Tài liệu Toán học quốc tế SASMO 2020 Lớp 3 - GEP Practice

 SASMO 2020, Primary 3 (Grade 3) Contest Questions 
 Primary 3 (Grade 3) – GEP Practice 
 2020 
 Contest Problems with Full Solutions 
Authors: 
 Henry Ong, BSc, MBA, CMA 
 Merlan Nagidulin, BSc 
© Singapore International Mastery Contests Centre (SIMCC) 
All Rights Reserved 
No part of this work may be reproduced or transmitted by any means, electronic or mechanical, including 
photocopying and recording, or by any information or retrieval system, without the prior permission of the 
publisher. SASMO 2020, Primary 3 (Grade 3) Contest Questions 
Alice’s first day in Caterpillar Club was Tuesday. She wants to throw a party on her 
40th day in the club. If Alice attends the club every day, on which day of the week 
will the party be? 
 A. Friday 
 B. Saturday 
 C. Sunday 
 D. Monday 
 E. None of the above 
Question 5 
How many multiples of 6 are there between 14 and 100? 
 A. 16 
 B. 15 
 C. 14 
 D. 13 
 E. None of the above 
 SASMO 2020, Primary 3 (Grade 3) Contest Questions 
Question 8 
What is the missing number in the sequence below? 
 1, 3, 7, 15, 31, ___ 
 A. 63 
 B. 47 
 C. 57 
 D. 59 
 E. None of the above 
Question 9 
If the four-digit number 3P78 is divisible by 3, how many possible values are there 
for P? 
 A. 4 
 B. 3 
 C. 5 
 D. 10 
 E. None of the above 
 SASMO 2020, Primary 3 (Grade 3) Contest Questions 
Question 12 
Alex, John and Sam went to buy oranges. Alex paid $20, John paid $15, and Sam 
only paid $5. They bought 120 oranges altogether. They divided them in proportion 
to the amount of money each of them had paid. How many oranges did John get? 
 A. 15 
 B. 30 
 C. 45 
 D. 60 
 E. None of the above 
Question 13 
A tank filled with 200 litres of water weighs 350 kg. The same tank filled with 150 
litres of water weighs 315 kg. What is the weight of the empty tank? 
 A. 120 kg 
 B. 150 kg 
 C. 165 kg 
 D. 210 kg 
 E. None of the above 
 SASMO 2020, Primary 3 (Grade 3) Contest Questions 
Section B (Correct answer – 4 points| Incorrect or No answer – 0 points) 
When an answer is a 1-digit number, shade “0” for the tens, hundreds and thousands place. 
Example: if the answer is 7, then shade 0007 
When an answer is a 2-digit number, shade “0” for the hundreds and thousands place. 
Example: if the answer is 23, then shade 0023 
When an answer is a 3-digit number, shade “0” for the thousands place. 
Example: if the answer is 785, then shade 0785 
When an answer is a 4-digit number, shade as it is. 
Example: if the answer is 4196, then shade 4196 
Question 16 
What is the sum of the first 30 numbers of the following pattern? 
 50, 49, 48, 47, 46  
Question 17 
If you increase the length of a rectangle by 12 cm, you will get a rectangle with a 
perimeter of 38 cm. What is the perimeter of the original rectangle? 
 SASMO 2020, Primary 3 (Grade 3) Contest Questions 
Question 20 
Study the picture below. 
Find the value of . 
Question 21 
If the area of the rectangle is 96 cm2, what is the area (in cm2) of the shaded 
region? 
 SASMO 2020, Primary 3 (Grade 3) Contest Questions 
Question 24 
The numbers 2, 5, 8, 11, 14, 17 and 20 can be placed in the 7 circles below such 
that the sum along each straight line is the same and each number can only be used 
once. What is the largest possible value of this sum? 
Question 25 
In the following, all the different letters stand for different digits. 
 P P P 
 + Q Q Q 
 R Q Q R 
Find the value of the 4-digit number RQQR. 
 END OF PAPER SASMO 2020, Primary 3 (Grade 3) Contest Solutions 
Question 4 
Every one week or 7 days later will return to the same day. 
Alice wants to throw a party on her 40th day or 39 days later in the club. 
39 ÷ 7 = 5R4 means that 39 days later will be 5 weeks and 4 days after Tuesday. 
5 weeks after Tuesday is still Tuesday and 4 days after Tuesday is Saturday. 
Answer: (B) 
Question 5 
From 1 to 100, there are 16 (100 ÷ 6 = 16푅4) multiples of 6. 
