M DUL Leow Yong Wei Yogeswari Konnasagan LENGKAP Year KSSR Semakan 5 Assessment Formative and Summative Digital Resources Mathematics • Info • Simulation • Gamified Quiz Wordwall Let's Grasp! Formative Assessment PBD Module Idea Starter Mathematics Comics Booster Zone UASA Section B PAK-21 HOTS & i-THINK Summative Assessment Summative Practices Ujian Pertengahan Sesi Akademik (UPSA) Ujian Akhir Sesi Akademik (UASA) Answers PdPc- friendly Features
1 © Penerbitan Pelangi Sdn. Bhd. 1. Numbers Numerals 326 451 Words Three hundred twenty-six thousand four hundred and fifty-one 2. Each digit of a given number has place value and digit value respectively. Digit 3 2 6 4 5 1 Place value hundred thousands ten thousands thousands hundreds tens ones Digit value 300 000 20 000 6 000 400 50 1 3. Partition 306 452 based on place values and digit values. Place value 3 hundred thousands + 0 ten thousands + 6 thousands + 4 hundreds + 5 tens + 2 ones Digit value 300 000 + 6 000 + 400 + 50 + 2 4. Compare the numbers Case 1: 62 753 5 digits 162 753 6 digits Case 2: 821 409 861 409 5. Number arrangement Arrange 361 094, 361 940, 316 094, 316 940 based on a certain order. Ascending order → arrange the numbers from the smallest value to the largest value from left to right → 316 094, 316 940, 361 094, 361 940 Descending order → arrange the numbers from the largest value to the smallest value from left to right → 361 940, 361 094, 316 940, 316 094 6. Number sequence 207 916 , m , 412 335 , n , 927 008 (a) For the value of m, accept any numbers between 207 916 and 412 335. Example: 345 678 (b) For the value of n, accept any numbers between 412 335 and 927 008. Example: 765 432 6-digit number is larger than 5-digit number. → 162 753 is larger than 62 753 → 162 753 > 62 753 or 62 753 < 162 753 60 000 > 20 000 → 861 409 is larger than 821 409 Let’s Grasp! What is the largest six-digit number that can be formed using the digits 3, 2, 6, 4, 5, and 1? Whole Numbers and Operations 1 UNIT Learning Area: Numbers and Operations
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 2 7. Identify and complete the number pattern. Example: 318 402 , 418 402 , 518 402 , n , 718 402 Identify the pattern 318 402 , 418 402 , 518 402 , n , 718 402 The number pattern is in ascending order by one hundred thousands. n = 518 402 + 100 000 = 618 402 8. A prime number must follow these conditions: (i) Greater than 1. (ii) Can only be divided by 1 and itself. There are 25 prime numbers within 100. 9. Estimate the number of marbles in container P based on the number of marbles in container Q as reference. The number of marbles in container P is about half of the number of marbles in container Q. Thus, the number of marbles in container P is approximately 200 marbles. 10. +1 to the digit at the place value to be rounded off (underlined) if the digit to the right (circled) is equal to 5, 6, 7, 8 or 9 +0 to the digit at the place value to be rounded off (underlined) if the digit to the right (circled) is equal to 0, 1, 2, 3, 4 Example 1: 6 +0 8 3 5 4 2 (Round off to the nearest ten thousand) 6 8 0 0 0 0 Example 2: +1 4 7 1 0 3 5 (Round off to the nearest hundred thousand) 5 0 0 0 0 0 11. Mixed operations Case 1: Without brackets (a) Addition and multiplication: Solve the multiplication first and then the addition (b) Subtraction and multiplication: Solve the multiplication first and then subtraction (c) Addition and division: Solve the division first and then addition (d) Subtraction and division: Solve the division first and then subtraction Case 2: With brackets Solve the operation in the brackets first and followed by the other operations. 12. Using unknown Case 1: Determine the unknown value in multiplication number sentence 4 × a = 20 b × 6 = 18 a = 20 ÷ 4 b = 18 ÷ 6 a = 5 b = 3 Case 2: Determine the unknown value in division number sentence 56 ÷ c = 8 d ÷ 9 = 10 c = 56 ÷ 8 d = 10 × 9 c = 7 d = 90 +100 000 +100 000 all digits on the right side of the underlined digit are changed to zero all digits on the right side of the underlined digit are changed to zero Prime numbers within 100. 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 Container P Container Q 400 marbles Info Order of operations
Mathematics Year 5 Unit 1 Whole Numbers and Operations 3 © Penerbitan Pelangi Sdn. Bhd. PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: PBD Module Number value Textbook: 1 – 9 LS 1.1.1 State numbers up to 1 000 000: (i) Read any number in words. (ii) Say any number in numerals. (iii) Write numbers in numerals and words. 1.1 1. Read and mark (✓) for the correct number. LS 1.1.1(i) (a) Three hundred twelve thousand four hundred and fifty-eight 302 458 312 458 ✓ (b) Nine hundred four thousand seven hundred and eighty-four 904 784 940 784 ✓ (c) Seven hundred thirty thousand six hundred and six 730 606 730 660 ✓ (d) Two hundred forty-five thousand eight hundred and eleven 245 801 245 811 ✓ 2. Say and match the number to the words. LS 1.1.1(ii) (a) 402 305 • • Four hundred two thousand and thirty-five (b) 420 350 • • Four hundred two thousand three hundred and five (c) 402 035 • • Four hundred twenty thousand five hundred and three (d) 420 503 • • Four hundred twenty thousand three hundred and fifty 3. Write the following numbers in numerals. LS 1.1.1(iii) (a) 135 629 One hundred thirty-five thousand six hundred twenty-nine (b) 316 007 Three hundred sixteen thousand and seven (c) 571 803 Five hundred seventy-one thousand eight hundred and three (d) 835 414 Eight hundred thirty-five thousand four hundred and fourteen 4. Write the following numbers in words. LS 1.1.1(iii) (a) 281 364 (b) 416 263 (c) 708 164 (d) 919 080
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 4 PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: LS 1.1.2 Determine the value of numbers up to 1 000 000: (i) State the place value and digit value of any number. (ii) Write any numbers in extended notation based on place value and digit value. 5. Write the place value and the digit value for the underlined digits. LS 1.1.2(i) PL 3 Number (a) 139 425 (b) 280 716 (c) 403 691 (d) 623 517 (e) 900 863 Place value Digit value 6. Partition the following numbers based on the place value. LS 1.1.2(ii) PL 3 (a) 219 367 = (b) 572 413 (c) 608 192 7. Partition the following numbers based on the digit value. LS 1.1.2(ii) PL 3 (a) 439 815 = 400 000 + 30 000 + 9 000 + 800 + 10 + 5 (b) 167 238 100 000 60 000 7 000 200 30 8 (c) 370 514 300 000 70 000 0 500 10 4
