Finding the Unknown | IT

Question 6

Is it possible to make a matchstick arrangement that appears in this sequence using exactly 200 sticks?

Question diagram 1
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Solution
Understand the Question
  • The number of matchsticks in each figure forms an arithmetic pattern: 3,5,7,9,3, 5, 7, 9, \dots
  • The general formula for the number of matchsticks MM in an arrangement of nn triangles is M=2n+1M = 2n + 1.
  • For an arrangement with 200200 matchsticks to be possible, nn must be a whole number.

Step 1 · Find the General Rule for the Pattern

Diagram 1

Counting the matchsticks for each arrangement:

  • For 11 triangle: M=3M = 3
  • For 22 triangles: M=3+2=5M = 3 + 2 = 5
  • For 33 triangles: M=5+2=7M = 5 + 2 = 7
  • For 44 triangles: M=7+2=9M = 7 + 2 = 9

The number of matchsticks increases by 22 each time.

General formula for nn triangles: M=2×n+1=2n+1M = 2 \times n + 1 = 2n + 1

Step 2 · Check if 200 Matchsticks Are Possible

Substitute M=200M = 200 into the formula: 200=2n+1200 = 2n + 1

Solve for nn:

2001=2n199=2n\begin{aligned} 200 - 1 &= 2n \\ 199 &= 2n \end{aligned} n=1992n=99.5\begin{aligned} n &= \dfrac{199}{2} \\[0.6em] n &= 99.5 \end{aligned}

Since the number of triangles nn must be a positive integer, n=99.5n = 99.5 is not valid.

Answer

No, it is not possible to make an arrangement in this sequence using exactly 200200 sticks.

Common Mistakes
  • Odd vs. Even Parity: The formula M=2n+1M = 2n + 1 always results in an odd number for any integer nn. Since 200200 is an even number, it can never be formed in this sequence.
  • Fractional Triangle Count: Overlooking the fact that nn represents the number of geometric shapes and must be a whole number, not a decimal.

More questions in IT

Q1

Find the unknown weights in the following cases:

Q2

Context: Finding the Unknown

Q. Discuss the answers with your classmates. Give reasons why you think your answer is right.

Q3

Find the unknown weight of the sack in the following cases. In Fig. 7.10, all the sacks have the same weight.

[Hint: If we remove equal weights from both the plates, will the weighing scale still be balanced? Remove one sack from each plate for Fig. 7.10.]

Q4

Jasmine decides to make a matchstick arrangement that appears in this sequence, using exactly 99 sticks. What will be the position number of this arrangement in the sequence?

Q5

Can you find ways to get the value of nn, such that 2n+1=992n + 1 = 99?

Q6

Is it possible to make a matchstick arrangement that appears in this sequence using exactly 200 sticks?

Q7

For the weighing scale problems in figures 7.6, 7.7, 7.8, 7.9, 7.10, and 7.11, frame equations by using letter-numbers to denote the unknown weight.

Q8

Context: For the problem in Fig. 7.6, let us denote the weight of one fried egg as ee. Since each slice of bread is 2, we have 2+2+2=62 + 2 + 2 = 6 on one side and e+ee + e on the other side. Since they are equal, we have 6=e+e6 = e + e, or 2e=62e = 6.

For the problem in Fig. 7.7, star=4\text{star} = 4, and we can denote the weight of one ring\text{ring} as yy. So, we have 16 on one side, and 4+2y4 + 2y on the other side. Thus, we have the equation 4+2y=164 + 2y = 16.

Q. Solve the equations that you frame and check if you get the same value for the unknown weight as you got previously.

Q9

Context: For the problem in Fig. 7.6, let us denote the weight of one fried egg as ee. Since each slice of bread is 22, we have 2+2+2=62 + 2 + 2 = 6 on one side and e+ee + e on the other side. Since they are equal, we have: 6=e+e, or6 = e + e, \text{ or} 2e=62e = 6

For the problem in Fig. 7.7, [flower] =4= 4, and we can denote the weight of one [donut] as yy. So, we have 1616 on one side, and 4+2y4 + 2y on the other side. Thus, we have the equation: 4+2y=164 + 2y = 16

Q. Frame 5 equations. Find methods to solve them.

Q10

Context: Consider the equation 2n+1=992n + 1 = 99.

