Telangana SCERT Solution Class X (10) Maths Chapter 4 Pair of linear equations in two variables Exercise 4.2
Pair of linear equations in two variables
Exercise – 4.2
(Q1) Form a pair of linear equations for each of the following problems and find their solutions.
(1) The ratio of incomes of two persons is 9:7 and the ratio of their expenditures is 4:3. If each of them manages to save RS 2000 per month then find their monthly income.
=> Solution:
First we have to form equation form given condition.
Let the two persons are ‘A’ and ‘B’
∴ Ratio of their income is 9:7
∴ Income of A becomes ₹9x and
Income of B incomes ₹7x
∴ The ratio of their expenditures is 4:3
Let the expenditure of A = RS 4y and then expenditure of B = 3y
If their savings is RS 2000 per month.
∴ A’ savings of RS 2000 and ‘B’ also savings of RS 2000
[generally => income – expenditure = savings]
A => 9x – 4y = 2000 ——– (1)
B => 7x – 3y = 2000 ——– (2)
Multiplying equ∩ (1) by 3 on both sides, we get
3 X 9x – 3 X 4y = 3 X 2000
27x – 12y = 6000 —– (3)
Multiplying equ∩ (2 4 on both sides, we get
4 X 7x – 4 X 3y = 4 X 2000
28x – 12y = 80000 ——– (4)
Subtracting equ∩ (3) from (4) we get
28x – 12y = 8000
– 27x – 12y = -6000
x = 2000
∴ 9x = 9 X 2000 = 18000
7x = 7 X 2000 = 14000
∴ Person A monthly is RS 18000 and
Person B monthly income is RS 14000
(2) The sum of a twp digit number and the number obtained by reversing the digit is 66. If the digits of the number by 2, find the number. How many such numbers are here?
=> Solution:
Let, the digits in unit place be ‘x’ and the digit in take place by ‘y’. (x>y)
Then the number = 10x + x
By reversing the digits then the number = 10x + y
But the sum of have two numbers is 66
i.e. 10y + x + 10x + y = 66
=> 11y + 11x = 66
=> 11 (y + x) = 66
∴ y = 18
Substitute y = 18 in equation (1),
x + 12 X 18 = 256
x + 216 = 256
x = 256 – 216
∴ x = 40
(i) Charge for journey
Fixed charge = RS 40 for 3 kms
Charge per km = RS 18 per kms.
(ii) Person need to pay for travelling 25 kms
= 40 + (25 – 3) X 18
= RS 436
(5) A fraction will be equal to 4/5 if 1 is added to both numerator and denominator. If, however, 5 is substracted from both numerator and denominator the fraction will be equal to 1/2 what is fraction?
=> Solution:
Let the fraction be x/y
From first condition,
X+1/y+1 = 4/5
5 (x+1) = 4 (y+1)
5x + 5 = 4y + 4
5x – 4y = 4 – 5
5x – 4y = – 1 —— (1)
From second condition,
x-5/y-5 = 1/2
2 (x – 5) = 1(y – 5)
2x – 10 = y – 5
2x – y = – 5 + 10
2x – y = 5 —— (2)
Multiply equ∩ (2) by 4, we get
4 X 2x – 4xy = 4 X 5
8x – 4y = 20 —— (3)
x = 60
Substituting x = 60 in equation (2), we get
X + y = 100
60 + y = 100
Y = 100 – 60
Y = 40
∴ The speed of the first car = 60 kmph and the speed of the second car = 40 kmph.
(7) Two angles are complementary. The larger angel is 30 less than twice the measure of the smaller angel. Find the measure of each angel.
=> Solution:
Let the larger angel be x0 and the smaller angel be y0.
We know that, the sum of the two complementary angels = 900
x0 + y0 = 900 —– (1)
And the larger angel is 30 less than the twice the measure of the smaller angel.
i.e. x0 = 2y0 – 30
∴ x0 – 2y0 = 30 —– (2)
(10) You have RS 12,000 /- saved amount and wants to invest it in two schemes yielding 10% and 15% interest. How much amount should be invested in each schemes so that you should get overall 12% interest.
=> Solution:
Amount available for investment = 12000.
Let, x be the amount to be invested in scheme yielding 10% return.
Therefore, amount to be invested in scheme yielding 15% = (12000 – x)
Desired return = 12000 X 12%
= 12000 X 12/100
= 1440
So, that, 1440 = 10% X x + 15% X (12000 – x)
= 10/100 X x + 15/100 X (12000 – x)
= 1/10 x + 15/100 X 12000 -15/100 x
= 1/10 x + 1800 -15/100 x
= 1/10 x -15/100 x + 1800
= – 0.05x + 1800
1440 – 1800 = -0.05x
-360 = -0.05x
∴ x = -360/-0.05
X = 7200
∴ 12000 – x = 12000 – 7200
= 4800
Here is your solution of Telangana SCERT Class 10 Math Chapter 4 Pair of linear equations in two variables Exercise 4.2
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