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Hardest GCSE Maths Questions

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Question: Solve the quadratic equation 2x² - 8x + 6 = 0.

Answer: x = 1 or x = 3.

Explanation: Factorising the equation or using the quadratic formula leads to the solutions x = 1 and x = 3.

Question: If y is directly proportional to x and y = 20 when x = 5, find y when x = 10.

Answer: y = 40.

 Explanation: Direct proportionality means y/x = constant. Given y = 20 when x = 5, the constant is 4. Thus, for x = 10, y = 10 × 4 = 40.

Question: A linear graph passes through the points (2, -3) and (4, 1). Find the equation of the line.

Answer: y = 2x – 7.

Explanation: First, calculate the slope (m) using (y₂ – y₁)/(x₂ – x₁) = (1 – (-3))/(4 – 2) = 4/2 = 2. Then, use one point to solve for y-intercept (b): -3 = 2(2) + b, so b = -7. Thus, the equation is y = 2x – 7.

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Question: Show that the difference between the squares of any two consecutive numbers is always an odd number.

Answer and Explanation:

Let the consecutive numbers be n and n+1.

The square of n is n^2.

The square of n+1 is (n+1)^2 which equals n^2 + 2n + 1.

The difference between the squares is (n^2 + 2n + 1) – n^2 which simplifies to 2n + 1.

Since 2n represents an even number, adding 1 results in an odd number.

Therefore, the difference is always odd.

Question: A ladder of 10 meters length leans against a wall with its bottom 6 meters from the wall. Find the height the ladder reaches on the wall up to two decimal places.

Answer and Explanation:

Using the right triangle formed by the ladder, the ground, and the wall:

The ladder’s length (hypotenuse) is 10 meters, and the distance from the wall (one leg) is 6 meters.

Apply the Pythagorean theorem: a^2 + b^2 = c^2, where a = 6 meters, and c = 10 meters.

Solve for b (height up the wall): b^2 = c^2 – a^2 = 10^2 – 6^2 = 100 – 36 = 64.

So, b = 8 meters.

The ladder reaches 8 meters up the wall.

Question: Solve the quadratic equation 2x^2 - 4x - 6 = 0.

Answer and Explanation:

Use the quadratic formula x = (-b ± sqrt(b^2 – 4ac)) / (2a) with a = 2, b = -4, and c = -6.

Discriminant is D = b^2 – 4ac = 16 + 48 = 64.

Solve for x: x = (4 ± sqrt(64)) / 4 = (4 ± 8) / 4.

Therefore, x = 3 or x = -1.

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Question: Calculate the sum of the first 20 terms of an arithmetic sequence with the first term 5 and a common difference of 3.

Answer and Explanation:

The sum S_n of the first n terms is given by S_n = n/2 [2a + (n – 1)d], where a = 5, d = 3, and n = 20.

So, S_20 = 20/2 [2(5) + 19(3)] = 10[10 + 57] = 10 * 67 = 670.

The sum of the first 20 terms is 670.

Question: A circle has a chord 8 cm long, located 6 cm from the center. Find the radius of the circle.

Answer and Explanation:

The perpendicular from the center to the chord bisects it, creating two 4 cm segments.

This forms a right-angled triangle with sides 4 cm, 6 cm, and the radius as the hypotenuse.

Apply the Pythagorean theorem: r^2 = 4^2 + 6^2 = 16 + 36 = 52.

Thus, r = sqrt(52).

Therefore, the radius of the circle is sqrt(52) cm.

Question: Find the equation of the line that passes through the points (3, 2) and (-1, -6).

Answer and Explanation:

First, calculate the slope (m) of the line using the formula m = (y2 – y1) / (x2 – x1), where the points are (x1, y1) = (3, 2) and (x2, y2) = (-1, -6).

So, m = (-6 – 2) / (-1 – 3) = -8 / -4 = 2.

Next, use the point-slope form of the equation of a line, y – y1 = m(x – x1), with m = 2 and one of the points, say (3, 2).

Substitute m, x1, and y1 into the equation: y – 2 = 2(x – 3).

Expanding and simplifying, y – 2 = 2x – 6, so y = 2x – 4.

Therefore, the equation of the line passing through the points (3, 2) and (-1, -6) is y = 2x – 4.

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