Mathematics Year 5 Unit 1 Whole Numbers and Operations 5 © Penerbitan Pelangi Sdn. Bhd. PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: LS 1.1.2 Determine the value of numbers up to 1 000 000: (iii) Compare the value of two numbers. 8. Compare and colour the smallest value in blue. LS 1.1.2(iii) PL 3 (a) 51 672 315 024 (b) 219 650 96 520 (c) 461 320 416 203 (d) 630 419 630 941 9. Compare and colour the largest value in purple. LS 1.1.2(iii) PL 3 (a) 14 326 164 023 (b) 731 028 37 802 (c) 800 416 800 164 (d) 268 031 268 310 10. Compare both numbers and fill in the blanks with “larger than” or “smaller than”. LS 1.1.2(iii) PL 3 (a) 537 461 lebih besar daripada 537 164 (b) 301 295 lebih kecil daripada 310 952 11. Using all the number cards below, form a six-digit smallest number and a six-digit largest number. LS 1.1.2(iii) PL 3 5 0 4 7 0 1 (a) Largest number : 754 100 (b) Smallest number : 100 457
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 6 PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: LS 1.1.2 Determine the value of numbers up to 1 000 000: (iv) Arrange numbers in ascending and descending order (v) Complete any number sequence in ascending and descending order 12. Arrange the following numbers according to the given order. LS 1.1.2(iv) PL 3 (a) 132 156 123 651 132 615 123 156 132 165 Ascending order: , , , , (b) 476 013 467 103 476 310 467 130 476 301 Descending order: , , , , (c) 915 321 591 312 591 213 915 132 591 123 Ascending order: , , , , (d) 681 304 681 043 681 034 681 403 681 430 Descending order: , , , , 13. Based on the following number sequence, determine the possible values of h and k in ascending order. LS 1.1.2(v) PL 3 h 205 631 372 146 668 007 k h = 164 325 k = 741 936 14. Based on the following number sequence, determine the possible values of m and n in descending order. LS 1.1.2(v) PL 3 741 026 m 552 961 n 132 460 m = 654 132 n = 249 615
Mathematics Year 5 Unit 1 Whole Numbers and Operations 7 © Penerbitan Pelangi Sdn. Bhd. PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: PBD Module Prime numbers Textbook: 10 – 11 LS 1.2.1 Identify prime numbers within 100. 1.2 15. State the characteristics of prime numbers. LS 1.2.1 PL 2 (a) lebih besar daripada nombor 1 (b) hanya boleh dibahagi dengan 1 dan diri sendiri 16. List all the prime numbers within the given numbers. LS 1.2.1 PL 2 (a) 10 until 20 11, 13, 17, 19 (b) 42 until 62 43, 47, 53, 59, 61 17. Determine whether the following number is a prime number. LS 1.2.1 PL 2 (a) Number Characteristics ✓/✗ 27 Greater than 1. 3 Can only be divided by itself. 7 Can be divided by other numbers. 3 27 is a prime number 7 (b) Number Characteristics ✓/✗ 37 Greater than 1. 3 Can only be divided by itself. 3 Can be divided by other numbers. 7 37 is a prime number 3 (c) Number Characteristics ✓/✗ 47 Greater than 1. 3 Can only be divided by itself. 3 Can be divided by other numbers. 7 47 is a prime number 3 (d) Number Characteristics ✓/✗ 57 Greater than 1. 3 Can only be divided by itself. 7 Can be divided by other numbers. 3 57 is a prime number 7 (e) Number Characteristics ✓/✗ 61 Greater than 1. 3 Can only be divided by itself. 3 Can be divided by other numbers. 7 61 is a prime number 3 (f) Number Characteristics ✓/✗ 15 Greater than 1. 3 Can only be divided by itself. 7 Can be divided by other numbers. 3 15 is a prime number 7
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 8 PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: PBD Module Number patterns Textbook: 12 – 13 LS 1.5.1 Identify patterns in given number series in ascending and descending order by ones up to tens, hundreds, thousands, ten thousands and hundred thousands. LS 1.5.2 Complete various number patterns that are given in ascending and descending order 1.5 18. Identify the number pattern for the following series of numbers. LS 1.5.1 PL 3 (a) 260 419 , 260 409 , 260 399 , 260 389 , 260 379 menurun sepuluh-sepuluh (b) 437 290 , 437 295 , 437 300 , 437 305 , 437 310 menaik lima-lima 19. Complete the following number sequences according to a certain number pattern. LS 1.5.2 PL 3 (a) 641 200 , 641 100 , 641 000 , 640 900 , 640 800 (b) 385 910 , 385 914 , 385 918 , 385 922 , 385 926 20. Complete the following number lines. State the number pattern. LS 1.5.1 LS 1.5.2 PL 3 (a) 523 165 523 175 523 185 523 195 523 205 523 215 Pattern: menaik sepuluh-sepuluh (b) 720 364 710 364 700 364 690 364 680 364 670 364 Pattern: menurun sepuluh ribu-sepuluh ribu (c) 295 132 295 140 295 148 295 156 295 164 295 172 Pattern: menaik lapan-lapan (d) 800 913 800 910 800 907 800 904 800 901 800 898 Pattern: menurun tiga-tiga
Mathematics Year 5 Unit 1 Whole Numbers and Operations 9 © Penerbitan Pelangi Sdn. Bhd. PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: PBD Module Estimation Textbook: 14 – 15 LS 1.3.1 Estimate quantity based on given reference set and justify the answer. 1.3 21. Write the estimated quantities based on the reference set of L. LS 1.3.1 PL 3 K L 600 sweets M (a) Jar K : 900 biji (b) Jar M : 300 biji PBD Module Rounding off numbers Textbook: 16 – 18 LS 1.4.1 Round off whole numbers up to the nearest hundred thousand. LS 1.4.2 Identify numbers that can be represented by a rounded off number up to the nearest hundred thousand. 1.4 22. Round off the following numbers. LS 1.4.1 PL 3 Number nearest hundred nearest thousand nearest ten thousand nearest hundred thousand (a) 326 185 326 200 326 000 330 000 300 000 (b) 486 307 486 300 486 000 490 000 500 000 (c) 572 618 572 600 573 000 570 000 600 000 (d) 709 535 709 500 710 000 710 000 700 000 (e) 841 352 841 400 841 000 840 000 800 000 23. Fill in the blanks with the correct answers. LS 1.4.2 PL 3 i-THINK Circle Map (a) Numbers that become 530 000 when rounded off to the nearest ten thousand. 534 184 525 123 527 094 532 368 530 826 529 114 (b) Numbers that become 800 000 when rounded off to the nearest hundred thousand. 847 329 750 694 771 425 824 118 800 932 793 854 i-THINK Map
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 10 PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: PBD Module Basic operations Textbook: 19 – 37 LS 1.6.1 Solve addition number sentences up to five numbers involving numbers up to six digits with sum within 1 000 000. 1.6 24. Solve the following additions. LS 1.6.1 PL 3 (a) 64 231 + 5 648 = 69 879 6 4 2 3 1 + 5 6 4 8 6 9 8 7 9 (b) 26 509 + 153 475 = 179 984 1 2 6 5 0 9 + 1 5 3 4 7 5 1 7 9 9 8 4 (c) 598 047 + 302 615 = 900 662 1 1 1 5 9 8 0 4 7 + 3 0 2 6 1 5 9 0 0 6 6 2 (d) 394 165 + 6 132 + 78 648 = 478 945 1 1 3 9 4 1 6 5 + 6 1 3 2 4 0 0 2 9 7 1 1 4 0 0 2 9 7 + 7 8 6 4 8 4 7 8 9 4 5 (e) 472 851 + 196 024 + 132 429 = 801 304 1 4 7 2 8 5 1 + 1 9 6 0 2 4 6 6 8 8 7 5 1 1 1 1 1 6 6 8 8 7 5 + 1 3 2 4 2 9 8 0 1 3 0 4 (f) 7 812 + 18 435 + 291 875 + 43 029 = 361 151 1 2 2 1 2 7 8 1 2 1 8 4 3 5 2 9 1 8 7 5 + 4 3 0 2 9 3 6 1 1 5 1 (g) 52 306 + 684 021 + 225 963 + 37 705 = 999 995 1 1 1 1 5 2 3 0 6 6 8 4 0 2 1 2 2 5 9 6 3 + 3 7 7 0 5 9 9 9 9 9 5 (h) 103 740 + 2 482 + 531 828 + 314 + 92 716 = 731 080 1 1 3 1 2 1 0 3 7 4 0 2 4 8 2 5 3 1 8 2 8 3 1 4 + 9 2 7 1 6 7 3 1 0 8 0 (i) 46 073 + 84 257 + 3 824 + 5 736 + 847 764 = 987 654 1 2 2 2 2 4 6 0 7 3 8 4 2 5 7 3 8 2 4 5 7 3 6 + 8 4 7 7 6 4 9 8 7 6 5 4
