations and substituting into the more complex ones.
  4. Check the solution by substituting back into all original conditions.

These problems are a practical application of algebra and logical reasoning, where understanding how to model real statements into equations is just as important as solving them.

Concept Concept Concept Concept Concept Concept Concept Concept Concept

Problem:

The sum of the three numbers is 51. If the first number is divided by the second, the quotient is 2 and the remainder 5; but if the second number is divided by the third, the quotient is 3 and the remainder 2. Determine the numbers.

Number Problems | Algebra – Problem 1: – Diagram Number Problems | Algebra – Problem 1: – Diagram Number Problems | Algebra – Problem 1: – Diagram

See images:

Number Problems | Algebra – Problem 1: – Diagram Number Problems | Algebra – Problem 1: – Diagram Number Problems | Algebra – Problem 1: – Diagram Number Problems | Algebra – Problem 1: – Diagram

Problem:

If the numerator of a certain fraction is increased by 2 and the denominator is increased by 1, the resulting fraction equals 1/2. If, however, the numerator is increased by 1 and the denominator decreased by 2, the resulting fraction equals 3/5 . Find the fraction.

Number Problems | Algebra – Problem 2: – Diagram Number Problems | Algebra – Problem 2: – Diagram Number Problems | Algebra – Problem 2: – Diagram

Number Problems | Algebra – Problem 2: – Diagram Number Problems | Algebra – Problem 2: – Diagram Number Problems | Algebra – Problem 2: – Diagram Number Problems | Algebra – Problem 2: – Diagram

Problem:

The ten's digit of a certain number is 3 less than the unit's digit. If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3. What is the original number?

Number Problems | Algebra – Problem 3: – Diagram Number Problems | Algebra – Problem 3: – Diagram Number Problems | Algebra – Problem 3: – Diagram

See images:

Number Problems | Algebra – Problem 3: – Diagram Number Problems | Algebra – Problem 3: – Diagram Number Problems | Algebra – Problem 3: – Diagram Number Problems | Algebra – Problem 3: – Diagram

Problem:

A number is expressed by 3 digits, which are in arithmetic progression. If the number is divided by the sum of the digits, the quotient is 26 and if 198 is to be added to the number, the digits will be reversed. Find the ten's digit.

Number Problems | Algebra – Problem 4: – Diagram Number Problems | Algebra – Problem 4: – Diagram Number Problems | Algebra – Problem 4: – Diagram

See images:

Number Problems | Algebra – Problem 4: – Diagram Number Problems | Algebra – Problem 4: – Diagram Number Problems | Algebra – Problem 4: – Diagram Number Problems | Algebra – Problem 4: – Diagram

Problem:

The sum of the ages of the three brothers is 63. If their ages are consecutive integers, what is the age of the eldest brother?

Number Problems | Algebra – Problem 5: – Diagram Number Problems | Algebra – Problem 5: – Diagram Number Problems | Algebra – Problem 5: – Diagram

See images:

Number Problems | Algebra – Problem 5: – Diagram Number Problems | Algebra – Problem 5: – Diagram Number Problems | Algebra – Problem 5: – Diagram Number Problems | Algebra – Problem 5: – Diagram

Problem:

Refer to the image shown:

Number Problems | Algebra – Problem 6: – Diagram Number Problems | Algebra – Problem 6: – Diagram Number Problems | Algebra – Problem 6: – Diagram

See images:

Number Problems | Algebra – Problem 6: – Diagram Number Problems | Algebra – Problem 6: – Diagram Number Problems | Algebra – Problem 6: – Diagram Number Problems | Algebra – Problem 6: – Diagram

Problem:

Refer to the image shown:

Number Problems | Algebra – Problem 7: – Diagram Number Problems | Algebra – Problem 7: – Diagram Number Problems | Algebra – Problem 7: – Diagram

See images:

Number Problems | Algebra – Problem 7: – Diagram Number Problems | Algebra – Problem 7: – Diagram Number Problems | Algebra – Problem 7: – Diagram Number Problems | Algebra – Problem 7: – Diagram

Problem:

Refer to the image shown:

Number Problems | Algebra – Problem 8: – Diagram Number Problems | Algebra – Problem 8: – Diagram Number Problems | Algebra – Problem 8: – Diagram

See images:

Number Problems | Algebra – Problem 8: – Diagram Number Problems | Algebra – Problem 8: – Diagram Number Problems | Algebra – Problem 8: – Diagram Number Problems | Algebra – Problem 8: – Diagram
-->
Scroll to zoom

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q57

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

The sum of the three numbers is 51. If the first number is divided by the second, the quotient is 2 and the remainder 5; but if the second number is divided by the third, the quotient is 3 and the remainder 2. Determine the numbers.

