NCERT Class 10 Maths Exercise 3.3 Solution – Chapter 3

Solution:

1. $$\text{Equations: }x+y=5\cdots (1),2x-3y=4\cdots (2)$$

   We have 2 equations. Equation 1 is x + y = 5 and equation 2 is 2x – 3y = 4.

2. $$\mathbf{\text{Elimination}}\mathbf{\text{Method:}}(1)\times 2\Rightarrow 2x+2y=10\cdots (3)$$

   First use the elimination method. Multiply equation 1 by 2, so that both equations have 2x and x can be removed.

3. $$(3)-(2):(2x+2y)-(2x-3y)=10-4\Rightarrow 5y=6\Rightarrow y=\frac{6}{5}$$

   Now subtract equation 2 from equation 3. 2x – 2x cancels and we get 5y = 6, so y = 6/5.

4. $$y=\frac{6}{5}\text{ in }(1):x+\frac{6}{5}=5\Rightarrow x=5-\frac{6}{5}=\frac{25-6}{5}=\frac{19}{5}$$

   Put the value of y in equation 1. x = 5 – 6/5 = 19/5.

5. $$\mathbf{\text{Substitution}}\mathbf{\text{Method:}}\text{From }(1):x=5-y\cdots (4)$$

   Now use the substitution method. From equation 1, x = 5 – y.

6. $$\text{Put }(4)\text{ in }(2):2(5-y)-3y=4\Rightarrow 10-2y-3y=4\Rightarrow 5y=6\Rightarrow y=\frac{6}{5}$$

   Put this value of x in equation 2. 2(5 – y) – 3y = 4. Simplifying gives 5y = 6, so y = 6/5.

7. $$y=\frac{6}{5}\text{ in }(4):x=5-\frac{6}{5}=\frac{19}{5}\therefore x=\frac{19}{5},y=\frac{6}{5}$$

   Put the value of y in equation 4. We get x = 19/5. Both methods give the same answer, so our solution is correct.

Solution:

1. $$3x+4y=10\cdots (1),2x-2y=2\cdots (2)$$

   We have 2 equations. The first is 3x + 4y = 10 and the second is 2x – 2y = 2. We will call them equation 1 and equation 2.

2. $$\text{Elimination Method: Make coefficients of }y\text{ equal}$$

   First solve by the elimination method. We will remove y, so the coefficients of y must be made equal in both equations.

3. $$(1)\times 1\Rightarrow 3x+4y=10\cdots (3),(2)\times 2\Rightarrow 4x-4y=4\cdots (4)$$

   Multiply equation 1 by 1, so it stays the same. Multiply equation 2 by 2 to get 4x – 4y = 4.

4. $$(3)+(4):7x=14\Rightarrow x=2$$

   Now add equations 3 and 4. y cancels and we get 7x = 14, so x = 2.

5. $$x=2\text{ in }(1):3(2)+4y=10\Rightarrow 4y=4\Rightarrow y=1$$

   Put x = 2 in equation 1. 6 + 4y = 10, so 4y = 4, which means y = 1.

6. $$\text{Substitution Method: From }(2):2x-2y=2\Rightarrow x-y=1\Rightarrow x=1+y$$

   Now the substitution method. Divide equation 2 by 2 to get x – y = 1, which means x = 1 + y.

7. $$\text{Sub in }(1):3(1+y)+4y=10\Rightarrow 3+7y=10\Rightarrow y=1,x=2$$

   Put x = 1 + y in equation 1. 3 + 3y + 4y = 10, so 7y = 7, which means y = 1 and x = 2. Both methods give the same answer!

8. $$x=2,y=1$$

   So the final answer is x = 2 and y = 1. Do not forget to clearly write both methods in the board exam.

Solution:

1. $$\text{Eq 1: }3x-5y-4=0\Rightarrow 3x-5y=4$$

   Write the first equation in standard form to get 3x – 5y = 4.

2. $$\text{Eq 2: }9x=2y+7\Rightarrow 9x-2y=7$$

   Rearrange the second equation too, to get 9x – 2y = 7.

3. $$\begin{array}{r}9x-15y=12\\ -(9x-2y=7)\\ -13y=5\end{array}$$

   For the elimination method, multiply equation 1 by 3 so that the coefficients of x are equal, then subtract equation 2. 9x – 9x cancels, -15y – (-2y) = -13y, and 12 – 7 = 5.

4. $$y=\frac{5}{-13}=\frac{-5}{13}$$

   So y = 5/(-13), which means y = -5/13.

5. $$3x-5\left(\frac{-5}{13}\right)=4\Rightarrow 3x+\frac{25}{13}=4\Rightarrow 3x=4-\frac{25}{13}=\frac{27}{13}$$

   Now put the value of y in equation 1: 3x + 25/13 = 4, so 3x = 4 – 25/13 = 27/13.

6. $$x=\frac{27}{13\times 3}=\frac{27}{39}=\frac{9}{13}$$

   Divide both sides by 3 to get x = 27/39, which simplifies to x = 9/13.

7. $$\text{Substitution: From Eq 1, }x=\frac{4+5y}{3}$$

   Now use the substitution method. From equation 1, write x in terms of y: x = (4 + 5y)/3.

8. $$9\left(\frac{4+5y}{3}\right)-2y=7\Rightarrow 3(4+5y)-2y=7\Rightarrow 12+15y-2y=7\Rightarrow 13y=-5\Rightarrow y=\frac{-5}{13}$$

   Put this in equation 2: 9 × (4 + 5y)/3, which is 3(4 + 5y) – 2y = 7. Solving gives 13y = -5, so y = -5/13 is confirmed, and x = 9/13 also comes out the same.

Solution:

1. $$\frac{x}{2}+\frac{2y}{3}=-1\cdots (1),x-\frac{y}{3}=3\cdots (2)$$

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

2. $$\text{Eq (1)}\times 6:3x+4y=-6\cdots (3),\text{Eq (2)}\times 3:3x-y=9\cdots (4)$$

   Multiply equation 1 by 6 to get 3x + 4y = -6. Multiply equation 2 by 3 to get 3x – y = 9.

3. $$\text{Eq (3)}-\text{Eq (4)}:(3x+4y)-(3x-y)=-6-9$$

   Now for the elimination method, subtract equation 4 from equation 3. 3x cancels.

4. $$5y=-15⟹y=-3$$

   We get 5y = -15, so y = -3.

5. $$\text{Put }y=-3\text{ in Eq (4)}:3x-(-3)=9⟹3x=6⟹x=2$$

   Put y = -3 in equation 4. 3x + 3 = 9, so 3x = 6, which means x = 2.

6. $$\mathbf{\text{Substitution}}\mathbf{\text{Method:}}\text{From Eq (4): }x=\frac{9+y}{3}$$

   Now solve by the substitution method too. Find the value of x from equation 4.

7. $$\text{Put in Eq (3): }3\cdot \frac{9+y}{3}+4y=-6⟹9+y+4y=-6⟹5y=-15⟹y=-3$$

   Put this value of x in equation 3. 9 + 5y = -6, so 5y = -15, which confirms y = -3.

8. $$x=2,y=-3$$

   Both methods give the same answer: x = 2 and y = -3. This is our final answer.

Solution:

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

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

2. $$\frac{x+1}{y-1}=1\Rightarrow x+1=y-1\Rightarrow x-y=-2\cdots (1)$$

   The first condition: add 1 to the numerator and subtract 1 from the denominator, and the fraction becomes 1. Cross multiply to get equation 1: x – y = -2.

3. $$\frac{x}{y+1}=\frac{1}{2}\Rightarrow 2x=y+1\Rightarrow 2x-y=1\cdots (2)$$

   The second condition: add 1 to the denominator only, and the fraction becomes 1/2. Cross multiply to get equation 2: 2x – y = 1.

4. $$\begin{array}{r}2x-y=1\cdots (2)\\ -(x-y=-2)\cdots (1)\\ x\phantom{-y}=3\end{array}$$

   Now use the elimination method. Subtract equation 1 from equation 2. y cancels and we get x = 3.

5. $$x=3\text{ put in }(1):3-y=-2\Rightarrow y=5$$

   Put x = 3 in equation 1: 3 – y = -2, so y = 5.

6. $$\text{Verification: }\frac{3+1}{5-1}=\frac{4}{4}=1\mathrm{✓}\frac{3}{5+1}=\frac{3}{6}=\frac{1}{2}\mathrm{✓}$$

   Check both conditions. In the first, 4/4 = 1, which is correct! In the second, 3/6 = 1/2, which is also correct.

7. $$\therefore \text{The fraction}={\displaystyle \frac{x}{y}}={\displaystyle \frac{3}{5}}$$

   So our final answer is: the fraction is 3/5.

Solution:

1. $$\text{Let Nuri’s age}=x,\text{Sonu’s age}=y$$

   First, choose the variables. Let Nuri’s present age be x and Sonu’s present age be y.

2. $$\text{5 years ago: }(x-5)=3(y-5)$$

   5 years ago, Nuri’s age was x – 5 and Sonu’s age was y – 5. At that time Nuri was three times as old as Sonu, so x – 5 = 3(y – 5).

3. $$x-5=3y-15⟹x-3y=-10\cdots (1)$$

   Simplify this. x – 5 = 3y – 15, and rearranging gives x – 3y = -10. This is our equation 1.

4. $$\text{10 years later: }(x+10)=2(y+10)$$

   Now think about 10 years later. Nuri’s age will be x + 10 and Sonu’s will be y + 10. At that time Nuri will be twice as old as Sonu, so x + 10 = 2(y + 10).

5. $$x+10=2y+20⟹x-2y=10\cdots (2)$$

   Simplify this. x + 10 = 2y + 20, and rearranging gives x – 2y = 10. This is our equation 2.

6. $$\text{Subtract (1) from (2): }(x-2y)-(x-3y)=10-(-10)$$

   Now use the elimination method. Subtract equation 1 from equation 2. x – x cancels.

7. $$y=20⟹x=2(20)+10=50$$

   After subtracting, we get y = 20. Now put the value of y in equation 2: x = 2 × 20 + 10 = 50.

