NCERT Class 10 Maths Exercise 3.2 Solution – Chapter 3

Solution:

1. $$\text{Two friends have total }14\text{ coins; one has }4\text{ more than other.}$$

   Think of real life: 2 friends have 14 coins together, and one has 4 more than the other. How many does each one have? This is the real-life version of this question.

2. $$x+y=14\cdots (1),x-y=4\cdots (2)$$

   We have 2 equations. Equation 1 is x + y = 14 and equation 2 is x – y = 4.

3. $$\text{From }(1):x=14-y$$

   In the substitution method, we first write one variable in terms of the other. From equation 1 we get x = 14 – y.

4. $$\text{Substitute in }(2):(14-y)-y=4$$

   Now put this value of x in equation 2. This gives 14 – y – y = 4.

5. $$14-2y=4⟹2y=10⟹y=5$$

   Simplify: 14 – 2y = 4, so 2y = 10, which means y = 5.

6. $$\text{Substitute }y=5\text{ in }x=14-y:x=14-5=9$$

   Now put the value of y back in the equation: x = 14 – 5, so x = 9.

7. $$\text{Verification: }(9+5=14\mathrm{✓}),(9-5=4\mathrm{✓})$$

   Always check! 9 + 5 = 14 is correct, and 9 – 5 = 4 is also correct. Both equations are satisfied.

8. $$\therefore x=9,y=5$$

   So the final answer is x = 9 and y = 5. This is what you should write in the board exam.

Solution:

1. $$s-t=3\cdots (1),{\displaystyle \frac{s}{3}}+{\displaystyle \frac{t}{2}}=6\cdots (2)$$

   We have 2 equations. The first is s – t = 3 and the second is s/3 + t/2 = 6.

2. $$\text{From }(1):s=t+3\cdots (3)$$

   In the substitution method, we first write one variable in terms of the other. From equation 1 we get s = t + 3.

3. $${\displaystyle \frac{(t+3)}{3}}+{\displaystyle \frac{t}{2}}=6$$

   Now put s = t + 3, which is equation 3, in equation 2.

4. $${\displaystyle \frac{2(t+3)+3t}{6}}=6⟹2t+6+3t=36$$

   Multiply both sides by the LCM, 6. This gives 2(t + 3) + 3t = 36.

5. $$5t=36-6=30⟹t=6$$

   Combine like terms: 5t = 30, so t = 6.

6. $$s=t+3=6+3=9$$

   Now put the value of t in equation 3: s = 6 + 3, so s = 9.

7. $$\text{Verify: }s-t=9-6=3\mathrm{✓},{\displaystyle \frac{9}{3}}+{\displaystyle \frac{6}{2}}=3+3=6\mathrm{✓}$$

   Always check in the board exam. Putting the values in both equations satisfies both, so the answer is correct.

8. $$s=9,t=6$$

   So our final answer is s = 9 and t = 6.

Solution:

1. $$3x-y=3\cdots (1)9x-3y=9\cdots (2)$$

   We have 2 equations. Equation 1 is 3x – y = 3 and equation 2 is 9x – 3y = 9.

2. $$\text{Divide equation (2) by }3:3x-y=3\cdots (2^{\mathrm{\prime }})$$

   Now divide equation 2 by 3. This gives 3x – y = 3, which is exactly the same as equation 1!

3. $$\text{Equation (1)}\equiv \text{Equation (2’)}\Rightarrow \text{Both equations are identical}$$

   Both equations are identical. It is the same line written in 2 different ways.

4. $$\text{From (1): }y=3x-3$$

   Let us try the substitution method. From equation 1, write y as y = 3x – 3.

5. $$\text{Substitute in (2): }9x-3(3x-3)=9\Rightarrow 9x-9x+9=9\Rightarrow 9=9$$

   Now put this value in equation 2. All the x terms cancel and we get 9 = 9, which is always true.

6. $$9=9\Rightarrow \text{True for ALL values of }x$$

   9 = 9 is always true. This means any value of x works, so there are infinite solutions.