From 1 to 14, there are 2 (6 and 12) multiples of 6. 
Thus, there are 16 − 2 = ퟒ multiples of 6 between 14 and 100. 
Answer: (C) 
Question 6 
Let us count the cubes on each stack from left to right. 
There are 3 cubes in the 1st stack. 
There are 3 cubes in the 2nd stack. 
There are 3 cubes in the 3rd stack. 
There are 6 cubes in the 4th stack. 
There are 5 + 4 + 2 + 1 + 3 = 15 cubes in the 5th stack. 
In total, there are 3 + 3 + 3 + 6 + 15 = cubes. 
Answer: (B) 
Question 7 
Emily arrived at the cinema 17 minutes after 3.55 pm, which is 4.12 pm. There are 2 
hours or 120 minutes from 1.47 pm to 3.47 pm. There are 25 minutes from 3.47 pm 
to 4.12 pm. Thus, Emily’s journey from her house to the cinema was 120 + 25 = 145 
minutes long. 
Answer: (E) 
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 SASMO 2020, Primary 3 (Grade 3) Contest Solutions 
Question 11 
The heaviest boy is 53 kg heavy, which is an odd-numbered weight. 
Mason’s weight is an even number, so he is not the heaviest. 
Joshua’s weight is a multiple of 5 which is 45 kg. So, he is not the heaviest. 
Christopher is not the heaviest as per the statement. 
Thus, the remaining boy, Mateo is the heaviest. 
Answer: (D) 
Question 12 
They paid altogether $20 + $15 + $5 = $40 for 120 oranges. 
$40 → 120 oranges 
$1 → 120 ÷ 4 = 3 oranges 
John paid $15 → 15 × 3 = 45 oranges 
Answer: (C) 
Question 13 
 Tank + 200 litres of water = 350 kg 
 Tank + 150 litres of water = 315 kg 
Subtracting the equations above: 50 litres of water = 35 kg 
 150 litres of water = 35 kg × 3 = 105 kg 
 Tank + 150 litres of water = Tank + 105kg = 315 kg 
 Thus, Tank = 315 kg − 105 kg = 퐤퐠 
Answer: (D) 
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 SASMO 2020, Primary 3 (Grade 3) Contest Solutions 
Question 16 
We notice that the 2nd number is (50 − 1), the third number is 50 − 2 and so on. 
Hence the 30th number is 50 − 29 = 21 and the sum is 
50 + 49 + 48   + 23 + 22 + 21 
= (50 + 21) + (49 + 22) + (48 + 23) + ⋯ + (36 + 35) 
= 71 × (30 ÷ 2) 
= 71 × 15 
= . 
Answer: 1065 
Question 17 
A rectangle has 2 lengths and 2 widths. 
When the length of the rectangle is increased by 12 cm for each side, its perimeter is 
increased by 12 × 2 = 24 cm. 
So, original perimeter + 24 cm = 38 cm 
Working backwards: original perimeter = 38 cm − 24 cm = ퟒ 퐜퐦 
Answer: 14 
Question 18 
1-part: 7 triangles 
2-part: 10 triangles 
3-part: 6 triangles 
4-part: 5 triangles 
6-part: 2 triangles 
Total number of triangles = 7 + 10 + 6 + 5 + 2 = 
Answer: 30 
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 SASMO 2020, Primary 3 (Grade 3) Contest Solutions 
Question 22 
It is given that 
 $29 = 2 boxes of chocolates + 3 packets of sweets + 4 baskets of fruits
 = 2 boxes + 3 packets + 4 baskets
 = 2 boxes + 2 packets + 1 packet + 1 basket + 3 baskets
 = 2 × (1 boxes + 1 packet) + (1 packet + 1 basket) + 3 baskets
 = 2 × $4 + $6 + 3 baskets = $14 + 3 baskets. 
Hence 
 3 baskets = $29 − 14 = $15 표 1 basket = $15 ÷ 3 = 5 
 1 푒푡 = $6 − $5 = $1 푛 1 표 = $4 − $1 = $ . 
Answer: $3 
Question 23 
To obtain the largest possible whole number, Diana needs to construct as many 
digits as possible. The digit ‘1’ requires the least number of matchsticks which is 2. 
The digit ‘7’ requires the second least number of matchsticks which is 3. Thus, the 
largest possible number that she can construct using exactly 17 matchsticks is 
71,111,111. 
The number 71,111,111 contains 8 digits. 
Answer: 8 
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