What is the value of x?

  1. 33
  2. 4
  3. 14
  4. 16

What is the value of y?

  1. 14
  2. 33
  3. 4
  4. 16

What is the value of z?

  1. 4
  2. 14
  3. 33
  4. 16

Part 1.

Let the three numbers be $x$, $y$, and $z$. From the division statements:
$x=2y+5$
$y=3z+2$
$x+y+z=51$
Substitute into the sum:
$(2(3z+2)+5)+(3z+2)+z=51$
$10z+11=51$, so $z=4$, $y=14$, and $x=33$.
$\boxed{x=33}$

Part 2.

From $y=3z+2$ and $z=4$:
$y=3(4)+2=14$
$\boxed{y=14}$

Part 3.

After substitution, $10z+11=51$, so:
$z=4$
$\boxed{z=4}$

Question Bank: q92

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

One number is five less than the other number. If their sum is 135, what are the numbers?

Answer:

  1. 65 & 70
  2. 70 & 75
  3. 60 & 65
  4. 75 & 80
Let the larger number be $x$. The smaller is $x-5$. Their sum is:
$x+(x-5)=135$
$2x=140$, so $x=70$ and $x-5=65$.
$\boxed{65\ \&\ 70}$

Question Bank: q121

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

Twice the middle digit of a three-digit number is the sum of the other two. If the number is divided by the sum of its digits, the answer is 56 and the remainder is 12. If the digits are reversed, the number becomes smaller by 594. Find the number.

Answer:

  1. 852
  2. 567
  3. 258
  4. 741
Let digits be $a, b, c$ (hundreds, tens, units).
$(1)\; 2b = a + c$ (b is arithmetic mean)
$(2)\; 100a+10b+c = 56(a+b+c)+12$
Simplify (2): $44a - 46b - 55c = 12 \Rightarrow$ after substituting $a=2b-c$: $14b - 33c = 4$
Testing integers: $c=2, b=5 \Rightarrow a = 2(5)-2 = 8$
Verify: $852 \div 15 = 56$ remainder $12$ ✓
$\boxed{852}$

Question Bank: q122

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

The product of two numbers is 1400. If three is subtracted from each number, their product becomes 1175. Find the bigger number.

Answer:

  1. 50
  2. 28
  3. 32
  4. 40
Let the two numbers be $x$ and $y$.
$xy = 1400 \quad (1)$
$(x-3)(y-3) = 1175 \Rightarrow xy - 3(x+y) + 9 = 1175$
$ 1400 - 3(x+y) = 1166 \Rightarrow x+y = 78 \quad (2)$
Quadratic: $t^2 - 78t + 1400 = 0 \Rightarrow t = \frac{78 \pm \sqrt{6084-5600}}{2} = \frac{78 \pm 22}{2}$
$t = 50 \text{ or } 28$
$\boxed{\text{Bigger number} = 50}$

Question Bank: q123

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

The product of three consecutive integers is 9240. Find the third integer.

Answer:

  1. 22
  2. 20
  3. 21
  4. 23
Let three consecutive integers be $n, n+1, n+2$.
$n(n+1)(n+2) = 9240$
Try $n = 20$: $20 \times 21 \times 22 = 9240$ ✓
$\boxed{\text{Third integer} = 22}$

Question Bank: q124

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

Ten less than four times a certain number is 14. Determine the number.

Answer:

  1. 6
  2. 4
  3. 5
  4. 7
$4x - 10 = 14$
$4x = 24$
$\boxed{x = 6}$

Question Bank: q125

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

The denominator of a certain fraction is three more than twice the numerator. If 7 is added to both terms of the fraction, the resulting fraction is 3/5. Find the original fraction.

Answer:

  1. 5/13
  2. 8/5
  3. 13/5
  4. 3/5
Let numerator $= n$, denominator $= 2n+3$.
Adding 7 to both: $\frac{n+7}{2n+10} = \frac{3}{5}$
$5(n+7) = 3(2n+10) \Rightarrow 5n+35 = 6n+30 \Rightarrow n = 5$
$\text{Denominator} = 2(5)+3 = 13$
$\boxed{\frac{5}{13}}$

Question Bank: q126

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

Three times the first of the three consectuve odd integers is three more than twice the third. Find the third integer.