7. $$\text{Geometrically: Both lines are COINCIDENT (same line)}$$

   On a graph, both equations show the same line. That is why there are infinitely many solutions.

8. $$\text{Infinitely Many Solutions: }y=3x-3,x\in \mathbb{R}$$

   Final answer: this pair of linear equations has infinitely many solutions, where y = 3x – 3 and x can be any real number.

Solution:

1. $$\begin{gathered} 0.2x+0.3y=1.3\cdots (1) \\ 0.4x+0.5y=2.3\cdots (2) \end{gathered}$$

   We have 2 equations. They have decimals, so let us clear those first.

2. $$\begin{gathered} \text{Multiply (1) by 10: }2x+3y=13\cdots (3) \\ \text{Multiply (2) by 10: }4x+5y=23\cdots (4) \end{gathered}$$

   Multiply both equations by 10, and all the decimals go away. Now the work is easier.

3. $$\text{From (3): }2x=13-3y⟹x=\frac{13-3y}{2}$$

   In the substitution method, we write one variable in terms of the other. From equation 3, find the value of x.

4. $$\text{Substitute in (4): }4\cdot \frac{13-3y}{2}+5y=23$$

   Now put this value of x in equation 4.

5. $$\begin{gathered} 2(13-3y)+5y=23 \\ 26-6y+5y=23 \\ -y=23-26=-3 \\ y=3 \end{gathered}$$

   Open the bracket and combine like terms. -y = -3, so y = 3.

6. $$x=\frac{13-3(3)}{2}=\frac{13-9}{2}=\frac{4}{2}=2$$

   Put y = 3 back in the expression for x. We get x = 2.

7. $$\begin{gathered} \text{Check in (1): }0.2(2)+0.3(3)=0.4+0.9=1.3\mathrm{✓} \\ \text{Check in (2): }0.4(2)+0.5(3)=0.8+1.5=2.3\mathrm{✓} \end{gathered}$$

   Always check in the board exam. Putting x = 2 and y = 3 in both original equations makes both sides equal, so the answer is correct.

8. $$\boxed{{\displaystyle x=2,y=3}}$$

   So the final answer is x = 2 and y = 3. Write it in a box in the board exam.

Solution:

1. $$\sqrt{2}x+\sqrt{3}y=0\cdots (1)$$

   The first equation is √2x + √3y = 0. We will call it equation 1.

2. $$\sqrt{3}x-\sqrt{8}y=0\cdots (2)$$

   The second equation is √3x – √8y = 0. We will call it equation 2.

3. $$\text{From (1): }x=\frac{-\sqrt{3}y}{\sqrt{2}}$$

   In the substitution method, we first write one variable in terms of the other. From equation 1, we get x = -(√3/√2)y.

4. $$\sqrt{3}\left(\frac{-\sqrt{3}y}{\sqrt{2}}\right)-\sqrt{8}y=0$$

   Now put this value of x in equation 2.

5. $$\frac{-3y}{\sqrt{2}}-\sqrt{8}y=0⟹\frac{-3y}{\sqrt{2}}-2\sqrt{2}y=0$$

   √3 × √3 = 3, and √8 can be written as 2√2. So the equation becomes -3y/√2 – 2√2y = 0.

6. $$\frac{-3y-2\sqrt{2}\cdot \sqrt{2}y}{\sqrt{2}}=0⟹\frac{-3y-4y}{\sqrt{2}}=0$$

   Now multiply by √2 to simplify the numerator. Since 2√2 × √2 = 4, we get (-3y – 4y)/√2 = 0.

7. $$-7y=0⟹y=0$$

   The numerator gives -7y = 0, so y = 0.

8. $$x=\frac{-\sqrt{3}\times 0}{\sqrt{2}}=0⟹x=0$$

   Put y = 0 back in the expression for x, and we also get x = 0. So the final answer is x = 0 and y = 0.

Solution:

1. $$\frac{3x}{2}-\frac{5y}{3}=-2\text{…(1)},\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\text{…(2)}$$

   We have 2 equations. First simplify them to remove the fractions.

2. $$\text{Eq (1)}\times 6:9x-10y=-12\text{…(3)}$$

   Multiply equation 1 by 6, the LCM of 2 and 3. This gives 9x – 10y = -12.

3. $$\text{Eq (2)}\times 6:2x+3y=13\text{…(4)}$$

   Multiply equation 2 by 6 too. This gives 2x + 3y = 13.

4. $$\text{From (4): }x=\frac{13-3y}{2}$$

   In the substitution method, find x from equation 4: x = (13 – 3y)/2.

5. $$9\cdot \frac{13-3y}{2}-10y=-12⟹\frac{117-27y}{2}-10y=-12$$

   Put this value in equation 3: 9(13 – 3y)/2 – 10y = -12.

6. $$117-27y-20y=-24⟹-47y=-141⟹y=3$$

   Multiply both sides by 2: 117 – 47y = -24. So -47y = -141, which means y = 3.

7. $$x=\frac{13-3(3)}{2}=\frac{13-9}{2}=\frac{4}{2}=2$$

   Put y = 3 back: x = (13 – 9)/2 = 2.

8. $$\therefore x=2,y=3$$

   So our final answer is x = 2 and y = 3. Write it clearly in a box in the board exam.

Solution:

1. $$\text{Eq 1: }2x+3y=11\text{Eq 2: }2x-4y=-24$$

   We have 2 equations. Both have 2x, so we will use subtraction.

2. $$(2x+3y)-(2x-4y)=11-(-24)$$

   Subtract equation 2 from equation 1. 2x – 2x cancels.

3. $$7y=35⟹y=5$$

   On the left, 3y – (-4y) = 7y, and on the right, 11 + 24 = 35. So y = 5.

4. $$2x+3(5)=11⟹2x=11-15=-4⟹x=-2$$

   Now put y = 5 in equation 1. 2x + 15 = 11, so 2x = -4, which means x = -2.

5. $$\text{Verify: }2(-2)-4(5)=-4-20=-24\mathrm{✓}$$

   Always check in the board exam. Put x = -2 and y = 5 in equation 2. The answer is correct.

6. $$y=mx+3⟹5=m(-2)+3$$

   Now put x = -2 and y = 5 in y = mx + 3.

7. $$5-3=-2m⟹2=-2m⟹m=-1$$

   5 – 3 = 2, and 2 = -2m, so m = -1.

Solution:

1. $$\text{Let the two numbers be }x\text{ and }y,\text{ where }x>y$$

   First, let the two numbers be x and y, where x is the bigger number.

2. $$x-y=26\cdots (1)x=3y\cdots (2)$$

   The question gives 2 equations. First, the difference of the numbers is 26. Second, one number is 3 times the other.

3. $$\text{Substitute }x=3y\text{ in equation (1):}3y-y=26$$

   Now use the substitution method. Put x = 3y from equation (2) in equation (1).

4. $$2y=26⟹y=\frac{26}{2}=13$$

   2y = 26. Dividing both sides by 2 gives y = 13.

5. $$x=3y=3\times 13=39$$

   Now put the value of y in equation (2): x = 3 × 13, so x = 39.

6. $$\text{Verification: }x-y=39-13=26\mathrm{✓}x=3\times 13=39\mathrm{✓}$$

   Always write the check in the board exam. The difference is 26 and 39 is 3 times 13, so both conditions are satisfied.

7. $$\therefore x=39,y=13$$

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   So the two numbers are 39 and 13. This is our final answer.

Solution:

1. $$\text{e.g., a straight road bent at a point}\Rightarrow x+y={180}^{\circ }$$

   In real life, supplementary angles form when a straight line bends at a point, like a turn on a straight road. The angles on both sides always add up to 180 degrees.

2. $$\text{Let larger angle}=x,\text{smaller angle}=y$$

   First, choose the variables. Let the bigger angle be x and the smaller angle be y.

3. $$\text{Equation 1: }x+y=180(\text{supplementary angles})$$

   Supplementary angles add up to 180 degrees, so the first equation is x + y = 180.

4. $$\text{Equation 2: }x-y=18(\text{larger exceeds smaller by 18})$$

   The second condition says the bigger angle is 18 degrees more than the smaller one, so x – y = 18.

