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Fukuoka Entrance Mathematics study materials
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Fukuoka Entrance Mathematics practice questions

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Question 1Numbers and Algebraic Expressions

Calculate the value of (-6) + (+13) - (-5).

  1. 24
  2. 12 Correct answer.
  3. -12
  4. 2

Explanation

Subtracting a negative number becomes addition, so -(-5) = +5. The expression becomes -6 + 13 + 5. Computing from the left: -6 + 13 = 7, then 7 + 5 = 12.

Question 2Numbers and Algebraic Expressions

Calculate the value of (-4)² × 3 - 20 ÷ (-5).

  1. 52 Correct answer.
  2. -52
  3. -28
  4. 44

Explanation

Follow the order of operations. First (-4)² = 16, then 16 × 3 = 48, and 20 ÷ (-5) = -4. Finally, 48 - (-4) = 48 + 4 = 52.

Question 3Numbers and Algebraic Expressions

Calculate 2/3 - 3/5.

  1. 1/15 Correct answer.
  2. -1/15
  3. 19/15
  4. -1/2

Explanation

Use the common denominator 15: 2/3 = 10/15 and 3/5 = 9/15. Then 10/15 - 9/15 = 1/15. The result is positive because 10/15 is larger than 9/15.

Question 4Numbers and Algebraic Expressions

Evaluate the expression 3a - 2b when a = 5 and b = -2.

  1. -11
  2. -19
  3. 11
  4. 19 Correct answer.

Explanation

Substitute the values: 3(5) - 2(-2) = 15 - (-4). Subtracting a negative becomes addition, so 15 + 4 = 19.

Question 5Numbers and Algebraic Expressions

Simplify the expression 4(x - 2y) - 3(x - 5y).

  1. 7x + 7y
  2. x - 7y
  3. x + 7y Correct answer.
  4. x - 23y

Explanation

Distribute: 4(x - 2y) = 4x - 8y and -3(x - 5y) = -3x + 15y. Combine like terms: (4x - 3x) + (-8y + 15y) = x + 7y.

Question 6Numbers and Algebraic Expressions

Solve the equation 4x + 3 = 2x - 9.

  1. x = -2
  2. x = 6
  3. x = -6 Correct answer.
  4. x = -3

Explanation

Move the x terms to the left and constants to the right: 4x - 2x = -9 - 3, so 2x = -12. Dividing by 2 gives x = -6. Check: 4(-6) + 3 = -21 and 2(-6) - 9 = -21.

Question 7Numbers and Algebraic Expressions

Solve the equation 0.3x - 1.2 = 0.9.

  1. x = -1
  2. x = 0.7
  3. x = 3
  4. x = 7 Correct answer.

Explanation

Add 1.2 to both sides: 0.3x = 2.1. Divide by 0.3: x = 2.1 ÷ 0.3 = 7. Multiplying both sides by 10 first (3x - 12 = 9) gives the same result.

Question 8Numbers and Algebraic Expressions

Simplify √48, writing it in the form a√b with b as small as possible.

  1. 2√12
  2. 4√3 Correct answer.
  3. 3√16
  4. 8√3

Explanation

Factor out the largest perfect square: 48 = 16 × 3, so √48 = √16 × √3 = 4√3. The radicand 3 has no perfect-square factors left.

Question 9Numbers and Algebraic Expressions

Calculate √3 × √27.

  1. 3√3
  2. 9 Correct answer.
  3. √30
  4. 81

Explanation

Use √a × √b = √(ab): √3 × √27 = √(3 × 27) = √81 = 9. Alternatively, √27 = 3√3, so √3 × 3√3 = 3 × 3 = 9.

Question 10Numbers and Algebraic Expressions

Solve the equation x² = 36.

  1. x = ±18
  2. x = ±6 Correct answer.
  3. x = 6
  4. x = 36

Explanation

Taking square roots of both sides gives x = ±√36 = ±6. Both 6² = 36 and (-6)² = 36, so a positive number has two square roots, one positive and one negative.

Question 11Numbers and Algebraic Expressions

Simplify (2x + 3y)/3 - (x - y)/2.

  1. (x + 9y)/6 Correct answer.
  2. (7x + 9y)/6
  3. (x + 9y)/5
  4. (x + 3y)/6

Explanation

Use the common denominator 6: [2(2x + 3y) - 3(x - y)]/6. Expand the numerator: 4x + 6y - 3x + 3y. Combine: x + 9y, so the result is (x + 9y)/6.