Mathematics Year 5 Unit 1 Whole Numbers and Operations 11 © Penerbitan Pelangi Sdn. Bhd. PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: LS 1.6.2 Solve subtraction number sentences up to three numbers within 1 000 000. 25. Solve the following subtractions. LS 1.6.2 PL 3 (a) 8 729 – 513 = 8 216 8 7 2 9 – 5 1 3 8 2 1 6 (b) 6 120 – 1 358 = 4 762 10 11 5 0 1 10 6 1 2 0 – 1 3 5 8 4 7 6 2 (c) 46 132 – 7 863 = 38 269 15 10 12 3 5 0 2 12 4 6 1 3 2 – 7 8 6 3 3 8 2 6 9 (d) 72 140 – 23 651 = 48 489 11 10 13 6 1 0 3 10 7 2 1 4 0 – 2 3 6 5 1 4 8 4 8 9 (e) 319 207 – 94 575 = 224 632 11 2 11 8 1 10 3 1 9 2 0 7 – 9 4 5 7 5 2 2 4 6 3 2 (f) 800 100 – 564 382 = 235 718 10 7 9 9 0 9 10 8 0 0 1 0 0 – 5 6 4 3 8 2 2 3 5 7 1 8 (g) 58 620 – 3 721 – 4 936 = 49 963 15 11 7 5 1 10 5 8 6 2 0 – 3 7 2 1 5 4 8 9 9 13 4 3 18 5 4 8 9 9 – 4 9 3 6 4 9 9 6 3 (h) 60 741 – 9 872 – 16 385 = 34 484 16 13 5 9 6 3 11 6 0 7 4 1 – 9 8 7 2 5 0 8 6 9 4 10 7 16 5 0 8 6 9 – 1 6 3 8 5 3 4 4 8 4 (i) 476 123 – 85 905 – 99 541 = 290 677 3 17 5 11 1 13 4 7 6 1 2 3 – 8 5 9 0 5 3 9 0 2 1 8 18 11 2 8 9 1 11 3 9 0 2 1 8 – 9 9 5 4 1 2 9 0 6 7 7 (j) 900 000 – 178 234 – 289 758 = 432 008 8 9 9 9 9 10 9 0 0 0 0 0 – 1 7 8 2 3 4 7 2 1 7 6 6 11 6 1 11 5 16 7 2 1 7 6 6 – 2 8 9 7 5 8 4 3 2 0 0 8
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 12 PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: LS 1.6.3 Solve multiplication number sentences of any number up to six digits with a number up to two digits 100 and 1 000 with product up to 1 000 000. 26. Solve the following multiplications. LS 1.6.3 PL 3 (a) 2 × 4 213 = 8 426 4 2 1 3 × 2 8 4 2 6 (b) 5 × 29 016 = 145 080 1 4 3 2 9 0 1 6 × 5 1 4 5 0 8 0 (c) 139 618 × 7 = 977 326 2 6 4 1 5 1 3 9 6 1 8 × 7 9 7 7 3 2 6 (d) 12 × 6 636 = 79 632 6 6 3 6 × 1 2 1 3 2 7 2 + 6 6 3 6 0 7 9 6 3 2 (e) 15 × 9 763 = 146 445 9 7 6 3 × 1 5 4 8 8 1 5 + 9 7 6 3 0 1 4 6 4 4 5 (f) 25 × 8 750 = 218 750 8 7 5 0 × 2 5 4 3 7 5 0 + 1 7 5 0 0 0 2 1 8 7 5 0 (g) 36 × 25 987 = 935 532 2 5 9 8 7 × 3 6 1 5 5 9 2 2 + 7 7 9 6 1 0 9 3 5 5 3 2 (h) 20 976 × 41 = 860 016 2 0 9 7 6 × 4 1 2 0 9 7 6 + 8 3 9 0 4 0 8 6 0 0 1 6 (i) 17 234 × 54 = 930 636 1 7 2 3 4 × 5 4 6 8 9 3 6 + 8 6 1 7 0 0 9 3 0 6 3 6 27. Solve the following multiplications. LS 1.6.3 PL 3 i-THINK Bridge Map equals to Relating factor 10 × 46 245 as as 100 × 3 917 1 000 × 804 (a) 462 450 (b) 391 700 (c) 804 000 equals to Relating factor 64 300 (d) 100 × 643 (e) 10 × 42 755 (f) 1 000 × 150 as as 427 550 150 000 i-THINK map
Mathematics Year 5 Unit 1 Whole Numbers and Operations 13 © Penerbitan Pelangi Sdn. Bhd. PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: LS 1.6.4 Solve division number sentences of any number within 1 000 000 with a number up to two digits, 100 and 1 000. 28. Solve the following divisions. LS 1.6.4 PL 3 (a) 64 082 ÷ 2 = 32 041 3 2 0 4 1 2 6 4 0 8 2 – 6 0 4 – 4 0 0 – 0 0 8 – 8 0 2 – 2 0 (b) 523 404 ÷ 9 = 58 156 5 8 1 5 6 9 5 2 3 4 0 4 – 4 5 7 3 – 7 2 1 4 – 9 5 0 – 4 5 5 4 – 5 4 0 (c) 906 015 ÷ 11 = 82 365 8 2 3 6 5 11 9 0 6 0 1 5 – 8 8 2 6 – 2 2 4 0 – 3 3 7 1 – 6 6 5 5 – 5 5 0 29. Solve the following divisions. LS 1.6.4 PL 3 (a) 429 615 ÷ 6 = 71 602 baki 3 7 1 6 0 2 6 4 2 9 6 1 5 – 4 2 0 9 – 6 3 6 – 3 6 0 1 – 0 1 5 – 1 2 3 (b) 707 070 ÷ 24 = 29 461 baki 6 2 9 4 6 1 24 7 0 7 0 7 0 – 4 8 2 2 7 – 2 1 6 1 1 0 – 9 6 1 4 7 – 1 4 4 3 0 – 2 4 6 (c) 846 260 ÷ 49 = 17 270 baki 30 1 7 2 7 0 49 8 4 6 2 6 0 – 4 9 3 5 6 – 3 4 3 1 3 2 – 9 8 3 4 6 – 3 4 3 3 0 – 0 3 0 30. Solve the following divisions. LS 1.6.4 PL 3 i-THINK Bridge Map equals to Relating factor 436 820 ÷ 10 as as 219 500 ÷ 100 715 060 ÷ 1 000 (a) 43 682 (b) 2 195 (c) 715 baki 60 i-THINK Map
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 14 PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: PBD Module Using unknown Textbook: 38 – 39 LS 1.8.1 Determine the value of an unknown in multiplication number sentences involving one multiplication operation with the product up to 1 000 000. LS 1.8.2 Determine the value of an unknown in division number sentences involving any number with a number up to two digits, 100 and 1 000 within 1 000 000. 1.8 31. Determine the value of unknown for the following mathematical sentences. LS 1.8.1 LS 1.8.2 PL 1 PL 3 (a) 3 × f = 6 f = 6 ÷ 3 f = 2 (b) g × 4 = 24 g = 24 ÷ 4 g = 6 (c) 10 × k = 70 k = 70 ÷ 10 k = 7 (d) m × 11 = 132 m = 132 ÷ 11 m = 12 (e) 12 × n = 108 n = 108 ÷ 12 n = 9 (f) p × 15 = 90 p = 90 ÷ 15 p = 6 (g) 6 ÷ r = 2 r = 6 ÷ 2 r = 3 (h) s ÷ 4 = 9 s = 9 × 4 s = 36 (i) v ÷ 5 = 12 v = 12 × 5 v = 60 (j) 80 ÷ w = 10 w = 80 ÷ 10 w = 8 (k) x ÷ 11 = 13 x = 13 × 11 x = 143 (l) 144 ÷ y = 12 y = 144 ÷ 12 y = 12 32. Determine the value of p. LS 1.8.1 PL 1 PL 3 Mathematics Comic The product of p and 1 000 is four hundred twenty-five thousand. What is the value of p ? Empat ratus dua puluh lima. Write the value of p in words in the blank space above. p × 1 000 = 425 000 p = 425 000 ÷ 1 000 p = 425 Mathematics Comic
Mathematics Year 5 Unit 1 Whole Numbers and Operations 15 © Penerbitan Pelangi Sdn. Bhd. PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: PBD Module Mixed operations Textbook: 40 – 49 LS 1.7.1 Calculate mixed operations within 1 000 000 with and without brackets: (i) Addition and multiplication, (ii) Subtraction and multiplication, 1.7 33. Determine the order of the mixed operations. Then, solve it. LS 1.7.1(i), (ii) PL 2 PL 3 (a) 198 716 + 14 × 3 721 = 250 810 Lakukan operasi darab terlebih dahulu diikuti dengan operasi tambah. 3 7 2 1 × 1 4 1 4 8 8 4 + 3 7 2 1 0 5 2 0 9 4 1 1 1 1 1 9 8 7 1 6 + 5 2 0 9 4 2 5 0 8 1 0 (b) 5 327 × 6 + 34 981 = 66 943 Lakukan operasi darab terlebih dahulu diikuti dengan operasi tambah. 5 3 2 7 × 6 3 1 9 6 2 1 1 3 1 9 6 2 + 3 4 9 8 1 6 6 9 4 3 (c) (74 706 + 6 129) × 7 = 565 845 Lakukan operasi di dalam kurungan terlebih dahulu diikuti dengan operasi dari kiri ke kanan. 1 1 7 4 7 0 6 + 6 1 2 9 8 0 8 3 5 5 5 2 3 8 0 8 3 5 × 7 5 6 5 8 4 5 (d) 12 × 69 456 – 108 426 = 725 046 Lakukan operasi darab terlebih dahulu diikuti dengan operasi tolak. 6 9 4 5 6 × 1 2 1 3 8 9 1 2 + 6 9 4 5 6 0 8 3 3 4 7 2 2 13 6 12 8 3 3 4 7 2 – 1 0 8 4 2 6 7 2 5 0 4 6 (e) 591 000 – 16 × 14 832 = 353 688 Lakukan operasi darab terlebih dahulu diikuti dengan operasi tolak. 1 4 8 3 2 × 1 6 8 8 9 9 2 + 1 4 8 3 2 0 2 3 7 3 1 2 10 8 0 9 9 10 5 9 1 0 0 0 – 2 3 7 3 1 2 3 5 3 6 8 8 (f) 13 × (253 914 – 184 356) = 904 254 Lakukan operasi di dalam kurungan terlebih dahulu diikuti dengan operasi dari kiri ke kanan. 6 9 5 5 8 × 1 3 2 0 8 6 7 4 + 6 9 5 5 8 0 9 0 4 2 5 4 14 10 1 4 13 8 0 14 2 5 3 9 1 4 – 1 8 4 3 5 6 6 9 5 5 8
