Download [PDF] 2014 CAIE P2 Mark Schemes 1112 Mathematics Specimen Cambridge Lower Secondary Checkpoint

File Information

Filename:	[PDF] 2014 CAIE P2 Mark Schemes 1112 Mathematics Specimen Cambridge Lower Secondary Checkpoint.pdf
Filesize:	149.98 KB
Uploaded:	30/07/2021 02:24:14
Keywords:	2014 caie p2 mark schemes 1112 mathematics specimen cambridge lower secondary checkpoint pdf
Description:	Download file or read online CAIE Cambridge primary checkpoint specimen paper Mathematics 1112/02/SM/14 for examination from 2014 mark schemes paper 2 - Cambridge Assessment International Education.
Downloads:	51

File Preview

Download Urls

Short Page Link

https://www.edufilestorage.com/2Rx

Full Page Link

https://www.edufilestorage.com/2Rx/PDF_2014_CAIE_P2_Mark_Schemes_1112_Mathematics_Specimen_Cambridge_Lower_Secondary_Checkpoint.pdf

HTML Code

<a href="https://www.edufilestorage.com/2Rx/PDF_2014_CAIE_P2_Mark_Schemes_1112_Mathematics_Specimen_Cambridge_Lower_Secondary_Checkpoint.pdf" target="_blank" title="Download from eduFileStorage.com"><img src="https://www.edufilestorage.com/cache/plugins/filepreviewer/1007/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg"/></a>

Forum Code

[url=https://www.edufilestorage.com/2Rx/PDF_2014_CAIE_P2_Mark_Schemes_1112_Mathematics_Specimen_Cambridge_Lower_Secondary_Checkpoint.pdf][img]https://www.edufilestorage.com/cache/plugins/filepreviewer/1007/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg[/img][/url]

[PDF] 2014 CAIE P2 Mark Schemes 1112 Mathematics Specimen Cambridge Lower Secondary Checkpoint.pdf | Plain Text

This document consists of 9 printed pages and 1 blank page. IB14 1112_02_SM/3RP © UCLES 2014 [Turn over Cambridge International Examinations Cambridge Checkpoint MATHEMATICS 1112/02 Paper 2 For Examination from 2014 SPECIMEN MARK SCHEME MAXIMUM MARK: 50

2 © UCLES 2014 1112/02/SM/14 Question 1 Part Mark Answer Further Information 2 grams metres litres Accept abbreviations. Award 1 mark for any two correct. Accept m 3 or cubic metres rather than litres. Total 2 Question 2 Part Mark Answer Further Information 1 7 Total 1 Question 3 Part Mark Answer Further Information 1 15 n + 21 Total 1 Question 4 Part Mark Answer Further Information (a) 1 68 (b) 1 9.7 (142…) Follow through from (a) as their (a) ÷ 7. Accept 10 if working is seen. Total 2

3 © UCLES 2014 1112/02/SM/14 [Turn over Question 5 Part Mark Answer Further Information 2 ($)87.20 Award 1 method mark for either 43.2 or 44 seen. or 7.20 × 6 + 8.80 × 5 seen. Total 2 Question 6 Part Mark Answer Further Information 1 560 (cm 3) Total 1 Question 7 Part Mark Answer Further Information 2 ( n = ) 9.5 or equivalent Award 1 mark for a correct first step that reduces the number of terms, e.g. 2n – 5 = 14 4 n = 2 n + 19 (or better) Total 2 Question 8 Part Mark Answer Further Information 2 1 mark for 3 correct. Grade 7 11 18 16 45 Grade 8 22 19 34 75 Total 33 37 50 120 Total 2

4 © UCLES 2014 1112/02/SM/14 Question 9 Part Mark Answer Further Information (a) 1 4, 3 (b) 2 –2 –3 –1 1 2 3 4 2 3 4 5 6 –1 –2 –35 y x = 1 0 x A B C 1 –4 Award 1 mark for a correct reflection in an incorrect vertical mirror line or y = 1. (c) 1 5, 0 (d) 1 2 (cm 2) Total 5 Question 10 Part Mark Answer Further Information 2 ($)24 Award 1 mark for 8 seen. or 9 72 × 3 seen. Total 2

