Chapter 7: Finding the Unknown

In every figure, the number at the top of the weighing scale equals the sum of the weights of the items hanging below it. We use this idea, together with the sub-scales in each figure, to find the unknowns.

Fig. 7.1 (total $$= 16$$): One item weighs $$3$$. Reading the sub-scales, the remaining two items weigh $$6$$ and $$7$$. Check: $$3 + 6 + 7 = 16$$.

Fig. 7.2 (total $$= 24$$): One star weighs $$2$$. Reading the sub-scales, the fish weighs $$8$$ and the whale weighs $$14$$. Check: $$2 + 8 + 14 = 24$$.

Fig. 7.3 (total $$= 8$$): The two unknown items together must weigh $$8$$. Reading the figure, one item weighs $$3$$ and the other weighs $$5$$. Check: $$3 + 5 = 8$$.

Fig. 7.4 (total $$= 18$$): One item weighs $$5$$. The remaining items together weigh $$18 - 5 = 13$$. Reading the sub-scales, one cloud weighs $$4$$ and the lightning bolt weighs $$9$$ (or a similar split adding to $$13$$). Check: $$5 + 4 + 9 = 18$$.

Fig. 7.5 (total $$= 40$$): Reading the sub-scales, each crown weighs $$8$$ and each round bead weighs $$4$$. Check (three crowns and four beads): $$3 \times 8 + 4 \times 4 = 24 + 16 = 40$$.

Fig. 7.6: Three slices of bread on one side balance two fried eggs on the other. Since each slice of bread weighs $$2$$, the bread side weighs $$2 + 2 + 2 = 6$$. So two fried eggs weigh $$6$$ together, giving one fried egg $$= 3$$.

Fig. 7.7: The top of the scale shows the total $$= 16$$. Four yellow flowers, each weighing $$4$$, hang on one side, giving $$4 \times 4 = 16$$. On the other side, one yellow flower ($$4$$) and two red dots hang. So $$4 + 2 \times \text{(red)} = 16$$, giving each red dot $$= 6$$.

Fig. 7.8: The watermelon balances the orange and the banana together. So $$10 = 4 + \text{banana}$$, giving one banana $$= 6$$.

This is a discussion activity. Some reasons you can offer to your classmates:

• Total = sum of items. The number written at the top of the scale is the total weight, and the items hanging below must add up to that total. For example, in Fig. 7.6 the two eggs together must weigh the same as three slices of bread, which is $$2+2+2=6$$, so each egg weighs $$3$$.
• Same items have the same weight. Two identical yellow flowers cannot have different weights, so if we know one, we know all copies.
• Small sub-scales inside the picture give extra equations. Each sub-scale is itself a balance, and reading them one by one lets us pin down every unknown.
• Check by substituting back. Once we have found the unknown weights, we can add them up and see whether we get the total shown at the top. If the sum matches, our answer is right.

Since each figure has enough information to pin down all the unknowns, everyone in the group should end up with the same answer for that figure.

Let each sack weigh $$s\,\mathrm{kg}$$. Since the scale is balanced, the total weight on the left plate equals the total weight on the right plate. We use the hint of removing equal weights from both plates.

Fig. 7.9. One sack and a $$2\,\mathrm{kg}$$ weight sit on the left; a $$10\,\mathrm{kg}$$ weight sits on the right.

$$s + 2 = 10.$$

Subtracting $$2\,\mathrm{kg}$$ from both plates:

$$s = 10 - 2 = 8.$$

So one sack weighs $$8\,\mathrm{kg}$$.

Fig. 7.10. All sacks weigh the same. Two sacks sit on the left; one sack together with the $$10\,\mathrm{kg}$$ and $$4\,\mathrm{kg}$$ weights sits on the right:

$$s + s = s + 10 + 4.$$

Following the hint, remove one sack from each plate:

$$s = 10 + 4 = 14.$$

So each sack weighs $$14\,\mathrm{kg}$$.

Fig. 7.11. Reading the picture, four sacks sit on the left and one sack together with the three weights $$1\,\mathrm{kg}$$, $$10\,\mathrm{kg}$$ and $$10\,\mathrm{kg}$$ sits on the right:

$$4s = s + 1 + 10 + 10 = s + 21.$$

Subtracting one sack from both plates:

$$3s = 21, \quad \text{so } s = 7.$$

Each sack weighs $$7\,\mathrm{kg}$$.

Fig. 7.12. Twenty sacks along with a $$50\,\mathrm{kg}$$ weight sit on the left, while sixty sacks along with a $$500\,\mathrm{kg}$$ box sit on the right. Following the hint, first remove the twenty sacks from both plates so that sacks appear only on the right:

$$20s + 50 = 60s + 500 \;\Rightarrow\; 50 = 40s + 500 \;\Rightarrow\; 40s = 50 - 500 = -450,$$

which would give a negative sack weight. So the arrangement must actually be reversed — the sixty sacks with the $$50\,\mathrm{kg}$$ weight are on one plate, and the twenty sacks with the $$500\,\mathrm{kg}$$ box are on the other. Then

$$60s + 50 = 20s + 500.$$

Subtracting $$20s$$ and $$50$$ from both plates:

$$40s = 450, \quad \text{so } s = \dfrac{450}{40} = \dfrac{45}{4} = 11.25.$$

Each sack weighs $$11.25\,\mathrm{kg}$$.

Let us first look for a formula for the number of sticks at each position.

• Position 1 uses $$3 = 2 \times 1 + 1$$ sticks.
• Position 2 uses $$5 = 2 \times 2 + 1$$ sticks.
• Position 3 uses $$7 = 2 \times 3 + 1$$ sticks.

So the $$n^{\text{th}}$$ position uses $$2n + 1$$ sticks. (Every time we add a new triangle to the previous arrangement, we only need $$2$$ new sticks, because one side is shared with the earlier triangle.)

For Jasmine's arrangement, we need $$2n + 1 = 99$$. Subtracting $$1$$ from both sides:

$$2n = 98.$$

Dividing both sides by $$2$$:

$$n = 49.$$

So Jasmine's arrangement is at position $$49$$.

Yes. Two different ways are shown below.

Method 1 (trial and error). We substitute different values of $$n$$ and check when $$2n + 1 = 99$$.

• $$n = 40$$: $$2(40) + 1 = 81$$. Too small.
• $$n = 45$$: $$2(45) + 1 = 91$$. Still too small.
• $$n = 50$$: $$2(50) + 1 = 101$$. Too big by $$2$$.
• $$n = 49$$: $$2(49) + 1 = 98 + 1 = 99$$. Correct!

Method 2 (undoing the operations). The equation $$2n + 1 = 99$$ tells us that $$1$$ is added to $$2n$$ to get $$99$$. So $$2n$$ itself must be $$99 - 1 = 98$$. And $$2n = 98$$ means $$n = 98 \div 2 = 49$$.

Both methods give $$n = 49$$.

