Question 1

We will use the principles of balanced mobiles and total weights to find the unknown values.

Step 1 — Calculate the weight of the red flower

We are given that the total weight of all objects in Fig. 7.1 is 16. We know that one green leaf weighs 3. There are 3 green leaves. Let us find the total weight of the green leaves.

$3 \times 3 = 9$

The total weight of the green leaves is 9. Let $F$ be the weight of one red flower. The total weight is the sum of the weights of the leaves and the flower.

$9 + F = 16$

$F = 16 - 9$

$F = 7$

$\boxed{7}$

Step 2 — Calculate the weights of the fish and submarine

We are given that the total weight of all objects in Fig. 7.2 is 24. The mobile is balanced. Let $S_w$ be the weight of one starfish. We are given $S_w = \textbf{2}$. Let $F_w$ be the weight of one fish. Let $Sub_w$ be the weight of one submarine.

The left side of the mobile has one starfish, then a fish, then another starfish. So, the left side weight is $S_w + F_w + S_w = 2S_w + F_w$. The right side of the mobile has one starfish, then a fish, then a submarine. So, the right side weight is $S_w + F_w + Sub_w$. Since the mobile is balanced, the weights on both sides are equal.

$2S_w + F_w = S_w + F_w + Sub_w$

$2S_w = S_w + Sub_w$

$S_w = Sub_w$

Since $S_w = \textbf{2}$, the weight of one submarine is 2. Now, let us use the total weight. The total weight is the sum of the weights on the left and right sides.

$(2S_w + F_w) + (S_w + F_w + Sub_w) = 24$

$3S_w + 2F_w + Sub_w = 24$

Let us substitute $S_w = \textbf{2}$ and $Sub_w = \textbf{2}$ into the equation.

$3(2) + 2F_w + 2 = 24$

$6 + 2F_w + 2 = 24$

$8 + 2F_w = 24$

$2F_w = 24 - 8$

$2F_w = 16$

$F_w = \frac{16}{2}$

$F_w = 8$

$\boxed{\text{Fish}=8, \text{Submarine}=2}$

Step 3 — Calculate the weights of the book and money bag

We are given that the total weight of all objects in Fig. 7.3 is 8. The mobile is balanced. Let $B$ be the weight of one red book. Let $M$ be the weight of one green money bag.

The left side of the mobile has 2 red books. The right side of the mobile has 2 green money bags. Since the mobile is balanced, the weights on both sides are equal.

$2B = 2M$

$B = M$

The total weight is the sum of the weights on the left and right sides.

$2B + 2M = 8$

Let us substitute $M$ with $B$ in the equation.

$2B + 2B = 8$

$4B = 8$

$B = \frac{8}{4}$

$B = 2$

Since $B = M$, the weight of one money bag is 2.

$\boxed{\text{Book}=2, \text{Money bag}=2}$

Step 4 — Calculate the weights of the cloud and lightning bolt

We are given that the total weight of all objects in Fig. 7.4 is 18. The mobile is balanced. Let $C$ be the weight of one cloud. Let $S$ be the weight of one sun. We are given $S = \textbf{5}$. Let $L$ be the weight of one lightning bolt.

The left side of the mobile has 3 clouds and 1 sun. So, the left side weight is $3C + S$. The right side of the mobile has 1 sun and 2 lightning bolts. So, the right side weight is $S + 2L$. Since the mobile is balanced, the weights on both sides are equal.

$3C + S = S + 2L$

$3C = 2L$

The total weight is the sum of the weights on the left and right sides.

$(3C + S) + (S + 2L) = 18$

$3C + 2S + 2L = 18$

Let us substitute $S = \textbf{5}$ into the equation.

$3C + 2(5) + 2L = 18$

$3C + 10 + 2L = 18$

$3C + 2L = 18 - 10$

$3C + 2L = 8$

Now we have two equations:

1. $3C = 2L$
2. $3C + 2L = 8$ Let us substitute $2L$ from equation (1) into equation (2).

$3C + 3C = 8$

$6C = 8$

$C = \frac{8}{6}$

$C = \frac{4}{3}$

Now, let us find $L$ using equation (1).

$3C = 2L$

$3\left(\frac{4}{3}\right) = 2L$

$4 = 2L$

$L = \frac{4}{2}$

$L = 2$

$\boxed{\text{Cloud}=4/3, \text{Lightning bolt}=2}$

Step 5 — Calculate the weights of the crown, water drop, and diamond

We are given that the total weight of all objects in Fig. 7.5 is 40. The mobile is balanced. We assume all weights are integers. Let $K$ be the weight of one crown. Let $W$ be the weight of one water drop. Let $D$ be the weight of one diamond.