5 © UCLES 2014 1112/02/SM/14 [Turn over Question 11 Part Mark Answer Further Information (a) 1 All 3 correct for the mark. x 0 2 4 6 y 2 6 10 14 (b) 1 y x 1 0 234567 –12 4 6 8 10 12 14 –2 Follow through from (a) provided line drawn is straight. Line must extend at least from x = 0 to x = 6. (c) 1 x = 2.5 (accept 2.4 – 2.6 inclusive) y = 7 (accept 6.9 – 7.1 inclusive) Both correct for one mark. Follow through from (b) if there is a single intersection. Total 3 Question 12 Part Mark Answer Further Information 1 5.455 Total 1 Question 13 Part Mark Answer Further Information 1 An answer that implies that each angle in a regular pentagon is 108° (or that the sum of the angles is 540°) or An answer that implies that each exterior angle in a regular pentagon is 72°. Total 1

6 © UCLES 2014 1112/02/SM/14 Question 14 Part Mark Answer Further Information 1 y(y – 8) Total 1 Question 15 Part Mark Answer Further Information (a) 1 22 (minutes) (b) 1 9 (minutes) (c) 1 27 (minutes) Total 3 Question 16 Part Mark Answer Further Information 2 6.25(%) Award 1 mark for 500 ÷ 8000 or 0.0625 or 93.75 or 1 – 0.9375 Total 2 Question 17 Part Mark Answer Further Information 1 In correct order. (x + 3) ( x – 4 ) = x 2 – x – 12 Total 1

7 © UCLES 2014 1112/02/SM/14 [Turn over Question 18 Part Mark Answer Further Information 1 92 900 000 (miles) Total 1 Question 19 Part Mark Answer Further Information (a) 1 ($)310 (accept answers between 307 and 313 inclusive) (b) 1 Puts a ring around $530 and gives a suitable reason (e.g. it fits in with the other points) or if a line of best fit is referred to. Total 2 Question 20 Part Mark Answer Further Information 4 1 mark for correct answer of x = 7.1 1 mark for at least 4 correct trials. 1 mark for evidence of ‘improvement’ (trials getting closer to 7.1) 1 mark for a trial of 7.15 Total 4

8 © UCLES 2014 1112/02/SM/14 Question 21 Part Mark Answer Further Information (a) 1 All correct for 1 mark. 1 2 3 4 1 1 2 3 4 2 2 4 6 8 3 3 6 9 12 4 4 8 12 16 (b) 1 4 3 or 16 12 or equivalent Follow through for their number of even numbers ÷ 16 Total 2 Question 22 Part Mark Answer Further Information 3 20 (years) Award 1 mark for correctly converting 2 tonnes to 2000 kg. (seen or implied) Award 1 mark for (‘their 2000’-105) ÷ 95 or Award 2 marks for answer of 19 (with working) or 19.9 Total 3

9 © UCLES 2014 1112/02/SM/14 Question 23 Part Mark Answer Further Information 3 Any answer that corrects to 5.0 to 1dp .e.g. 5.0462 or 5.05 Award 2 marks for an answer of 2.5… or Award 1 mark for A = πr 2 seen or evidence of its use. Award 1 mark for r 2 = π 20 (= 6.3…) Total 3 Question 24 Part Mark Answer Further Information 3 8.65(68…) (cm) Award 3 marks for 8.7 or 8.66 (cm) or 3 + 32 Award 2 marks for sight of 5.65(68) or 32 or equivalent e.g. 4 2 Award 1 mark for 4 2 + 4 2 = 32 or Any attempt at using Pythagoras’ theorem. Total 3

10 © UCLES 2014 1112/02/SM/14 BLANK PAGE