Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

To do:

We have to form the pair of linear equations and find their solutions by any algebraic method.

(i) Let the fixed charge and the daily charge for mess be $x$ and $y$ respectively.

When student A takes food for 20 days, he has to pay Rs. 1000 as hostel charges.

This implies,

$x + 20y = 1000$.....(i)

When student B takes food for 26 days, he has to pay Rs. 1180 as hostel charges.

This implies,

$x + 26y = 1180$.....(ii)

Subtracting equation (i) from equation (ii), we get,

$(x+26y)-(x+20y)=1180-1000$

$x-x+26y-20y=180$

$6y=180$

$y=\frac{180}{6}$

$y=30$

Substituting $y=30$ in equation (i), we get,

$x+20(30)=1000$

$x+600=1000$

$x=1000-600$

$x=400$

The fixed charge is Rs. 400 and the cost of food per day is Rs. 30.    

(ii) Let the numerator and denominator of the original fraction be $x$ and $y$ respectively.

The original fraction$=\frac{x}{y}$

The fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator. 

This implies,

New fraction$=\frac{x-1}{y}$

According to the question,

$\frac{x-1}{y}=\frac{1}{3}$

$3(x-1)=1(y)$    (On cross multiplication)

$3x-3=y$

$y=3x-3$.....(i)

When 8 is added to the denominator, it becomes $\frac{1}{4}$.

This implies,

$\frac{x}{y+8}=\frac{1}{4}$

$4(x)=1(y+8)$    (On cross multiplication)

$4x=y+8$

$4x-y-8=0$

$4x-(3x-3)-8=0$     (From (i))

$4x-3x+3-8=0$

$x-5=0$

$x=5$

$\Rightarrow y=3(5)-3$

$y=15-3$

$y=12$

Therefore, the original fraction is $\frac{5}{12}$.    

(iii) Let the number of questions answered correctly be $x$ and the number of questions answered incorrectly be $y$.

This implies,

 The total number of questions $=x+y$.

In the first case, 3 marks were awarded for each right answer and $-1$ for every wrong answer. 

According to the question,

$40=3x+(-1)y$

$y=3x-40$.....(i)

In the second case, 4 marks were awarded for each right answer and $-2$ for every wrong answer. 

According to the question,

$50=4x+(-2)y$

$2y=4x-50$

$2y=2(2x-25)$

$y=2x-25$.....(ii)

From (i) and (ii), we get,

$3x-40=2x-25$

$3x-2x=40-25$

$x=15$

This implies,

$y=2(15)-25$

$y=30-25$

$y=5$

$\Rightarrow x+y=15+5=20$

The total number of questions in the test is 20. 

(iv) We know that,

Distance$=$ Speed $\times$ Time.

Distance between the two places A and B $= 100\ km$.

Let the speed of the first car starting from A be $x\ km/hr$ and the speed of the second car starting from B be $y\ km/hr$.

Let the cars meet at point P when they are moving in the same direction and at point Q when they are moving in the opposite direction.

When they travel in the same direction, they meet in 5 hours.

Distance travelled by the first car in 5 hours $AP= 5\times x\ km=5x\ km$.

Distance travelled by the second car in 5 hours $BP= 5\times y\ km=5y\ km$.

$AP-BP=100$

$5x-5y=100$

$5(x-y)=5\times20$

$x-y=20$.....(i)

When they travel in the opposite direction, they meet in 1 hour.

Distance travelled by the first car in 1 hour $AQ= 1\times x\ km=x\ km$.

Distance travelled by the second car in 1 hour $BQ= 1\times y\ km=y\ km$.

$AQ+BQ=AB$

$x + y = 100$….(ii)

Adding equations (i) and (ii), we get,

$x-y+x+y=20+100$

$2x = 120$

$x = \frac{120}{2}$

$x=60$

Substituting $x=60$ in equation (ii), we get,

$60+y=100$

$y = 100-60$

$y = 40$

Therefore, the speed of the first car is $60\ km/hr$ and the speed of the second car is $40\ km/hr$.   

(v) Let the original length of the rectangle be $l$ and the breadth be $b$. 

Area of the original rectangle $=lb$.

In the first case, the length is reduced by 5 units and breadth is increased by 3 units, the area of the rectangle is reduced by 9 square units. 

New length $=l-5$

New breadth $=b+3$

The area formed by the new rectangle $=(l-5)(b+3)$ sq. units

According to the question,

$(l-5)(b+3)=lb-9$

$lb-5b+3l-15=lb-9$

$3l-5b=15-9$

$3l-5b=6$.....(i)

In the second case, the length is increased by 3 units and breadth is increased by 2 units, the area is increased by 67 sq. units.

New length $=l+3$

New breadth $=b+2$

The area formed by the new rectangle $=(l+3)(b+2)$ sq. units

According to the question,

$(l+3)(b+2)=lb+67$

$lb+2l+3b+6=lb+67$

$2l+3b=67-6$

$2l+3b=61$.....(ii)

Subtracting $2\times(i)$ from $3\times(ii)$, we get,

$3(2l+3b)-2(3l-5b)=3(61)-2(6)$

$6l-6l+9b+10b=183-12$

$19b=171$

$b=\frac{171}{19}$

$b=9$

$2l+3(9)=61$    (From (ii))

$2l=61-27$

$2l=34$

$l=\frac{34}{2}$

$l=17$

The dimensions of the rectangle are $17$ units (length) and $9$ units (breadth).   

Tutorialspoint

Updated on: 10-Oct-2022

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