Singapore Math Level 3A & 3B – Challenge Questions - 12 Questions
CHALLENGE QUESTIONS OF 3A
Friends, we have not presented the Singapore Math Level
3A Challenge Questions exactly as they are in the textbook but have presented
them in a new way, i.e., slightly differently. However, the essence has been
kept the same, so let's focus on solving them.
Q(1-25): Solve the following questions or story
problems.
1. Look at the table below, which shows the amount Alex
saved from Sunday to Tuesday.
|
Sunday
|
Monday
|
Tuesday
|
|
$6
|
$12
|
$18
|
Based on the pattern in the table, how much money will
Alex save by Saturday?
Solution(1):
Here, 12÷6=2; 18÷6=3.
Sunday to Saturday = 7 days.
So, Alex will save by Saturday
= $6×1+$6×2+$6×3+$6×4+$6×5+$6×6+$6×7
= $6 + $12 + $18 + $24 + $30 + $36 + $42
= $168 (Answer)
2. Complete the pattern with the correct numbers.
3,4,7,11,18,29,47, __, ___,____.
Solution(2):
Here, 3+4=7; 4+7=11;…….
So, 47+29 = 76; 76+47 = 123; 123+76 = 199
The completed pattern is 3,4,7,11,18,29,47,76,123,199.
3. At a gathering, 6 handshakes took place. If each
student shook hands with another student once, how many students were at the
gathering?
Solution(3):
M shook hands with N, O, and P. (3 handshakes)
N shook hands with O and P. (2 handshakes)
O shook hands with P. (1 handshake)
3 + 2 + 1 = 6 handshakes;
∵ 4
students were at the gathering.
4. Write 2-digit numbers by using digits 1, 2, and 3
where the digits in each number cannot be repeated. From the numbers you have
written, select all the 2-digit numbers that are divisible by 4.
Solution(4):
According to the second condition, 12 and 32 are
divisible by 4.
Find Z.
Solution(5):
Let the digits be M, N, O and P.
|
M
|
N
Smallest |
O
Largest |
P
|
(O+ P) – (M + N) = 8
M + N + O + P = 26
Use the guess-and-check method.
O + P = 9 + 8 = 17
M +N = 5 + 4 = 9
17 – 9 = 8
5 + 4 + 9 + 8 = 26
Z is 5,498.
Another
Solution:
Let the digits be M, N, O and P.
|
M
|
N
Smallest |
O
Largest |
P
|
(M+N) = (O+P) – 8
Or, (M+N) – (O+P) = -8
Or, (O+P) – (M+N) = 8 ……(i)
M+N+O+P = 26 ……(ii)
By adding (i) and (ii), we get.
2(O+P)=8+26
Or, 2(O+P)=34
Or, O+P = 34/2
Or, O+P = 17
Or, 9 + 8 = 17 [∵O
is the largest digit]
Now,
M+N+17 = 26 [By substituting the value.]
Or, M+N = 26 – 17
Or, M+N = 9
Or, 5+4 = 9 [∵N
is smallest digit and Z is less than 6,000 ]
∵ Z = 5498 (Ans.)
6. A number lies between 20 and 39. If you divide it by
5, you get a remainder of 2. If you divide it by 6, you also get a remainder of
2. What is the number?
Solution(6):
In the 1st condition, the numbers are:
5×4+2 = 20+2 = 22;
5×5+2 = 25+2 = 27;
5×6+2 = 30+2 = 32;
5×7+2 = 35+2 = 37.
In the 2nd condition, the numbers are:
6×4+2 = 24+2 = 26;
6×5+2 = 30+2 = 32;
6×6+2 = 36+2 = 38.
Therefore,
32 ÷ 5 = 6 R 2,
32 ÷ 6 = 5 R 2.
The number is 32.
7. Complete the pattern with the correct numbers.
2, 6, 10, 15, 20, 24, , __, ____, 38, 42, ___,____.
Solution(7):
Here,
2+4=6;
6+4=10;
10+5=15;
15+5=20;
20+4=24;
24+4=28;
28+5=33;
32+5=38;
38+4=42;
42+4=46;
46+5=51;
The completed pattern is: 2, 6, 10, 15, 20, 24, , 28, 33,
38, 42, 46, 51.
8. A group of 6 students shook hands with one another
at a conference. Each student shook hands with another student once. How many
handshakes were take place?
Solution(8):
1st student exchanged handshakes with 5
other student.
2nd student exchanged handshakes with 4
other student.
3rd student exchanged handshakes with 3
other student.
4th student exchanged handshakes with 2
other student.
5th student exchanged handshakes with 1
other student.
5 +4 +3 +2 +1 =15
15 handshakes were exchanged.
9. When a box of apples is shared among 3 girls, there
is 1 apple left. When the box of apples is shared among 4 girls, there is 1
apple left. How many apples are there in the box? (Assume that the number of
apples is not greater than 20.)
Solution(9):
First,
3×1+1=4;
3×2+1=7;
3×3+1=10;
3×4+1=13;
3×5+1=16;
3×6+1=19.
Second,
4×1+1=5;
4×2+1=9;
4×3+1=13;
4×4+1=17.
This means,
13 ÷ 3 = 4 R 1,
13 ÷ 4 = 3 R 1.
There are 13 apples in the box.
10. The last digit of a number is the same as the 1st
digit. The 1st digit is 1 less than the 2nd digit. The
sum of all its digits is 4. Find the 3-digit number and it will be an odd
number.
Solution(10):
If 3 digits are x,y and z.
So,
x=z
y-x=1
x+y+z=4
Using the guess-and-check method:
2 – 1 = 1
1 + 2 + 1 = 4
The 3-digit odd number is 121.
Let’s solve this using a different method:
If 3 digits are x,y and z.
So,
x=z
y-x=1
or, y=1+x
x+y+z=4
or, x+1+x+x=4
or, 3x+1=4
or, 3x=4-1
or, 3x=3
or, x=3/3=1
∵z=1
y=1+1=2
The 3-digit odd number is 121.
11. The table below shows the number of workers and the
number of days required to complete a task. Find the number of days needed for
10 workers to finish the same task.
|
No.
of workers
|
No.
of days needed
|
|
50
|
32
|
|
40
|
44
|
|
30
|
56
|
|
20
|
68
|
|
10
|
?
|
Solution(11):
Here,
44-32=12; 56-44=12;……..
Therefore,
32+12=44;
44+12=56;
56+12=68;
68+12=80;
10 workers need 80 days to finish the same task.
12. Suppose a magazine with 100 pages is opened. The
sum of the facing page numbers of the magazine is divisible by 5. The quotient
is equal to the product of 7 and 3. What are the facing page numbers?
Solution(12):
3 × 7 = 21
21 × 5 = 105
The sum of the facing page numbers of the magazine =
105.
52 + 53 = 105
The facing page numbers are: 52 and 53.
Also: