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5.1H Writing a Ratio as a Fraction or Linear Function: Higher Tier Practice

EDEXCEL 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A recipe uses ingredients in the ratio 5:3:2 for flour:butter:sugar. Work out the fraction of the total mixture that is sugar.

(Total for Question 1 is 2 marks)

The lengths of three pieces of ribbon are in the ratio 7:4:9. The longest piece is 63 cm. Work out the length of the shortest piece.

(Total for Question 2 is 3 marks)

A mixture contains acid and water in the ratio a:(2a + 3). If the total volume is 120 ml, express the volume of acid as a function of a, then find the acid volume when a = 3.

(a)Express the volume of acid, V ml, as a function of a.(2)

(b)Find V when a = 3. Give your answer in ml, simplified.(2)

(Total for Question 3 is 4 marks)

A photographer sets the ratio of red to blue light intensity to be (2x + 1):(x - 1). The total intensity is fixed at 450 units. Given x > 1, find x and state the intensities of red and blue light.

(Total for Question 4 is 5 marks)

Write the linear function y = mx + c that maps the ratio x:(3x + 4) to the fraction of x in the total, expressed as a function of x for x not equal to -4/3. Simplify the function if possible.

(Total for Question 5 is 5 marks)

Show that if two quantities are in the ratio (k + 2):(2k - 1) and their sum is S, then the first quantity equals S*(k+2)/(3k+1). Then, for k = 2 and S = 240, find the two quantities.

(a)Show that the first quantity equals S*(k+2)/(3k+1).(3)

(b)For k = 2 and S = 240 find the two quantities.(3)

(Total for Question 6 is 6 marks)

A town council divides its budget between parks and libraries in the ratio 5:(t + 3). The total budget is 1600 units. The council then changes the split so that the ratio of parks to libraries becomes 3:2, while keeping the total budget at 1600 units and keeping the actual amount allocated to parks exactly the same as before. Find t.

(Total for Question 7 is 8 marks)

Show that for positive x the expression f(x) = (4x + 6)/(2x + 3) is equal to 2 for all x except where it is not defined. Then explain what this means in terms of the ratio 4x+6 : 2x+3.

(a)Show that f(x) = (4x + 6)/(2x + 3) = 2 for all x where the expression is defined.(6)

(b)Explain what this implies about the ratio 4x+6 : 2x+3 for all x where it is defined.(4)

(Total for Question 8 is 10 marks)

A chemist uses volumes in the ratio (x + 1):(2x - 5). For which value of x are the two volumes equal? By how much does the first volume exceed the second when x = 8?

(a)Find the value of x for which the two volumes are equal.(3)

(b)When x = 8 and the total volume is 660 ml, find how many ml the second volume exceeds the first by.(4)

(Total for Question 9 is 7 marks)

Mark scheme · 5.1H Writing a Ratio as a Fraction or Linear Function: Higher Tier Practice

Question 1

M1 write fraction sugar = 2/(5+3+2) or equivalent method
A1 2/10 = 1/5 cao
Answer:1/5 (one fifth). Reason: sugar = 2 out of 10 total, so 2/10 = 1/5, cao.
Working check: 5+3+2=10 so sugar fraction = 2/10 = 1/5.

Question 2

M1 use scale factor: 63 corresponds to 9 (the largest part) so factor = 63/9 = 7
M1 identify shortest part is 4 (smallest of 7, 4, 9) and multiply by factor: 4 * 7
A1 28 cm cao
Answer:28 cm.
Working check: scale factor = 63/9 = 7, shortest part is 4, shortest = 4 * 7 = 28.

Question 3

(a) M1 method: fraction acid = a/(a + (2a+3)) = a/(3a+3)
(a) A1 V = 120 * a/(3a+3) or equivalent simplified form cao
(a) Answer:V(a) = 120 * a/(3a + 3).
Working check: total ratio = a + (2a+3) = 3a+3 so acid fraction = a/(3a+3).
(b) M1 substitute a = 3 into V = 120 * a/(3a+3) to get 120 * 3/(9+3)
(b) A1 30 ml cao
(b) Answer:30 ml.
Working check: V = 120 * 3/12 = 120 * 1/4 = 30 ml.

Question 4

M1 form equation using total intensity: (2x+1)+(x-1)=450
M1 simplify to 3x = 450
M1 solve x = 150
A1 x = 150 cao
A1 red = 2(150)+1 = 301 units, blue = 150-1 = 149 units cao
Answer:x = 150; red intensity = 301 units; blue intensity = 149 units.
Working check: (2x+1)+(x-1)=3x=450 so x=150. Red=2(150)+1=301, blue=150-1=149. Red+blue=301+149=450, consistent.

Question 5

M1 form fraction x/(x + 3x + 4) = x/(4x+4)
M1 simplify to x/(4(x+1)) or 1/4 * x/(x+1)
M1 express as linear function y = (1/4) * x/(x+1) and note not defined at x = -1 and x = -4/3? (identify domain)
A1 state simplified expression y = x/(4x+4) cao
A1 state domain restriction x not equal to -1 and note function form; accept equivalent simplified forms cao
Answer:y = x/(4x + 4). Simplified form y = x/(4(x + 1)). Domain: x not equal to -1.
Working check: total = x + (3x+4) = 4x+4 so fraction of x = x/(4x+4).

Question 6

(a) M1 write fraction first = (k+2)/((k+2)+(2k-1))
(a) M1 simplify denominator: (k+2)+(2k-1)=3k+1
(a) A1 conclude first = S*(k+2)/(3k+1) cao
(a) Answer:First = S*(k+2)/(3k+1).
Working check: sum of ratio parts = k+2 + 2k-1 = 3k+1 so fraction = (k+2)/(3k+1) and multiply by S.
(b) M1 substitute k = 2 into formula: first = 240*(4)/(7) or equivalent
(b) M1 calculate first = 960/7 and second = 240 - first or use ratio parts 4:3 to get integer values? note
(b) A1 first = 960/7 and second = 240*(3)/(7) = 720/7 cao
(b) Answer:First = 960/7; second = 720/7.
Working check: ratio for k=2 is 4:3 so parts sum 7. First = 240 * 4/7 = 960/7, second = 240 * 3/7 = 720/7.

Question 7

M1 express original parts: parks = 5p, libraries = (t+3)p so total = (t+8)p = 1600
M1 after change ratio becomes 3:2 so parks = 3q, libraries = 2q and total =5q =1600 so q = 320
M1 parks amount is unchanged: 5p = 3q so p = 3q/5
M1 substitute q=320 to get p = 3*320/5 = 192
M1 use total (t+8)p =1600 so (t+8)*192 = 1600
M1 solve (t+8) = 1600/192 = 25/3? show steps
M1 find t = 1600/192 - 8 = (25/3) - 8 = (25 - 24)/3 = 1/3
A1 t = 1/3 cao
Answer:t = 1/3.
Working check: total 1600 = 5p + (t+3)p = (t+8)p. New ratio 3:2 gives q = 320 and parks 3q = 960 so p = 960/5 = 192. Then t+8 = 1600/192 = 25/3 so t = 25/3 - 8 = 1/3.

Question 8

(a) M1 attempt to divide numerator by denominator or factor: 4x+6 = 2(2x+3)
(a) M1 write (4x+6)/(2x+3) = 2(2x+3)/(2x+3)
(a) M1 cancel (2x+3) provided 2x+3 not zero
(a) M1 conclude f(x) = 2 for all x with 2x+3 not equal to 0
(a) M1 state excluded value x = -3/2
(a) A1 complete proof f(x)=2 and domain note cao
(a) Answer:f(x) = 2 for all x except x = -3/2, where f is undefined.
Working check: 4x+6 = 2(2x+3) so f(x) = 2(2x+3)/(2x+3) = 2, provided 2x+3 not zero.
(b) M1 state that ratio simplifies to 2:1 for all allowed x
(b) M1 mention that both numerator and denominator scale by same factor so ratio constant
(b) M1 note excluded x = -3/2 where denominator is zero
(b) A1 clear statement: 4x+6 : 2x+3 = 2 : 1 for all x except x = -3/2 cao
(b) Answer:The ratio 4x+6 : 2x+3 is equivalent to 2 : 1 for every x where defined. It is constant because numerator = 2 times denominator; x = -3/2 is excluded since denominator is zero.
Working check: dividing both parts by (2x+3) gives 2:1.

Question 9

(a) M1 set x+1 = 2x - 5 and form equation
(a) M1 solve: -x = -6 so x = 6
(a) A1 x = 6 cao
(a) Answer:x = 6.
Working check: x+1 = 2x-5 gives x = 6.
(b) M1 substitute x = 8 to get ratio parts 9 : 11
(b) M1 total parts = 20 so part value = 660/20 = 33
(b) M1 compute difference second - first = (11 - 9)*33 = 2*33
(b) A1 66 ml cao (second is 11*33 = 363, first is 9*33 = 297, so second exceeds first by 66 ml)
(b) Answer:66 ml. When x = 8 the ratio is 9:11, total parts 20 so unit = 33 ml. First = 297 ml, second = 363 ml, so the second exceeds the first by 66 ml.
Working check: 660/20 = 33, 9*33 = 297, 11*33 = 363, difference = 66 ml.

Writing a Ratio as a Fraction or Linear Function: Higher Tier Practice - Worksheets, Questions and Revision