Finding the Unknown | IT

Question 1

Find the unknown weights in the following cases:

Solution

Understand the Question

A hanging mobile remains balanced when the total weight on the left arm equals the total weight on the right arm: $\text{Left Arm Weight} = \text{Right Arm Weight}$
If the total weight of a balanced mobile is given, each arm supports exactly half of the total weight: $\text{Left Arm Weight} = \text{Right Arm Weight} = \dfrac{\text{Total Weight}}{2}$
We set up linear equations from these balance conditions and given values to find each unknown weight.

(i) Find the weight of the red flower in Fig. 7.1.

Step 1 · Calculate the Weight of the Red Flower

Given total weight $= 16$ and each green leaf weighs $3$ . Diagram 1

$\text{Total weight of 3 leaves} = 3 \times 3 = 9$

Let $F$ be the weight of the red flower:

\begin{aligned} 9 + F &= 16 \\ F &= 16 - 9 \\ &= 7 \end{aligned}

Answer

(i) $\text{Red flower} = 7$

(ii) Find the weights of the fish and submarine in Fig. 7.2.

Step 1 · Calculate the Weights of the Fish and Submarine

Given total weight $= 24$ and starfish weight $S_w = 2$ . Diagram 2

Let $F_w$ be the weight of a fish and $Sub_w$ be the weight of a submarine.

Since the mobile is balanced:

\begin{aligned} 2S_w + F_w &= S_w + F_w + Sub_w \\ 2S_w &= S_w + Sub_w \\ Sub_w &= S_w = 2 \end{aligned}

Using the total weight equation:

\begin{aligned} (2S_w + F_w) + (S_w + F_w + Sub_w) &= 24 \\ 3S_w + 2F_w + Sub_w &= 24 \end{aligned}

Substitute $S_w = 2$ and $Sub_w = 2$ :

\begin{aligned} 3(2) + 2F_w + 2 &= 24 \\ 8 + 2F_w &= 24 \\ 2F_w &= 16 \\ F_w &= 8 \end{aligned}

Answer

(ii) $\text{Fish} = 8, \quad \text{Submarine} = 2$

(iii) Find the weights of the book and money bag in Fig. 7.3.

Step 1 · Calculate the Weights of the Book and Money Bag

Given total weight $= 8$ . Diagram 3

Let $B$ be the weight of one book and $M$ be the weight of one money bag.

Since the mobile is balanced:

\begin{aligned} 2B &= 2M \\ B &= M \end{aligned}

Using the total weight equation:

\begin{aligned} 2B + 2M &= 8 \\ 2B + 2B &= 8 \\ 4B &= 8 \\ B &= 2 \end{aligned}

Since $B = M$ , $M = 2$ .

Answer

(iii) $\text{Book} = 2, \quad \text{Money bag} = 2$

(iv) Find the weights of the cloud and lightning bolt in Fig. 7.4.

Step 1 · Calculate the Weights of the Cloud and Lightning Bolt

Given total weight $= 18$ and sun weight $S = 5$ . Diagram 4

Let $C$ be the weight of a cloud and $L$ be the weight of a lightning bolt.

Since the mobile is balanced:

\begin{aligned} 3C + S &= S + 2L \\ 3C &= 2L \quad \dots (1) \end{aligned}

Using the total weight:

\begin{aligned} (3C + S) + (S + 2L) &= 18 \\ 3C + 2S + 2L &= 18 \\ 3C + 2(5) + 2L &= 18 \\ 3C + 2L &= 8 \quad \dots (2) \end{aligned}

Substitute $2L = 3C$ into equation (2):

\begin{aligned} 3C + 3C &= 8 \\ 6C &= 8 \\ C &= \dfrac{8}{6} = \dfrac{4}{3} \end{aligned}

Now substitute $C = \dfrac{4}{3}$ into equation (1):

\begin{aligned} 3\left(\dfrac{4}{3}\right) &= 2L \\ 4 &= 2L \\ L &= 2 \end{aligned}

Answer

(iv) $\text{Cloud} = \dfrac{4}{3}, \quad \text{Lightning bolt} = 2$

(v) Find the weights of the crown, water drop, and diamond in Fig. 7.5.

