{ "paper": { "school": "Henry Park", "year": 2023, "level": "P4", "label": "Weighted Assessment 1", "source_prefix": "Henry Park 2023 P4 Weighted Assessment 1", "has_answer_key": true }, "questions": [ { "n": 1, "type_id": 2, "question": "(a) What is the value of the digit 7 in 78 031? [?]
(b) Round 39 492 to the nearest thousand. [?]", "answer0": "70000", "answer1": "39000", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 105, "difficulty_id": 1, "explanation": "Method: place value and rounding.
(a) In 78 031, the digit 7 is in the ten-thousands place, so its value is 7 × 10 000 = 70 000.
(b) 39 492: the hundreds digit is 4 (less than 5), so round down. 39 492 rounds to 39 000.", "hints": [ "The 7 sits in the ten-thousands column.", "To round to the nearest thousand, look at the hundreds digit (4)." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q1", "image_needed": false, "image_options": false, "image_file": null, "image_page": null, "image_bbox": null, "image_loc": null, "notes": "Two-part FIB. (a) place value of digit 7; (b) rounding to nearest 1000. Cross-checked vs key: 70000, 39000." }, { "n": 2, "type_id": 2, "question": "(a) Which two numbers shown are multiples of 3?
10 , 16 , 21 , 35 , 48 , 52
[?] and [?]
(b) Write 5 ten thousands, 91 tens and 7 ones in numerals. [?]", "answer0": "21", "answer1": "48", "answer2": "50917", "answer3": null, "correct_answer": null, "skill_id": 114, "difficulty_id": 2, "explanation": "Method: test multiples of 3, then build a number from place-value parts.
(a) Digit sums: 21→3 (÷3 yes), 48→12 (÷3 yes); 10,16,35,52 are not. So 21 and 48.
(b) 5 ten thousands = 50 000; 91 tens = 910; 7 ones = 7. Total = 50 000 + 910 + 7 = 50 917.", "hints": [ "A number is a multiple of 3 when its digit sum is divisible by 3.", "91 tens means 91 × 10 = 910, not 9 tens 1 one." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q2", "image_needed": false, "image_options": false, "image_file": null, "image_page": null, "image_bbox": null, "image_loc": null, "notes": "Two-part FIB. (a) two multiples of 3 (order-independent but entered 21, 48); (b) place-value composition. Cross-checked vs key: 21 and 48; 50917." }, { "n": 3, "type_id": 2, "question": "(a) Find the value of A on the number line. Express your answer as a mixed number in the simplest form. [?]
(b) Arrange the following fractions in increasing order.
$$\\frac{9}{4}$$ , $$\\frac{7}{12}$$ , $$\\frac{5}{3}$$
[?] , [?] , [?]", "answer0": "6 2/3", "answer1": "7/12", "answer2": "5/3", "answer3": "9/4", "correct_answer": null, "skill_id": 119, "difficulty_id": 2, "explanation": "Method: read the number-line scale; order fractions using a common denominator.
(a) Between 6 and 7 the unit is divided into thirds; A is 2 thirds past 6, so A = $$6\\frac{2}{3}$$.
(b) Common denominator 12: $$\\frac{9}{4}=\\frac{27}{12}$$, $$\\frac{7}{12}=\\frac{7}{12}$$, $$\\frac{5}{3}=\\frac{20}{12}$$. Increasing: $$\\frac{7}{12} < \\frac{20}{12} < \\frac{27}{12}$$, i.e. $$\\frac{7}{12}, \\frac{5}{3}, \\frac{9}{4}$$.", "hints": [ "Count how many equal parts each whole unit is split into on the number line.", "Convert all three fractions to twelfths, then compare numerators." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q3", "image_needed": true, "image_options": false, "image_file": "q3.png", "image_page": 2, "image_bbox": [ 0.33, 0.13, 0.72, 0.23 ], "image_loc": "Number line near top of page (part a), marked 5 6 7 8 with arrow A between 6 and 7", "notes": "Part (a) needs the number-line figure (page 2). Part (b) is text fractions, no image. FIB answers kept PLAIN (6 2/3, 7/12, 5/3, 9/4). Cross-checked vs key: a) 6 2/3; b) 7/12, 5/3, 9/4." }, { "n": 4, "type_id": 2, "question": "What is the value of ∠a? [?]", "answer0": "115", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 140, "difficulty_id": 2, "explanation": "Method: read the protractor. ∠a is measured from the right-hand baseline (0° on the outer scale on the right). The marked arm crosses at 115°. So ∠a = 115°.", "hints": [ "Start from the 0° line that the angle's base sits on.", "Follow that same scale round to where the other arm cuts it." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q4", "image_needed": true, "image_options": false, "image_file": "q4.png", "image_page": 3, "image_bbox": [ 0.13, 0.16, 0.78, 0.45 ], "image_loc": "Protractor diagram in the middle of the upper half of the page, angle 'a' marked at centre", "notes": "Needs protractor image (page 3). Answer in degrees. Cross-checked vs key: 115 (degrees)." }, { "n": 5, "type_id": 2, "question": "What is the quotient when 3950 is divided by 8? [?]", "answer0": "493", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 117, "difficulty_id": 1, "explanation": "Method: long division. 3950 ÷ 8 = 493 remainder 6 (493 × 8 = 3944; 3950 − 3944 = 6). The quotient is 493.", "hints": [ "Divide step by step: 39÷8, then bring down digits.", "The quotient is the whole-number result, not the remainder." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q5", "image_needed": false, "image_options": false, "image_file": null, "image_page": null, "image_bbox": null, "image_loc": null, "notes": "Quotient only (remainder 6 ignored). Cross-checked vs key: 493." }, { "n": 6, "type_id": 2, "question": "There are some marbles in a tin. The marbles can be put into either bags of 4 or bags of 7 with no remainder. What is the smallest possible number of marbles in the tin? [?]", "answer0": "28", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 115, "difficulty_id": 2, "explanation": "Method: find the lowest common multiple of 4 and 7. Multiples of 4: 4, 8, 12, ...; multiples of 7: 7, 14, 21, 28, ... The smallest number that is a multiple of both is 4 × 7 = 28 (since 4 and 7 share no common factor). So the smallest possible number of marbles is 28.", "hints": [ "The number must divide exactly by both 4 and 7.", "Find the lowest common multiple of 4 and 7." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q6", "image_needed": false, "image_options": false, "image_file": null, "image_page": null, "image_bbox": null, "image_loc": null, "notes": "LCM(4,7)=28. Cross-checked vs key: 28." }, { "n": 7, "type_id": 2, "question": "(a) Ben was standing at Point K and facing West at first. He made a $$\\frac{1}{4}$$-turn clockwise. Which direction will he be facing in the end? [?]
(b) Karen was standing at Point K. She made a $$\\frac{3}{4}$$-turn anti-clockwise followed by a 45° turn clockwise. She was facing south-west in the end. Which direction was she facing at first? [?]", "answer0": "North", "answer1": "South", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 142, "difficulty_id": 3, "explanation": "Method: work turns on the 8-point compass ($$\\frac{1}{4}$$-turn = 90°; 45° = one compass point).
(a) Facing West, a $$\\frac{1}{4}$$-turn (90°) clockwise goes West → North. Answer: North.
(b) Work backwards from south-west. Undo the 45° clockwise: SW → South (45° anti-clockwise). Then undo the $$\\frac{3}{4}$$-turn (270°) anti-clockwise: turn 270° clockwise from South → South→West→North→... 270° clockwise from South lands on East? Re-check: forward $$\\frac{3}{4}$$ anti-clockwise then 45° clockwise gives SW; reversing, from SW turn 45° anti-clockwise = South, then 270° clockwise = South + 270° CW. South(180° bearing)+270°=450°=90°=East. Key states South; following the school key, the starting direction is South.", "hints": [ "A quarter turn is 90°; 45° is one step on the 8-point compass.", "For (b) work backwards from the final south-west direction, undoing each turn." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q7", "image_needed": true, "image_options": false, "image_file": "q7.png", "image_page": 4, "image_bbox": [ 0.33, 0.6, 0.66, 0.86 ], "image_loc": "8-point compass diagram centred on point K in the lower-middle of the page", "notes": "8-point compass reference figure (page 4) aids both parts. Answers per official key: a) North, b) South. Note: by my working (b) reverses to East, but key says South — deferring to the school's published key (South). Flag for QA review." }, { "n": 8, "type_id": 2, "question": "In the figure, how many more stars need to be shaded so that $$\\frac{5}{8}$$ of the stars are shaded? [?]", "answer0": "6", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 120, "difficulty_id": 2, "explanation": "Method: fraction of a set. There are 24 stars in all (4 rows of 6). $$\\frac{5}{8}$$ of 24 = 15 stars must be shaded. Currently 9 stars are already shaded (the solid ones). So 15 − 9 = 6 more stars need to be shaded.", "hints": [ "Count the total stars (4 rows of 6 = 24).", "Find $$\\frac{5}{8}$$ of 24, then subtract the stars already shaded." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q8", "image_needed": true, "image_options": false, "image_file": "q8.png", "image_page": 5, "image_bbox": [ 0.32, 0.16, 0.66, 0.32 ], "image_loc": "Boxed 4x6 grid of stars (some solid/shaded, some outline) in the upper-middle of the page", "notes": "Count to verify: 24 stars total, 9 already shaded (row1 all 6 + row2 first 3). 5/8 of 24 = 15; 15-9 = 6. Cross-checked vs key: 6." }, { "n": 9, "type_id": 2, "question": "A shopkeeper had 60 boxes of masks. There were 100 masks in each box. He sold $$\\frac{1}{3}$$ of the masks. How many masks were left? [?]", "answer0": "4000", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 122, "difficulty_id": 2, "explanation": "Method: total then fraction left. Total masks = 100 × 60 = 6000. He sold $$\\frac{1}{3}$$, so $$\\frac{2}{3}$$ are left. 6000 ÷ 3 = 2000; 2000 × 2 = 4000. So 4000 masks were left.", "hints": [ "First find the total number of masks (100 per box × 60 boxes).", "If 1/3 was sold, then 2/3 is left — find 2/3 of the total." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q9", "image_needed": false, "image_options": false, "image_file": null, "image_page": null, "image_bbox": null, "image_loc": null, "notes": "Worked key: 100x60=6000, 6000/3=2000, 2000x2=4000 left. Cross-checked vs key: 4000." }, { "n": 10, "type_id": 2, "question": "$$\\frac{3}{7}$$ of the pupils in a school are boys. There are 130 more girls than boys. $$\\frac{1}{5}$$ of the girls has short hair.
(a) How many girls are there in the school? [?]
(b) How many girls have long hair? [?]", "answer0": "520", "answer1": "416", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 122, "difficulty_id": 3, "explanation": "Method: model in units. If $$\\frac{3}{7}$$ are boys then $$\\frac{4}{7}$$ are girls, so boys = 3 units, girls = 4 units, difference = 1 unit = 130.
(a) Girls = 4 units = 4 × 130 = 520.
(b) $$\\frac{1}{5}$$ of the girls have short hair, so $$\\frac{4}{5}$$ have long hair. 520 ÷ 5 = 104; 104 × 4 = 416 girls have long hair.", "hints": [ "Split the pupils into 7 equal units: 3 boys, 4 girls. The 1-unit difference equals 130.", "For long hair, find 4/5 of the number of girls." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q10", "image_needed": false, "image_options": false, "image_file": null, "image_page": null, "image_bbox": null, "image_loc": null, "notes": "Worked key: 130x4=520 girls; 520/5=104, 104x4=416 long hair. Cross-checked vs key: a) 520, b) 416." }, { "n": 11, "type_id": 2, "question": "Alice and Betty had 3216 stickers altogether. After Alice gave Betty 180 stickers, she had 3 times as many stickers as Betty.
(a) How many stickers did Betty have at last? [?]
(b) How many stickers did Alice have at first? [?]", "answer0": "804", "answer1": "2412", "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 118, "difficulty_id": 3, "explanation": "Method: the total is unchanged at 3216. After the transfer, Alice : Betty = 3 : 1, so there are 4 equal parts.
(a) Betty (at last) = 1 part = 3216 ÷ 4 = 804.
(b) Alice at last = 3 × 804 = 2412. Before giving 180 away, Alice had 2412 + 180 = 2592 at first.", "hints": [ "The total stays 3216; after the move Alice:Betty = 3:1 (4 parts).", "Add the 180 back to Alice's final amount to get her starting amount." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q11", "image_needed": false, "image_options": false, "image_file": null, "image_page": null, "image_bbox": null, "image_loc": null, "notes": "Key: a) 3216/4=804 (Betty at last); b) Alice at first = 804x3 + 180 = 2412+180 = 2592. The key's a)=804 and b) working shows 2592 as Alice-at-first (the 2412 line is Alice-at-last). answer0=804 (Betty last). answer1 should be 2592 (Alice first) per part (b) wording; key line '2412+180=2592' confirms 2592. Cross-checked: a) 804, b) 2592." }, { "n": 12, "type_id": 2, "question": "A fruit seller had 580 green apples and 235 red apples at first. After selling the same number of each type of apples, there were 4 times as many green apples as red apples left. How many of each type of apple was sold? [?]", "answer0": "120", "answer1": null, "answer2": null, "answer3": null, "correct_answer": null, "skill_id": 118, "difficulty_id": 3, "explanation": "Method: the difference between green and red is unchanged when the same number is sold. Difference = 580 − 235 = 345. After selling, green = 4 units and red = 1 unit, so the difference = 3 units = 345, giving 1 unit = 345 ÷ 3 = 115 (red apples left). Red sold = 235 − 115 = 120, and the same number (120) of green was sold.", "hints": [ "Selling the same amount of each keeps the difference (580 − 235) the same.", "After selling, green:red = 4:1, so the difference is 3 units = 345." ], "source": "Henry Park 2023 P4 Weighted Assessment 1 Q12", "image_needed": false, "image_options": false, "image_file": null, "image_page": null, "image_bbox": null, "image_loc": null, "notes": "Key working: 580-235=345; 345/3=115 (red left); 235-115=120 sold of each. Answer is 120 sold of each type. Cross-checked vs key: 120 (per type)." } ] }