5. $$\text{From Eq 1: }x=180-y$$

   Now use the substitution method. From equation 1, x = 180 – y.

6. $$(180-y)-y=18\Rightarrow 180-2y=18$$

   Put this value in equation 2. 180 – y – y = 18, which means 180 – 2y = 18.

7. $$2y=162\Rightarrow y=81$$

   2y = 162. Dividing both sides by 2 gives y = 81 degrees.

8. $$x=180-81=99$$

   Now find x. x = 180 – 81, so x = 99 degrees.

9. $$\therefore x={99}^{\circ },y={81}^{\circ }$$

   So our final answer is: the bigger angle is 99 degrees and the smaller angle is 81 degrees.

Solution:

1. $$\text{Shop bill: }7\times (\text{bat price})+6\times (\text{ball price})=3800$$

   This question is based on real-life shopping. At a sports shop, different quantities give different bills, and we have to find the price of each item.

2. $$\text{Let cost of 1 bat}=x\text{ (Rs.)},\text{cost of 1 ball}=y\text{ (Rs.)}$$

   First, choose the variables. Let the cost of one bat be Rs. x and the cost of one ball be Rs. y.

3. $$\begin{gathered} 7x+6y=3800\cdots (1) \\ 3x+5y=1750\cdots (2) \end{gathered}$$

   The two conditions give us 2 equations. The first is 7x + 6y = 3800 and the second is 3x + 5y = 1750.

4. $$\text{From (1): }7x=3800-6y⟹x=\frac{3800-6y}{7}$$

   In the substitution method, find x from equation 1: x = (3800 – 6y)/7.

5. $$3\times \frac{3800-6y}{7}+5y=1750$$

   Now put this value of x in equation 2: 3(3800 – 6y)/7 + 5y = 1750.

6. $$\frac{11400-18y}{7}+5y=1750⟹11400-18y+35y=12250⟹17y=850⟹y=50$$

   Multiply both sides by 7 to get 11400 – 18y + 35y = 12250. So 17y = 850, which means y = 50.

7. $$x=\frac{3800-6\times 50}{7}=\frac{3800-300}{7}=\frac{3500}{7}=500$$

   Put y = 50 in equation 1: x = (3800 – 300)/7 = 3500/7 = 500.

8. $$\text{Verification: }3(500)+5(50)=1500+250=1750\mathrm{✓}$$

   Always write the check in the board exam: 3 × 500 + 5 × 50 = 1750, which satisfies equation 2. It is correct!

9. $$\text{Cost of 1 bat}=\text{Rs. }500,\text{Cost of 1 ball}=\text{Rs. }50$$

   So the final answer is: one bat costs Rs. 500 and one ball costs Rs. 50.

Solution:

1. $$\text{Taxi fare}=\underset{\text{fixed}}{\underbrace{x}}+\underset{\text{distance charge}}{\underbrace{y\times d}}$$

   This question is directly linked to real life. When you book an Ola or Uber, there is a fixed base fare and the rest depends on distance, just like this formula.

2. $$\text{Let fixed charge}=\text{₹}x,\text{charge per km}=\text{₹}y$$

   First, choose the variables. Let the fixed charge be Rs. x and the charge per km be Rs. y.

3. $$x+10y=105\cdots (1)$$

   The total charge for 10 km is Rs. 105, so the first equation is x + 10y = 105.

4. $$x+15y=155\cdots (2)$$

   The total charge for 15 km is Rs. 155, so the second equation is x + 15y = 155.

5. $$\text{From (1):}x=105-10y$$

   In the substitution method, find x from equation 1: x = 105 – 10y.

6. $$(105-10y)+15y=155⟹105+5y=155⟹y=10$$

   Now put this value in equation 2: 105 + 5y = 155, so 5y = 50, which means y = 10.

7. $$x=105-10(10)=105-100=5$$

   Put the value of y back: x = 105 – 100, so x = 5.

8. $$\text{For 25 km:}5+25\times 10=5+250=\text{₹}255$$

   Find the charge for 25 km: 5 + 25 × 10 = Rs. 255.

9. $$x=\text{₹}5,y=\text{₹}10\text{per km},\text{25 km charge}=\text{₹}255$$

   So the final answer is: the fixed charge is Rs. 5, the charge per km is Rs. 10, and for 25 km you pay Rs. 255.

Solution:

1. $$\text{e.g. Scoring fraction in a game: }\frac{\text{goals scored}}{\text{total shots}}$$

   Fractions are very common in real life, like goals scored out of total shots in a game. If the same number is added to both, the fraction changes, just like here!

2. $$\text{Let numerator}=x,\text{denominator}=y,\text{fraction}={\displaystyle \frac{x}{y}}$$

   First, let the numerator be x and the denominator be y, so the fraction is x/y.

3. $$\frac{x+2}{y+2}=\frac{9}{11}⟹11(x+2)=9(y+2)⟹11x-9y=-4\cdots (1)$$

   The first condition says that adding 2 to both the numerator and the denominator gives 9/11. Cross multiply to get equation 1: 11x – 9y = -4.

4. $$\frac{x+3}{y+3}=\frac{5}{6}⟹6(x+3)=5(y+3)⟹6x-5y=-3\cdots (2)$$

   The second condition says that adding 3 to both gives 5/6. Cross multiply to get equation 2: 6x – 5y = -3.

5. $$\text{From (1): }11x-9y=-4⟹x=\frac{9y-4}{11}$$

   Now use the substitution method. From equation 1, x = (9y – 4)/11.

6. $$6\left(\frac{9y-4}{11}\right)-5y=-3⟹54y-24-55y=-33⟹-y=-9⟹y=9$$

   Put this value of x in equation 2, multiply both sides by 11 and simplify. This gives y = 9.

7. $$x=\frac{9(9)-4}{11}=\frac{81-4}{11}=\frac{77}{11}=7$$

   Now put y = 9 back in the expression for x. This gives x = 77/11, so x = 7.

8. $$\text{Verification: }\frac{7+2}{9+2}=\frac{9}{11}\mathrm{✓}\frac{7+3}{9+3}=\frac{10}{12}=\frac{5}{6}\mathrm{✓}$$

   Always check: both conditions are satisfied, so the answer is correct.

9. $$\therefore \text{The fraction}=\frac{x}{y}=\frac{7}{9}$$

   So the final answer is: the fraction is 7/9.

Solution:

1. $$\text{Age Puzzle: Father-Son age gaps}\Rightarrow \text{Linear Equations!}$$

   In real life, whenever we compare someone’s age in the past or future, it becomes a linear equation. This idea is also used in big exams like banking and SSC.

2. $$\text{Let Jacob’s present age}=x\text{ yrs, Son’s}=y\text{ yrs}$$

   First, choose the variables. Let Jacob’s present age be x and his son’s present age be y.

3. $$(x+5)=3(y+5)\Rightarrow x-3y=10\cdots (1)$$

   Five years hence means add 5 to both ages. Jacob’s age will be 3 times his son’s age, so x + 5 = 3(y + 5). This simplifies to x – 3y = 10.

4. $$(x-5)=7(y-5)\Rightarrow x-7y=-30\cdots (2)$$

   Five years ago means subtract 5 from both ages. Jacob’s age was 7 times his son’s age, so x – 5 = 7(y – 5). This simplifies to x – 7y = -30.

5. $$\text{From (1): }x=10+3y$$

   Now use the substitution method. From equation 1, x = 10 + 3y.

6. $$(10+3y)-7y=-30\Rightarrow -4y=-40\Rightarrow y=10$$

   Put this value in equation 2: 10 + 3y – 7y = -30, so -4y = -40, which means y = 10.

7. $$x=10+3(10)=10+30=40$$

   Put the value of y back: x = 10 + 3 × 10, so x = 40.

8. $$\text{Jacob’s present age}=40\text{ yrs},\text{Son’s present age}=10\text{ yrs}$$

   So Jacob’s present age is 40 years and his son’s present age is 10 years. This is our final answer!