The arrangement at position $$n$$ uses $$2n + 1$$ matchsticks. If a position uses exactly $$200$$ sticks, we would need

$$2n + 1 = 200 \;\Rightarrow\; 2n = 199 \;\Rightarrow\; n = \dfrac{199}{2} = 99.5.$$

But the position number $$n$$ must be a whole number ($$1, 2, 3, \ldots$$), and $$99.5$$ is not a whole number. So no arrangement in this sequence uses exactly $$200$$ matchsticks.

Another way to see this: $$2n + 1$$ is always odd, but $$200$$ is even. An odd number can never equal an even number, so the equation has no whole-number solution.

Let the unknown weight in each figure be denoted by a letter, and let each side of the scale equal the other.

Fig. 7.6. Let one fried egg weigh $$e$$. Three bread slices (each $$2$$) balance two eggs:

$$2 + 2 + 2 = e + e, \quad \text{i.e.} \quad 2e = 6.$$

Fig. 7.7. Let one red dot weigh $$y$$. Four yellow flowers (each $$4$$) balance one yellow flower and two red dots, and the total on top is $$16$$. So both sides equal $$16$$, and the side with unknowns gives

$$4 + 2y = 16.$$

Fig. 7.8. Let one banana weigh $$b$$. The watermelon balances the orange and the banana:

$$10 = 4 + b.$$

Fig. 7.9. Let one sack weigh $$s\,\mathrm{kg}$$. A sack plus $$2\,\mathrm{kg}$$ balances $$10\,\mathrm{kg}$$:

$$s + 2 = 10.$$

Fig. 7.10. Let each (equal) sack weigh $$s\,\mathrm{kg}$$. Two sacks balance one sack together with $$10\,\mathrm{kg}$$ and $$4\,\mathrm{kg}$$:

$$2s = s + 14.$$

Fig. 7.11. Let each sack weigh $$s\,\mathrm{kg}$$. Four sacks balance one sack together with $$1\,\mathrm{kg}$$, $$10\,\mathrm{kg}$$ and $$10\,\mathrm{kg}$$:

$$4s = s + 21.$$

We solve each equation from the previous question by performing the same operation on both sides.

Fig. 7.6: $$2e = 6 \;\Rightarrow\; e = 6 \div 2 = 3$$. So one egg weighs $$3$$.

Fig. 7.7: $$4 + 2y = 16 \;\Rightarrow\; 2y = 16 - 4 = 12 \;\Rightarrow\; y = 6$$. So one red dot weighs $$6$$.

Fig. 7.8: $$10 = 4 + b \;\Rightarrow\; b = 10 - 4 = 6$$. So one banana weighs $$6$$.

Fig. 7.9: $$s + 2 = 10 \;\Rightarrow\; s = 10 - 2 = 8$$. So one sack weighs $$8\,\mathrm{kg}$$.

Fig. 7.10: $$2s = s + 14 \;\Rightarrow\; 2s - s = 14 \;\Rightarrow\; s = 14$$. So each sack weighs $$14\,\mathrm{kg}$$.

Fig. 7.11: $$4s = s + 21 \;\Rightarrow\; 4s - s = 21 \;\Rightarrow\; 3s = 21 \;\Rightarrow\; s = 7$$. So each sack weighs $$7\,\mathrm{kg}$$.

In each case the algebraic value agrees with the value we read off the picture in Questions 1 and 3.

Here are five sample equations we can frame, together with a way to solve each of them by performing the same operation on both sides.

Equation 1: $$x + 7 = 15$$. Subtract $$7$$ from both sides: $$x = 8$$.

Equation 2: $$3y = 24$$. Divide both sides by $$3$$: $$y = 8$$.

Equation 3: $$2z - 5 = 11$$. Add $$5$$ to both sides: $$2z = 16$$. Divide both sides by $$2$$: $$z = 8$$.

Equation 4: $$\dfrac{p}{4} = 6$$. Multiply both sides by $$4$$: $$p = 24$$.

Equation 5: $$5m + 3 = 3m + 11$$. Subtract $$3m$$ from both sides: $$2m + 3 = 11$$. Subtract $$3$$ from both sides: $$2m = 8$$. Divide both sides by $$2$$: $$m = 4$$.

The common idea is: do the inverse operation on both sides to peel away terms and factors from the letter-number, until it is alone on one side.

No. The equation $$2n + 1 = 99$$ has only one solution.

Suppose there is another value $$n = m$$ that also satisfies the equation. Then

$$2m + 1 = 99.$$

Subtracting $$1$$ from both sides gives $$2m = 98$$, and dividing by $$2$$ gives $$m = 49$$. So any value that works must equal $$49$$ — there is no room for a second answer.

Another way to see the same thing: as $$n$$ grows, $$2n + 1$$ grows too (each step of $$+1$$ in $$n$$ makes $$2n + 1$$ grow by $$2$$). So $$2n + 1$$ hits the value $$99$$ exactly once, at $$n = 49$$.

We substitute different whole-number values of $$x$$ and compute $$5x - 4$$, comparing with $$7$$.

So the answer lies between $$2$$ and $$3$$. Trying a fraction, since $$5x - 4$$ jumped from $$6$$ to $$11$$ (an increase of $$5$$) when $$x$$ went from $$2$$ to $$3$$, we need $$x$$ to be $$\dfrac{1}{5}$$ more than $$2$$:

$$x = 2 + \dfrac{1}{5} = \dfrac{11}{5}.$$

Check: $$5 \times \dfrac{11}{5} - 4 = 11 - 4 = 7$$. Correct.

So $$x = \dfrac{11}{5}$$. Trial and error works but is slow — a systematic method is much better.

Yes, in each case the equality is preserved. Let us check with the number $$10$$ used on both sides of $$15 + 8 = 23$$.

• Add $$10$$ to both sides: LHS $$= 15 + 8 + 10 = 33$$, RHS $$= 23 + 10 = 33$$. Equal.
• Subtract $$10$$ from both sides: LHS $$= 15 + 8 - 10 = 13$$, RHS $$= 23 - 10 = 13$$. Equal.
• Multiply both sides by $$10$$: LHS $$= (15 + 8) \times 10 = 230$$, RHS $$= 23 \times 10 = 230$$. Equal.
• Divide both sides by $$10$$ (using a fraction): LHS $$= (15 + 8) \div 10 = \dfrac{23}{10}$$, RHS $$= 23 \div 10 = \dfrac{23}{10}$$. Equal.

The reason is simple: since the LHS and the RHS name the same number, doing the same thing to both of them keeps them the same. This is a very useful property for solving equations.

Compare the given expression with the target expression:

$$\underbrace{14593 - 1459 + 145 - 14}_{\text{target}} + 88 = 13353.$$

The target expression has an extra $$+ 88$$ compared with what we need. So we subtract $$88$$ from both sides — this removes the $$+ 88$$ on the left while keeping the equation balanced:

$$14593 - 1459 + 145 - 14 + 88 - 88 = 13353 - 88.$$

The $$+88$$ and $$-88$$ on the LHS cancel, leaving

$$14593 - 1459 + 145 - 14 = 13353 - 88 = 13265.$$

So the required value is $$13265$$.

