NCERT Exemplar Solutions Class 10 Maths Pair of Linear Equations — Exercise 3.4

NCERT Exemplar Problems, Class 10 Mathematics — Pair of Linear Equations in Two Variables, Exercise 3.4. All 13 questions solved.

The long-answer exercise splits cleanly. Questions 1 to 5 are geometric: draw or compute the lines, find where they cross, and work out an area. Questions 6 to 13 are word problems of the kind that carry full marks in a board paper — boats and streams, fares, discounts, investments, digits.

For the geometric half, the only technique is that a vertex of the figure is an intersection of two of the given lines (counting an axis as a line). Find all of them and the figure is determined; the area then comes from $\frac12 \times \text{base} \times \text{height}$ for a triangle, or $\frac12(a+b)h$ for a trapezium.

For the word problems, three setups cover almost everything here:

• Boats and streams. Downstream speed is $b + s$, upstream is $b – s$. Time is distance over speed, and the resulting equations are linear in $\frac{1}{b+s}$ and $\frac{1}{b-s}$ — not in $b$ and $s$. Substituting for those two reciprocals first is what keeps Question 8 manageable.
• Two-digit numbers. If the tens digit is $t$ and the units digit is $u$, the number is $10t + u$ — never $tu$.
• Swapped conditions. Questions 11 and 12 both give a second scenario with two rates exchanged. Adding and subtracting the two equations is almost always faster than elimination, because the swap makes the coefficient sums equal.

Key insight. In every word problem here, name the quantity the question asks for and write one equation per sentence of the problem. The sentences and the equations are in one-to-one correspondence; if you end up with fewer equations than unknowns, a sentence has been read but not used.

Question 1

Graphically, solve the following pair of equations: $2x + y = 6$ and $2x – y + 2 = 0$. Find the ratio of the areas of the two triangles formed by the lines representing these equations with the $x$-axis and the lines with the $y$-axis.

Solution. The two lines meet where both hold. Adding them,

$$(2x + y) + (2x – y) = 6 + (-2) \implies 4x = 4 \implies x = 1, \quad y = 4$$

so the solution is $(1, 4)$ — the point marked on the graph.

Triangle with the $x$-axis. Its third side lies along $y = 0$, so its other two vertices are the $x$-intercepts: $2x + y = 6$ gives $(3, 0)$ and $2x – y = -2$ gives $(-1, 0)$. Base $= 3 – (-1) = 4$; height $= $ the $y$-coordinate of the apex $= 4$.

$$\text{Area}_1 = \frac12 \times 4 \times 4 = 8 \text{ square units}$$

Triangle with the $y$-axis. Now the $y$-intercepts: $(0, 6)$ and $(0, 2)$. Base $= 6 – 2 = 4$; height $=$ the $x$-coordinate of the apex $= 1$.

$$\text{Area}_2 = \frac12 \times 4 \times 1 = 2 \text{ square units}$$

$$\text{Ratio} = 8 : 2 = 4 : 1$$

Question 2

Determine, graphically, the vertices of the triangle formed by the lines $y = x$, $3y = x$ and $x + y = 8$.

Solution. Each vertex is where two of the three lines meet, so take them in pairs.

• $y = x$ and $3y = x$. Substituting, $3y = y$, so $y = 0$ and $x = 0$: the vertex is $(0, 0)$. Both lines pass through the origin, which is why the triangle has a vertex there.
• $y = x$ and $x + y = 8$. Substituting, $2x = 8$, so $x = 4$ and $y = 4$: the vertex is $(4, 4)$.
• $3y = x$ and $x + y = 8$. Substituting $x = 3y$, $3y + y = 8$, so $y = 2$ and $x = 6$: the vertex is $(6, 2)$.

On graph paper: $y = x$ is the line through $(0,0)$ and $(4,4)$; $3y = x$ is the shallower line through $(0,0)$ and $(6,2)$; $x + y = 8$ joins $(8,0)$ to $(0,8)$ and cuts the other two at the points found above.

Question 3

Draw the graphs of the equations $x = 3$, $x = 5$ and $2x – y – 4 = 0$. Also find the area of the quadrilateral formed by the lines and the $x$-axis.

