iGCSE Mathematics (0580) :E1.8 Use the four rules for calculations.iGCSE Style Questions Paper 4

Question

(a) Factorise 121y2– m2.

Answer/Explanation

Ans: (11y − m) (11y + m) final answer

(b) Write as a single fraction in its simplest form.

\(\frac{4}{3x-5}+\frac{x+2}{x-1}\)

Answer/Explanation

Ans: \(\frac{3x^{2}+5x-14}{(3x-5)(x-1)}\) final answer

(c) Solve the equation.
3x2+ 2x – 7 = 0
Show all your working and give your answers correct to 2 decimal places.

Answer/Explanation

Ans: \(\frac{-2\pm \sqrt{2^{2}-4(3)(-7)}}{2\times 3}\)

− 1.90      1.23 final answers

(d) In this part, all lengths are in centimetres.

ABCD is a trapezium with area 15cm2.
(i) Show that 2x2+ 5x – 12 = 0.

Answer/Explanation

Ans: \(\frac{1}{2}(x+4+3x+2)(x+1)=15\)

4x2 + 4x + 6x + 6 = 30  or    2x2 + 2x + 3x + 3 = 15

2x2 + 5x – 12 = 0

(ii) Solve the equation 2x2+ 5x – 12 = 0.

Answer/Explanation

Ans: \(1.5 or \frac{3}{2}, – 4\)

(iii) Write down the length of AB.

Answer/Explanation

Ans: \(6.5 or \frac{13}{2}\)

Question

(a)\(s=ut+\frac{1}{2}at^{2}\)
Find the value of s when u = 5.2 , t = 7 and a = 1.6
s = …………………………………………
(b) Simplify.
(i) 3a-5b-a+2b
………………………………………….
(ii)\(\frac{5}{3x}\times \frac{9x}{20}\)
………………………………………….
(c) Solve
(i)\( \frac{15}{x}=-3\)
x = …………………………………………
(ii) 4(5-3x)=23
x = …………………………………………
(d) Simplify.
\((27x^{9})^{\frac{2}{3}}\)
………………………………………….
(e) Expand and simplify.
(3x – 5y)(2x + y)
………………………………………….

Answer/Explanation

(a) The formula is:
\(s = ut + (\frac{1}{2})at^2\)
Substituting u = 5.2, t = 7, and a = 1.6, we get:
\(s = (5.2)(7) + (\frac{1}{2})(1.6)(7)^2\)
Simplifying this expression, we get:
s = 36.4 + 39.2
Therefore, the value of s is:
s = 75.6
(b)(i)3a-a- 5b+2b=(3a-a)+(-5b+2b)=2a-3b
Therefore, the simplified expression is:
3a-5b-a+2b=2a-3b
(b)(ii)\(\frac{5}{3x}\times \frac{9x}{20}=\frac{\left ( 5\times 9x \right )}{\left ( 3x\times 20 \right )}=\frac{45x}{60x}\)
Simplifying further by canceling out the common factor of x in the numerator and denominator,
\(\frac{45x}{60x}=\frac{3}{4}\)
(c)(i)\( \frac{15}{x}=-3\)
Multiplying both sides by x, we get
15 = -3x
Dividing both sides by -3, we get
x = -5
(ii) 4(5 – 3x) = 23
20 – 12x = 23
-12x=3
\(x=-\frac{1}{4} \)or -0.25
(d) To simplify the expression \(\left ( 27x^{9} \right )^{\frac{2}{3}}\),we can use the property of exponents that says \((a^b)^c = a^(b\times c)\), and simplify the base and exponent separately
\((27x^{9})^{\frac{2}{3}} = 27^{\frac{2}{3}}\times(x^{9})^{\frac{2}{3}}=(3^{3})^{\frac{2}{3}}\times x^{3\times 2}=9x^{6}\)
(e)To expand the given expression, we can use the distributive property of multiplication,
(3x – 5y)(2x + y) = 3x(2x + y) – 5y(2x + y)
Now, we can use the distributive property again to simplify each of these products,
=(3x.2x) + (3x.y)-(5y.2x)-(5y.y)
\(= 6x^2+3xy-10xy-5y^2\)
\(= 6x^2 -7xy-5y^2\)
Therefore, the expanded and simplified expression is \(6x^{2} – 7xy – 5y^{2}\)

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