Question 2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction? (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. (iv) Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Class 10 - Math - Pair of Linear Equations in Two Variables Page 57
Answers
Step 1:First multiply both equation by some suitable non zero constant to make the coefficient of one variable( either x or y) numerically equal.
Step 2:Then add or subtract one equation from the other so that one variable get eliminated . if you get an equation in one variable go to step 3.
Step 3:solve the equation in one variable x or y so obtained to get its value.
Step 4:Substitute this value of x(or y) in either of the original equation to get the value of the Other variable.
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Solution:
(i) Let the fraction be x/y
1st case:
A.T.Q⇒x + 1/y – 1 = 1
⇒x + 1 =y – 1
⇒ x – y = -2 ….............................................. (i)
2nd case:
A.T.Q
⇒x/y+1 = 1/2
⇒2 x= y+1
⇒ 2 x – y = 1 …......................................................... (ii)
Subtracting equation (i) from equation (ii), we get
2 x – y = 1
x – y = -2
- + +
------------------------
x = 3
⇒x = 3 …........................................................ (iii)
Putting this value in equation (i), we get
3 – y = -2
–y = -5
y = 5
Hence, the fraction is x/y = 3/5
(ii) Let present age of Nuri = x
&present age of Sonu = y
5 years ago,
Nuri's age= (x-5) yr
sonu's age= (y-5) yr
A.T.Q,
⇒(x – 5) = 3(y – 5)
⇒(x – 5) = 3y – 15
⇒ x – 3y = -15+5
⇒x-3y = -10…........................................................ (i)
After 10 years
Nuri's age= (x+10) yr
Sonu's age= (y+10) yr
A.T.Q
⇒(x + 10y) = 2(y + 10)
⇒x + 10y = 2y + 20
⇒x – 2y = 20 - 10
⇒x – 2y = 10 ........................................................ii
Subtracting equation (i) from equation (ii), we get
⇒(x – 2y) - (x-3y) = 10+10
⇒-2y+3y=20
⇒y = 20 …................................................................................ (iii)
Putting this value in equation (i), we get
x – 60 = -10
x = 50
Hence, age of Nuri = 50 years and age of Sonu = 20 years.
(iii) Let the unit digit and tens digits of the number be x and y
Then, number = 10y + x
Number after reversing the digits = 10x + y
A.T.Q
⇒x + y = 9 …....................................... (i)
⇒9(10y + x) = 2(10x + y)
⇒90y +9x = 20x +2y
⇒88y = 11x
⇒x= 88y/11= 8y
x = 8 y...................................................................ii
On substituting x= 8 y in eq. ii on eq i
⇒x+y=9
⇒8y+y=9
⇒9y = 9
⇒y = 1
Putting the value in equation (ii), we get
⇒x=8y = 8×1 = 8
⇒x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.
(iv) Let the number of Rs 50 notes and Rs 100 notes be x and y
A.T.Q,
x + y = 25 …........................................................................... (i)
50x + 100y = 2000 ….................................................. (ii)
Multiplying equation (i) by 50, we get
50x + 50y = 1250 … (iii)
Subtracting equation (iii) from equation (ii), we get
⇒(50x - 50 x ) - (100y -50y) = 2000 - 1250
⇒50y = 750
⇒y = 15
Putting this value in equation (i),
⇒x + y = 25
⇒x +15=25
⇒x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.
(v) Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y
A.T.Q
x + 4y = 27 … (i)
x + 2y = 21 … (ii)
Subtracting equation (ii) from equation (i), we get
⇒(x- x ) - (4y- 2y) = 27-21
⇒2y = 6
⇒y = 3 … (iii)
Putting in equation (i), we get
⇒x + 12 =27
⇒x = 15
Hence, fixed charge = Rs 15 and Charge per day = Rs 3.
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Answer:
Step 1:First multiply both equation by some suitable non zero constant to make the coefficient of one variable( either x or y) numerically equal.
Step 2:Then add or subtract one equation from the other so that one variable get eliminated . if you get an equation in one variable go to step 3.
Step 3:solve the equation in one variable x or y so obtained to get its value.
Step 4:Substitute this value of x(or y) in either of the original equation to get the value of the Other variable.
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Solution:
(i) Let the fraction be x/y
1st case:
A.T.Q⇒x + 1/y – 1 = 1
⇒x + 1 =y – 1
⇒ x – y = -2 ….............................................. (i)
2nd case:
A.T.Q
⇒x/y+1 = 1/2
⇒2 x= y+1
⇒ 2 x – y = 1 …......................................................... (ii)
Subtracting equation (i) from equation (ii), we get
2 x – y = 1
x – y = -2
- + +
------------------------
x = 3
⇒x = 3 …........................................................ (iii)
Putting this value in equation (i), we get
3 – y = -2
–y = -5
y = 5
Hence, the fraction is x/y = 3/5
(ii) Let present age of Nuri = x
&present age of Sonu = y
5 years ago,
Nuri's age= (x-5) yr
sonu's age= (y-5) yr
A.T.Q,
⇒(x – 5) = 3(y – 5)
⇒(x – 5) = 3y – 15
⇒ x – 3y = -15+5
⇒x-3y = -10…........................................................ (i)
After 10 years
Nuri's age= (x+10) yr
Sonu's age= (y+10) yr
A.T.Q
⇒(x + 10y) = 2(y + 10)
⇒x + 10y = 2y + 20
⇒x – 2y = 20 - 10
⇒x – 2y = 10 ........................................................ii
Subtracting equation (i) from equation (ii), we get
⇒(x – 2y) - (x-3y) = 10+10
⇒-2y+3y=20
⇒y = 20 …................................................................................ (iii)
Putting this value in equation (i), we get
x – 60 = -10
x = 50
Hence, age of Nuri = 50 years and age of Sonu = 20 years.
(iii) Let the unit digit and tens digits of the number be x and y
Then, number = 10y + x
Number after reversing the digits = 10x + y
A.T.Q
⇒x + y = 9 …....................................... (i)
⇒9(10y + x) = 2(10x + y)
⇒90y +9x = 20x +2y
⇒88y = 11x
⇒x= 88y/11= 8y
x = 8 y...................................................................ii
On substituting x= 8 y in eq. ii on eq i
⇒x+y=9
⇒8y+y=9
⇒9y = 9
⇒y = 1
Putting the value in equation (ii), we get
⇒x=8y = 8×1 = 8
⇒x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.
(iv) Let the number of Rs 50 notes and Rs 100 notes be x and y
A.T.Q,
x + y = 25 …........................................................................... (i)
50x + 100y = 2000 ….................................................. (ii)
Multiplying equation (i) by 50, we get
50x + 50y = 1250 … (iii)
Subtracting equation (iii) from equation (ii), we get
⇒(50x - 50 x ) - (100y -50y) = 2000 - 1250
⇒50y = 750
⇒y = 15
Putting this value in equation (i),
⇒x + y = 25
⇒x +15=25
⇒x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.
(v) Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y
A.T.Q
x + 4y = 27 … (i)
x + 2y = 21 … (ii)
Subtracting equation (ii) from equation (i), we get
⇒(x- x ) - (4y- 2y) = 27-21
⇒2y = 6
⇒y = 3 … (iii)
Putting in equation (i), we get
⇒x + 12 =27
⇒x = 15
Hence, fixed charge = Rs 15 and Charge per day = Rs 3.
Step-by-step explanation: