Algebra Play | IT

Question 11

In the following grids, find the values of the shapes and fill in the empty squares:

Question diagram 1
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Solution
Understand the Question
  • Each row and column in a puzzle grid represents a sum of the shapes contained in it.
  • We can translate each row of shapes into a linear equation by assigning a variable to each unique shape.
  • Solving the pair of simultaneous linear equations gives the numerical value of each shape.
  • Once the shape values are determined, we substitute them into the remaining rows and columns to find the missing totals.

Step 1 · Find Shape Values in First Grid

Let the blue square be SS and the red circle be CC.Diagram 1

From Row 1:

S+S+C=272S+C=27…(1)\begin{aligned} S + S + C &= 27 \\ 2S + C &= 27 \quad \dots (1) \end{aligned}

From Row 2:

C+C+S=21S+2C=21…(2)\begin{aligned} C + C + S &= 21 \\ S + 2C &= 21 \quad \dots (2) \end{aligned}

Subtract equation (2) from equation (1):

(2S+C)−(S+2C)=27−21S−C=6S=C+6…(3)\begin{aligned} (2S + C) - (S + 2C) &= 27 - 21 \\ S - C &= 6 \\ S &= C + 6 \quad \dots (3) \end{aligned}

Substitute equation (3) into equation (2):

(C+6)+2C=213C+6=213C=21−63C=15C=153=5\begin{aligned} (C + 6) + 2C &= 21 \\ 3C + 6 &= 21 \\ 3C &= 21 - 6 \\ 3C &= 15 \\ C &= \dfrac{15}{3} = 5 \end{aligned}

Substitute C=5C = 5 into equation (3): S=5+6=11S = 5 + 6 = 11

Thus, Red Circle(C)=5\text{Red Circle} (C) = 5 and Blue Square(S)=11\text{Blue Square} (S) = 11.

Step 2 · Find Missing Row Total in First Grid

Diagram 2

Row 3 consists of a red circle, a blue square, and a red circle:

C+S+C=5+11+5=21\begin{aligned} C + S + C &= 5 + 11 + 5 \\ &= 21 \end{aligned}

Step 3 · Find Shape Values in Second Grid

Let the blue circle be BC\text{BC} and the purple diamond be PD\text{PD}.Diagram 3

From Row 1:

BC+PD+PD=18BC+2PD=18…(4)\begin{aligned} \text{BC} + \text{PD} + \text{PD} &= 18 \\ \text{BC} + 2\text{PD} &= 18 \quad \dots (4) \end{aligned}

From Row 2:

PD+BC+BC=152BC+PD=15…(5)\begin{aligned} \text{PD} + \text{BC} + \text{BC} &= 15 \\ 2\text{BC} + \text{PD} &= 15 \quad \dots (5) \end{aligned}

Multiply equation (5) by 22:

2×(2BC+PD)=2×154BC+2PD=30…(6)\begin{aligned} 2 \times (2\text{BC} + \text{PD}) &= 2 \times 15 \\ 4\text{BC} + 2\text{PD} &= 30 \quad \dots (6) \end{aligned}

Subtract equation (4) from equation (6):

(4BC+2PD)−(BC+2PD)=30−183BC=12BC=123=4\begin{aligned} (4\text{BC} + 2\text{PD}) - (\text{BC} + 2\text{PD}) &= 30 - 18 \\ 3\text{BC} &= 12 \\ \text{BC} &= \dfrac{12}{3} = 4 \end{aligned}

Substitute BC=4\text{BC} = 4 into equation (5):

2(4)+PD=158+PD=15PD=15−8=7\begin{aligned} 2(4) + \text{PD} &= 15 \\ 8 + \text{PD} &= 15 \\ \text{PD} &= 15 - 8 = 7 \end{aligned}

Thus, Blue Circle(BC)=4\text{Blue Circle} (\text{BC}) = 4 and Purple Diamond(PD)=7\text{Purple Diamond} (\text{PD}) = 7.

Step 4 · Find Missing Row Total in Second Grid

Diagram 4

Row 3 consists of a purple diamond, a blue circle, and a blue circle:

PD+BC+BC=7+4+4=15\begin{aligned} \text{PD} + \text{BC} + \text{BC} &= 7 + 4 + 4 \\ &= 15 \end{aligned}

Step 5 · Find Missing Column Totals in Second Grid

Diagram 5

For Column 1: BC+PD+PD=4+7+7=18\text{BC} + \text{PD} + \text{PD} = 4 + 7 + 7 = 18

For Column 2: PD+BC+BC=7+4+4=15\text{PD} + \text{BC} + \text{BC} = 7 + 4 + 4 = 15

For Column 3: PD+BC+BC=7+4+4=15\text{PD} + \text{BC} + \text{BC} = 7 + 4 + 4 = 15

Answer
  • Blue square: 1111
  • Red circle: 55
  • Blue circle: 44
  • Purple diamond: 77
  • Grid 1 Missing Row Total: 2121
  • Grid 2 Missing Row Total: 1515
  • Grid 2 Missing Column Totals: 18,15,1518, 15, 15
Common Mistakes
  • Distinguishing Similar Shapes: Mixing up the red circle from Grid 1 with the blue circle from Grid 2.
  • Sign Errors in Elimination: Forgetting to distribute the negative sign to all terms when subtracting one equation from another (e.g., (2S+C)−(S+2C)=S−C(2S + C) - (S + 2C) = S - C, not S+3CS + 3C).
  • Swapping Variables: Assigning the value of SS to CC or BC\text{BC} to PD\text{PD} during final substitution.

More questions in IT

Q1

Context: Think of a number trick:

  1. Think of a number.
  2. Double it.
  3. Add four.
  4. Divide by two.
  5. Subtract the original number you thought of.

Q. I predict you get 2. Am I right? Try it out with different starting numbers. Do you always end up with the same value, 2? Why?

Q2

Context: Consider the following number trick:

  1. Think of a number.
  2. Double it.
  3. Add four.
  4. Divide by two.
  5. Subtract the original number you thought of. (This trick always results in 2).

Q. How would you change this game to make the final answer 3? What about 5?

Q3

Context: Consider the following number trick:

  1. Think of a number.
  2. Double it.
  3. Add four.
  4. Divide by two.
  5. Subtract the original number you thought of. (This trick always results in 22).

Q. Can you come up with more complicated steps that always lead to the same final value?

Q4

Context: In the date trick, the final answer is given by 100M+165+D100M + 165 + D, where MM is the month and DD is the day.

Q. Find the dates if the final answers are the following:

(i) 1269

(ii) 394

(iii) 296

Q5

Context: In the date trick, the final answer is given by 100M+165+D100M + 165 + D, where MM is the month and DD is the day.

Q. Can you change the steps in this trick and still find the original date? Instead of subtracting 165 from the final answer, you might have to subtract some other number.

Q6

Try to devise your own 'Think of a Number' trick.

Q7

Use the same rule to fill these pyramids:

Q8

Fill the following pyramids:

Q9

What is the relationship between the numbers in the bottom row and the number at the top?

Let us start with the simplest pyramid.

Q10

What about a pyramid with three rows?

Using letter numbers for the bottom row, we can write an expression for the top row.

Q11

In the following grids, find the values of the shapes and fill in the empty squares:

Q12

Context: 6.5 The Largest Product

Q. Fill the digits 2, 3, and 5 in □□×□\square\square \times \square, using each digit once. What is the largest product possible?

Q13

Context: Mukta's trick: Choose a 2-digit number of different digits, reverse the digits to get another number, find their difference, and divide the result by 9.

Q. If we choose other 2-digit numbers, and follow the steps, will there always be no remainder?

Q14

Context: Suppose a two-digit number is abab. When it is reversed, the new number is baba. If b>ab > a, the difference is ba−ab=9(b−a)ba - ab = 9(b - a), which is divisible by 9.

Q. Can you work out what happens if a>ba > b?

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