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Date	November 2018	Marks available	1	Reference code	18N.1.SL.TZ0.10
Level	SL	Paper	1	Time zone	no time zone
Command term	Identify	Question number	10	Adapted from	N/A

Question

The following method, calcBMI() accepts person’s height (H) in metres (m) and weight (W) in kilograms (kg) and returns their Body Mass Index (BMI).

calcBMI(H, W)
  X = H * H
  B = W / X
  return B
endcalcBMI

Boris weighs 104 kg and is 2.00 m tall. His BMI can be calculated by calling method calcBMI() as follows

BorisBMI = calcBMI(2.00, 104).

A person can belong to one of the following four weight categories:

The data about a group of adults and their height measurement (in metres) and weight measurement (in kg) is held in three one-dimensional arrays.

Where
NAME is a one-dimensional array holding names (currently sorted in alphabetical order).
WEIGHT is a one-dimensional array holding weight measurement in kilograms.
HEIGHT is a one-dimensional array holding height measurement in metres.

For example,
NAME[0] is Annie.
Her weight measurement is 52.40 kg and can be found in WEIGHT[0].
HEIGHT[0] is 1.56 which represents Annie’s height measurement in metres.

State the value of variable BorisBMI.

[1]

Use pseudocode to construct an algorithm which accepts a person’s BMI and outputs the weight category the person belongs to.

[4]

State the name of the person whose height is held in HEIGHT[3].

[1]

Identify one reason why a binary search algorithm cannot be used to find the name of person whose height is given.

[1]

d.i.

Describe how the name of person whose height is given could be output.

[2]

d.ii.

Construct an algorithm which will output the names of all the people whose BMI is greater than this group’s average BMI.

You should call method calcBMI() in your answer.

[6]

Markscheme

Award [1 max].

26.0;

Award [4 max].

Award [1] for outputting 'underweight' and correct condition in if statement.
Award [1] for outputting 'normal weight' and correct logical expression (18.5 < B < 25.0).
Award [1] for outputting 'overweight ' and correct logical expression (25.0 < B < 30.0).
Award [1] for outputting 'obese' (30.0 and greater than 30.0).
Award [1] for using if-else.

Example answer 1:

category(B)
   if B < 18.5
     output('underweight')
   else if B < 25.0
     output('normal weight')
   else if B < 30.0
     output('overweight')
   else
     output('obese')
   end if
end category

Example answer 2:

category(B)
  if B < 18.5
    output('underweight')
  end if
  if B >= 18.5 and B < 25.0
    output('normal weight')
  end if
  if B >= 25.0 and B < 30.0
    output('overweight')
  end if
  if B >= 30.0
    output('obese')
  end if
end category

Award [1 max].

Paul;

Award [1 max].

Binary search can be applied only on sorted array / array HEIGHT is not sorted so binary search cannot be used;

d.i.

Award [1] for identifying each stage in the process of printing out a person’s name up to [2 max].

Linear (sequential) search could be used to find the position (array index) of a given height measurement in array HEIGHT;
And the name in array NAME at found position should be outputted;

d.ii.

Award [6 max].

Award [3 max] marks for calculating average.
Award [1] for initialization and changing the sum (of all BMIs) within the loop.
Award [1] for a correct loop.
Award [1] for correct parameters in the call of method calcBMI().
Award [1] for calculating average for the 30 persons (sum of all BMIs over 30).

Award [3 max] marks for displaying appropriate names after calculating the group’s average BMI.
Award [1] for an if statement within the correct loop.
Award [1] for condition in if (comparing the BMI of the person with the average BMI).
Award [1] for the correct output [1] for correct parameters in the method call
(calcBMI(HEIGHT[K], WEIGHT[K]).

Example answer:

sum = 0
loop for K from 0 to 29
  sum = sum + calcBMI(HEIGHT[K], WEIGHT[K])
end loop
average = sum/30

loop for K from 0 to 29
  if calcBMI(HEIGHT[K], WEIGHT[K]) > average then
    output(NAME[K])
  end if
end loop

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

Syllabus sections

Topic 4: Computational thinking, problem-solving and programming » 4.2 Connecting computational thinking and program design

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18M.1.SL.TZ0.12a:

The collection DATA contains the following data:

2, 4, 1, −2, −4, 1, 0

Consider the following pseudocode:

COUNTER = 0
SUM = 0
DATA.resetNext()
loop for X from 0 to 6
  if DATA.getNext() > 0
    ARRAY[X] = DATA.getNext()
    COUNTER = COUNTER + 1
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  end if
end loop
output SUM/COUNTER

Trace the pseudocode using the table below:

18N.1.SL.TZ0.7:

Construct a trace table for the following algorithm

A = 3
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loop while B >= A
  A = A + 1
  output(B − A)
  B = B − 1
end loop

19N.1.SL.TZ0.7:

Construct a trace table for the following algorithm.

K = 1
N = 1
M = 2
loop while K < 5
   output(N,M)
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21M.1.SL.TZ0.8:

List the output from the given algorithm for the following input.

2, 6, 8, 9, 12, 15, 18, 20

loop for Count from 0 to 7
    input NUMBER
    if NUMBER div 2 = NUMBER / 2 then
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19N.1.SL.TZ0.10a:

Consider the following algorithm.

N = 372
X = N DIV 100
Y = X + 10 * (N MOD 100 DIV 10)
Z = Y + (N MOD 10) * 100

Determine the values of variables X, Y, and Z after execution of this algorithm. Show your working.

19M.2.SL.TZ0.4b:
Using the rules and initial values above, construct the pseudocode that would enable the area of the sea ice and the sea level rise to be calculated if there was an increase of 0.04 °C in the ocean surface temperature.

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Topic 4: Computational thinking, problem-solving and programming

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