Solution. $x = 3$ and $x = 5$ are vertical lines; $2x – y – 4 = 0$ is $y = 2x – 4$, which crosses the $x$-axis at $(2, 0)$. The four vertices of the enclosed figure are:

• $x = 3$ meets the $x$-axis at $(3, 0)$;
• $x = 5$ meets the $x$-axis at $(5, 0)$;
• $x = 5$ meets $y = 2x – 4$ at $y = 10 – 4 = 6$, giving $(5, 6)$;
• $x = 3$ meets $y = 2x – 4$ at $y = 6 – 4 = 2$, giving $(3, 2)$.

The two vertical sides are parallel, so this is a trapezium with parallel sides of lengths $2$ and $6$, a distance $5 – 3 = 2$ apart.

$$\text{Area} = \frac12 \left(a + b\right)h = \frac12 (2 + 6)(2) = 8 \text{ square units}$$

Question 4

The cost of $4$ pens and $4$ pencil boxes is ₹100. Three times the cost of a pen is ₹15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.

Solution. Let a pen cost ₹$x$ and a pencil box ₹$y$. The two sentences give one equation each:

$$4x + 4y = 100, \qquad 3x = y + 15$$

The first simplifies to $x + y = 25$. Substituting $y = 3x – 15$ from the second:

$$x + 3x – 15 = 25 \implies 4x = 40 \implies x = 10, \quad y = 15$$

Check: $4(10) + 4(15) = 40 + 60 = 100$ ✓, and $3(10) = 30 = 15 + 15$ ✓.

Question 5

Determine, algebraically, the vertices of the triangle formed by the lines $3x – y = 3$, $2x – 3y = 2$ and $x + 2y = 8$.

Solution. Three lines, three pairs, three vertices.

$3x – y = 3$ and $2x – 3y = 2$. From the first, $y = 3x – 3$. Substituting:

$$2x – 3(3x – 3) = 2 \implies 2x – 9x + 9 = 2 \implies -7x = -7 \implies x = 1, \ y = 0$$

$3x – y = 3$ and $x + 2y = 8$. Again $y = 3x – 3$:

$$x + 2(3x – 3) = 8 \implies 7x = 14 \implies x = 2, \ y = 3$$

$2x – 3y = 2$ and $x + 2y = 8$. From the second, $x = 8 – 2y$:

$$2(8 – 2y) – 3y = 2 \implies 16 – 7y = 2 \implies y = 2, \ x = 4$$

Question 6

Ankita travels $14$ km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels $2$ km by rickshaw, and the remaining distance by bus. On the other hand, if she travels $4$ km by rickshaw and the remaining distance by bus, she takes $9$ minutes longer. Find the speed of the rickshaw and of the bus.

Solution. Let the rickshaw travel at $r$ km/h and the bus at $b$ km/h. Time is distance over speed, so the two journeys give

$$\frac{2}{r} + \frac{12}{b} = \frac12 \qquad\text{and}\qquad \frac{4}{r} + \frac{10}{b} = \frac12 + \frac{9}{60} = \frac{13}{20}$$

(the $9$ minutes converted to $\frac{9}{60} = \frac{3}{20}$ of an hour). These are not linear in $r$ and $b$, but they are linear in the reciprocals. Put $u = \frac1r$ and $v = \frac1b$:

$$2u + 12v = \frac12, \qquad 4u + 10v = \frac{13}{20}$$

Multiply the first by $2$ and subtract:

$$4u + 24v = 1 \quad\text{minus}\quad 4u + 10v = \frac{13}{20} \implies 14v = \frac{7}{20} \implies v = \frac{1}{40}$$

So $b = 40$. Then $2u = \frac12 – \frac{12}{40} = \frac12 – \frac{3}{10} = \frac{1}{5}$, giving $u = \frac{1}{10}$ and $r = 10$.

Question 7

A person, rowing at the rate of $5$ km/h in still water, takes thrice as much time in going $40$ km upstream as in going $40$ km downstream. Find the speed of the stream.

Solution. Let the stream flow at $s$ km/h. Rowing upstream the speeds subtract, downstream they add:

$$\text{upstream speed} = 5 – s, \qquad \text{downstream speed} = 5 + s$$

“Thrice as much time” over the same distance $40$ km:

$$\frac{40}{5 – s} = 3 \times \frac{40}{5 + s}$$

The $40$s cancel, leaving $5 + s = 3(5 – s)$:

$$5 + s = 15 – 3s \implies 4s = 10 \implies s = 2.5$$

Check: upstream $2.5$ km/h takes $16$ hours for $40$ km; downstream $7.5$ km/h takes $\frac{16}{3}$ hours — a factor of three ✓. Note that $s < 5$, as it must be for upstream travel to be possible at all.

