SELINA Solution Class 9 Simultaneous (Linear) Equation (Including problems ) Exercise 6F

Question 1

Five years ago, A's age was four times the age of B. Five years hence, A’s age will be twice the age of B. Find their preset ages.

Sol:

Let present age of A = x years
And present age of B = y years

According to the question,
Five years ago,
x - 5 = 4(y - 5)
x - 4y = -15                        ...(1)

Five years later,
x + 5 = 2(y + 5)
x + 5 = 2y + 10
x - 2y = 5                          ....(2)

Now subtracting (1) from (2)
       x - 2y = 5
   -  x - 4y = - 15  
       -    +     +      
            2y = 20
             y = 10

From (1)
x - 4 (10) = -15
x = 25
Present ages of A and B are 25 years and 10 years respectively.

Question 2

A is 20 years older than B. 5 years ago, A was 3 times as old as B. Find their present ages.

Sol:

Let A’s presentage be x years
and B’s present age be y years

According to the question
x = y + 20
x - y = 20                            ...(1)

Five years ago,
x - 5 = 3(y - 5)
x - 5 = 3y - 15
x  - 3y = - 10                      ...(2)

Subtracting (1) and (2),
      x - 3y = - 10
  -   x - y  = 20    
     -    +    -        
         - 2y = - 30
             y = 15

From (1)
x = 15 + 20
x = 35
Thus, present ages of A and B are 35 years and 15 years.

Question 3

Four years ago, a mother was four times as old as her daughter. Six years later, the mother will be two and a half times as old as her daughter at that time. Find the present ages of the mother and her daughter.

Sol:

Let the present age of the mother be x years.
and the present age of the daughter be y year.
4 yrs ago age = (x - 4)
mother's age = 4(y - 4) = 4y - 16
According to the equation,
x - 4 = 4( y - 4 )
⇒ x - 4 = 4y - 16
⇒  x - 4y = - 12                ...(1)
And,
6 yrs later daughter's age = x + 6
mother's age= ${\displaystyle 2\frac{1}{2}(y+6)}$
x + 6 = ${\displaystyle 2\frac{1}{2}}$( y + 6 )
⇒ x + 6 = ${\displaystyle \frac{5}{2}}$y + 15
⇒ x - ${\displaystyle \frac{5}{2}}$ y = 9              ...(2)

Solving (1) and (2), We get
y = 14 and x = 44
Hence, the present age of the mother is 44 years
and the present age of the daughter is 14 years.

Question 4

The age of a man is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children at that time. Find the present age of the man.

Sol:

Let the present age of the man be x years.
and let the sum of the ages of his two children be y years.
According to the question,
x = 2y                            ...(1)
and x + 20 = y + 40      ....(2) [ Since he has two children ]

Solving (1) and (2), We get,
2y + 20 = y + 40
⇒ y = 20
So, x = 2y ⇒ x = 40.
Hence, the present age of the man is 40 years.

Question 5

The annual incomes of A and B are in the ratio 3 : 4 and their annual expenditure are in the ratio 5 : 7. If each Rs. 5000; find their annual incomes.

Sol:

Let A’s annual in come = Rs.x
and B’s annual income = Rs. y

According to the question,
${\displaystyle \frac{x}{y}=\frac{3}{4}}$
4x - 3y = 0                   ...(1)
and,
${\displaystyle \frac{x-5000}{y-5000}=\frac{5}{7}}$
7x - 5y = 10000            ...(2)

Multiplying equation no. (1) by 7 and (2) by 4.
28x - 21y = 0                ....(3)
28x - 20y = 40,000       .....(4)

Subtracting equation (4) from (3)
     28x - 21y = 0
-   28x - 20y = 40,000 
    -      +         -           
            - y = - 40,000
              y = 40,000

From (1)
4x - 3(40,000) = 0
x = 30,000
Thus, A’s income in Rs. 30,000 and B’s income is Rs. 40,000.

Question 6

In an examination, the ratio of passes to failures was 4 : 1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. Find the number of students who appeared for the examination.

Sol:

Let the no. of pass candidates be x
and the no. of fail candidates be y.

