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Date November 2010 Marks available 1 Reference code 10N.1.sl.TZ0.14
Level SL only Paper 1 Time zone TZ0
Command term Write down Question number 14 Adapted from N/A

Question

The length of a square garden is (x + 1) m. In one of the corners a square of 1 m length is used only for grass. The rest of the garden is only for planting roses and is shaded in the diagram below.

 

The area of the shaded region is A .

Write down an expression for A in terms of x.

[1]
a.

Find the value of x given that A = 109.25 m2.

[3]
b.

The owner of the garden puts a fence around the shaded region. Find the length of this fence.

[2]
c.

Markscheme

(x + 1)2 – 1  or  x2 + 2x     (A1)     (C1)

[1 mark]

a.

(x + 1)2 – 1 = 109.25     (M1)

x2 + 2x 109.25 = 0     (M1)


Notes: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for correctly removing the brackets.


OR

(x + 1)2 1 = 109.25     (M1)

x + 1 = \(\sqrt {110.25} \)     (M1)


Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for taking the square root of both sides.


OR

(x + 1)2 – 10.52 = 0     (M1)

(x 9.5) (x + 11.5) = 0     (M1)


Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for factorised left side of the equation.


x = 9.5     (A1)(ft)     (C3)


Note: Follow through from their expression in part (a).

The last mark is lost if x is non positive.

If the follow through equation is not quadratic award at most (M1)(M0)(A1)(ft).

 

[3 marks]

b.

4 × (9.5 + 1) = 42 m     (M1)(A1)(ft)     (C2)


Notes: Award (M1) for correct method for finding the length of the fence. Accept equivalent methods.

 

[2 marks]

c.

Examiners report

Some candidates were able to answer this question correctly, but the majority experienced difficulty in finding the correct expression for the area of the shaded region. Those who showed working could then be awarded follow through marks for correctly equating their expressions to the given area and for their found value of x. Many candidates also could not find the perimeter of the shaded region in part c) even though they had found the value of x correctly.

a.

Some candidates were able to answer this question correctly, but the majority experienced difficulty in finding the correct expression for the area of the shaded region. Those who showed working could then be awarded follow through marks for correctly equating their expressions to the given area and for their found value of x. Many candidates also could not find the perimeter of the shaded region in part c) even though they had found the value of x correctly.

b.

Some candidates were able to answer this question correctly, but the majority experienced difficulty in finding the correct expression for the area of the shaded region. Those who showed working could then be awarded follow through marks for correctly equating their expressions to the given area and for their found value of x. Many candidates also could not find the perimeter of the shaded region in part c) even though they had found the value of x correctly.

c.

Syllabus sections

Topic 1 - Number and algebra » 1.0 » Basic manipulation of simple algebraic expressions, including factorization and expansion

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