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8. $$\therefore \text{Nuri’s age}=50\text{ years},\text{Sonu’s age}=20\text{ years}$$

   So the final answer is: Nuri’s present age is 50 years and Sonu’s present age is 20 years.

Solution:

1. $$\text{Lock code: tens}=x,\text{units}=y,\text{number}=10x+y$$

   Imagine your bag has a 2-digit lock code whose digits add up to 9. In the same way, in real life we make unknown digits into variables and solve equations.

2. $$\text{Let unit digit}=y,\text{tens digit}=x$$

   First, choose the variables. Let the digit in the tens place be x and the digit in the units place be y.

3. $$\text{Original number}=10x+y,\text{Reversed number}=10y+x$$

   So the original two-digit number is 10x + y, and the number with reversed digits is 10y + x.

4. $$x+y=9\cdots (1)$$

   The first condition says the sum of the digits is 9, so equation 1 is x + y = 9.

5. $$9(10x+y)=2(10y+x)⟹90x+9y=20y+2x⟹88x-11y=0⟹8x-y=0\cdots (2)$$

   The second condition says 9 times the original number equals 2 times the reversed number. Simplifying gives equation 2: 8x – y = 0.

6. $$\text{Add (1) and (2):}(x+y)+(8x-y)=9+0⟹9x=9⟹x=1$$

   Now use the elimination method. Add equation 1 and equation 2. y cancels and we get 9x = 9, which means x = 1.

7. $$\text{Substitute }x=1\text{ in (1):}1+y=9⟹y=8$$

   Put x = 1 in equation 1: 1 + y = 9, so y = 8.

8. $$\text{Check: }9\times 18=162=2\times 81\mathrm{✓}$$

   Let us check: 9 × 18 = 162, and 2 × 81 is also 162. Both conditions are satisfied, so it is correct.

9. $$\text{The number}=10x+y=10(1)+8=\boxed{{\displaystyle 18}}$$

   So our final answer is: the two-digit number is 18.

Solution:

1. $$\text{Bank cashier problem}\Rightarrow \text{2 unknowns, 2 conditions}\Rightarrow \text{Linear pair}$$

   This is a real-life bank situation. Whenever the count of 2 different things and their total value are given, they always form a pair of linear equations.

2. $$\text{Let }x=\text{no. of ₹}50\text{ notes},y=\text{no. of ₹}100\text{ notes}$$

   First, choose the variables. Let x be the number of Rs. 50 notes and y be the number of Rs. 100 notes.

3. $$x+y=25\cdots (1)$$

   Meena got 25 notes in total, so the first equation is x + y = 25.

4. $$50x+100y=2000\cdots (2)$$

   The total amount is Rs. 2000, so the second equation is 50x + 100y = 2000.

5. $$\text{Simplify (2): }x+2y=40\cdots (3)$$

   Divide all of equation 2 by 50. It simplifies to x + 2y = 40. This makes the calculation easier.

6. $$\text{(3) }-\text{(1): }(x+2y)-(x+y)=40-25$$

   Now use the elimination method. Subtract equation 1 from equation 3.

7. $$y=15$$

   x – x cancels, and 2y – y = 40 – 25, so y = 15.

8. $$x+15=25\Rightarrow x=10$$

   Put the value of y in equation 1: x + 15 = 25, so x = 10.

9. $$\therefore \text{ ₹}50\text{ notes}=10,\text{₹}100\text{ notes}=15$$

   So the final answer is: Meena got 10 notes of Rs. 50 and 15 notes of Rs. 100.

Solution:

1. $$\text{OTT plans: base fee + per-extra-episode charge}$$

   This idea is very common in real life. An OTT platform has a fixed base subscription and a separate charge for extra content, exactly like this.

2. $$\text{Let fixed charge (3 days)}=x,\text{ extra/day}=y$$

   First, choose the variables. Let the fixed charge for the first 3 days be Rs. x, and the charge for each extra day after that be Rs. y.

3. $$\text{Saritha: }x+4y=27\cdots (1)$$

   Saritha kept the book for 7 days, which is 4 extra days after the 3 fixed days. So x + 4y = 27.

4. $$\text{Susy: }x+2y=21\cdots (2)$$

   Susy kept the book for 5 days, which is 2 extra days after the 3 fixed days. So x + 2y = 21.

5. $$\text{Subtract (2) from (1): }(x+4y)-(x+2y)=27-21$$

   Now use the elimination method. Subtract equation 2 from equation 1. x – x cancels.

6. $$2y=6⟹y=3$$

   The left side has 2y and the right side has 6, so y = Rs. 3 per extra day.

7. $$x+2(3)=21⟹x=21-6=15$$

   Put the value of y in equation 2: x + 6 = 21, so x = Rs. 15.

8. $$\text{Check: }15+4(3)=27\mathrm{✓},15+2(3)=21\mathrm{✓}$$

   Check by putting the values in both equations. Both are satisfied, so the answer is correct.

9. $$\text{Fixed charge}=\text{Rs. }15,\text{Extra charge/day}=\text{Rs. }3$$

   Final answer: the library’s fixed charge is Rs. 15 and the charge for each extra day is Rs. 3.