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 16 PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: LS 1.7.1 Calculate mixed operations within 1 000 000 with and without brackets: (iii) Addition and division, (iv) Subtraction and division. 34. Determine the order of the mixed operations. Then, solve it. LS 1.7.1(iii), (iv) PL 2 PL 3 (a) 9 852 ÷ 2 + 316 942 = 321 868 Lakukan operasi bahagi terlebih dahulu diikuti dengan operasi tambah. 4 9 2 6 2 9 8 5 2 – 8 1 8 – 1 8 0 5 – 4 1 2 – 1 2 0 1 1 4 9 2 6 + 3 1 6 9 4 2 3 2 1 8 6 8 (b) (260 491 + 494 140) ÷ 8 = 94 328 baki 7 Lakukan operasi di dalam kurungan terlebih dahulu diikuti dengan operasi dari kiri ke kanan. 1 1 2 6 0 4 9 1 + 4 9 4 1 4 0 7 5 4 6 3 1 9 4 3 2 8 8 7 5 4 6 3 1 – 7 2 3 4 – 3 2 2 6 – 2 4 2 3 – 1 6 7 1 – 6 4 7 (c) 410 520 – 369 538 ÷ 13 = 382 094 Lakukan operasi bahagi terlebih dahulu diikuti dengan operasi tambah. 2 8 4 2 6 13 3 6 9 5 3 8 – 2 6 1 0 9 – 1 0 4 5 5 – 5 2 3 3 – 2 6 7 8 – 7 8 0 10 11 3 0 10 4 1 10 4 1 0 5 2 0 – 2 8 4 2 6 3 8 2 0 9 4 (d) 653 950 ÷ (100 – 75) = 26 158 Lakukan operasi di dalam kurungan terlebih dahulu diikuti dengan operasi dari kiri ke kanan. 100 – 75 = 25 2 6 1 5 8 25 6 5 3 9 5 0 – 5 0 1 5 3 – 1 5 0 3 9 – 2 5 1 4 5 – 1 2 5 2 0 0 – 2 0 0 0
Mathematics Year 5 Unit 1 Whole Numbers and Operations 17 © Penerbitan Pelangi Sdn. Bhd. PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: PBD Module Problem solving Textbook: 50 – 59 LS 1.9.1 Solve problems involving whole numbers up to 1 000 000 in daily situations. LS 1.9.2 Solve daily problems involving basic operations and mixed operations within 1 000 000. 1.9 35. Solve the following problems. Applications Daily LS 1.9.1 LS 1.9.2 (a) The diagram shows a number card. 476 320 Hamid needs five more number cards. The value of each card is added with 10 000 from its previous number. Arrange all the number cards in descending order. PL 4 (b) The table shows the number of residents in four towns. Town E F G H Number of residents 351 682 276 340 249 865 360 918 Which of the towns will have 300 000 residents when it is rounded off to the nearest hundred thousand? PL 4 (c) There are 124 562 packs of chocolate samples will be distributed in a food exhibition. Each visitor will receive 3 packs of the chocolate samples. How many visitors will receive the chocolate samples? Is there a surplus in the samples? If yes, how many? PL 5
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 18 PERFORMANCE LEVEL () 1 2 3 4 5 6 Date: LS 1.9.2 Solve daily problems involving basic operations and mixed operations within 1 000 000. LS 1.9.3 Solve multiplication and division problems in daily situations involving one unknown. 36. Solve the following problems. Applications Daily LS 1.9.2 LS 1.9.3 (a) The table shows the number of hand sanitisers sold in two weeks. Week Number of bottles First 16 450 Second 7 times the number in the first week Calculate the total number of hand sanitisers sold in the two weeks. PL 5 Jumlah cecair pensanitasi tangan yang telah dijual dalam dua minggu = 16 450 + 7 × 16 450 = 16 450 + 115 150 = 131 600 (b) The diagram shows the number of bottles of orange juice produced by two factories. APEX factory FINAX factory 18 650 bottles of orange juice Calculate the number of bottles of orange juice produced by both factories. PL 5 Jumlah jus oren yang telah dihasilkan = 18 650 ÷ 5 + 18 650 = 3 730 + 18 650 = 22 380 botol (c) 100 000 storybooks will be distributed to h number of schools equally. Each school receives 100 storybooks. What is the value of h? PL 6 Applying 100 000 ÷ h = 100 h = 100 000 ÷ 100 h = 1 000
Mathematics Year 5 Unit 1 Whole Numbers and Operations 19 © © Penerbitan Pelangi Sdn. Bhd Penerbitan Pelangi Sdn. Bhd.. Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 Teacher’s Signature: Date: UASA Section B (a) MK Publishing Company produced 320 516 reference books and storybooks in a particular year. The number of reference books produced is 17 840 less than the number of storybooks. Calculate the number of reference books produced by the company. Bilangan buku rujukan = (320 516 − 17 840) ÷ 2 = 302 676 ÷ 2 = 151 338 (b) The diagram shows the conversation between two workers of MK Publishing Company. Our company sold 54 090 books in June. The number of books sold in July is 3 840 more than June. Mr. Muthu Mr. Goh Calculate the total number of books sold in both months. Bilangan buku yang dijual pada bulan Julai Jumlah bilangan buku yang dijual = 54 090 + 3 840 = 54 090 + 57 930 = 57 930 = 112 020 (c) MK Publishing Company distributed 65 700 books in January. The number of books distributed by PT Publishing Company in that month is 4 times the number of MK Publishing Company. What is the number of books distributed by PT Publishing Company? Bilangan buku yang diagihkan oleh Syarikat Penerbitan PT = 65 700 × 4 = 262 800 (d) There are 249 150 comic books in the store owned by MK Publishing Company. About 100 boxes of comic books where each box contains 50 comics have been sent to the bookstore. Calculate the balance of comic books that are still in the store. Baki komik yang masih ada di dalam stor = 249 150 − 100 × 50 = 249 150 − 5 000 = 244 150 Booster Zone Booster Zone
© Penerbitan Pelangi Sdn. Bhd. 20 A Section 1. The diagram shows a number card. 76 195 (a) State the place value of the underlined digit. Puluh (b) Round off the number to the nearest ten thousand. 80 000 [2 marks] 2. 42 × m = 84 thousand Calculate the value of m. 42 × m = 84 ribu 42 × m = 84 000 m = 84 000 ÷ 42 m = 2 000 [2 marks] 3. 912 870 ÷ (513 − 506) = 912 870 ÷ (513 − 506) = 912 870 ÷ 7 = 130 410 [2 marks] 4. A company distributed storybooks to 65 schools where each school received 250 storybooks. The company still has 1 300 storybooks in the store. Calculate the total number of storybooks owned by the company before distribution. Bilangan buku cerita sebelum pengagihan = (250 × 65) + 1 300 = 16 250 + 1 300 = 17 550 [2 marks] 5. The table shows the number of visitors visiting the robotics exhibition in three months. Month Number of visitors April 24 160 May 3 times the number in April June 2 times the number in May Calculate the number of visitors in June. Bilangan pengunjung pada bulan Mei = 24 160 × 3 = 72 480 Bilangan pengunjung pada bulan Jun = 72 480 × 2 = 144 960 [2 marks] SUMMATIVE PRACTICE 1