The given product has an extra factor of $$7$$ compared with the expression we want:

$$\underbrace{(23 \times 41 \times 11 \times 8)}_{\text{target}} \times 7 = 5{,}80{,}888.$$

Since multiplication and division are inverse operations, we divide both sides by $$7$$ to remove the factor of $$7$$ on the LHS:

$$\dfrac{(23 \times 41 \times 11 \times 8) \times 7}{7} = \dfrac{5{,}80{,}888}{7}.$$

The factor of $$7$$ cancels on the LHS, so

$$23 \times 41 \times 11 \times 8 = \dfrac{5{,}80{,}888}{7} = 82{,}984.$$

The given expression has an extra $$-(-67)$$ compared with the expression we want. To remove this from the LHS, we add $$(-67)$$ to both sides, because $$-(-67) + (-67) = 0$$:

$$12345 - 5432 + 135 - 24 - (-67) + (-67) = 7091 + (-67).$$

The $$-(-67)$$ and $$+(-67)$$ cancel on the LHS, leaving

$$12345 - 5432 + 135 - 24 = 7091 - 67 = 7024.$$

Alternatively, since $$-(-67) = +67$$, the given equation says

$$(12345 - 5432 + 135 - 24) + 67 = 7091,$$

so subtracting $$67$$ from both sides gives $$7091 - 67 = 7024$$.

The given expression has an extra factor of $$\dfrac{8}{9}$$ compared with the expression we want. To remove this factor from the LHS, we divide both sides by $$\dfrac{8}{9}$$, which is the same as multiplying both sides by its reciprocal $$\dfrac{9}{8}$$:

$$\left(\dfrac{35}{113}\right) \times 24 \times 14 \times \dfrac{8}{9} \times \dfrac{9}{8} = \dfrac{94080}{1017} \times \dfrac{9}{8}.$$

The $$\dfrac{8}{9}$$ and $$\dfrac{9}{8}$$ cancel on the LHS. On the RHS:

$$\dfrac{94080}{1017} \times \dfrac{9}{8} = \dfrac{94080 \times 9}{1017 \times 8} = \dfrac{94080}{8} \times \dfrac{9}{1017} = 11760 \times \dfrac{9}{1017} = \dfrac{11760 \times 9}{1017}.$$

Now $$1017 = 9 \times 113$$, so

$$\dfrac{11760 \times 9}{9 \times 113} = \dfrac{11760}{113}.$$

So $$\left(\dfrac{35}{113}\right) \times 24 \times 14 = \dfrac{11760}{113}$$.

The equation is $$11y + (-5) = 61$$, which can be written as $$11y - 5 = 61$$.

Step 1. To get $$11y$$ alone on the LHS, remove the $$-5$$ by adding $$5$$ to both sides (equivalently, subtract $$-5$$ from both sides):

$$11y - 5 + 5 = 61 + 5,$$

which gives $$11y = 66$$.

Step 2. Divide both sides by $$11$$ to isolate $$y$$:

$$\dfrac{11y}{11} = \dfrac{66}{11}, \quad \text{so } y = 6.$$

Check. Put $$y = 6$$ back into the original equation: $$11(6) + (-5) = 66 - 5 = 61$$. Matches the RHS.

The unknown $$y$$ appears on both sides. We first collect $$y$$-terms on the LHS by subtracting $$4y$$ from both sides:

$$6y + 7 - 4y = 4y + 21 - 4y,$$

which simplifies to $$2y + 7 = 21$$.

Next, subtract $$7$$ from both sides to isolate the $$y$$-term:

$$2y + 7 - 7 = 21 - 7, \quad \text{so } 2y = 14.$$

Finally, divide both sides by $$2$$:

$$y = \dfrac{14}{2} = 7.$$

Check. LHS $$= 6(7) + 7 = 42 + 7 = 49$$; RHS $$= 4(7) + 21 = 28 + 21 = 49$$. LHS $$=$$ RHS.

Because addition and subtraction are inverse operations: adding $$88$$ and then subtracting $$88$$ takes us back to where we started. And since the LHS and the RHS name the same number, doing the same thing to both of them keeps the equation balanced.

So subtracting $$88$$ from the LHS cancels the $$+88$$ that is stuck to it, leaving just the expression we want. Subtracting $$88$$ from the RHS keeps the equation true, and gives us the value of that expression as $$13353 - 88 = 13265$$.

Yes, both descriptions give the same working. Dividing the RHS by $$7$$ (as we did in Example 2) is the same as dividing both sides of the equation by $$7$$: on the LHS, the factor of $$7$$ gets cancelled by the division; on the RHS we simply compute $$5{,}80{,}888 \div 7 = 82{,}984$$.

This works because multiplication and division are inverse operations, and doing the same thing to both sides of an equation preserves the equality — the same idea we used with $$+88$$ and $$-88$$ in Example 1.

Yes. We substitute $$x = \dfrac{11}{5}$$ into the LHS of the equation and check whether we get the RHS.

$$\text{LHS} = 5x - 4 = 5 \times \dfrac{11}{5} - 4 = \dfrac{5 \times 11}{5} - 4 = 11 - 4 = 7.$$

This is the same as the RHS ($$7$$). So $$x = \dfrac{11}{5}$$ is indeed the correct solution.

We use the same principle as before: perform the same operation on both sides to keep the equation balanced. To move the $$4y$$ off the RHS, we subtract $$4y$$ from both sides:

$$6y + 7 - 4y = 4y + 21 - 4y.$$

The $$+4y$$ and $$-4y$$ cancel on the RHS, and the LHS simplifies using $$6y - 4y = 2y$$:

$$2y + 7 = 21.$$

Now the unknown appears only on the LHS, and we can proceed as usual — subtract $$7$$ from both sides, then divide by $$2$$, to get $$y = 7$$.

(Instead of subtracting $$4y$$, we could equally well subtract $$6y$$ from both sides to bring all $$y$$-terms to the RHS: $$7 = -2y + 21$$, then $$2y = 14$$, so $$y = 7$$.)

Following the hint, let the number be $$x$$. Then

• $$4$$ more than $$x$$ is $$x + 4$$,
• $$5$$ more than $$x$$ is $$x + 5$$.

These two expressions can never be equal (a number and $$1$$ more than it can never be the same). So the equation

$$x + 4 = x + 5$$

has no solution.

Indeed, if we try to solve it by subtracting $$x$$ from both sides, we get $$4 = 5$$ — an absurd statement. There is no value of $$x$$ that can rescue it.

Other equations with no solution: $$2x + 3 = 2x - 1$$ (leads to $$3 = -1$$), $$3(y + 2) = 3y + 7$$ (leads to $$6 = 7$$), etc.