Question 12Numbers and Algebraic Expressions

Expand (x + 4)(x - 6).

  1. x² + 2x - 24
  2. x² - 10x - 24
  3. x² - 2x - 24 Correct answer.
  4. x² - 2x + 24

Explanation

Use the expansion pattern (x + a)(x + b) = x² + (a + b)x + ab. Here a = 4 and b = -6, so a + b = -2 and ab = -24. The result is x² - 2x - 24.

Question 13Numbers and Algebraic Expressions

Expand (3x + 2)².

  1. 9x² + 4
  2. 3x² + 12x + 4
  3. 9x² - 12x + 4
  4. 9x² + 12x + 4 Correct answer.

Explanation

Use (a + b)² = a² + 2ab + b² with a = 3x and b = 2. Then (3x)² = 9x², 2(3x)(2) = 12x, and 2² = 4. The result is 9x² + 12x + 4.

Question 14Numbers and Algebraic Expressions

Factor x² + 5x - 14.

  1. (x + 7)(x - 2) Correct answer.
  2. (x - 2)(x - 7)
  3. (x + 14)(x - 1)
  4. (x - 7)(x + 2)

Explanation

Find two numbers whose sum is +5 and product is -14: those are +7 and -2. So x² + 5x - 14 = (x + 7)(x - 2). Expanding confirms x² - 2x + 7x - 14 = x² + 5x - 14.

Question 15Numbers and Algebraic Expressions

Factor 2x² - 18 completely.

  1. 2(x + 3)(x - 3) Correct answer.
  2. 2(x + 3)²
  3. 2(x - 3)²
  4. 2(x² - 9)

Explanation

First factor out the common factor 2: 2(x² - 9). Then factor the difference of squares: x² - 9 = (x + 3)(x - 3). The complete factorization is 2(x + 3)(x - 3).

Question 16Numbers and Algebraic Expressions

Solve the simultaneous equations x + y = 10 and x - y = 4.

  1. x = 3, y = 7
  2. x = 7, y = -3
  3. x = 6, y = 4
  4. x = 7, y = 3 Correct answer.

Explanation

Add the two equations to eliminate y: 2x = 14, so x = 7. Substitute back: 7 + y = 10, so y = 3. Check in the second equation: 7 - 3 = 4.

Question 17Numbers and Algebraic Expressions

Solve the simultaneous equations 2x + 3y = 12 and x - y = 1.

  1. x = 0, y = 4
  2. x = 6, y = 0
  3. x = 3, y = 2 Correct answer.
  4. x = 2, y = 3

Explanation

From the second equation, x = y + 1. Substitute into the first: 2(y + 1) + 3y = 12, so 5y + 2 = 12 and y = 2. Then x = 2 + 1 = 3. Check: 2(3) + 3(2) = 12 and 3 - 2 = 1.

Question 18Numbers and Algebraic Expressions

At a shop near Dazaifu Tenmangu, buying 2 umegae-mochi and 1 cup of tea costs 700 yen, while buying 1 umegae-mochi and 2 cups of tea costs 500 yen. Find the price of one umegae-mochi.

  1. 300 yen Correct answer.
  2. 200 yen
  3. 350 yen
  4. 250 yen

Explanation

Let m be the mochi price and t the tea price. Then 2m + t = 700 and m + 2t = 500. Double the first equation: 4m + 2t = 1400. Subtract the second: 3m = 900, so m = 300. Check: t = 700 - 600 = 100, and 300 + 2(100) = 500.

Question 19Numbers and Algebraic Expressions

Solve the quadratic equation x² - 5x + 6 = 0.

  1. x = -2, x = -3
  2. x = -1, x = -6
  3. x = 1, x = 6
  4. x = 2, x = 3 Correct answer.

Explanation

Factor the left side: find two numbers with sum -5 and product +6, which are -2 and -3. So (x - 2)(x - 3) = 0, giving x = 2 or x = 3. Check: 2² - 5(2) + 6 = 0.

Question 20Numbers and Algebraic Expressions

Solve the quadratic equation x² + 2x - 8 = 0.

  1. x = -2, x = -4
  2. x = 2, x = 4
  3. x = 2, x = -4 Correct answer.
  4. x = -2, x = 4

Explanation

Find two numbers with sum +2 and product -8: those are +4 and -2. So (x + 4)(x - 2) = 0, giving x = -4 or x = 2. Check: 2² + 2(2) - 8 = 4 + 4 - 8 = 0.

Question 21Numbers and Algebraic Expressions

Solve the quadratic equation x² - 4x + 1 = 0.

  1. x = 2 ± 2√3
  2. x = 4 ± √3
  3. x = -2 ± √3
  4. x = 2 ± √3 Correct answer.

Explanation

The equation does not factor over integers, so use the quadratic formula: x = [4 ± √(16 - 4)]/2 = [4 ± √12]/2 = [4 ± 2√3]/2 = 2 ± √3. Dividing both terms of the numerator by 2 is the key final step.

Question 22Numbers and Algebraic Expressions

Calculate √12 + √27.

  1. 6√3
  2. √39
  3. 5√6
  4. 5√3 Correct answer.

Explanation

Simplify each radical first: √12 = 2√3 and √27 = 3√3. Then 2√3 + 3√3 = 5√3, since like radicals combine in the same way as like terms.

Question 23Numbers and Algebraic Expressions

Calculate (√5 + 2)(√5 - 2).

  1. -1
  2. 3
  3. 9
  4. 1 Correct answer.

Explanation

Use the difference-of-squares pattern (a + b)(a - b) = a² - b². Here a² = (√5)² = 5 and b² = 4, so the result is 5 - 4 = 1. The middle terms -2√5 and +2√5 cancel.

Question 24Numbers and Algebraic Expressions

Calculate √50 - √8.

  1. 3√2 Correct answer.
  2. 4
  3. 7√2
  4. √42

Explanation

Simplify each radical: √50 = 5√2 and √8 = 2√2. Then 5√2 - 2√2 = 3√2. Like radicals combine by subtracting their coefficients.

Question 25Numbers and Algebraic Expressions

Rationalize the denominator and simplify 6/√3.

  1. 3√2
  2. 6√3
  3. 2
  4. 2√3 Correct answer.

Explanation

Multiply numerator and denominator by √3: 6/√3 = 6√3/(√3 × √3) = 6√3/3 = 2√3. Rationalizing removes the radical from the denominator.

Question 26Numbers and Algebraic Expressions

Expand and simplify (x + 3)(x - 3) - (x - 2)².

  1. 4x - 5
  2. -13
  3. -4x - 13
  4. 4x - 13 Correct answer.

Explanation

Expand each part: (x + 3)(x - 3) = x² - 9 and (x - 2)² = x² - 4x + 4. Subtract: x² - 9 - (x² - 4x + 4) = x² - 9 - x² + 4x - 4 = 4x - 13. The x² terms cancel.

Question 27Numbers and Algebraic Expressions

Given a + b = 6 and ab = 4, find the value of a² + b².

  1. 32
  2. 20
  3. 36
  4. 28 Correct answer.

Explanation

Use the identity (a + b)² = a² + 2ab + b², rearranged as a² + b² = (a + b)² - 2ab. Substitute: 6² - 2(4) = 36 - 8 = 28. This avoids needing to find a and b individually.

Question 28Numbers and Algebraic Expressions

Solve the equation (x - 1)² = 2x + 6.

  1. x = 5, x = 1
  2. x = 5, x = -1 Correct answer.
  3. x = 4, x = -1
  4. x = -5, x = 1

Explanation

Expand the left side: x² - 2x + 1 = 2x + 6. Move everything to one side: x² - 4x - 5 = 0. Factor: (x - 5)(x + 1) = 0, so x = 5 or x = -1. Check x = 5: (5 - 1)² = 16 and 2(5) + 6 = 16.

Question 29Numbers and Algebraic Expressions

Given x = √3 + 1, find the value of x² - 2x + 3.

  1. 2
  2. 4 + 2√3
  3. 3
  4. 5 Correct answer.

Explanation

Rewrite the expression using x - 1: x² - 2x + 3 = (x - 1)² + 2. Since x = √3 + 1, we have x - 1 = √3, so (x - 1)² = 3. The value is 3 + 2 = 5. Direct substitution gives the same result but with more work.

Question 30Linear Equations and Functions

y is proportional to x, and y = 12 when x = 4. Find the value of y when x = 7.

  1. 19
  2. 21 Correct answer.
  3. 28
  4. 16

Explanation

A proportional relationship has the form y = ax. From 12 = 4a, the constant is a = 3, so y = 3x. When x = 7, y = 3 × 7 = 21.