Question 8

A motor boat can travel $30$ km upstream and $28$ km downstream in $7$ hours. It can travel $21$ km upstream and return in $5$ hours. Find the speed of the boat in still water and the speed of the stream.

Solution. Let the boat do $b$ km/h in still water and the stream $s$ km/h. Substitute for the two reciprocal speeds at once:

$$u = \frac{1}{b – s} \quad (\text{upstream}), \qquad d = \frac{1}{b + s} \quad (\text{downstream})$$

“Return” in the second journey means $21$ km upstream and $21$ km downstream. The two conditions become linear:

$$30u + 28d = 7, \qquad 21u + 21d = 5$$

Divide the second by $21$: $u + d = \frac{5}{21}$, so $u = \frac{5}{21} – d$. Substituting into the first:

$$30\left(\frac{5}{21} – d\right) + 28d = \frac{50}{7} – 2d = 7 \implies 2d = \frac{50}{7} – 7 = \frac17 \implies d = \frac{1}{14}$$

So $b + s = 14$. And $u = \frac{5}{21} – \frac{1}{14} = \frac{10 – 3}{42} = \frac16$, so $b – s = 6$.

Adding and subtracting: $b = 10$ and $s = 4$.

Question 9

A two-digit number is obtained by either multiplying the sum of the digits by $8$ and then subtracting $5$ or by multiplying the difference of the digits by $16$ and then adding $3$. Find the number.

Solution. Let the tens digit be $t$ and the units digit $u$, so the number itself is $10t + u$. Each of the two descriptions gives an equation.

$$10t + u = 8(t + u) – 5 \implies 10t + u = 8t + 8u – 5 \implies 2t – 7u = -5$$ $$10t + u = 16(t – u) + 3 \implies 10t + u = 16t – 16u + 3 \implies -6t + 17u = 3$$

Multiply the first by $3$ and add to the second:

$$6t – 21u = -15 \quad\text{plus}\quad -6t + 17u = 3 \implies -4u = -12 \implies u = 3$$

Then $2t = 7(3) – 5 = 16$, so $t = 8$ and the number is $83$.

Check: the digit sum is $11$, and $8(11) – 5 = 83$ ✓; the digit difference is $5$, and $16(5) + 3 = 83$ ✓.

Question 10

A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from station A to B costs ₹2530. Also, one reserved first class ticket and one reserved first class half ticket from A to B costs ₹3810. Find the full first class fare from station A to B, and also the reservation charges for a ticket.

Solution. Let the full fare be ₹$F$ and the reservation charge ₹$R$. A full reserved ticket costs $F + R$; a half reserved ticket costs $\frac{F}{2} + R$ — the reservation is not halved, which is the sentence the question turns on.

$$F + R = 2530$$ $$\left(F + R\right) + \left(\frac{F}{2} + R\right) = 3810$$

Substituting the first into the second:

$$2530 + \frac{F}{2} + R = 3810 \implies \frac{F}{2} + R = 1280$$

Subtracting this from $F + R = 2530$:

$$\frac{F}{2} = 1250 \implies F = 2500, \quad R = 2530 – 2500 = 30$$

Question 11

A shopkeeper sells a saree at $8\%$ profit and a sweater at $10\%$ discount, thereby getting a sum ₹1008. If she had sold the saree at $10\%$ profit and the sweater at $8\%$ discount, she would have got ₹1028. Find the cost price of the saree and the list price (price before discount) of the sweater.

Solution. Let the saree’s cost price be ₹$x$ and the sweater’s list price ₹$y$ — the question names these two quantities precisely, and mixing up cost price with selling price is the usual source of error.

A profit of $8\%$ on the saree sells it for $1.08x$; a discount of $10\%$ on the sweater sells it for $0.90y$. So

$$1.08x + 0.90y = 1008$$ $$1.10x + 0.92y = 1028$$

Clear the decimals by multiplying by $100$, then eliminate $y$: multiply the first by $92$ and the second by $90$:

$$9936x + 8280y = 92\,736$$ $$9900x + 8280y = 92\,520$$

Subtracting, $36x = 216$, so $x = 600$. Then $0.90y = 1008 – 648 = 360$, giving $y = 400$.

Check the second scenario: $1.10(600) + 0.92(400) = 660 + 368 = 1028$ ✓.