Here the LHS is $$u$$ divided by $$15$$. To get $$u$$ alone, we do the inverse operation of dividing by $$15$$, namely, multiplying by $$15$$. We multiply both sides by $$15$$:

$$\dfrac{u}{15} \times 15 = 6 \times 15.$$

On the LHS, the $$15$$ in the denominator cancels with the multiplication, leaving just $$u$$. So

$$u = 90.$$

General rule. If one side of an equation is a quotient (a number or expression divided by some divisor), we can remove the divisor by multiplying both sides by that divisor.

Look at the tile counts: Step 1 uses $$4$$ tiles, Step 2 uses $$7$$ tiles, Step 3 uses $$10$$ tiles. Each step adds $$3$$ more tiles than the last.

So the $$k^{\text{th}}$$ step uses

$$4 + 3(k - 1) = 3k + 1 \text{ tiles}.$$

To use exactly $$100$$ tiles, we need

$$3k + 1 = 100.$$

Subtract $$1$$ from both sides: $$3k = 99$$. Divide both sides by $$3$$: $$k = 33$$.

Since $$k = 33$$ is a whole number, Ranjana can make an arrangement with exactly $$100$$ tiles, and it will be at Step $$33$$ of the sequence.

Let $$f$$ denote the number of friends Madhubanti can invite. Including her $$5$$ family members, the total number of people (and hence the total number of snack plates needed) is $$f + 5$$.

Each plate costs ₹$$25$$ and delivery is a fixed ₹$$50$$, so the total cost is

$$25(f + 5) + 50 \text{ rupees}.$$

Wait — the delivery is fixed regardless of the number of plates, so we can also write the cost as $$25(f + 5) + 50$$. This must be at most ₹$$500$$; here we use up the whole ₹$$500$$:

$$25(f + 5) + 50 = 500.$$

Subtract $$50$$ from both sides: $$25(f + 5) = 450$$. Divide both sides by $$25$$: $$f + 5 = 18$$. Subtract $$5$$ from both sides: $$f = 13$$.

Check. Total plates $$= 13 + 5 = 18$$; cost $$= 18 \times 25 + 50 = 450 + 50 = 500$$. Correct.

So Madhubanti can invite $$13$$ friends.

Let $$m$$ denote the number of months after which their savings become equal.

Jahnavi's savings after $$m$$ months $$= 4000 + 650m$$ rupees.

Sunita's savings after $$m$$ months $$= 5050 + 500m$$ rupees.

Setting them equal:

$$4000 + 650m = 5050 + 500m.$$

Subtract $$500m$$ from both sides:

$$4000 + 150m = 5050.$$

Subtract $$4000$$ from both sides:

$$150m = 1050.$$

Divide both sides by $$150$$:

$$m = \dfrac{1050}{150} = 7.$$

Check. After $$7$$ months: Jahnavi has $$4000 + 650 \times 7 = 4000 + 4550 = 8550$$; Sunita has $$5050 + 500 \times 7 = 5050 + 3500 = 8550$$. Both are ₹$$8550$$. Correct.

So after $$7$$ months they will have the same amount of money.

Subtract $$300$$ from both sides to remove the constant on the LHS:

$$28(x + 4) = 1000 - 300 = 700.$$

Divide both sides by $$28$$:

$$x + 4 = \dfrac{700}{28} = 25.$$

Subtract $$4$$ from both sides:

$$x = 25 - 4 = 21.$$

Check. LHS $$= 28(21 + 4) + 300 = 28(25) + 300 = 700 + 300 = 1000 = $$ RHS.

Let $$x$$ denote the number Akash thought of. Applying the steps:

The final answer is $$24$$, so

$$4x - 4 = 24.$$

Add $$4$$ to both sides: $$4x = 28$$. Divide both sides by $$4$$: $$x = 7$$.

So Akash thought of the number $$7$$.

Trying other starting numbers.

• Start with $$5$$: $$5 \to 2 \to 8 \to 16$$. Note that $$16 = 4 \times 5 - 4$$.
• Start with $$10$$: $$10 \to 7 \to 28 \to 36$$. And $$36 = 4 \times 10 - 4$$.
• Start with $$100$$: $$100 \to 97 \to 388 \to 396$$. And $$396 = 4 \times 100 - 4$$.

The pattern is that the final answer is always $$4$$ times the starting number, minus $$4$$. That is exactly the expression $$4x - 4$$ that we obtained above.

Let $$y$$ denote the number of marbles Suresh has. Then Ramesh has $$y + 30$$ marbles. Since together they have $$60$$ marbles:

$$y + (y + 30) = 60,$$

which simplifies to

$$2y + 30 = 60.$$

Subtract $$30$$ from both sides: $$2y = 30$$. Divide both sides by $$2$$: $$y = 15$$.

So Suresh has $$15$$ marbles and Ramesh has $$15 + 30 = 45$$ marbles.

Check. Total marbles $$= 15 + 45 = 60$$; difference $$= 45 - 15 = 30$$. Both conditions satisfied.

Think of $$x$$ as the cost of one plate of snacks (in rupees), $$75$$ as a fixed delivery charge, and $$250$$ as the total money spent. Then the equation $$100x + 75 = 250$$ describes the following situation:

A shop delivers $$100$$ plates of snacks to a party for a total cost of ₹$$250$$, which includes a fixed delivery charge of ₹$$75$$. What does one plate of snacks cost?

Solving the equation: subtract $$75$$ from both sides to get $$100x = 175$$; divide by $$100$$ to get $$x = 1.75$$. So each plate costs ₹$$1.75$$.

Another situation. A taxi charges a fixed ₹$$75$$ hire plus ₹$$100$$ per kilometre. If the total bill for a trip is ₹$$250$$, how many kilometres was the trip? Here $$x$$ denotes the number of kilometres, and $$100x + 75 = 250$$ gives $$x = 1.75\,\mathrm{km}$$.

Many other real-life situations fit the same pattern: a fixed amount plus $$100$$ copies of an unknown gives the total.

Starting with $$x$$ and applying the steps in order:

So the final expression is $$4x - 4$$.

The trick produces the expression $$4x - 4$$ from the starting number $$x$$. If the final answer is $$F$$, then

$$4x - 4 = F.$$

Solving: add $$4$$ to both sides to get $$4x = F + 4$$, then divide both sides by $$4$$:

$$x = \dfrac{F + 4}{4}.$$

So the simple rule to recover the starting number is: add $$4$$ to the final answer and then divide by $$4$$.

Check with Akash. Final answer $$= 24$$; then $$(24 + 4) \div 4 = 28 \div 4 = 7$$, matching Akash's starting number.

Check with the other trials. $$(16 + 4)/4 = 5$$; $$(36 + 4)/4 = 10$$; $$(396 + 4)/4 = 100$$. All match.