Mathematics Year 5 Unit 1 Whole Numbers and Operations 21 © Penerbitan Pelangi Sdn. Bhd. B Section 1. (a) Happy Bakery produced 52 731 packets of various flavored breads in January. The owner of the bakery predicts that the production of bread will increase by 100 packets per month for the following months. Calculate the number of breads produced in the fourth month. +100 +100 +100 52 731 52 831 52 931 53 031 Maka, penghasilan roti pada bulan keempat ialah sebanyak 53 031 bungkus. [2 marks] (b) The table shows the number of breads sold by Happy Bakery according to flavor. Flavor Chocolate Pandan Coconut Number of breads 24 095 16 703 21 846 (i) Form a five-digit number by using all the underlined digits value. 20 706 (ii) If the price of a packet of bread is RM5, calculate the sale of all three flavors. Jumlah bilangan roti Hasil jualan = 24 095 + 16 703 + 21 846 = 62 644 × RM5 = 62 644 = RM313 220 [4 marks] (c) Happy Bakery received an order of 6 300 packets of bread. All of the breads were put in a few boxes. If each box contains 100 packets of bread, how many boxes are needed by Happy Bakery? Bilangan kotak yang diperlukan = 6 300 ÷ 100 = 63 buah kotak [3 marks] (d) Happy Bakery supplies 3 250 packets of bread to several stalls daily. Happy Bakery does not supply bread for 4 days in April. Calculate the number of breads supplied by Happy Bakery in April. Bilangan roti yang dibekalkan dalam bulan April = 3 250 × (30 − 4) = 3 250 × 26 = 84 500 [3 marks] Gamified Quiz 1
Mathematics Year 5 Unit 1 Whole Numbers and Operations © Penerbitan Pelangi Sdn. Bhd. 22 Date: PAK–21 ACTIVITY The teacher prepares one or two sets of number cards with digits 0 until 9 and put them into a box as shown in the diagram below. 3 0 1 9 4 6 2 7 5 8 Instruction: Pupils are divided into groups. The boxes are distributed to each group and each group draws six number cards. Each group will form questions based on what was learnt in Unit 1. Their final work can be written on a A3 paper or in the form of MS PowerPoint slides. Each group will then present their activity. Example of the activity: 1. Example of the chosen number. 4 9 6 0 3 1 2. The largest 6-digit number: 964 310 The smallest 5-digit number: 10 346 3. 964 310 written in words: Nine hundred sixty-four thousand three hundred and ten 4. Partition 964 310 based on the digit value: 900 000 + 60 000 + 4 000 + 300 + 10 5. The place value of digit 9 in the number 964 310 is hundred thousands. 6. Prime number: 3 7. 964 310 rounded off to the nearest ten thousand: 960 000 8. 964 310 + 10 346 = 974 656 9. 964 310 – 10 346 = 953 964 PAK-21 ACTIVITY 1 UNIT Whole Numbers and Operations Fun-and-Pick
23 © Penerbitan Pelangi Sdn. Bhd. 1. Solving fraction and mixed number. Case 1: 1 × 2 5 = 1 × 2 5 = 2 5 Case 2: 3 × 1 3 4 = 3 × 7 4 = 21 4 = 5 1 4 convert the answer to a mixed number convert to improper fraction Case 3: 1 2 × 1 3 = 1 × 1 2 × 3 = 1 6 numerator × numerator denominator × denominator Case 4: 1 1 2 × 1 1 6 = 3 2 × 7 6 = 7 4 = 1 3 4 reduce 3 and 6 by multiple of 3 1 2 2. Rounding off decimal numbers Round off 3.7258 to 1 decimal place 2 decimal places 3 decimal places 3.7258 → 3.7 3.7258 → 3.73 3.7258 → 3.726 3. Addition and subtraction of decimals 2 . 3 0 + 4 . 1 8 6 . 4 8 add zero at the empty place value to make calculation easier the decimal points are arranged in one straight line 9 . – 1 . 2 3 4 8 . 2 3 4 Common mistake: (a) 2.3 + 4.18 = (b) 9 – 1.234 = 8 9 9 10 9 . 0 0 0 – 1 . 2 3 4 7 . 7 6 6 +0 +1 +1 Let’s Grasp! 84.61 – 55.2 + 37.095 = Fractions, Decimals and Percentages 2 UNIT Learning Area: Numbers and Operations
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages © Penerbitan Pelangi Sdn. Bhd. 24 4. Multiplication of decimals (a) 3 × 4.58 = (b) 10 × 0.456 = 4.56 1 2 4 . 5 8 × 3 1 3 . 7 4 regroup when the total multiplication is 10 or more the decimal point of the answer is based on the number of decimal places in the question when a decimal number is multiplied with 10, 100 or 1 000, the decimal point will be moved once, twice or thrice to the right based on the “number of zeroes” 5. Division of decimals (a) 6.3 ÷ 2 = (b) 14.58 ÷ 10 = 1.458 when a decimal number is divided with 10, 100 or 1 000, the decimal point will be moved once, twice or thrice to the left based on the “number of zeroes” place the decimal point in the answer according to the decimal point in the question 3 . 1 5 2 6 . 3 0 – 6 0 3 – 2 1 0 – 1 0 0 add 0 to continue dividing if there is a remainder 6. Converting mixed number to percentage and vice versa. (a) A mixed number can be converted to a percentage based on the following steps. (i) Multiply the whole number and the fraction with 100% separately. Example 1: 2 3 5 = (2 × 100%) + ( 3 5 × 100%) = 200% + (3 × 20%) = 200% + 60% = 260% (ii) Conversion table that must be memorised: 1 2 1 4 3 4 1 5 2 5 3 5 4 5 1 10 50% 25% 75% 20% 40% 60% 80% 10% (b) Percentage can be converted to a mixed number based on the following steps. (i) Divide the percentage with 100% separately. (ii) Simplify the fraction if needed. Example 2: 325% = 300% + 25% = 300% 100% + 25% 100% = 3 1 4 20 1 4 3 1 7. Quantity of percentages (a) 20% of 50 = (b) 120% of 50 = = 20 100 × 50 = 120 100 × 50 = 10 = 60 value of percentage (20%) < 100%, answer < 50 value of percentage (120%) > 100%, answer > 50 Info Percentages 1
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages 25 © Penerbitan Pelangi Sdn. Bhd. Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 PBD Module Fractions Textbook: 65 – 69 LS 2.1.1 Multiply fractions of two numbers involving whole numbers, proper fractions and mixed numbers. 2.1 1. Solve the following multiplications. LS 2.1.1 PL 1 PL 3 (a) 2 × 3 7 = 2 × 3 7 = 6 7 (b) 7 8 × 2 = 7 × 2 8 = 14 8 = 1 6 8 = 1 3 4 (c) 2 1 4 × 5 = 9 4 × 5 = 9 × 5 4 = 45 4 = 11 1 4 (d) 1 6 × 4 7 = 1 × 4 6 × 7 = 4 42 = 2 21 (e) 2 9 × 3 4 = 2 × 3 9 × 4 = 6 36 = 1 6 (f) 3 1 2 × 3 4 = 7 2 × 3 4 = 7 × 3 2 × 4 = 21 8 = 2 5 8 (g) 5 6 × 1 1 2 = 5 6 × 3 2 = 5 × 3 6 × 2 = 15 12 = 1 1 4 (h) 1 2 5 × 4 1 2 = 7 5 × 9 2 = 7 × 9 5 × 2 = 63 10 = 6 3 10 (i) 1 1 3 × 2 1 6 = 4 3 × 13 6 = 4 × 13 3 × 6 = 52 18 = 2 8 9