According to the question,
${\displaystyle \frac{x}{y}=\frac{y}{1}}$
x - 4y = 0                                     ...(1)
and
${\displaystyle \frac{x-20}{y-10}=\frac{5}{1}}$
x - 5y = -30                                 ...(2)

Subtracting equation (2) from (1),
    x - 4y = 0
-   x - 5y = - 30  
    -    +       +     
          y = 30

From (1)
- 4(30) = 0
x = 120

Total students appeared = x +  y = 120 + 30 = 150.

Question 7

A and B both the have some pencils. If A gives 10 pencils to B, then B will have twice as many as A. And if B gives 10 pencils to A, then they will have the same number of pencils. How many pencils does each have ?

Sol:

Let the numberof pencils with A = x
and the number of pencils with B = y.

If A gives 10 pencils to B,
y + 10 = 2( x - 10 )
2x - y = 30                                    ...(1)

If B gives to pencils to A
y - 10 = x + 10 
x - y = - 20                                   ...(2)

Subtracting equation (1) and (2),
     x  -  y = - 20 
-   2x - y = 30    
     - x = - 50
        x = 50

From (1)
2(50) - y = 30
y = 70
Thus, A has 50 pencils and B has 70 pencils.

Question 8

1250 persons went to sea a circus-show. Each adult paid Rs. 75 and each child paid Rs. 25 for the admission ticket. Find the number of adults and number of children, if the total collection from them amounts to Rs. 61,250.

Sol:

Let the number of adults = x
and the number of children = y

According to the question,
x + y = 1250                                ...(1)
75x + 25y = 61250                      ...(2)

Subtracting equation (2) from (1)
       3x + y = 2450
-       x + y = 1250  
     -     -        -        
       2x = 1200
         x = 600

From (1)
600 + y = 1250
y = 650
Thus, number of adults = 600 and the number of children = 650.

Question 9

Two articles A and B are sold for Rs. 1,167 making 5% profit on A and 7% profiton A and 7% profit on B. IF the two articles are sold for Rs. 1,165, a profit of 7% is made on A and a profit of 5% is made on B. Find the cost prices of each article.

Sol:

Let the cost price of article A = Rs. x
and the cost price of articles B = Rs. y

According to the question,
( x + 5% of x ) +( y + 7% of y ) = 1167

${\displaystyle [x+\frac{5}{100}x]+[y+\frac{7}{100}y]=1167}$

${\displaystyle \frac{21x}{20}+\frac{107y}{100}=1167}$

105x + 107y = 116700                     ...(1)
and,
`[107x]/100 + [105y]/100 = 1165
107x + 105y = 116,500                ...(2)

Adding (1) and (2)
    105x + 107y = 116700
+  107x + 105y = 116500  
     212x + 212y = 233200
Dividing by 212,
x + y = 1100                               ...(3)

subtracting (2) from (1)
    105x + 107y = 116700 
-   107x + 105y = 116500  
     -        -            -             
    - 2x + 2y = 200
Dividing by 2,
- x + y = 100                              ...(4)

Adding equation (4) and (3)
       - x + y = 100
   +    x + y = 1100  
              2y = 1200
                y = 600

from (3)
x +600 = 1100
x = 500
Thus, cost price of article A is Rs. 500 and that of article B is Rs. 600.

Question 10

Pooja and Ritu can do a piece of work in ${\displaystyle 17\frac{1}{7}}$ days. If one day work of Pooja be three fourth of one day work of Ritu’ find in how many days each will do the work alone.

Sol:

Let Pooja’s 1 day work = ${\displaystyle \frac{1}{x}}$

and Ritu's 1 day work = ${\displaystyle \frac{1}{y}}$
According the question,
${\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{7}{120}}$                   ....(1)
and
${\displaystyle \frac{1}{x}=\frac{3}{4}.\frac{1}{y}}$

${\displaystyle y=\frac{3}{4}x}$                                ....(2)

Using the value of y from (2) in (1)
${\displaystyle \frac{1}{x}+\frac{4}{3x}=\frac{7}{120}}$

${\displaystyle \frac{1}{x}\left(\frac{7}{3}\right)=\frac{7}{120}}$
x = 40

From (2)
${\displaystyle y=\frac{3}{4}\left(40\right)}$
y = 30

Pooja will complete the work in 40 days and Ritu will complete the work in 30 days.