Answer:

  1. 15
  2. 11
  3. 13
  4. 9
Let consecutive odd integers be $n, n+2, n+4$.
$3n = 2(n+4)+3 \Rightarrow 3n = 2n+11 \Rightarrow n = 11$
Third integer: $11+4$
$\boxed{= 15}$

Question Bank: q470

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

The ten’s digit of a certain number is 3 less than the unit’s digit. If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3. What is the original number?

Answer:

  1. 47
  2. 58
  3. 14
  4. 36
Let the units digit be $u$ and the tens digit be $u-3$. The number is:
$10(u-3)+u=11u-30$
Sum of digits is $2u-3$. Given quotient 4 remainder 3:
$11u-30=4(2u-3)+3$
$11u-30=8u-9$
$u=7$
The number is $47$.
$\boxed{47}$

Question Bank: q719

MSTE - Algebra / Number Problems / Engr. Janclyde Espinosa (Clidez)

What is the value of k if the sum of odd consecutive positive integers from 1 to k equals 441?

  1. 41
  2. 21
  3. 30
  4. 29
The sum of the first $n$ odd positive integers is $n^2$. If the sum from 1 to $k$ is 441:
$n^2=441$, so $n=21$. The 21st odd integer is:
$k=2n-1=41$
$\boxed{41}$

Question Bank: t75

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

Which of the following must be true for the product of $(x + 1)(x)(x - 1)$ when $x$ is a positive integer.

  1. it is either odd or even
  2. it is always divisible by 3
  3. it can be negative
  4. it always is divisible by 4
The factors $(x-1)$, $x$, and $(x+1)$ are three consecutive integers. In any three consecutive integers, one factor must be divisible by 3.
Therefore their product is always divisible by 3.
$\boxed{\text{it is always divisible by 3}}$

Question Bank: t196

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The sum of seven consecutive integers is zero. What is the smallest integer?

  1. -4
  2. -1
  3. -3
  4. -2
For seven consecutive integers with sum zero, the middle integer must be zero.
The integers are $-3,-2,-1,0,1,2,3$.
The smallest is $-3$.
$\boxed{-3}$

Question Bank: t235

MSTE - Algebra / Algebra Fundamentals / Gemini mapped Chapter 1 to 3

The seven-digit telephone number of a certain town starts with $350$. If the last four digits cannot begin or end with zero, how many telephone numbers can be assigned?

  1. 7290
  2. 10000
  3. 6561
  4. 8100
The number starts with 350, so only the last four digits vary.
The first of the last four digits cannot be zero: 9 choices.
The two middle digits can be any digit: 10 choices each.
The last digit cannot be zero: 9 choices.
Total $=9\times10\times10\times9=8100$.
$\boxed{8100}$

Question Bank: w4

MSTE - Algebra / Number Problems / MSTE May 2019

During the 2006 Winter Olympic Games in Torino, Italy, the total number of gold medals won by Germany, Canada, and the United States were three consecutive odd integers. Of these three countries, Germany won the most gold medals and Canada won the fewest. If the sum of the first integer, twice the second integer, and four times the third integer is 69, find the number of gold medals won by each country.

  1. Germany: 11; U.S: 9, and Canada: 7
  2. Germany: 12; U.S: 10, and Canada: 8
  3. Germany: 13; U.S: 11, and Canada: 9
  4. Germany: 14; U.S: 12, and Canada: 10
Let Canada $= x$, U.S. $= x+2$, Germany $= x+4$ (consecutive odd integers, common difference 2).
$x + 2(x+2) + 4(x+4) = 69$
$7x + 20 = 69 \Rightarrow x = 7$
Canada $= 7$, U.S. $= 9$, $\boxed{\text{Germany} = 11}$

Question Bank: w65

MSTE - Algebra / Number Problems / MSTE November 2019

Some Civil Engineering students decide to have a parade during Engineering Week. When the students form ranks of 3 abreast, two are left over. In ranks of 5, four students are left over. In ranks of 7, six are left over. In ranks of 11, ten are left over. What is the least possible number of students in the parade?

  1. 1362
  2. 1154
  3. 986
  4. 1028
Each remainder is exactly one less than its divisor, so $N+1$ is divisible by 3, 5, 7, and 11:
$N + 1 = 3 \times 5 \times 7 \times 11 = 1155$
$N = 1154$
Check: $1154 = 3(384)+2 = 5(230)+4 = 7(164)+6 = 11(104)+10$.
$\boxed{N = 1154}$