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages © Penerbitan Pelangi Sdn. Bhd. 26 Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 PBD Module Decimals Textbook: 70 – 79 LS 2.2.1 Round off decimals up to three decimal places. 2.2 2. Round off the following decimals according to the given place value. LS 2.2.1 PL 2 one decimal place two decimal places three decimal places (a) 0.2915 0.3 0.29 0.292 (b) 1.3266 1.3 1.33 1.327 (c) 2.0874 2.1 2.09 2.087 (d) 3.5678 3.6 3.57 3.568 (e) 4.9139 4.9 4.91 4.914 3. Complete these. LS 2.2.1 PL 2 i-THINK Bridge Map round off to 1 decimal place Relating factor 0.4293 kg as as as 2.1564 kg 4.8652 kg 6.3571 kg (a) 0.4 kg (b) 2.2 kg (c) 4.9 kg (d) 6.4 kg round off to 2 decimal places Relating factor 1.075 l as as as 3.4068 l 5.9176 l 7.2405 l (e) 1.08 l (f) 3.41 l (g) 5.92 l (h) 7.24 l round off to 3 decimal places Relating factor 8.1357 m as as as 6.3618 m 4.6854 m 2.7749 m (i) 8.136 m (j) 6.362 m (k) 4.685 m (l) 2.775 m i-THINK Map
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages 27 © Penerbitan Pelangi Sdn. Bhd. Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 LS 2.2.2 Solve mixed operations number sentence involving addition and subtraction of decimals up to three decimal places. 4. Solve the following mixed operations. LS 2.2.2 PL 1 PL 3 (a) 1.5 + 7.65 – 3.9 = 5.25 1 1 . 5 0 + 7 . 6 5 9 . 1 5 8 11 9 . 1 5 – 3 . 9 0 5 . 2 5 (b) 5.2 – 1.38 + 2.7 = 6.52 11 4 1 10 5 . 2 0 – 1 . 3 8 3 . 8 2 1 3 . 8 2 + 2 . 7 0 6 . 5 2 (c) 4.36 – 2.8 + 5.991 = 7.551 3 13 4 . 3 6 – 2 . 8 0 1 . 5 6 1 1 1 . 5 6 0 + 5 . 9 9 1 7 . 5 5 1 (d) 3.2 + 4.97 – 6.265 = 1.905 1 3 . 2 0 + 4 . 9 7 8 . 1 7 7 11 6 10 8 . 1 7 0 – 6 . 2 6 5 1 . 9 0 5 (e) 2.49 + 8.5 – 1.564 = 9.426 1 2 . 4 9 + 8 . 5 0 1 0 . 9 9 0 10 8 10 1 0 . 9 9 0 – 1 . 5 6 4 9 . 4 2 6 (f) 9.1 – 7.265 + 5.05 = 6.885 10 8 0 9 10 9 . 1 0 0 – 7 . 2 6 5 1 . 8 3 5 1 . 8 3 5 + 5 . 0 5 0 6 . 8 8 5
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages © Penerbitan Pelangi Sdn. Bhd. 28 Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 LS 2.2.3 Multiply decimals up to three decimal places with numbers up to two digits, 100 and 1 000. LS 2.2.4 Divide decimals with numbers up to two digits, 100, 1 000, with quotient up to three decimal places. 5. Calculate the following. LS 2.2.3 LS 2.2.4 PL 1 PL 3 (a) 4 × 1.63 = 6.52 2 1 1 . 6 3 × 4 6 . 5 2 (b) 6.423 × 5 = 32.115 2 1 1 6 . 4 2 3 × 5 3 2 . 1 1 5 (c) 15 × 4.132 = 61.98 4 . 1 3 2 × 1 5 2 0 6 6 0 + 4 1 3 2 0 6 1 . 9 8 0 (d) 86.048 ÷ 2 = 43.024 4 3 . 0 2 4 2 8 6 . 0 4 8 – 8 0 6 – 6 0 0 – 0 0 4 – 4 0 8 – 8 0 (e) 26.3 ÷ 20 = 1.315 1 . 3 1 5 20 2 6 . 3 0 0 – 2 0 6 3 – 6 0 3 0 – 2 0 1 0 0 – 1 0 0 0 (f) 431.145 ÷ 15 = 28.743 2 8 . 7 4 3 15 4 3 1 . 1 4 5 – 3 0 1 3 1 – 1 2 0 1 1 1 – 1 0 5 6 4 – 6 0 4 5 – 4 5 0 6. Complete these. LS 2.2.3 LS 2.2.4 PL 3 i-THINK Bridge Map equals to Relating factor 10 × 0.267 as as 100 × 0.085 0.0461 × 1 000 (a) 2.67 (b) 8.5 (c) 46.1 equals to Relating factor 5.8 ÷ 10 as as 47.1 ÷ 100 31.2 ÷ 1 000 (d) 0.58 (e) 0.471 (f) 0.0312 i-THINK Map
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages 29 © Penerbitan Pelangi Sdn. Bhd. Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 PBD Module Percentages Textbook: 80 – 83 LS 2.3.1 Convert mixed numbers to percentages and vice versa. 2.3 7. Write the percentages based on the given fractions. LS 2.3.1 PL 2 i-THINK Bubble Map Percentages (a) 40% (b) 50% (c) 225% (e) 65% (d) 125% (a) 2 5 = 2 × 20 5 × 20 = 40 100 = 40% (e) 13 20 = 13 × 5 20 × 5 = 65 100 = 65% (b) 1 2 = 1 × 50 2 × 50 = 50 100 = 50% (c) 2 1 4 = 2 1 + 1 4 = 2 × 100 1 × 100 + 1 × 25 4 × 25 = 200 100 + 25 100 = 200% + 25% = 225% (d) 1 2 8 = 1 1 4 = 1 1 + 1 4 = 1 × 100 1 × 100 + 1 × 25 4 × 25 = 100 100 + 25 100 = 125% i-THINK map
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages © Penerbitan Pelangi Sdn. Bhd. 30 Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 LS 2.3.1 Convert mixed numbers to percentages and vice versa. LS 2.3.2 Calculate quantity of percentage up to more than 100% and vice versa. 8. Convert the following mixed numbers to percentages. LS 2.3.1 PL 2 (a) 3 1 2 = 7 2 = 7 2 × 100% = 350% (b) 1 3 4 = 7 4 = 7 4 × 100% = 175% (c) 4 1 5 = 21 5 = 21 5 × 100% = 420% 9. Convert the following percentages to mixed numbers. LS 2.3.1 PL 2 (a) 140% = 140 100 = 1 4 10 = 1 2 5 (b) 230% = 230 100 = 2 3 10 (c) 480% = 480 100 = 4 8 10 = 4 4 5 10. Calculate the quantity based on the given percentage. LS 2.3.2 PL 3 (a) 20% of 100 = 20 100 × 100 = 20 (b) 45% of 240 = 45 100 × 240 = 108 (c) 5% of 90 cakes = 5 100 × 90 = 4.5 biji kek (d) 140% of RM120 = 140 100 × RM120 = RM168 (e) 125% of 360 kg = 125 100 × 360 kg = 450 kg (f) 180% of 4 l = 180 100 × 4 l = 7.2 l 11. Find the percentage for the following quantities. LS 2.3.2 PL 3 (a) 78 men of 260 people = 78 260 × 100% = 30% (b) RM288 of RM120 = 288 120 × 100% = 240% (c) 153 g of 60 g = 153 60 × 100% = 255%
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages 31 © Penerbitan Pelangi Sdn. Bhd. Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 PBD Module Problem solving Textbook: 84 – 87 LS 2.4.1 Solve daily problems involving fractions, decimals and percentages. 2.4 12. Solve the following problems. Applications Daily LS 2.4.1 (a) The diagram shows five equal boxes. There are several boxes that are shaded in reds as per shown in the diagram. Auni shades 1 2 of the unshaded boxes with green. Calculate the fraction of the box that is shaded with green from the whole diagram. PL 4 Pecahan petak yang dilorekkan dengan warna hijau = 1 2 × 2 5 = 1 5 (b) The diagram shows a conversation between two brothers. Hasif Naim Our sister has poured 2 3 5 l of corn juice into several glasses. The original volume of the corn juice is 1 1 2 times the volume of corn juice that has been poured into the glasses. Calculate the original volume of the corn juice prepared by their sister. PL 5 Isi padu asal air jagung = 1 1 2 × 2 3 5 l = 3 9 10 l
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages © Penerbitan Pelangi Sdn. Bhd. 32 Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 LS 2.4.1 Solve daily problems involving fractions, decimals and percentages. 13. Solve the following problems. Applications Daily LS 2.4.1 (a) The diagram shows the distance between location E and location F. F G E 82.16 km Given that the distance from E to F is 5 times the distance from F to G. Calculate the distance from F to G. PL 5 Jarak dari F ke G = 82.16 km ÷ 5 = 16.432 km (b) The total number of reference books in a library is 3 000. 60% of the total books are Mathematics reference books. (i) Calculate the number of Mathematics reference books in the library. PL 4 Bilangan buku rujukan Matematik = 60% daripada 3 000 = 60 100 × 3 000 = 1 800 (ii) 20% of the Mathematics reference book has been borrowed. How many Mathematics reference books are still left in the library? Routine NonPL 6 Peratus buku rujukan Matematik yang tinggal = 100% – 20% = 80% Bilangan buku rujukan Matematik yang tinggal = 80 100 × 1 800 = 1 440
