NCERT Solutions Maths Class 10 Exercise 3.4
1.
Solve the following pair of linear equations by the elimination method and the
substitution method:
(i) x + y = 5 and 2x – 3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x − 5y –
4 = 0 and 9x = 2y + 7 (IV)
Solution: (i) x + y = 5 … (1)
2x – 3y = 4 … (2)
Elimination
Method:
Multiplying equation (1) by 3, we get
3x + 3y = 15 … (3)
2x −
3y = 4 … (2)
Adding
equations (2) and (3), we get
5x = 19
⇒ x = 19/5
Putting
the value of x in equation (1), we get
19/5 + y = 5
⇒ y = 5 – 19/5 = 6/5
Therefore, x = 19/5 and y = 6/5.
Substitution Method:
x + y = 5 … (1)
2x −
3y = 4 … (2)
From
equation (1), we get
x = 5 − y
Putting
this value of x in equation (2), we
get
2(5 − y) − 3y = 4
⇒ 10 − 2y −
3y = 4
⇒ 5y = 6
⇒ y = 6/5
Putting
the value of y in equation (1), we
get
x = 5 – 6/5 = 19/5
Therefore, x = 19/5 and y = 6/5.
(ii) 3x
+ 4y = 10 … (1)
2x
– 2y = 2 … (2)
Elimination Method:
Multiplying equation (2) by 2, we get
4x −
4y = 4 … (3)
3x + 4y = 10 … (1)
Adding
equations (3) and (1), we get
7x = 14
⇒ x = 2
Putting
the value of x in equation (1), we get
3(2) + 4y = 10
⇒ 4y = 10 – 6 = 4
⇒ y = 1
Therefore, x = 2 and y = 1.
Substitution Method:
3x + 4y = 10 … (1)
2x −
2y = 2 … (2)
From
equation (2), we get
2x = 2 + 2y
⇒ x = 1 + y … (3)
Putting
this value of x in equation (1), we get
3(1 + y) + 4y = 10
⇒ 3 + 3y + 4y = 10
⇒ 7y = 7
⇒ y = 1
Putting the value of y in equation (3), we
get
x = 1 + 1 = 2
x = 2
Therefore, x = 2 and y = 1.
(iii) 3x − 5y – 4 = 0 … (1)
9x = 2y + 7… (2)
Elimination Method:
Multiplying equation (1) by 3, we get
9x −
15y – 12 = 0 … (3)
Equation (2) can be written as
9x −
2y – 7 = 0 … (2)
Subtracting
equation (2) from equation (3), we get
−13y – 5 = 0
⇒ −13y = 5
⇒ y = −5/13
Putting
the value of y in equation (1), we get
3x – 5(−5/13) − 4 = 0
⇒ 3x = 4 – 25/13 = (52 –
25)/13 = 27/13
⇒ x = 27/39 = 9/13
Therefore, x = 9/13 and y = −5/13.
Substitution Method:
3x − 5y – 4 = 0 … (1)
9x = 2y + 7… (2)
Equation (2) can be written as
9x – 2y = 7… (2)
From equation (1), we have
3x = 4 + 5y
⇒ x = (4 + 5y)/3
Putting
this value of x in equation (2), we get
9(4 + 5y)/3 −
2y = 7
⇒ 12 + 15y −
2y = 7
⇒ 13y = −5
⇒ y = −5/13
Putting
the value of y in equation (1), we get
3x – 5(−5/13) = 4
⇒ 3x = 4 – 25/13 = (52 – 25)/13
= 27/13
⇒ x = 27/39 = 9/13
Therefore, x = 9/13 and y = −5/13.
(iv)
Multiplying equation (2) by 2, we get
Adding equations (3) and (1), we get
5x/2
= 5
⇒ x = 2
Putting
the value of x in equation (2), we get
2
– y/3 = 3
⇒ y = −3
Therefore, x = 2 and y = −3.
Substitution Method:
From equation (2), we have
Putting this value of x in equation (1), we get
⇒ 5y + 9 = −6
⇒ 5y = −15
⇒ y = −3
Putting
the value of y in equation (1), we get
⇒ x = 2
Therefore, x = 2 and y = −3.
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 ½ 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 a 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.
Solution:
(i) Let
the numerator of the required fraction be x and the
denominator be y.
According
to the given conditions, we have
⇒ x + 1 = y – 1 and 2x = y + 1
⇒ x – y = −2
… (1)
and 2x – y = 1… (2)
x = 3
Putting
the value of x in equation (1), we get
3
– y = − 2
−y = −5
⇒ y = 5
Therefore, the
required fraction is 3/5.
(ii) Let the present age of Nuri be x years and the present age of Sonu be y years.
5 years ago, the age
of Nuri = (x – 5) years
5 years ago, the age
of Sonu = (y – 5) years
According to the first
condition, we have
(x − 5) = 3(y −
5)
⇒ x – 5 = 3y – 15
⇒ x −
3y = −10 …… (1)
10 years later, the age of Sonu = (y + 10) years
According
to the second condition, we have
(x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x −
2y = 10 …… (2)
Subtracting equation (1) from equation (2), we get
y = 10 − (−10) = 20 years
Putting the value of y in equation (1), we get
x – 3(20) = −10
⇒ x – 60 = −10
⇒ x = 50 years
Therefore,
the present age of Nuri is 50 years and the present age of Sonu is 20 years.
(iii) Let the digit at the tens place be x and the digit at the ones place be y.
Number = 10x
+ y
Number obtained by
reversing the order of the digits = 10y + x
According to the first condition, we have
x + y = 9 …… (1)
According to the second condition, we have
9 × Number = 2 × Number obtained by reversing the
order of the digits
9(10x + y) = 2(10y + x)
⇒ 90x + 9y = 20y + 2x
⇒ 88x = 11y
⇒ 8x = y
⇒ 8x – y = 0 …… (2)
Adding equations (1) and (2),
we get
9x = 9
⇒ x = 1
Putting the value of x in equation (1), we get
1 + y = 9
⇒ y = 9 – 1 = 8
Therefore, the number = 10x + y = 10(1) + 8 =
10 + 8 = 18
(iv) Let the number of Rs 100 notes be x and the number of Rs 50 notes be y.
According
to the given conditions, we have
x + y = 25 …… (1)
and 100x + 50y = 2000
⇒ 2x + y = 40 …… (2)
Subtracting equation (2) from
equation (1), we get
−x = −15
⇒ x = 15
Putting the value of x in
equation (1), we get
15 + y = 25
⇒ y = 25 – 15 = 10
Therefore,
the number of Rs 100 notes is 15 and the number of Rs 50 notes is 10.
(v) Let the fixed charge for 3 days be Rs x and the additional charge for each day thereafter be
Rs y.
According
to given condition, we have
x + 4y = 27 …… (1)
x + 2y = 21 …… (2)
Subtracting equation (2) from
equation (1), we get
2y = 6
⇒ y = 3
Putting the value of y in equation (1),
we get
x + 4(3) = 27
⇒ x = 27 – 12 = 15
Therefore, the fixed charge for 3 days
is Rs 15 and the additional charge for each day thereafter is Rs 3.