Finding Common Ground | FIO

Question 15

The length, width, and height of a box are 12 cm12\text{ cm}, 18 cm18\text{ cm}, and 36 cm36\text{ cm} respectively. Which of the following sized cubes can be packed in this box without leaving gaps?

(a) 9 cm9\text{ cm}

(b) 6 cm6\text{ cm}

(c) 4 cm4\text{ cm}

(d) 3 cm3\text{ cm}

(e) 2 cm2\text{ cm}

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • For cubes to completely fill a rectangular box without leaving any gaps, the side length of the cube must divide the length (12 cm12\text{ cm}), width (18 cm18\text{ cm}), and height (36 cm36\text{ cm}) exactly.
  • Therefore, a cube of side length ss can be packed without gaps if and only if ss is a common factor of 1212, 1818, and 3636.
  • We find all common factors of 1212, 1818, and 3636, and check which given options are in that list.

Step 1 · Find the Common Factors of the Dimensions

Given dimensions of the box: Length L=12 cm,Width W=18 cm,Height H=36 cm\text{Length } L = 12\text{ cm}, \quad \text{Width } W = 18\text{ cm}, \quad \text{Height } H = 36\text{ cm}Diagram 1

Listing factors of each dimension:

  • Factors of 12:1,2,3,4,6,12\text{Factors of } 12: 1, 2, 3, 4, 6, 12
  • Factors of 18:1,2,3,6,9,18\text{Factors of } 18: 1, 2, 3, 6, 9, 18
  • Factors of 36:1,2,3,4,6,9,12,18,36\text{Factors of } 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

The common factors of 1212, 1818, and 3636 are: 1,2,3,61, 2, 3, 6

Step 2 · Check the Given Options

A cube of side length ss can be packed without gaps if ss is a common factor of 12,18,12, 18, and 3636:

  • (a) 9 cm9\text{ cm}: 99 does not divide 1212 exactly     \implies Cannot be packed

  • (b) 6 cm6\text{ cm}:

12÷6=2,18÷6=3,36÷6=612 \div 6 = 2, \quad 18 \div 6 = 3, \quad 36 \div 6 = 6

66 divides all three dimensions     \implies Can be packed

  • (c) 4 cm4\text{ cm}: 44 does not divide 1818 exactly     \implies Cannot be packed

  • (d) 3 cm3\text{ cm}:

12÷3=4,18÷3=6,36÷3=1212 \div 3 = 4, \quad 18 \div 3 = 6, \quad 36 \div 3 = 12

33 divides all three dimensions     \implies Can be packed

  • (e) 2 cm2\text{ cm}:
12÷2=6,18÷2=9,36÷2=1812 \div 2 = 6, \quad 18 \div 2 = 9, \quad 36 \div 2 = 18

22 divides all three dimensions     \implies Can be packed

Answer

The cubes that can be packed without leaving gaps are (b) 6 cm6\text{ cm}, (d) 3 cm3\text{ cm}, and (e) 2 cm2\text{ cm}.

Common Mistakes
  • Volume Divisibility Error: Thinking that if the total volume of the box is divisible by the volume of the cube (s3s^3), the cubes will fit without gaps. The side length must divide each individual dimension (L,W,HL, W, H) separately.
  • Checking Only One Dimension: Assuming a cube fits if its side divides only one or two dimensions (e.g., 9 cm9\text{ cm} divides 1818 and 3636, but fails for 1212; 4 cm4\text{ cm} divides 1212 and 3636, but fails for 1818).

More questions in FIO

Q1

List all the factors of the following numbers:

(a) 90

(b) 105

(c) 132

(d) 360 (this number has 24 factors)

(e) 840 (this number has 32 factors)

Q2

Find the common factors and the HCF of the following numbers:

(a) 50, 60

(b) 140, 275

(c) 77, 725

(d) 370, 592

(e) 81, 243

Q3

How do we directly find the HCF without listing all the factors?

Q4

Find the HCF of the following numbers:

(a) 24, 180

(b) 42, 75, 24

(c) 240, 378

(d) 400, 2500

(e) 300, 800

Q5

Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: 72=6×1272 = 6 \times 12 and 144=8×18144 = 8 \times 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not?

Q6

Find the LCM of the following numbers:

(a) 30, 72

(b) 36, 54

(c) 105, 195, 65

(d) 222, 370

Q7

Make a general statement about the HCF for the following pairs of numbers. You could consider examples before coming up with general statements. Look for possible explanations of why they hold.

(a) Two consecutive even numbers

(b) Two consecutive odd numbers

(c) Two even numbers

(d) Two consecutive numbers

(e) Two co-prime numbers

Share your observations with the class.

Q8

The LCM of 3 and 24 is 24 (it is one of the two given numbers).

(a) Find more such number pairs where the LCM is one of the two numbers.

(b) Make a general statement about such numbers. Describe such number pairs using algebra.

Q9

Make a general statement about the LCM for the following pairs of numbers. You could consider examples before coming up with these general statements. Look for possible explanations of why they hold.

(a) Two multiples of 3

(b) Two consecutive even numbers

(c) Two consecutive numbers

(d) Two co-prime numbers

Q10

In the two rows below, colours repeat as shown. When will the blue stars meet next?

Q11

(a) Is 5×7×11×115 \times 7 \times 11 \times 11 a multiple of 5×7×7×11×25 \times 7 \times 7 \times 11 \times 2?

(b) Is 5×7×11×115 \times 7 \times 11 \times 11 a factor of 5×7×7×11×25 \times 7 \times 7 \times 11 \times 2?

Q12

Find the HCF and LCM of the following (state your answers in the form of prime factorisations):

(a) 3×3×5×7×73 \times 3 \times 5 \times 7 \times 7 and 12×7×1112 \times 7 \times 11

(b) 45 and 36

Q13

Find two numbers whose HCF is 1 and LCM is 66.

Q14

A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka.)

Q15

The length, width, and height of a box are 12 cm12\text{ cm}, 18 cm18\text{ cm}, and 36 cm36\text{ cm} respectively. Which of the following sized cubes can be packed in this box without leaving gaps?

(a) 9 cm9\text{ cm}

(b) 6 cm6\text{ cm}

(c) 4 cm4\text{ cm}

(d) 3 cm3\text{ cm}

(e) 2 cm2\text{ cm}

Q16

Among the numbers below, which is the largest number that perfectly divides both 306 and 36?

(a) 36

(b) 612

(c) 18

(d) 3

(e) 2

(f) 360

Q17

Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.

Q18

Children are playing ‘Fire in the Mountain’. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially?

(a) 72

(b) 90

(c) 45

(d) 3

(e) 36

(f) None of these

Q19

Tick the correct statement(s). The LCM of two different prime numbers (m,nm, n) can be:

(a) Less than both numbers

(b) In between the two numbers

(c) Greater than both numbers

(d) Less than m×nm \times n

(e) Greater than m×nm \times n

Q20

A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?

Q21

What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?

Q22

Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together 815\dfrac{8}{15}, 120\dfrac{1}{20}, 736\dfrac{7}{36}, 1163\dfrac{11}{63} and 121\dfrac{1}{21}. What do you get? How can we find this sum efficiently?

← Back to Finding Common Ground