Question 31Linear Equations and Functions

y is inversely proportional to x, and y = 6 when x = 2. Find the value of y when x = 4.

  1. 3 Correct answer.
  2. 2
  3. 12
  4. 8

Explanation

Inverse proportion means y = a/x with xy = a. From 6 = a/2, the constant is a = 12. When x = 4, y = 12/4 = 3. Doubling x halves y.

Question 32Linear Equations and Functions

For the function y = 2x - 5, find the value of y when x = -3.

  1. 1
  2. -11 Correct answer.
  3. -1
  4. 11

Explanation

Substitute x = -3: y = 2(-3) - 5 = -6 - 5 = -11. Multiplying 2 by a negative x gives a negative product, which then combines with -5.

Question 33Linear Equations and Functions

Find the slope of the line passing through the points (1, 3) and (3, 9).

  1. 2
  2. 3 Correct answer.
  3. 6
  4. 1/3

Explanation

Slope = (change in y) ÷ (change in x) = (9 - 3)/(3 - 1) = 6/2 = 3. The y-values increase by 6 while the x-values increase by 2.

Question 34Linear Equations and Functions

A train departs Hakata Station and travels 240 km at a constant speed in exactly 3 hours. What is its speed in km per hour?

  1. 60 km/h
  2. 80 km/h Correct answer.
  3. 72 km/h
  4. 90 km/h

Explanation

Speed = distance ÷ time = 240 ÷ 3 = 80 km/h. Distance traveled is proportional to time when the speed is constant, with the constant of proportionality equal to the speed.

Question 35Linear Equations and Functions

Which of the following equations represents y being inversely proportional to x?

  1. y = 12x²
  2. y = 12x
  3. y = x + 12
  4. y = 12/x Correct answer.

Explanation

Inverse proportion has the form y = a/x, equivalently xy = a, where a is a nonzero constant. The equation y = 12/x matches this form with a = 12: as x doubles, y halves.

Question 36Quadratic Functions and Equations

For the function y = 3x², find the value of y when x = -2.

  1. -12
  2. 12 Correct answer.
  3. -36
  4. 36

Explanation

Substitute x = -2: y = 3 × (-2)² = 3 × 4 = 12. The square of a negative number is positive, so y is positive whenever x is nonzero.

Question 37Linear Equations and Functions

A linear function has slope 3, and its graph passes through the point (2, 5). Find its equation.

  1. y = 3x + 5
  2. y = 3x + 1
  3. y = 3x - 1 Correct answer.
  4. y = 2x + 1

Explanation

A line with slope 3 has the form y = 3x + b. Substitute the point (2, 5): 5 = 3(2) + b, so b = 5 - 6 = -1. The equation is y = 3x - 1. Check: at x = 2, y = 6 - 1 = 5.

Question 38Linear Equations and Functions

Find the coordinates of the intersection point of the lines y = -x + 4 and y = 2x - 5.

  1. (3, -1)
  2. (3, 1) Correct answer.
  3. (-3, 7)
  4. (1, 3)

Explanation

Set the expressions equal: -x + 4 = 2x - 5, so 9 = 3x and x = 3. Then y = -3 + 4 = 1. The intersection is (3, 1). Check in the second line: 2(3) - 5 = 1.

Question 39Quadratic Functions and Equations

The graph of y = ax² passes through the point (2, -12). Find the value of y when x = 3.

  1. -18
  2. -27 Correct answer.
  3. 27
  4. -12

Explanation

Substitute the given point: -12 = a(2²) = 4a, so a = -3 and the function is y = -3x². At x = 3, y = -3 × 9 = -27.

Question 40Linear Equations and Functions

A taxi in Hakata charges a base fare of 500 yen plus 100 yen for each kilometer traveled. What is the total fare for an 8 km ride?

  1. 1,400 yen
  2. 900 yen
  3. 1,300 yen Correct answer.
  4. 800 yen

Explanation

The fare is a linear function: y = 100x + 500, where x is the distance in km. For x = 8, y = 100 × 8 + 500 = 800 + 500 = 1,300 yen.

Question 41Linear Equations and Functions

For the linear function y = -2x + 7, find the rate of change as x increases from 1 to 4.

  1. -6
  2. -2 Correct answer.
  3. 7
  4. 2

Explanation

For any linear function y = ax + b, the rate of change (change in y ÷ change in x) is always the slope a, regardless of the interval. Here a = -2. Verify: y goes from 5 to -1, a change of -6 over an x-change of 3, and -6 ÷ 3 = -2.