Step 1 · Calculate the Weights of Crown, Water Drop, and Diamond

Given total weight $= 40$ . Let $K$ be the weight of a crown, $W$ of a water drop, and $D$ of a diamond. Diagram 5

Since the mobile is balanced:

\begin{aligned} 3K + W &= K + 2W + 2D \\ 2K &= W + 2D \quad \dots (1) \end{aligned}

From the total weight:

\begin{aligned} (3K + W) + (K + 2W + 2D) &= 40 \\ 4K + 3W + 2D &= 40 \quad \dots (2) \end{aligned}

Substitute $2D = 2K - W$ into equation (2):

\begin{aligned} 4K + 3W + (2K - W) &= 40 \\ 6K + 2W &= 40 \\ 3K + W &= 20 \implies W = 20 - 3K \end{aligned}

Substitute $W = 20 - 3K$ into equation (1):

\begin{aligned} 2K &= (20 - 3K) + 2D \\ 5K - 20 &= 2D \\ D &= \dfrac{5K - 20}{2} \end{aligned}

For positive integer weights, $K$ must satisfy $4 < K < \dfrac{20}{3}$ :

If $K = 5$ : $W = 5$ , $D = \dfrac{5}{2} = 2.5$ (not an integer)
If $K = 6$ : $W = 2$ , $D = \dfrac{10}{2} = 5$ (integers)

Therefore, $K = 6$ , $W = 2$ , and $D = 5$ .

Answer

(v) $\text{Crown} = 6, \quad \text{Water drop} = 2, \quad \text{Diamond} = 5$

(vi) Find the weight of the egg in Fig. 7.6.

Step 1 · Calculate the Weight of the Egg

Given toast weight $T = 2$ . Let $E$ be the weight of one egg. Diagram 6

Since the mobile is balanced:

\begin{aligned} 3T &= 2E \\ 3(2) &= 2E \\ 6 &= 2E \\ E &= 3 \end{aligned}

Answer

(vi) $\text{Egg} = 3$

(vii) Find the weight of the 'O' shape in Fig. 7.7.

Step 1 · Calculate the Weight of the 'O' Shape

Given 'X' shape weight $X = 4$ . Let $O$ be the weight of one 'O' shape. Diagram 7

Since the mobile is balanced:

\begin{aligned} 3X &= 2O \\ 3(4) &= 2O \\ 12 &= 2O \\ O &= 6 \end{aligned}

Answer

(vii) $\text{'O' shape} = 6$

(viii) Find the weight of the banana in Fig. 7.8.

Step 1 · Calculate the Weight of the Banana

Given watermelon weight $W_m = 10$ and orange weight $O_r = 4$ . Let $B_n$ be the weight of one banana. Diagram 8

Since the mobile is balanced:

\begin{aligned} W_m + O_r &= B_n \\ 10 + 4 &= B_n \\ B_n &= 14 \end{aligned}

Answer

(viii) $\text{Banana} = 14$

Common Mistakes

Total Weight vs. Arm Weight: Forgetting that in a balanced mobile with total weight $W$ , each side weighs $\dfrac{W}{2}$ . For example, in Fig. 7.1, the total weight $16$ includes all leaves and the flower combined.
Integer Constraint: In problems with multiple unknowns (like Fig. 7.5), ensure all derived weights are positive integers.
Incorrect Multipliers: Confusing the count of items on each side of the mobile.

More questions in IT

Find the unknown weights in the following cases:

Context: Finding the Unknown

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

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.]

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?

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

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

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.

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

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

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

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

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

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

Q10

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

Q. Can this equation have any other solution?

Q11

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

Q12

Consider an equation $15 + 8 = 23$ . 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 $23 \times 41 \times 11 \times 8 \times 7 = 5,80,888$ . To find the value of the expression $23 \times 41 \times 11 \times 8$ , we can divide $5,80,888$ by $7$ .

Q. Is this the same as dividing both sides by $7$ , which removes the factor $7$ 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 $3k + 1$ which gives the number of tiles needed to make an arrangement in Step $k$ . To check whether an arrangement is possible using 100 tiles at some Step $k$ , we can solve the equation: $3k + 1 = 100$ . Find the value of $k$ .

Q15

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

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

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.

Think of a number.
Subtract 3 from the number.
Multiply the result by 4.
Add 8 to the product.
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:

\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 $y$ , then the number of marbles with Ramesh is $y + 30$ . Since the total number of marbles is 60, we have the equation: $2y + 30 = 60$

Q. Use this to find both the unknowns.

Q19

Write equations whose solution is $y = 5$ . 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 $5x + 4 = 3x + 8$ and $3x - 6 = 2x + 4$ .

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 $Ax + B = Cx + D$ is $x = \dfrac{D - B}{A - C}$ .

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

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?