NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 3: Pair of Linear Equations in Two Variables

Exercise 3.4

Step 1: Write both equations in slope form (y = ...).

Equation (1): \(2x + y = 6\)

Subtract \(2x\) from both sides:

\(y = 6 - 2x\)

Equation (2): \(2x - y + 2 = 0\)

Subtract 2 from both sides:

\(2x - y = -2\)

Add \(y\) to both sides:

\(2x = y - 2\)

So, \(y = 2x + 2\).

Step 2: Find the point of intersection.

At the intersection, both y’s are equal:

\(6 - 2x = 2x + 2\)

Bring terms together:

\(6 - 2 = 2x + 2x\)

\(4 = 4x\)

So, \(x = 1\).

Now put \(x = 1\) in equation (1):

\(y = 6 - 2(1)\)

\(y = 6 - 2 = 4\).

So, intersection point is \((1, 4)\).

Step 3: Find intercepts on the x-axis.

For equation (1): \(y = 0\).

So, \(2x = 6\Rightarrow x = 3\).

Point = \((3, 0)\).

For equation (2): \(y = 0\).

So, \(2x + 2 = 0\Rightarrow 2x = -2\Rightarrow x = -1\).

Point = \((-1, 0)\).

Step 4: Area of triangle with x-axis.

Base = distance between \((3, 0)\) and \((-1, 0)\) = 4.

Height = y-coordinate of intersection point = 4.

Area = \(\tfrac{1}{2} \times 4 \times 4 = 8\).

Step 5: Find intercepts on the y-axis.

For equation (1): \(x = 0\).

So, \(y = 6\). Point = \((0, 6)\).

For equation (2): \(x = 0\).

So, \(y = 2\). Point = \((0, 2)\).

Step 6: Area of triangle with y-axis.

Base = distance between \((0, 6)\) and \((0, 2)\) = 4.

Height = x-coordinate of intersection point = 1.

Area = \(\tfrac{1}{2} \times 4 \times 1 = 2\).

Step 7: Find ratio of areas.

Ratio = Area with x-axis : Area with y-axis

= \(8 : 2 = 4 : 1\).

Step 1: Find the intersection of \(y = x\) and \(3y = x\).

From the first line: \(y = x\).

From the second line: \(x = 3y\).

Put \(y = x\) into \(x = 3y\):

\(x = 3x\).

Subtract \(x\) from both sides:

\(x - 3x = 0\).

\(-2x = 0\).

So, \(x = 0\).

Then, \(y = 0\).

First vertex is \((0,0)\).

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Step 2: Find the intersection of \(y = x\) and \(x + y = 8\).

From the first line: \(y = x\).

Substitute \(y = x\) into \(x + y = 8\):

\(x + x = 8\).

\(2x = 8\).

So, \(x = 4\).

Since \(y = x\), we get \(y = 4\).

Second vertex is \((4,4)\).

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Step 3: Find the intersection of \(3y = x\) and \(x + y = 8\).

From the first line: \(x = 3y\).

Substitute into \(x + y = 8\):

\(3y + y = 8\).

\(4y = 8\).

So, \(y = 2\).

Now, \(x = 3y = 3 \times 2 = 6\).

Third vertex is \((6,2)\).

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Final Answer:

The vertices of the triangle are \((0,0)\), \((4,4)\), and \((6,2)\).

Step 1: Rewrite the slant line equation.

The line is given as \(2x - y - 4 = 0\).

Move terms around: \(y = 2x - 4\).

So, this is a straight line which goes upward (slope = 2) and cuts the y-axis at \(-4\).

Step 2: Understand the other two lines.

The equations \(x = 3\) and \(x = 5\) are vertical lines. They are parallel to the y-axis.

Step 3: Think about the region we want.

The x-axis is \(y = 0\).

So, the closed shape is formed between:

• The vertical line at \(x=3\),
• The vertical line at \(x=5\),
• The slant line \(y=2x-4\),
• And the x-axis \(y=0\).

This shape is a trapezium (a quadrilateral with one pair of parallel sides).

Step 4: Find the heights of the trapezium.

We plug in x-values into the slant line \(y=2x-4\):

At \(x=3\): \(y = 2(3) - 4 = 6 - 4 = 2\).

At \(x=5\): \(y = 2(5) - 4 = 10 - 4 = 6\).

So the slant line is at height 2 above the x-axis at \(x=3\), and at height 6 above the x-axis at \(x=5\).