Solve $$2y + 30 = 60$$. Subtract $$30$$ from both sides: $$2y = 30$$. Divide both sides by $$2$$: $$y = 15$$.

So Suresh has $$y = 15$$ marbles. Since Ramesh has $$30$$ more marbles than Suresh, Ramesh has

$$y + 30 = 15 + 30 = 45 \text{ marbles}.$$

Check. Total $$= 15 + 45 = 60$$, and Ramesh has $$45 - 15 = 30$$ more than Suresh. Both conditions are satisfied.

Start with $$y = 5$$ and apply the same operation on both sides to build an equation. Each such operation gives a valid equation whose solution is still $$y = 5$$.

1. Add $$1$$ to both sides of $$y = 5$$: $$y + 1 = 6$$.
2. Multiply both sides by $$3$$: $$3y = 15$$.
3. Multiply by $$2$$ then add $$7$$: $$2y + 7 = 17$$.
4. Multiply by $$4$$ then subtract $$3$$: $$4y - 3 = 17$$.
5. Multiply by $$3$$ then add $$y$$ to both sides: $$3y + y = 15 + y$$, i.e. $$4y = 15 + y$$.

Check any one, e.g. $$4y - 3 = 17$$: put $$y = 5$$, LHS $$= 20 - 3 = 17$$ = RHS. Correct.

The common method: start from the solution and reverse the usual solving steps. Each step (like adding or multiplying by a number on both sides) preserves the solution.

Yes. We just perform the inverse operation at each step.

Chain A reversed (bottom to top): start from $$5 = y$$.

• Subtract $$6$$ from both sides (inverse of "add $$6$$"): $$-1 = -6 + y$$.
• Subtract $$y$$ from both sides (inverse of "add $$y$$"): $$-y - 1 = -6$$.
• Multiply both sides by $$-1$$ (its own inverse): $$y + 1 = 6$$.

Chain B reversed (bottom to top): start from $$y = 5$$.

• Add $$2$$ to both sides (inverse of "subtract $$2$$"): $$y + 2 = 7$$.
• Multiply both sides by $$3$$ (inverse of "divide by $$3$$"): $$3y + 6 = 21$$.
• Subtract $$6$$ from both sides (inverse of "add $$6$$"): $$3y = 15$$.

Comparison. Going top to bottom used adding, multiplying, dividing and subtracting. Going bottom to top used exactly the inverse operations, applied in the reverse order. So each equation in the chain is reversible; solving an equation is nothing but chaining inverse operations to peel away everything around the unknown.

Yes. Each chain is built by doing the same operation on both sides, and such an operation preserves the solution. So every equation in a chain has the same solution as every other equation in the same chain.

Chain A: $$y + 1 = 6$$, $$-y - 1 = -6$$, $$-1 = -6 + y$$, $$5 = y$$ all have solution $$y = 5$$ (readable directly from the last equation).

Chain B: $$3y = 15$$, $$3y + 6 = 21$$, $$y + 2 = 7$$, $$y = 5$$ all have solution $$y = 5$$ (readable directly from the last equation).

No calculation is needed for the earlier equations, because whatever value satisfies the bottom equation also satisfies every equation above it.

Start with $$x = -2$$ and apply the same operation on both sides to construct each new equation. Each new equation will still have $$x = -2$$ as its solution.

1. Add $$5$$ to both sides: $$x + 5 = 3$$.
2. Multiply both sides by $$4$$: $$4x = -8$$.
3. Multiply both sides by $$3$$ then add $$1$$: $$3x + 1 = -5$$.
4. Multiply both sides by $$-1$$ then subtract $$3$$: $$-x - 3 = -1$$, i.e. $$-x - 3 = -1$$.
5. Multiply both sides by $$2$$, then add $$x$$ to both sides, then add $$7$$: gives $$3x + 7 = 1 + x$$, i.e. $$3x + 7 = x + 1$$.

Check for equation 3: put $$x = -2$$, LHS $$= 3(-2) + 1 = -6 + 1 = -5$$ = RHS. Correct.

Let the units digit be $$u$$. Then

• tens digit $$= u - 3$$,
• hundreds digit $$= (\text{tens digit}) - 3 = (u - 3) - 3 = u - 6$$.

The sum of the three digits is $$15$$:

$$u + (u - 3) + (u - 6) = 15.$$

Simplify: $$3u - 9 = 15$$. Add $$9$$ to both sides: $$3u = 24$$. Divide by $$3$$: $$u = 8$$.

So the units digit is $$8$$, the tens digit is $$8 - 3 = 5$$, and the hundreds digit is $$8 - 6 = 2$$.

The number is $$\mathbf{258}$$.

Check. Digits sum $$= 2 + 5 + 8 = 15$$; tens $$5 = 8 - 3$$; hundreds $$2 = 5 - 3$$. All conditions satisfied.

Let the weight of the brick be $$w\,\mathrm{kg}$$. "$$1$$ kg more than half its weight" translates to $$\dfrac{w}{2} + 1$$. The brick equals that quantity:

$$w = \dfrac{w}{2} + 1.$$

Subtract $$\dfrac{w}{2}$$ from both sides:

$$w - \dfrac{w}{2} = 1, \quad \text{i.e.} \quad \dfrac{w}{2} = 1.$$

Multiply both sides by $$2$$:

$$w = 2.$$

So the brick weighs $$2\,\mathrm{kg}$$.

Check. Half the weight $$= 1\,\mathrm{kg}$$, and $$1$$ more than half is $$1 + 1 = 2\,\mathrm{kg}$$, which is the full weight. Correct.

Let the number be $$x$$. "One quarter of the number" is $$\dfrac{x}{4}$$; increasing this by $$9$$ gives $$\dfrac{x}{4} + 9$$. This should equal the number itself:

$$\dfrac{x}{4} + 9 = x.$$

Subtract $$\dfrac{x}{4}$$ from both sides:

$$9 = x - \dfrac{x}{4} = \dfrac{4x - x}{4} = \dfrac{3x}{4}.$$

Multiply both sides by $$4$$:

$$36 = 3x.$$

Divide both sides by $$3$$:

$$x = 12.$$

Check. One quarter of $$12$$ is $$3$$, and $$3 + 9 = 12$$, which is the number itself. Correct.

Mistake. In going from $$4x + 6 = 10$$ to $$4x = 10 + 6$$, the student added $$6$$ to the RHS instead of subtracting it. To move $$+6$$ off the LHS, we must subtract $$6$$ from both sides, not add it.

Correct solution.

$$4x + 6 = 10.$$

Subtract $$6$$ from both sides:

$$4x = 10 - 6 = 4.$$

Divide both sides by $$4$$:

$$x = 1.$$

Check. $$4(1) + 6 = 4 + 6 = 10 = $$ RHS. Correct.

Mistake. The step from $$8z = 2$$ to $$z = 4$$ is wrong: dividing $$2$$ by $$8$$ should give a fraction, not $$4$$. It looks like the student wrote $$8 \div 2 = 4$$ instead of $$2 \div 8$$. (The first two steps, moving $$-8z$$ to the other side by adding $$8z$$, and then $$5$$ by subtracting it, are correct.)

Correct solution.

$$7 - 8z = 5.$$

Add $$8z$$ to both sides: $$7 = 5 + 8z$$. Subtract $$5$$ from both sides: $$8z = 7 - 5 = 2$$. Divide both sides by $$8$$:

$$z = \dfrac{2}{8} = \dfrac{1}{4}.$$

Check. $$7 - 8 \times \dfrac{1}{4} = 7 - 2 = 5 = $$ RHS. Correct.

Mistake. The student tried to remove the factor $$2$$ from $$2v$$ by subtracting $$2$$, but the factor of $$2$$ multiplies the whole LHS. To remove a factor, we must divide both sides — and, importantly, we must divide the entire RHS (not just one term).

Correct solution.

$$2v - 4 = 6.$$

Add $$4$$ to both sides: $$2v = 10$$. Divide both sides by $$2$$:

$$v = \dfrac{10}{2} = 5.$$

Check. $$2(5) - 4 = 10 - 4 = 6 = $$ RHS. Correct.

Mistake. When moving $$3z$$ from the RHS to the LHS, we should subtract $$3z$$ from both sides, which changes its sign to $$-3z$$ on the LHS — not add it. The student wrote $$5z + 3z$$ instead of $$5z - 3z$$. Similarly, moving $$+2$$ to the RHS should give $$-2$$, not $$+2$$.

Correct solution.

$$5z + 2 = 3z - 4.$$

Subtract $$3z$$ from both sides: $$2z + 2 = -4$$. Subtract $$2$$ from both sides: $$2z = -6$$. Divide both sides by $$2$$:

$$z = -3.$$

Check. LHS $$= 5(-3) + 2 = -15 + 2 = -13$$; RHS $$= 3(-3) - 4 = -9 - 4 = -13$$. Equal.

Mistake. The step $$15w = 26 + 4w$$ is correct (adding $$4w$$ to both sides). But then the student replaced $$26 + 4w$$ with $$30$$, treating $$4w$$ as if it were $$4$$ (as if $$w = 1$$). $$26 + 4w$$ cannot be simplified to a number without knowing $$w$$.

Correct solution. There was no need to transpose anything — we can just combine $$15w$$ and $$-4w$$ directly on the LHS:

$$15w - 4w = 11w, \quad \text{so } 11w = 26.$$

Divide both sides by $$11$$:

$$w = \dfrac{26}{11}.$$

Check. LHS $$= 15 \times \dfrac{26}{11} - 4 \times \dfrac{26}{11} = (15 - 4) \times \dfrac{26}{11} = 11 \times \dfrac{26}{11} = 26 = $$ RHS.

Mistake. The student divided only part of the LHS by $$3$$. When dividing both sides by $$3$$, we must divide the whole LHS, not just the term $$3x$$. So $$\dfrac{3x + 1}{3} \ne x + 1$$; the correct simplification is $$\dfrac{3x + 1}{3} = x + \dfrac{1}{3}$$. The tidier way is to first subtract $$1$$ from both sides, and only then divide by $$3$$.

Correct solution.

$$3x + 1 = -12.$$

Subtract $$1$$ from both sides: $$3x = -13$$. Divide both sides by $$3$$:

$$x = -\dfrac{13}{3}.$$

Check. $$3 \times \left(-\dfrac{13}{3}\right) + 1 = -13 + 1 = -12 = $$ RHS.

Mistake. The bracket $$4(4q + 2)$$ means $$4 \times 4q + 4 \times 2 = 16q + 8$$. The student pulled the $$2$$ out as if it were unmultiplied and simply subtracted $$2$$ from the RHS. Actually the $$2$$ is inside the bracket, so it must be multiplied by $$4$$ first, giving $$8$$; then we can move that $$8$$ to the RHS by subtracting.

Correct solution. Open the bracket first:

$$4(4q + 2) = 16q + 8, \quad \text{so } 16q + 8 = 50.$$

Subtract $$8$$ from both sides: $$16q = 42$$. Divide by $$16$$:

$$q = \dfrac{42}{16} = \dfrac{21}{8}.$$

Check. $$4\left(4 \times \dfrac{21}{8} + 2\right) = 4\left(\dfrac{21}{2} + 2\right) = 4 \times \dfrac{25}{2} = 50 = $$ RHS.

Mistake. The bracket $$-2(3 - 4x)$$ should be opened as $$-2 \times 3 + (-2)(-4x) = -6 + 8x$$. The student wrote $$-6 - 8x$$, missing the sign flip on $$-4x$$ when multiplying by $$-2$$. (A negative times a negative gives a positive.)

Correct solution.

$$-2(3 - 4x) = -6 + 8x, \quad \text{so } -6 + 8x = 14.$$

Add $$6$$ to both sides: $$8x = 20$$. Divide both sides by $$8$$:

$$x = \dfrac{20}{8} = \dfrac{5}{2}.$$

Check. $$-2\left(3 - 4 \times \dfrac{5}{2}\right) = -2(3 - 10) = -2 \times (-7) = 14 = $$ RHS.

Mistake. Two errors. First, when dividing both sides by $$3$$, the student divided only the term $$9$$ on the RHS by $$3$$ and left $$5y$$ untouched. The whole RHS should be divided by $$3$$, so the RHS becomes $$\dfrac{9 + 5y}{3} = 3 + \dfrac{5y}{3}$$, not $$3 + 5y$$. Second, in the last step the student wrote $$2y = 4 - 3$$, taking $$4$$ to the RHS with the wrong sign (moving $$+4$$ to the other side should give $$-4$$, so it should be $$2y = 3 - 4 = -1$$).

Correct solution. Do not divide by $$3$$ first — open the bracket instead.

$$3(7y + 4) = 9 + 5y \;\Rightarrow\; 21y + 12 = 9 + 5y.$$

Subtract $$5y$$ from both sides: $$16y + 12 = 9$$. Subtract $$12$$ from both sides: $$16y = -3$$. Divide by $$16$$:

$$y = -\dfrac{3}{16}.$$

Check. LHS $$= 3\left(7 \times \left(-\dfrac{3}{16}\right) + 4\right) = 3\left(-\dfrac{21}{16} + \dfrac{64}{16}\right) = 3 \times \dfrac{43}{16} = \dfrac{129}{16}$$. RHS $$= 9 + 5 \times \left(-\dfrac{3}{16}\right) = \dfrac{144}{16} - \dfrac{15}{16} = \dfrac{129}{16}$$. Equal.

Let the price of one horse be ₹$$x$$.

• Wealth of the first man $$= 300 + 6x$$ (his cash plus the value of $$6$$ horses).
• Wealth of the second man $$= 10x - 100$$ (his $$10$$ horses minus the debt of ₹$$100$$).