The left side of the mobile has 3 crowns and 1 water drop. So, the left side weight is $3K + W$. The right side of the mobile has 1 crown, 2 water drops, and 2 diamonds. So, the right side weight is $K + 2W + 2D$. Since the mobile is balanced, the weights on both sides are equal.

$3K + W = K + 2W + 2D$

$2K = W + 2D \quad \text{(Equation 1)}$

The total weight is the sum of the weights on the left and right sides.

$(3K + W) + (K + 2W + 2D) = 40$

$4K + 3W + 2D = 40 \quad \text{(Equation 2)}$

From Equation 1, we can write $2D = 2K - W$. Let us substitute $2D$ into Equation 2.

$4K + 3W + (2K - W) = 40$

$6K + 2W = 40$

Let us divide the entire equation by 2.

$3K + W = 20 \quad \text{(Equation 3)}$

From Equation 3, we can express $W$ in terms of $K$.

$W = 20 - 3K$

Let us substitute this expression for $W$ into Equation 1.

$2K = (20 - 3K) + 2D$

$5K - 20 = 2D$

$D = \frac{5K - 20}{2}$

For weights to be positive, $W > 0$ and $D > 0$. $20 - 3K > 0 \implies K < 20/3 \implies K < 6.66...$ $\frac{5K - 20}{2} > 0 \implies 5K > 20 \implies K > 4$ So, $K$ must be an integer between 4 and 6.66.... The possible integer values for $K$ are 5 and 6. We also need $D$ to be an integer. If $K = 5$: $W = 20 - 3(5) = 5$. $D = \frac{5(5) - 20}{2} = \frac{5}{2} = 2.5$. This is not an integer. If $K = 6$: $W = 20 - 3(6) = 2$. $D = \frac{5(6) - 20}{2} = \frac{10}{2} = 5$. These are integers. So, the unique integer solution is $K = \textbf{6}$, $W = \textbf{2}$, $D = \textbf{5}$.

$\boxed{\text{Crown}=6, \text{Water drop}=2, \text{Diamond}=5}$

Step 6 — Calculate the weight of the egg

The mobile in Fig. 7.6 is balanced. Let $T$ be the weight of one toast. We are given $T = \textbf{2}$. Let $E$ be the weight of one egg.

The left side of the mobile has 3 toasts. So, the left side weight is $3T$. The right side of the mobile has 2 eggs. So, the right side weight is $2E$. Since the mobile is balanced, the weights on both sides are equal.

$3T = 2E$

Let us substitute $T = \textbf{2}$ into the equation.

$3(2) = 2E$

$6 = 2E$

$E = \frac{6}{2}$

$E = 3$

$\boxed{3}$

Step 7 — Calculate the weight of the 'O' shape

The mobile in Fig. 7.7 is balanced. Let $X$ be the weight of one 'X' shape. We are given $X = \textbf{4}$. Let $O$ be the weight of one 'O' shape.

The left side of the mobile has 3 'X' shapes. So, the left side weight is $3X$. The right side of the mobile has 2 'O' shapes. So, the right side weight is $2O$. Since the mobile is balanced, the weights on both sides are equal.

$3X = 2O$

Let us substitute $X = \textbf{4}$ into the equation.

$3(4) = 2O$

$12 = 2O$

$O = \frac{12}{2}$

$O = 6$

$\boxed{6}$

Step 8 — Calculate the weight of the banana

The mobile in Fig. 7.8 is balanced. Let $W_m$ be the weight of one watermelon. We are given $W_m = \textbf{10}$. Let $O_r$ be the weight of one orange. We are given $O_r = \textbf{4}$. Let $B_n$ be the weight of one banana.

The left side of the mobile has 1 watermelon and 1 orange. So, the left side weight is $W_m + O_r$. The right side of the mobile has 1 banana. So, the right side weight is $B_n$. Since the mobile is balanced, the weights on both sides are equal.

$W_m + O_r = B_n$

Let us substitute $W_m = \textbf{10}$ and $O_r = \textbf{4}$ into the equation.

$10 + 4 = B_n$

$14 = B_n$

$\boxed{14}$

Answer

Fig 7.1: Red flower = 7 Fig 7.2: Blue fish = 8 Fig 7.2: Grey submarine = 2 Fig 7.3: Red book = 2 Fig 7.3: Green money bag = 2 Fig 7.4: Cloud = 4/3 Fig 7.4: Lightning bolt = 2 Fig 7.5: Crown = 6 Fig 7.5: Water drop = 2 Fig 7.5: Diamond = 5 Fig 7.6: Egg = 3 Fig 7.7: 'O' shape = 6 Fig 7.8: Banana = 14