Step 5: Work out the distance between the vertical lines.

The vertical distance along the x-axis is \(5 - 3 = 2\) units.

Step 6: Apply the trapezium area formula.

Area of trapezium = \(\tfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{distance}\).

Here, the two parallel sides are the heights: 2 units and 6 units.

So, area = \(\tfrac{1}{2} \times (2 + 6) \times 2 = \tfrac{1}{2} \times 8 \times 2 = 8\).

Final Answer: The area of the quadrilateral is \(8\) square units.

Step 1: Assume variables.

Let the cost of one pen be \(p\) (in rupees).

Let the cost of one pencil box be \(b\) (in rupees).

Step 2: Form the first equation.

We are told that 4 pens and 4 pencil boxes cost Rs 100.

This means: \(4p + 4b = 100\).

Divide everything by 4 to make it simpler: \(p + b = 25\).

Step 3: Form the second equation.

We are told that three times the cost of a pen is Rs 15 more than the cost of a pencil box.

So: \(3p = b + 15\).

Or we can write it as: \(b = 3p - 15\).

Step 4: Solve the equations.

From the first equation: \(p + b = 25\).

Now replace \(b\) with \(3p - 15\) (from the second equation):

\(p + (3p - 15) = 25\).

That gives: \(4p - 15 = 25\).

Add 15 on both sides: \(4p = 40\).

Divide by 4: \(p = 10\).

Step 5: Find the cost of the pencil box.

Use \(b = 3p - 15\).

Put \(p = 10\): \(b = 3(10) - 15 = 30 - 15 = 15\).

Final Answer: Pen costs Rs 10 each, and Pencil box costs Rs 15 each.

We need to find the points where these three lines meet each other. Each pair of lines will intersect at one vertex of the triangle.

Step 1: Intersection of Line 1 (\(3x - y = 3\)) and Line 2 (\(2x - 3y = 2\))

From Line 1: \(3x - y = 3 \Rightarrow y = 3x - 3\).

Now substitute this value of \(y\) into Line 2:

\(2x - 3(3x - 3) = 2\)

\(2x - 9x + 9 = 2\)

\(-7x + 9 = 2\)

\(-7x = -7\)

\(x = 1\).

When \(x = 1\), \(y = 3(1) - 3 = 0\).

So, first vertex = (1, 0).

Step 2: Intersection of Line 2 (\(2x - 3y = 2\)) and Line 3 (\(x + 2y = 8\))

From Line 3: \(x + 2y = 8 \Rightarrow x = 8 - 2y\).

Now substitute this into Line 2:

\(2(8 - 2y) - 3y = 2\)

\(16 - 4y - 3y = 2\)

\(16 - 7y = 2\)

\(-7y = -14\)

\(y = 2\).

If \(y = 2\), then \(x = 8 - 2(2) = 4\).

So, second vertex = (4, 2).

Step 3: Intersection of Line 3 (\(x + 2y = 8\)) and Line 1 (\(3x - y = 3\))

From Line 3: \(x = 8 - 2y\).

Substitute this into Line 1:

\(3(8 - 2y) - y = 3\)

\(24 - 6y - y = 3\)

\(24 - 7y = 3\)

\(-7y = -21\)

\(y = 3\).

If \(y = 3\), then \(x = 8 - 2(3) = 2\).

So, third vertex = (2, 3).

Final Answer:

The triangle formed has vertices at \((1,0)\), \((4,2)\), and \((2,3)\).

Step 1: Assume the speeds.

Let the speed of the rickshaw be \(r\) km/h.

Let the speed of the bus be \(b\) km/h.

Step 2: Recall the formula for time.

\(\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}\)

Step 3: Write the first condition (Case 1).

Distance by rickshaw = 2 km, so time = \(\dfrac{2}{r}\).

Distance by bus = 12 km, so time = \(\dfrac{12}{b}\).

Total time = 30 minutes = \(\dfrac{1}{2}\) hour.

So, \(\dfrac{2}{r} + \dfrac{12}{b} = \dfrac{1}{2}\). … (1)

Step 4: Write the second condition (Case 2).

Distance by rickshaw = 4 km, so time = \(\dfrac{4}{r}\).

Distance by bus = 10 km, so time = \(\dfrac{10}{b}\).

Total time = 30 min + 9 min = 39 min = \(\dfrac{13}{20}\) hour.

So, \(\dfrac{4}{r} + \dfrac{10}{b} = \dfrac{13}{20}\). … (2)

Step 5: Make the equations easier.