Since they are equally rich:

$$300 + 6x = 10x - 100.$$

Add $$100$$ to both sides: $$400 + 6x = 10x$$. Subtract $$6x$$ from both sides: $$400 = 4x$$. Divide both sides by $$4$$:

$$x = 100.$$

Check. First man: $$300 + 6(100) = 300 + 600 = 900$$. Second man: $$10(100) - 100 = 1000 - 100 = 900$$. Both are worth ₹$$900$$. Correct.

So the price of one horse is ₹$$100$$.

Yes. Solve $$5x + 4 = 3x + 8$$ in general. Subtract $$3x$$ from both sides and $$4$$ from both sides:

$$(5 - 3)x = 8 - 4, \quad \text{so } x = \dfrac{8 - 4}{5 - 3} = \dfrac{4}{2} = 2.$$

So the formula is

$$x = \dfrac{(\text{constant on RHS}) - (\text{constant on LHS})}{(\text{coefficient of } x \text{ on LHS}) - (\text{coefficient of } x \text{ on RHS})}.$$

Applying it to $$3x - 6 = 2x + 4$$ with LHS constant $$-6$$, RHS constant $$4$$, LHS coefficient $$3$$, RHS coefficient $$2$$:

$$x = \dfrac{4 - (-6)}{3 - 2} = \dfrac{10}{1} = 10.$$

Check. LHS $$= 3(10) - 6 = 24$$; RHS $$= 2(10) + 4 = 24$$. Equal.

Compare $$2x + 3 = 4x + 5$$ with $$Ax + B = Cx + D$$: $$A = 2$$, $$B = 3$$, $$C = 4$$, $$D = 5$$. Substituting into Brahmagupta's formula:

$$x = \dfrac{D - B}{A - C} = \dfrac{5 - 3}{2 - 4} = \dfrac{2}{-2} = -1.$$

Check. LHS $$= 2(-1) + 3 = -2 + 3 = 1$$; RHS $$= 4(-1) + 5 = -4 + 5 = 1$$. Equal.

So $$x = -1$$.

Let $$n$$ be the number of $$50$$-rupee notes; then she also has $$n$$ notes of $$100$$ rupees.

Total amount $$= 50n + 100n = 150n$$ rupees, and this equals $$750$$:

$$150n = 750.$$

Divide both sides by $$150$$: $$n = \dfrac{750}{150} = 5$$.

Check. $$5$$ notes of ₹$$50$$ $$= 250$$ and $$5$$ notes of ₹$$100$$ $$= 500$$. Total $$= 250 + 500 = 750$$. Correct.

So Ranju has $$5$$ notes of ₹$$50$$ and $$5$$ notes of ₹$$100$$.

From the picture we can see $$4$$ blue dots that are not covered, and $$3$$ black blobs, each hiding the same number of dots. Let one blob cover $$b$$ dots.

Total dots $$= (\text{visible}) + 3 \times (\text{under each blob})$$:

$$4 + 3b = 25.$$

Subtract $$4$$ from both sides: $$3b = 21$$. Divide by $$3$$: $$b = 7$$.

Check. $$4 + 3(7) = 4 + 21 = 25$$. Correct.

So each blob covers $$7$$ blue dots.

Machine 1. Let the input be $$x$$. The two "$$\div 3$$" boxes give $$\dfrac{x}{3 \times 3} = \dfrac{x}{9}$$. So

$$\dfrac{x}{9} = 5 \;\Rightarrow\; x = 45.$$

Check: $$45 \to 15 \to 5$$. Correct.

Machine 2. Let the input be $$x$$. The two "$$-4$$" boxes give $$x - 4 - 4 = x - 8$$. So

$$x - 8 = -11 \;\Rightarrow\; x = -11 + 8 = -3.$$

Check: $$-3 \to -7 \to -11$$. Correct.

Let the distance travelled be $$k$$ kilometres. Total cost $$= $$ fixed fee $$+$$ per-km charge $$\times$$ distance:

$$800 + 20k = 2200.$$

Subtract $$800$$ from both sides: $$20k = 1400$$. Divide by $$20$$:

$$k = \dfrac{1400}{20} = 70.$$

Check. $$800 + 20(70) = 800 + 1400 = 2200$$. Correct.

So the taxi travelled $$70$$ km.

Let the smaller number be $$x$$. Then the larger number is $$3x$$, and their sum is

$$x + 3x = 76, \quad \text{so } 4x = 76.$$

Divide both sides by $$4$$: $$x = 19$$. The larger number is $$3 \times 19 = 57$$.

Check. $$19 + 57 = 76$$; $$57 = 3 \times 19$$. Both conditions satisfied.

The window is $$34\,\mathrm{cm}$$ tall. From this we subtract the fixed gaps at the top ($$3\,\mathrm{cm}$$) and the bottom ($$2\,\mathrm{cm}$$) to get the total space occupied by the equal internal gaps between the rods. From the figure, there are $$5$$ such equal gaps between rods. Let each gap be $$g\,\mathrm{cm}$$. Then

$$3 + 5g + 2 = 34, \quad \text{so } 5g + 5 = 34.$$

Subtract $$5$$ from both sides: $$5g = 29$$. Divide by $$5$$:

$$g = \dfrac{29}{5} = 5.8\,\mathrm{cm}.$$

Check. $$3 + 5 \times 5.8 + 2 = 3 + 29 + 2 = 34$$. Correct.

So the gap between two rods is $$5.8\,\mathrm{cm}$$.

Let the cost of one chocolate milkshake be ₹$$m$$. Then one fruit juice costs ₹$$(m - 15)$$. The total cost of $$4$$ juices and $$7$$ milkshakes is

$$4(m - 15) + 7m = 600.$$

Expand: $$4m - 60 + 7m = 600$$, so $$11m - 60 = 600$$. Add $$60$$: $$11m = 660$$. Divide by $$11$$: $$m = 60$$.

So a milkshake costs ₹$$60$$ and a juice costs ₹$$60 - 15 = 45$$.

Check. $$4(45) + 7(60) = 180 + 420 = 600$$. Correct.

From $$28p - 36 = 98$$, add $$36$$ to both sides: $$28p = 134$$. Divide both sides by $$14$$: $$2p = \dfrac{134}{14} = \dfrac{67}{7}$$. So $$14p = \dfrac{134}{2} = 67$$.

Value of $$14p - 19$$. $$14p - 19 = 67 - 19 = 48$$.

Value of $$28p - 38$$. Notice $$28p - 38 = (28p - 36) - 2 = 98 - 2 = 96$$.

Check with $$p$$. From $$28p = 134$$, $$p = \dfrac{134}{28} = \dfrac{67}{14}$$. Then $$14p = 67$$ and $$28p = 134$$, and both values ($$48$$ and $$96$$) agree.

The three angles of any triangle add up to $$180^{\circ}$$.