Q. Can this equation have any other solution?

Q11

Try solving 5x4=75x - 4 = 7 using trial and error.

Q12

Consider an equation 15+8=2315 + 8 = 23. If we add, subtract, multiply or divide the same number on both sides, will it still preserve the equality of LHS and RHS?

Q13

Context: It is known that 23×41×11×8×7=5,80,88823 \times 41 \times 11 \times 8 \times 7 = 5,80,888. To find the value of the expression 23×41×11×823 \times 41 \times 11 \times 8, we can divide 5,80,8885,80,888 by 77.

Q. Is this the same as dividing both sides by 77, which removes the factor 77 and leaves only the expression to be evaluated on the LHS?

Q14

Ranjana creates a sequence of arrangements with square tiles as shown below. Can she extend the sequence and make an arrangement using 100 tiles? If yes, which step in the sequence will it be?

We have the expression 3k+13k + 1 which gives the number of tiles needed to make an arrangement in Step kk. To check whether an arrangement is possible using 100 tiles at some Step kk, we can solve the equation: 3k+1=1003k + 1 = 100. Find the value of kk.

Q15

We have the expression 3k+13k + 1 which gives the number of tiles needed to make an arrangement in Step kk.

To check whether an arrangement is possible using 100 tiles at some Step kk, we can solve the equation: 3k+1=1003k + 1 = 100. Find the value of kk.

Q16

Context: Example 11: Riyaz created a math trick, which he tries out on his friend Akash.

Riyaz asked Akash to perform the following steps without revealing the answer to any of the intermediate steps.

  1. Think of a number.
  2. Subtract 3 from the number.
  3. Multiply the result by 4.
  4. Add 8 to the product.
  5. Reveal the final answer.

The final answer revealed by Akash was 24. Using this, Riyaz correctly figured out the starting number that Akash had thought of. Find this number.

Q. Try the steps using different numbers as the starting number. Do you see any relation between the starting number and final answer?

Q17

Context: Since Akash's final answer was 24, we have the equation:

4x4=244(x1)=24x1=6(Dividing both sides by 4)\begin{aligned} 4x - 4 &= 24 \\ 4(x - 1) &= 24 \\ x - 1 &= 6 \quad \text{(Dividing both sides by 4)} \end{aligned}

Thus, Akash thought of the number 7.

Can you think of a simple rule that you can use to get the starting number from the final answer?

Q18

Context: Ramesh and Suresh have 60 marbles between them. Ramesh has 30 more marbles than Suresh. If the number of marbles with Suresh is yy, then the number of marbles with Ramesh is y+30y + 30. Since the total number of marbles is 60, we have the equation: 2y+30=602y + 30 = 60

Q. Use this to find both the unknowns.

Q19

Write equations whose solution is y=5y = 5. Share the equations you made with each other and discuss the methods used.

Q20

Can you form a chain going from the bottom equation to the top? Compare the operations used when going from the top to the bottom and from the bottom to the top.

Q21

Without calculating, can you find the value of the unknown in each equation in the chains above?

[Hint: We have seen that the value that satisfies an equation also satisfies the new equation obtained by performing the same operation on both sides of the original equation.]

Q22

7.3 Mind the Mistake, Mend the Mistake

The following are some equations along with the steps used to solve them to find the value of the letter-number. Go through each solution and decide whether the steps are correct. If there is a mistake, describe the mistake, correct it and solve the equation.

Q26

Context: Consider the equations 5x+4=3x+85x + 4 = 3x + 8 and 3x6=2x+43x - 6 = 2x + 4.

Q. Can we come up with a formula to solve these equations? That is, for the first equation, can we perform some operations using 5, 4, 3, and 8 that will directly give us the solution? Using a similar method, can you solve the second equation using the numbers 3, -6, 2 and 4?

Q27

Context: Brahmagupta's formula for solving equations of the form Ax+B=Cx+DAx + B = Cx + D is x=DBACx = \dfrac{D - B}{A - C}.

Q. Using this formula can you solve this equation 2x+3=4x+52x + 3 = 4x + 5?

Q28

There are some children and donkeys on a beach. Together they have 28 heads and 80 feet. How many donkeys are there? How many children are there?

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