Question 42Quadratic Functions and Equations

For the function y = (1/2)x², find the rate of change as x increases from -4 to -2.

  1. -1
  2. -6
  3. -3 Correct answer.
  4. 3

Explanation

Compute the y-values: at x = -4, y = (1/2)(16) = 8; at x = -2, y = (1/2)(4) = 2. Rate of change = (2 - 8)/(-2 - (-4)) = -6/2 = -3. On the negative side of the parabola, y decreases as x increases toward 0.

Question 43Linear Equations and Functions

A school has a fixed supply of printing paper. Using 40 sheets per day, the supply lasts exactly 12 days. How many days will the same supply last if 30 sheets are used per day?

  1. 10 days
  2. 9 days
  3. 16 days Correct answer.
  4. 18 days

Explanation

The total supply is 40 × 12 = 480 sheets. Daily use and days lasted are inversely proportional, so at 30 sheets per day the supply lasts 480 ÷ 30 = 16 days.

Question 44Linear Equations and Functions

Find the equation of the line passing through the points (-1, 2) and (3, 10).

  1. y = (1/2)x + 5/2
  2. y = 2x + 4 Correct answer.
  3. y = 4x + 6
  4. y = 2x - 4

Explanation

Slope = (10 - 2)/(3 - (-1)) = 8/4 = 2. Using the point (3, 10): 10 = 2(3) + b, so b = 4. The equation is y = 2x + 4. Check the other point: 2(-1) + 4 = 2.

Question 45Quadratic Functions and Equations

The graph of y = ax² passes through the point (-2, 20). Find the value of a.

  1. a = 5 Correct answer.
  2. a = 5/2
  3. a = 10
  4. a = -5

Explanation

Substitute the point: 20 = a(-2)² = 4a, so a = 20 ÷ 4 = 5. Since (-2)² = 4 is positive and y = 20 is positive, a must be positive.

Question 46Linear Equations and Functions

A candle is 20 cm long and burns down at a constant rate of 2 cm per hour. After how many hours will the candle be 8 cm long?

  1. 4 hours
  2. 12 hours
  3. 10 hours
  4. 6 hours Correct answer.

Explanation

Let t be the hours elapsed. The length is y = 20 - 2t. Set 20 - 2t = 8, so 2t = 12 and t = 6. Check: after 6 hours, 20 - 12 = 8 cm remain.

Question 47Linear Equations and Functions

Find the x-coordinate of the point where the graph of y = (3/2)x - 2 crosses the x-axis.

  1. 2/3
  2. -2
  3. 3/2
  4. 4/3 Correct answer.

Explanation

The graph crosses the x-axis where y = 0. Set (3/2)x - 2 = 0, so (3/2)x = 2 and x = 2 × (2/3) = 4/3. The x-intercept is (4/3, 0).

Question 48Linear Equations and Functions

y is proportional to x, and y = -15 when x = 6. Find the value of y when x = -8.

  1. 30
  2. 12
  3. 20 Correct answer.
  4. -20

Explanation

The constant of proportionality is a = -15 ÷ 6 = -5/2, so y = (-5/2)x. When x = -8, y = (-5/2) × (-8) = 20. The product of two negatives is positive.

Question 49Quadratic Functions and Equations

Find all intersection points of the parabola y = x² and the line y = x + 2.

  1. (-2, 4) and (1, 3)
  2. (2, 2) and (-1, 1)
  3. (2, 4) and (1, 1)
  4. (2, 4) and (-1, 1) Correct answer.

Explanation

Set x² = x + 2, so x² - x - 2 = 0. Factor: (x - 2)(x + 1) = 0, giving x = 2 or x = -1. At x = 2, y = 4; at x = -1, y = 1. The points are (2, 4) and (-1, 1), and both satisfy the line equation too.

Question 50Quadratic Functions and Equations

For the function y = 2x² with the domain -1 ≤ x ≤ 2, find the range of y.

  1. 0 ≤ y ≤ 2
  2. 0 ≤ y ≤ 4
  3. 0 ≤ y ≤ 8 Correct answer.
  4. 2 ≤ y ≤ 8

Explanation

Because the domain includes x = 0, the minimum value of y is 2(0)² = 0, not the value at an endpoint. The maximum occurs at the endpoint farthest from 0: at x = 2, y = 2(4) = 8, while at x = -1, y = 2. So the range is 0 ≤ y ≤ 8.