Triangle 1. The three angles are $$y$$, $$y + 15$$ and $$y + 15$$. So

$$y + (y + 15) + (y + 15) = 180.$$

Simplify: $$3y + 30 = 180$$. Subtract $$30$$: $$3y = 150$$. Divide by $$3$$: $$y = 50$$. So the three angles are $$50^{\circ}$$, $$65^{\circ}$$ and $$65^{\circ}$$. Check: $$50 + 65 + 65 = 180$$.

Triangle 2. The three angles are $$x$$, $$x - 10$$ and $$x + 10$$. So

$$x + (x - 10) + (x + 10) = 180.$$

Simplify: $$3x = 180$$. Divide by $$3$$: $$x = 60$$. So the three angles are $$60^{\circ}$$, $$50^{\circ}$$ and $$70^{\circ}$$. Check: $$60 + 50 + 70 = 180$$.

Start with $$u = 6$$ and apply the same operation to both sides to build each new equation.

1. Add $$4$$ to both sides: $$u + 4 = 10$$.
2. Multiply both sides by $$5$$: $$5u = 30$$.
3. Multiply by $$2$$ then subtract $$3$$: $$2u - 3 = 9$$.
4. Multiply by $$3$$ and then add $$u$$ to both sides: $$3u + u = 18 + u$$, i.e. $$4u = u + 18$$.

Check the fourth: put $$u = 6$$; LHS $$= 24$$; RHS $$= 6 + 18 = 24$$. Equal.

Let the first person receive $$x$$.

• Second person: $$2 \times x = 2x$$.
• Third person: $$3 \times (2x) = 6x$$.
• Fourth person: $$4 \times (6x) = 24x$$.

Total:

$$x + 2x + 6x + 24x = 132.$$

Simplify: $$33x = 132$$. Divide both sides by $$33$$:

$$x = \dfrac{132}{33} = 4.$$

Check. The four amounts are $$4, 8, 24, 96$$ and $$4 + 8 + 24 + 96 = 132$$. Correct.

So the first person is given $$4$$.

Let the height of the giraffe be $$h$$ metres. Then "$$2\dfrac{1}{2}$$ m more than half its height" is $$\dfrac{h}{2} + \dfrac{5}{2}$$. Setting this equal to the height:

$$h = \dfrac{h}{2} + \dfrac{5}{2}.$$

Subtract $$\dfrac{h}{2}$$ from both sides:

$$h - \dfrac{h}{2} = \dfrac{5}{2}, \quad \text{i.e.} \quad \dfrac{h}{2} = \dfrac{5}{2}.$$

Multiply both sides by $$2$$:

$$h = 5.$$

So the giraffe is $$5\,\mathrm{m}$$ tall.

Check. Half of $$5$$ is $$2.5$$, and $$2.5 + 2.5 = 5\,\mathrm{m}$$. Correct.

Let the number be $$x$$. Then "$$x$$ increased by $$36$$" equals "ten times $$x$$":

$$x + 36 = 10x.$$

Subtract $$x$$ from both sides:

$$36 = 9x.$$

Divide both sides by $$9$$:

$$x = \dfrac{36}{9} = 4.$$

Check. $$4 + 36 = 40 = 10 \times 4$$. Correct.

Solve each equation. The path from Start to End is traced by following the solved value of each equation from one cell to the next.

• $$8x = 20 + 3x \Rightarrow 5x = 20 \Rightarrow x = 4$$ (worked example).
• $$-7 = 11 - 3x \Rightarrow 3x = 11 + 7 = 18 \Rightarrow x = 6$$.
• $$15 = 19 - 4x \Rightarrow 4x = 19 - 15 = 4 \Rightarrow x = 1$$.
• $$2x - 9 = -3 \Rightarrow 2x = -3 + 9 = 6 \Rightarrow x = 3$$.
• $$-2x = -42 \Rightarrow x = 21$$.
• $$2x + 3 = x + 5 \Rightarrow x = 5 - 3 = 2$$.
• $$8m + 8 = -72 \Rightarrow 8m = -80 \Rightarrow m = -10$$.
• $$2(x + 1) - 10 = 18 \Rightarrow 2(x + 1) = 28 \Rightarrow x + 1 = 14 \Rightarrow x = 13$$.
• $$2x + 5 = 3(x - 1) \Rightarrow 2x + 5 = 3x - 3 \Rightarrow 5 + 3 = 3x - 2x \Rightarrow x = 8$$.
• $$-4 = 16 - 5k \Rightarrow 5k = 16 + 4 = 20 \Rightarrow k = 4$$.
• $$2x - 9 = 3 - x \Rightarrow 3x = 12 \Rightarrow x = 4$$.
• $$30 = 4 - 50n \Rightarrow 50n = 4 - 30 = -26 \Rightarrow n = -\dfrac{26}{50} = -\dfrac{13}{25}$$.

The path from Start to End is built by matching each solved value with the arrow labelled by that value in the maze (for instance, the Start box gives $$x = 4$$ and takes the arrow labelled $$4$$; that leads to $$2x - 9 = -3$$, which gives $$x = 3$$ and hence the arrow labelled $$3$$; and so on).

Let $$d$$ be the number of donkeys and $$c$$ the number of children. Each has $$1$$ head, but a donkey has $$4$$ feet and a child has $$2$$ feet.

Head count: $$c + d = 28$$, so $$c = 28 - d$$.

Foot count: $$2c + 4d = 80$$. Substitute $$c = 28 - d$$:

$$2(28 - d) + 4d = 80.$$

Expand: $$56 - 2d + 4d = 80$$, i.e. $$56 + 2d = 80$$. Subtract $$56$$: $$2d = 24$$. Divide by $$2$$: $$d = 12$$. Then $$c = 28 - 12 = 16$$.

Check. Heads: $$16 + 12 = 28$$. Feet: $$16 \times 2 + 12 \times 4 = 32 + 48 = 80$$. Both match.

So there are $$12$$ donkeys and $$16$$ children on the beach.

Yes, the final answer is always $$8$$. Let us follow the steps algebraically with the starting number $$x$$.

Notice that the letter $$x$$ disappears after the "subtract the original number" step — every trace of your original choice is wiped out. So the final answer is always $$8$$, no matter what $$x$$ was.

Try it with $$x = 7$$: $$7 \to 14 \to 24 \to 12 \to 5 \to 8$$. And with $$x = 100$$: $$100 \to 200 \to 210 \to 105 \to 5 \to 8$$. Always $$8$$.

Making your own trick. Pick any number you would like the final answer to be, say $$12$$. Design steps so that $$x$$ cancels out and only $$12$$ remains. For example: think of $$x$$; multiply by $$3$$ to get $$3x$$; add $$9$$ to get $$3x + 9$$; divide by $$3$$ to get $$x + 3$$; subtract the original number to get $$3$$; add $$9$$ to get $$12$$. The final answer is always $$12$$.