Put \(u = \dfrac{1}{r}\) and \(v = \dfrac{1}{b}\).

Equation (1): \(2u + 12v = \dfrac{1}{2}\).

Equation (2): \(4u + 10v = \dfrac{13}{20}\).

Step 6: Solve the equations.

From (1): \(2u + 12v = \dfrac{1}{2}\) → Divide by 2 → \(u + 6v = \dfrac{1}{4}\). … (3)

From (2): \(4u + 10v = \dfrac{13}{20}\) → Divide by 2 → \(2u + 5v = \dfrac{13}{40}\). … (4)

Multiply (3) by 2: \(2u + 12v = \dfrac{1}{2}\). … (5)

Now subtract (4) from (5):

\((2u + 12v) - (2u + 5v) = \dfrac{1}{2} - \dfrac{13}{40}\)

\(7v = \dfrac{20}{40} - \dfrac{13}{40} = \dfrac{7}{40}\)

So, \(v = \dfrac{1}{40}\).

Step 7: Find bus speed.

Since \(v = \dfrac{1}{b}\), we get \(b = 40\) km/h.

Step 8: Find rickshaw speed.

Put \(v = \dfrac{1}{40}\) in (3):

\(u + 6 \times \dfrac{1}{40} = \dfrac{1}{4}\)

\(u + \dfrac{6}{40} = \dfrac{1}{4}\)

\(u + \dfrac{3}{20} = \dfrac{1}{4}\)

\(u = \dfrac{1}{4} - \dfrac{3}{20} = \dfrac{5}{20} - \dfrac{3}{20} = \dfrac{2}{20} = \dfrac{1}{10}\)

So, \(r = 10\) km/h.

Final Answer: Rickshaw speed = 10 km/h, Bus speed = 40 km/h.

Step 1: Let the speed of the stream be \(s\) km/h.

Step 2: When rowing upstream (against the current), the water slows the boat down. So, effective speed = (speed in still water – speed of stream) = \(5 - s\) km/h.

Step 3: When rowing downstream (with the current), the water helps the boat go faster. So, effective speed = (speed in still water + speed of stream) = \(5 + s\) km/h.

Step 4: Time is calculated as distance ÷ speed. - Time to go 40 km upstream = \(\dfrac{40}{5 - s}\). - Time to go 40 km downstream = \(\dfrac{40}{5 + s}\).

Step 5: It is given that upstream time is three times the downstream time. So, \(\dfrac{40}{5 - s} = 3 \times \dfrac{40}{5 + s}\).

Step 6: Cancel 40 from both sides: \(\dfrac{1}{5 - s} = \dfrac{3}{5 + s}\).

Step 7: Cross-multiply: \(5 + s = 3(5 - s)\).

Step 8: Simplify: \(5 + s = 15 - 3s\). Add \(3s\) on both sides → \(5 + 4s = 15\). Subtract 5 → \(4s = 10\). Divide by 4 → \(s = 2.5\).

Final Answer: The speed of the stream is 2.5 km/h.

Step 1: Understand the problem.

- When a boat goes upstream (against the flow of water), its speed becomes slower: \(u = b - s\).

- When a boat goes downstream (with the flow of water), its speed becomes faster: \(d = b + s\).

Here, \(b\) is the speed of the boat in still water, and \(s\) is the speed of the stream.

Step 2: Use the first condition (7 hours).

The boat goes 30 km upstream and 28 km downstream in 7 hours. Time = Distance ÷ Speed, so:

\(\dfrac{30}{u} + \dfrac{28}{d} = 7\) … (Equation 1)

Step 3: Use the second condition (5 hours).

The boat goes 21 km upstream and 21 km downstream in 5 hours. So:

\(\dfrac{21}{u} + \dfrac{21}{d} = 5\)

Simplify: \(21 \left( \dfrac{1}{u} + \dfrac{1}{d} \right) = 5\)

So: \(\dfrac{1}{u} + \dfrac{1}{d} = \dfrac{5}{21}\) … (Equation 2)

Step 4: Make substitution.

Let \(A = \dfrac{1}{u}\) and \(B = \dfrac{1}{d}\).

From Equation 1: \(30A + 28B = 7\).

From Equation 2: \(A + B = \dfrac{5}{21}\).

Step 5: Solve the equations.

Multiply Equation 2 by 28: \(28A + 28B = \dfrac{140}{21} = \dfrac{20}{3}\).

Now subtract this from Equation 1: \((30A + 28B) - (28A + 28B) = 7 - \dfrac{20}{3}\).

\(2A = \dfrac{21 - 20}{3} = \dfrac{1}{3}\).

So, \(A = \dfrac{1}{6}\). That means \(u = 6\).

From Equation 2: \(\dfrac{1}{6} + B = \dfrac{5}{21}\).

So, \(B = \dfrac{5}{21} - \dfrac{1}{6} = \dfrac{1}{14}\). That means \(d = 14\).

Step 6: Find boat and stream speed.

\(b = \dfrac{u + d}{2} = \dfrac{6 + 14}{2} = 10\).

\(s = \dfrac{d - u}{2} = \dfrac{14 - 6}{2} = 4\).

Final Answer: Boat in still water = 10 km/h, Stream = 4 km/h.

Step 1: Represent the digits

Let the tens digit be \(a\) and the units digit be \(b\).

So the number can be written as \(10a + b\).

Step 2: Form the first equation

The question says: "The number equals 8 times the sum of its digits minus 5".

Sum of digits = \(a + b\).

So: \(10a + b = 8(a + b) - 5\).

Simplify:

\(10a + b = 8a + 8b - 5\)

Bring all terms to one side:

\(10a - 8a + b - 8b = -5\)

\(2a - 7b = -5\)   → Equation (1)

Step 3: Form the second equation

The question also says: "The number equals 16 times the difference of its digits plus 3".

Difference of digits = \(a - b\).

So: \(10a + b = 16(a - b) + 3\).

Simplify:

\(10a + b = 16a - 16b + 3\)

Bring all terms to one side:

\(10a - 16a + b + 16b = 3\)

\(-6a + 17b = 3\)   → Equation (2)

Step 4: Solve the two equations

Equation (1): \(2a - 7b = -5\)

Equation (2): \(-6a + 17b = 3\)

Multiply Equation (1) by 3 to eliminate \(a\):

\(6a - 21b = -15\)

Add this to Equation (2):

\((-6a + 17b) + (6a - 21b) = 3 + (-15)\)

\(-4b = -12\)

So, \(b = 3\).

Step 5: Find \(a\)

Put \(b = 3\) into Equation (1):

\(2a - 7(3) = -5\)

\(2a - 21 = -5\)

\(2a = 16\)

\(a = 8\).

Step 6: Write the number

The tens digit is 8, and the units digit is 3.

So the number is 83.

Step 1: Let the full fare (without reservation charge) be \(F\). Let the reservation charge be \(R\).

Step 2: For a full reserved ticket:

Full fare + reservation charge = \(F + R = 2530\).

Step 3: For a half reserved ticket:

The fare will be half of full fare = \(\tfrac{F}{2}\).

Reservation charge is the same = \(R\).

So, one half reserved ticket = \(\tfrac{F}{2} + R\).

Step 4: For one full reserved + one half reserved ticket together:

\((F + R) + (\tfrac{F}{2} + R) = 3810\).

Simplify: \(F + R + \tfrac{F}{2} + R = 3810\).

So, \(\tfrac{3F}{2} + 2R = 3810\).

Step 5: From Step 2 we know: \(F + R = 2530\). So, \(F = 2530 - R\).

Step 6: Substitute \(F = 2530 - R\) in the equation \(\tfrac{3F}{2} + 2R = 3810\):

\(\tfrac{3}{2}(2530 - R) + 2R = 3810\).

\(3795 - 1.5R + 2R = 3810\).

\(3795 + 0.5R = 3810\).

Step 7: Subtract 3795 from both sides:

\(0.5R = 15\).

So, \(R = 30\).

Step 8: Put \(R = 30\) in \(F + R = 2530\):

\(F + 30 = 2530\).

So, \(F = 2500\).

Final Answer: Full fare = Rs 2500, Reservation charge = Rs 30.

Step 1: Assume variables

Let the cost price of the saree = \(S\).

Let the list price of the sweater = \(L\).

Step 2: Write equation for first case

The saree is sold at 8% profit. So selling price of saree = \(S + 8\%\text{ of }S = 1.08S\).

The sweater is sold at 10% discount. So selling price of sweater = \(L - 10\%\text{ of }L = 0.90L\).

Total money = Rs 1008.

Equation (1): \(1.08S + 0.90L = 1008\).

Step 3: Write equation for second case

Saree is sold at 10% profit. So selling price of saree = \(1.10S\).

Sweater is sold at 8% discount. So selling price of sweater = \(0.92L\).

Total money = Rs 1028.

Equation (2): \(1.10S + 0.92L = 1028\).

Step 4: Subtract the two equations

Equation (2) − Equation (1):

\((1.10S - 1.08S) + (0.92L - 0.90L) = 1028 - 1008\)

\(0.02S + 0.02L = 20\).

Divide both sides by 0.02: \(S + L = 1000\).

Step 5: Put this value in Equation (1)

From step 4, \(L = 1000 - S\).

Substitute into Equation (1):

\(1.08S + 0.90(1000 - S) = 1008\).

Simplify: \(1.08S + 900 - 0.90S = 1008\).

\(0.18S + 900 = 1008\).

\(0.18S = 108\).

\(S = 600\).

Step 6: Find \(L\)

\(L = 1000 - S = 1000 - 600 = 400\).

Final Answer: Saree cost price = Rs 600, Sweater list price = Rs 400.

Step 1: Assume the amounts.

Let the amount invested in Scheme A = \(x\).

Let the amount invested in Scheme B = \(y\).

Step 2: Write the first condition.

Interest from A = \(8\% \times x = 0.08x\).

Interest from B = \(9\% \times y = 0.09y\).

Total interest = 1860.

So, \(0.08x + 0.09y = 1860\). … (1)

Step 3: Write the second condition.

If the amounts were interchanged:

Interest from A = \(9\% \times x = 0.09x\).

Interest from B = \(8\% \times y = 0.08y\).

Total interest now = 1880 (20 more).

So, \(0.09x + 0.08y = 1880\). … (2)

Step 4: Simplify the equations.

Multiply both equations by 100 to remove decimals:

(1) becomes \(8x + 9y = 186000\).

(2) becomes \(9x + 8y = 188000\).

Step 5: Subtract the equations.

Subtract (1) from (2):

\((9x - 8x) + (8y - 9y) = 188000 - 186000\).

\(x - y = 2000\). … (3)

Step 6: Solve for one variable.

From (3): \(x = y + 2000\).

Step 7: Substitute back.

Put \(x = y + 2000\) in equation (1):

\(8(y + 2000) + 9y = 186000\).

\(8y + 16000 + 9y = 186000\).

\(17y + 16000 = 186000\).

\(17y = 170000\).

\(y = 10000\).

Step 8: Find \(x\).

\(x = y + 2000 = 10000 + 2000 = 12000\).

Final Answer:

Scheme A = Rs 12,000 and Scheme B = Rs 10,000.

Step 1: Assume numbers of bananas.

Let the number of bananas in Lot A = a.

Let the number of bananas in Lot B = b.

Step 2: Write cost for Lot A and B at first selling rates.

• Lot A: Rs 2 for 3 bananas → price of 1 banana = Rs \(\tfrac{2}{3}\).
• So cost of a bananas = \(\tfrac{2}{3}a\).
• Lot B: Re 1 each → cost of b bananas = \(1 \times b = b\).

Total cost = 400 → equation (1):

\(\tfrac{2}{3}a + b = 400\)

Step 3: Write cost at changed rates.

• Lot A: Re 1 each → cost = \(a\).
• Lot B: Rs 4 for 5 bananas → price of 1 banana = Rs \(\tfrac{4}{5}\).
• So cost of b bananas = \(\tfrac{4}{5}b\).

Total cost = 460 → equation (2):

\(a + \tfrac{4}{5}b = 460\)

Step 4: Remove fractions for easier solving.

• Multiply equation (1) by 3: \(2a + 3b = 1200\).
• Multiply equation (2) by 5: \(5a + 4b = 2300\).

Step 5: Solve the two equations.

We have:

\(2a + 3b = 1200\) … (i)

\(5a + 4b = 2300\) … (ii)

Multiply (i) by 5: \(10a + 15b = 6000\).

Multiply (ii) by 2: \(10a + 8b = 4600\).

Subtract: \((10a + 15b) - (10a + 8b) = 6000 - 4600\).

So, \(7b = 1400 → b = 200\).

Put \(b = 200\) in (i):

\(2a + 3(200) = 1200 → 2a + 600 = 1200\).

So, \(2a = 600 → a = 300\).

Step 6: Find total bananas.

Total bananas = \(a + b = 300 + 200 = 500\).

Answer: 500 bananas (Lot A: 300, Lot B: 200).