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.

Question

Homi · Accepted Answer

We will represent unknown weights with letters. We then set up equations based on balanced scales.

Step 1 — Figure 7.6: Bread and Eggs

Let us call the weight of one bread slice 'b'. Let us call the weight of one fried egg 'e'. The scale shows three bread slices. It balances three fried eggs. So, we can write the equation: $3b = 3e$ We know one bread slice weighs 2 units. Let us substitute $b = 2$ into our equation. $3 \times 2 = 3e$ $6 = 3e$ To find 'e', we divide both sides by 3. $e = \frac{6}{3}$

$\boxed{e = 2 \text{ units}}$

Diagram 1

Step 2 — Figure 7.7: X and O Shapes

Let us call the weight of one X shape 'x'. Let us call the weight of one O shape 'o'. The scale shows five X shapes. It balances three O shapes. So, we can write the equation: $5x = 3o$ We know one X shape weighs 4 units. Let us substitute $x = 4$ into our equation. $5 \times 4 = 3o$ $20 = 3o$ To find 'o', we divide both sides by 3. $o = \frac{20}{3}$

$\boxed{o = 6.67 \text{ units (approx)}}$

Diagram 2

Step 3 — Figure 7.8: Fruits

Let us call the weight of one watermelon 'w'. Let us call the weight of one orange 'o'. Let us call the weight of one banana 'b'. The left side has one watermelon and one banana. The right side has one orange and two bananas. So, we can write the equation: $w + b = o + 2b$ We know one watermelon weighs 10 units. We know one orange weighs 4 units. Let us substitute $w = 10$ and $o = 4$ . $10 + b = 4 + 2b$ To find 'b', we subtract 'b' from both sides. $10 = 4 + 2b - b$ $10 = 4 + b$ Now, we subtract 4 from both sides. $b = 10 - 4$

$\boxed{b = 6 \text{ units}}$

Diagram 3

Step 4 — Figure 7.9: Sacks and Weights

Let us call the weight of one sack 's'. The left pan has one sack and a 2 kg weight. The right pan has a 10 kg weight and a 2 kg weight. The scale is balanced. So, we can write the equation: $s + 2 = 10 + 2$ First, we simplify the right side. $s + 2 = 12$ To find 's', we subtract 2 from both sides. $s = 12 - 2$

$\boxed{s = 10 \text{ kg}}$

Diagram 4

Step 5 — Figure 7.10: Sacks and Weights

Let us call the weight of one sack 's'. The left pan has two sacks. The right pan has a 10 kg weight, a 4 kg weight, and one sack. The scale is balanced. So, we can write the equation: $2s = 10 + 4 + s$ First, we simplify the right side. $2s = 14 + s$ To find 's', we subtract 's' from both sides. $2s - s = 14$

$\boxed{s = 14 \text{ kg}}$

Diagram 5

Step 6 — Figure 7.11: Sacks and Weights

Let us call the weight of one sack 's'. The left pan has six sacks. The right pan has two 10 kg weights, one 1 kg weight, and two sacks. The scale is balanced. So, we can write the equation: $6s = 10 + 10 + 1 + 2s$ First, we simplify the right side. $6s = 21 + 2s$ To find 's', we subtract $2s$ from both sides. $6s - 2s = 21$ $4s = 21$ Now, we divide both sides by 4. $s = \frac{21}{4}$

$\boxed{s = 5.25 \text{ kg}}$

Diagram 6

Answer

(i) Fig. 7.6: Equation: $3b = 3e$ . Unknown egg weight: 2 units. (ii) Fig. 7.7: Equation: $5x = 3o$ . Unknown O shape weight: 20/3 units (approx 6.67 units). (iii) Fig. 7.8: Equation: $w + b = o + 2b$ . Unknown banana weight: 6 units. (iv) Fig. 7.9: Equation: $s + 2 = 10 + 2$ . Unknown sack weight: 10 kg. (v) Fig. 7.10: Equation: $2s = 10 + 4 + s$ . Unknown sack weight: 14 kg. (vi) Fig. 7.11: Equation: $6s = 10 + 10 + 1 + 2s$ . Unknown sack weight: 5.25 kg.

Question 7

Answer

More questions in IT

7.3 Mind the Mistake, Mend the Mistake