Question 12

Susan invested a certain amount of money in two schemes A and B, which offer interest at the rate of $8\%$ per annum and $9\%$ per annum, respectively. She received ₹1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received ₹20 more as annual interest. How much money did she invest in each scheme?

Solution. Let ₹$x$ go into scheme A and ₹$y$ into scheme B. Interchanging means $y$ earns $8\%$ and $x$ earns $9\%$.

$$0.08x + 0.09y = 1860$$ $$0.09x + 0.08y = 1880$$

Because the coefficients are swapped, adding and subtracting is much faster than elimination.

Adding: $0.17(x + y) = 3740$, so $x + y = 22\,000$. Subtracting the first from the second: $0.01x – 0.01y = 20$, so $x – y = 2000$.

Hence $x = 12\,000$ and $y = 10\,000$.

Check: $0.08(12000) + 0.09(10000) = 960 + 900 = 1860$ ✓, and interchanged, $1080 + 800 = 1880$ ✓ — exactly ₹20 more.

Question 13

Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of ₹2 for $3$ bananas and the second lot at the rate of ₹1 per banana, and got a total of ₹400. If he had sold the first lot at the rate of ₹1 per banana, and the second lot at the rate of ₹4 for $5$ bananas, his total collection would have been ₹460. Find the total number of bananas he had.

Solution. Let lot A hold $x$ bananas and lot B hold $y$. A rate of “₹2 for 3 bananas” is $\frac23$ rupees each, and “₹4 for 5 bananas” is $\frac45$ each.

$$\frac{2}{3}x + y = 400 \qquad\text{and}\qquad x + \frac{4}{5}y = 460$$

Clear the fractions — multiply the first by $3$ and the second by $5$:

$$2x + 3y = 1200, \qquad 5x + 4y = 2300$$

Eliminate $x$: multiply the first by $5$ and the second by $2$:

$$10x + 15y = 6000, \qquad 10x + 8y = 4600 \implies 7y = 1400 \implies y = 200$$

Then $2x = 1200 – 600 = 600$, so $x = 300$. The total is $300 + 200 = 500$.

Check: $\frac23(300) + 200 = 200 + 200 = 400$ ✓, and $300 + \frac45(200) = 300 + 160 = 460$ ✓.

Common mistakes

• Using the wrong height for the second triangle in Question 1. For the triangle standing on the $y$-axis, the “height” is the horizontal distance to the apex, which is its $x$-coordinate, $1$ — not $4$. The two triangles share an apex and have equal bases, so the whole ratio is decided by this one number.
• Forgetting that “return” means both ways in Question 8. “$21$ km upstream and return” is $21$ up plus $21$ down, giving $21u + 21d = 5$. Reading it as $21$ km in total makes the numbers ugly and the answer wrong.
• Halving the reservation charge in Question 10. A half ticket is half the fare plus the full reservation. The problem says so explicitly, and the whole calculation rests on it.
• Writing a two-digit number as $tu$ in Question 9. It is $10t + u$. This is the single most common error in digit problems, and it produces a quadratic where a linear pair was intended.
• Treating $8\%$ profit and $10\%$ discount as the same operation in Question 11. Profit adds to the cost price ($1.08x$); discount subtracts from the list price ($0.90y$). One goes up and one comes down.
• Solving Question 12 by elimination. It works, but the swapped coefficients make adding and subtracting far shorter — the sum gives $x + y$ and the difference gives $x – y$ directly. Look for this whenever two equations are mirror images.
• Converting minutes incorrectly in Question 6. Nine minutes is $\frac{9}{60} = \frac{3}{20}$ hours, because the speeds are in km/h. Mixing units is the commonest failure in this question.
• Not checking that the stream is slower than the boat. In Questions 7 and 8 a solution with $s \ge b$ would be physically impossible; both come out sensibly ($2.5 < 5$ and $4 < 10$), and noticing this is a free check on the algebra.

Practise next

• Exemplar Exercise 3.3 — shorter versions of the same word problems, plus the parameter questions this exercise assumes you have done.
• Exemplar Exercises 3.1 and 3.2 for the consistency conditions underlying Questions 1 to 5.
• NCERT textbook Exercise 3.3 — the book’s own long word problems, including a boat-and-stream question of exactly the shape of Question 8.
• NCERT textbook Exercise 3.1 for graphical work of the kind Questions 1 to 3 extend.