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages 33 © © Penerbitan Pelangi Sdn. Bhd Penerbitan Pelangi Sdn. Bhd.. Date: Teacher’s Signature: Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 UASA Section B (a) Mei Ling has 2 1 4 m of fabric. She used 4 5 of the fabric to make a shirt for her son. What is the length of fabric, in m, used by Mei Ling to make the shirt? Panjang kain yang digunakan = 4 5 × 2 1 4 m = 4 5 × 9 4 m = 9 5 m = 1 4 5 m (b) The table shows the length of fabric based on colour bought by Mei Ling. Fabric Length Red 29.7 m Blue 3.8 m more than red fabric Green 7.6 m less than blue fabric Calculate the length, in m, of green fabric bought by Mei Ling. Panjang kain berwarna hijau = 29.7 m + 3.8 m − 7.6 m = 33.5 m − 7.6 m = 25.9 m (c) Mei Ling received an order to sew 20 clothes. 40% of the clothes are t-shirts while, 50% of the remaining clothes are baju kurung. Find the number of baju kurung that Mei Ling need to sew. Bilangan baju kemeja = 40 100 × 20 = 8 Baki baju = 20 − 8 = 12 Bilangan baju kurung = 50 100 × 12 = 6 (d) Mei Ling keeps the unused fabric into two boxes P and Q. Given that the length of fabric in box P is 4.62 m. The length of fabric in box Q is 3 times the length of fabric in box P. Calculate the length of fabric in box Q. Round off the answer to one decimal place. Panjang kain di dalam kotak Q = 4.62 m × 3 = 13.86 m = 13.9 m Booster Zone Booster Zone
© Penerbitan Pelangi Sdn. Bhd. 34 A Section 1. 2 1 3 × 39 = 2 1 3 × 39 = 7 3 × 39 = 91 [2 marks] 2. The diagram shows a number card. 8.707 Round off the decimal number to two decimal places. 8.71 [1 mark] 3. 94.316 + – 27.38 = 125.233 What is the number that should be written in the above? 13 12 8 3 2 11 9 4 . 3 1 6 – 2 7 . 3 8 0 6 6 . 9 3 6 11 14 11 12 0 1 4 1 2 13 1 2 5 . 2 3 3 – 6 6 . 9 3 6 5 8 . 2 9 7 [2 marks] 4. 3 × 5.164 = 1 1 5 . 1 6 4 × 3 1 5 . 4 9 2 [2 marks] 5. Convert 325% to a mixed number. 325% = 325 100 = 3 25 100 = 3 1 4 [2 marks] 6. The diagram shows two rectangles of equal size. Each of the rectangle is divided into several equal parts. Calculate the total percentage of the shaded area in both rectangles. Pecahan kawasan berlorek = 1 4 5 Peratusan kawasan berlorek = 9 × 20 5 × 20 = 180 100 = 180% [2 marks] 7. Haikal jogs 43.05 km in a week. The distance Haikal jogs every day is the same. Calculate the distance, in km, Haikal’s daily jogging distance. Jarak Haikal berjoging setiap hari = 43.05 km ÷ 7 = 6.15 km [2 marks] SUMMATIVE PRACTICE 2
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages 35 © Penerbitan Pelangi Sdn. Bhd. B Section 1. (a) The table shows the number of chickens in chicken coops A and B owned by Encik Hisham. Coop A B Number of chickens 400 35% of the number of chickens in coop A Calculate the number of chickens in coop B. Bilangan ayam di dalam reban B = 35 100 × 400 = 140 ekor [2 marks] (b) Encik Hisham collected 480 eggs from his chicken farm in a particular day. 1 4 of the total eggs is grade A while the balance are grade B and grade C. If the number of Grade B eggs is 11 2 of the number of grade A eggs, calculate the number of grade C eggs collected by Encik Hisham. Bilangan telur gred A = 1 4 × 480 = 120 biji Bilangan telur gred B = 11 2 × 120 = 3 2 × 120 = 180 biji Bilangan telur gred C = 480 − 120 − 180 = 180 biji [5 marks]
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages © Penerbitan Pelangi Sdn. Bhd. 36 (c) There are 120 eggs in a basket. Encik Hisham found that 20% of the eggs were spoiled. Calculate the number of eggs that are still good. Bilangan telur yang telah rosak = 20 100 × 120 = 24 biji Bilangan telur yang masih elok = 120 − 24 = 96 biji [3 marks] (d) Encik Hisham supplied 820 eggs to a supermarket in January. If the number of eggs supplied in February is 3 4 of the number of eggs supplied in January, how many eggs were supplied in February? Bilangan telur yang dibekalkan pada bulan Februari = 3 4 × 820 = 615 biji [2 marks] Gamified Quiz 2
Mathematics Year 5 Unit 2 Fractions, Decimals and Percentages 37 © Penerbitan Pelangi Sdn. Bhd. Date: PERFORMANCE LEVEL () 1 2 3 4 5 6 The teacher prepares 12 mixed number cards and percentage cards for each group. The card received by each group will be recorded into a table as shown below. Instruction: The pupils are divided into several groups. Each group member will take turns to provide a response in terms of changing the mixed number to percentage and vice versa. Each group will then present the activity. No. Mixed number Percentage No. Mixed number Percentage 01 3 1 2 07 425% 02 540% 08 1 4 5 03 2 3 4 09 750% 04 180% 10 5 3 10 05 4 5 8 11 320% 06 910% 12 6 1 4 Example of the activity: No. Mixed number Percentage No. Mixed number Percentage 01 3 1 2 350% 07 4 1 4 425% 02 5 2 5 540% 08 1 4 5 180% 03 2 3 4 275% 09 7 1 2 750% 04 1 4 5 180% 10 5 3 10 530% 05 4 5 8 462.5% 11 3 1 5 320% 06 9 1 10 910% 12 6 1 4 625% PAK–21 ACTIVITY PAK-21 ACTIVITY 2 UNIT Fractions, Decimals and Percentages Round Table
© Penerbitan Pelangi Sdn. Bhd. 52 Mathematics Year 5 Ujian Pertengahan Sesi Akademik (UPSA) 1. The diagram shows three number cards. 380 422 70 458 199 600 (a) Form a six-digit number by using all the underlined digits value. 180 400 (b) Calculate the difference between the largest number and the smallest number. = 380 422 − 70 458 = 309 964 [3 marks] 2. 3 hundred thousand + 67 thousand + 18 hundred = 3 ratus ribu + 67 ribu + 18 ratus = 300 000 + 67 000 + 1 800 = 367 000 + 1 800 = 368 800 [2 marks] 3. A factory produces 445 700 pencils. 2 576 damaged pencils are discarded and the rest are put into 22 boxes equally. Calculate the number of pencil in each box Bilangan pensel yang terdapat di dalam setiap kotak = (445 700 − 2 576) ÷ 22 = 443 124 ÷ 22 = 20 142 [2 marks] 4. There are 62 seats in each row in a stadium. Calculate the total number of seats if the stadium has 839 rows. Jumlah bilangan tempat duduk = 839 × 62 = 52 018 [2 marks] 5. 7 9 × 9 14 = 7 9 × 9 14 = 7 14 = 1 2 [2 marks] 6. Round off 14.951 to two decimal places. 14.95 [1 mark] 7. 14.9 + 0.17 − 2.76 = 14.9 + 0.17 − 2.76 = 15.07 − 2.76 = 12.31 [2 marks] 8. Convert 33 5 to percentages. 33 5 = 3 1 + 3 5 = 3 × 100 1 × 100 + 3 × 20 5 × 20 = 300 100 + 60 100 = 300% + 60% = 360% [2 marks] Section A (26 marks) Answer all questions. 1 1 Ujian Pertengahan Sesi Akademik Score 50 (UPSA)
© Penerbitan Pelangi Sdn. Bhd. 54 Mathematics Year 5 Ujian Pertengahan Sesi Akademik (UPSA) Section B (24 marks) Answer all questions. 1. (a) The table shows the number of cupcakes sold by Maria’s bakery for three consecutive years. Year Number of cupcakes First 5 350 less than the second year Second 108 660 Third 240% of the first year Calculate the number of cupcakes sold on the third year. Bilangan kek cawan yang dijual pada tahun pertama = 108 660 − 5 350 = 103 310 biji Bilangan kek cawan yang dijual pada tahun ketiga = 240 100 × 103 310 = 247 944 biji [4 marks] (b) Maria’s bakery received an order of 1 426 cupcakes for a wedding. All of the cupcakes are put into several boxes. If each box contains 23 cupcakes, what is the number of boxes needed? Bilangan kotak yang diperlukan = 1 426 ÷ 23 = 62 buah kotak [2 marks] (c) Maria’s bakery workers prepared 186 chocolate-flavored cupcakes on a particular day. The number of orange-flavored cupcakes prepared on that day was 25 6 of the number of chocolate-flavored cupcakes. Calculate the number of orange-flavored cupcakes prepared on that day. Bilangan kek cawan berperisa oren = 25 6 × 186 = 17 6 × 186 = 527 biji [2 marks] 31 1
Mathematics Year 5 Ujian Akhir Sesi Akademik (UASA) © Penerbitan Pelangi Sdn. Bhd. 132 1. State the digit value of digit 2 in 420 935. 20 000 [1 mark] 2. The diagram shows two number cards. Form a prime number using both digits in the number cards. Nombor yang boleh dibentuk menggunakan kedua-dua nombor ialah 35 dan 53. 35 bukan nombor perdana kerana dapat dibahagi dengan 5 dan 7. Maka, nombor perdana yang dapat dibentuk ialah 53. [1 mark] 3. 184 400 ÷ 8 = 23 050 8 184 400 –16 24 –24 0 4 – 0 40 – 40 00 – 0 0 [2 marks] 4. Convert 5 3 4 to percentage. 5 3 4 = 5 1 + 3 4 = 5 × 100 1 × 100 + 3 × 25 4 × 25 = 500 100 + 75 100 = 500% + 75% = 575% [2 marks] 5. Encik Haikal bought a motorcycle for RM5 380. He paid a down payment of RM388 in cash and the balance in installments over 2 years. Calculate the monthly payment made by Encik Haikal. Harga motosikal selepas pendahuluan = RM5 380 − RM388 = RM4 992 Ansuran bulanan = RM4 992 ÷ (2 × 12) = RM4 992 ÷ 24 = RM208 [3 marks] Section A (26 marks) Answer all questions. Ujian Akhir Sesi Akademik Score 50 (UASA)
Mathematics Year 5 Ujian Akhir Sesi Akademik (UASA) 135 © Penerbitan Pelangi Sdn. Bhd. Section B (24 marks) Answer all questions. 1. (a) The diagram shows the mass of a watermelon. 4 080 g Convert the mass of the watermelon to kg. State the answer in fraction. 4 080 g = 4 080 1 000 kg = 4 80 1 000 kg = 4 2 25 kg [2 marks] (b) The price of 1 kg of watermelon is RM2.90. Aiman bought a few watermelons with the total mass of 12 kg and paid RM50.00. Calculate the balance that he received. Harga 12 kg tembikai = RM2.90 × 12 = RM34.80 Baki = RM50.00 – RM34.80 = RM15.20 [3 marks] (c) The diagram shows the ingredients needed to prepare watermelon juice.
A1 © Penerbitan Pelangi Sdn. Bhd. Mathematics Year 5 Answers 1 UNIT Whole Numbers and Operations 1. (a) 312 458 (b) 904 784 (c) 730 606 (d) 245 811 2. (a) (b) (c) (d) 3. (a) 135 629 (b) 316 007 (c) 571 803 (d) 835 414 4. (a) Two hundred eighty-one thousand three hundred and sixty-four (b) Four hundred sixteen thousand two hundred and sixty-three (c) Seven hundred eight thousand one hundred and sixty-four (d) Nine hundred nineteen thousand and eighty 5. (a) ten thousands, 30 000 (b) thousands, 0 (c) hundred thousands, 400 000 (d) tens, 10 (e) hundreds, 800 6. (a) 2 hundred thousands + 1 ten thousands + 9 thousands + 3 hundreds + 6 tens + 7 ones (b) 5 hundred thousands + 7 ten thousands + 2 thousands + 4 hundreds + 1 tens + 3 ones (c) 6 hundred thousands + 0 ten thousands + 8 thousands + 1 hundreds + 9 tens + 2 ones 7. (a) 400 000 + 30 000 + 9 000 + 800 + 10 + 5 (b) 100 000 + 60 000 + 7 000 + 200 + 30 + 8 (c) 300 000 + 70 000 + 0 + 500 + 10 + 4 8. (a) 51 672 (b) 96 520 (c) 416 203 (d) 630 419 9. (a) 164 023 (b) 731 028 (c) 800 416 (d) 268 310 10. (a) larger than (b) smaller than 11. (a) 754 100 (b) 100 457 12. (a) 123 156, 123 651, 132 156, 132 165, 132 615 (b) 476 310, 476 301, 476 013, 467 130, 467 103 (c) 591 123, 591 213, 591 312, 915 132, 915 321 (d) 681 430, 681 403, 681 304, 681 043, 681 034 13. h = 164 325 ; k = 741 936 (Any correct answers are accepted.) 14. m = 654 132 ; n = 249 615 (Any correct answers are accepted.) 15. (a) greater than 1 (b) can only be divided by 1 and itself 16. (a) 11, 13, 17, 19 (b) 43, 47, 53, 59, 61 17. (a) ✓ ✗ ✓ ✗ (b) ✓ ✓ 7 ✓ (c) ✓ 3 7 3 (d) ✓ ✗ ✓ ✗ (e) ✓ 3 7 3 (f) ✓ ✗ ✓ ✗ 18. (a) descending order by tens (b) ascending order by fives 19. (a) 641 100, 641 000, 640 900 (b) 385 910 ; 385 922 20. (a) 523 185 ; 523 205 ascending order by tens (b) 710 364 ; 670 364 descending order by ten thousands (c) 295 148 ; 295 156 ascending order by eights (d) 800 913 ; 800 898 descending order by threes 21. (a) 900 beads (b) 300 beads (Any correct answers are accepted) 22. (a) 326 200 ; 326 000 ; 330 000 ; 300 000 (b) 486 300 ; 486 000 ; 490 000 ; 500 000 (c) 572 600 ; 573 000 ; 570 000 ; 600 000 (d) 709 500 ; 710 000 ; 710 000 ; 700 000 (e) 841 400 ; 841 000 ; 840 000 ; 800 000 23. (a) Accept any numbers between 525 000 until 534 999. (b) Accept any numbers between 750 000 until 849 999. 24. (a) 69 879 (b) 179 984 (c) 900 662 (d) 478 945 (e) 801 304 (f) 361 151 (g) 999 995 (h) 731 080 (i) 987 654 ANSWERS
M DUL LENGKAP The M DUL LENGKAP series for Years 4, 5 and 6 is published specifically for Pentaksiran Bilik Darjah (PBD). Moreover, this series is formulated to fulfil the requirements of formative and summative assessments as outlined by the Malaysian Ministry of Education. The contents of this series are aligned with the Dokumen Standard Kurikulum dan Pentaksiran (DSKP) and textbook. All practices in this series are systematically arranged with extra features to assist pupils in mastering the lessons. Digital resources such as info, simulations and gamified quizzes are also included for a more effective learning experience. W.M: RM10.95 / E.M: RM11.65 PELANGI PelangiPublishing PelangiBooks PelangiBooks Subject / Year 4 5 6 Bahasa Melayu English Matematik Mathematics Sains Science Sejarah Pendidikan Islam Pelajaran Jawi Pendidikan Jasmani dan Pendidikan Kesihatan 4 5 6 MODUL LENGKAP PBD GENIUS PBD Year 5 KSSR Semakan Mathematics ISBN: 978-629-470-219-6 WRC495934