# CLASS-9SIMULTANEOUS LINEAR EQUATIONS - PROBLEMS ON MONEY

Problems on Money

Example.1)  The monthly income of A & B are in the ratio 2 : 5, and their monthly expenditures are in the ratio 1 : 3. If A saves \$ 4000 and B saves \$ 3000 per month, find the monthly income of each.

Ans.) As per the given condition, the monthly income of A & B are in the ratio 2 : 5, their monthly expenditures are in the ratio 1 : 3.

So, let the monthly income of A & B be \$ 2x and \$ 5x respectively and let their monthly expenditures be \$ y and \$ 3y respectively. Then,

Monthly saving of A = \$ (2x – y)

Monthly savings of B = \$ (5x – 3y)

As per the given condition –

(2x – y) = 4000  ………………(i)

(5x – 3y) = 3000  ………………(ii)

Multiply (i) by 5, and we get –

10x – 5y = 20000 ………………(iii)

Multiply (ii) by 2, and we get –

10x – 6y = 6000 ……………..(iv)

Now, we will subtract (iii) from (iv), and we get –

10x – 6y =  6000

10x – 5y = 20000

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- y = - 14000

Or,                    y = 14000

Now, we will substitute y in the equation (i)

(2x – y) = 4000

=>     2x – 14000 = 4000

=>     2x =  18000

=>      x  =  9000

So, A’s income =>  2x =  2 X 9000 =  \$ 18000

And, B’s income =>  5x = 5 X 9000 =  \$ 45000          (Ans.)

Example.2) Taxi charges in a city consist of fixed charges and the remaining depending upon the distance traveled in kilometers. If a person travels 70 km, he pays \$ 1130 and for traveling 100 km, he pays \$ 1550. Find the fixed charges and the rate per km.

Ans.)  Let the fixed charges be \$ x and the other charges by \$ y per km.

Then, x + 70y = 1130  ………………(i)

And,  x + 100y = 1550  ……………….(ii)

Now, we will subtract (i) from (ii), we get -

x + 100y = 1550

x +  70y = 1130

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30y  =  420

=>           y  =  14

Now, we will substitute the value of ‘y’ in (i), and we find –

x + 70y = 1130

=>      x  +  (70 X 14) = 1130

=>      x  =  1130 – 980 = 150

So, fixed charges = \$ 150 and rate = \$ 14 per km      (Ans.)

Example.3) A part of monthly hostel charges in a school 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 23 days, he has to pay \$ 4730 as hostel charges, where as a student B, who takes food for 30 days, pays \$ 5850 as hostel charges.  Find the fixed charges and the cost of food per day.

Ans.) Let the fixed charges per month be \$ x and cost of food per day be \$ y. Then,

x + 23y = 4730 ……………..(i)

and,     x + 30y = 5850 ……………….(ii)

On subtracting (i) from (ii), we get –

x + 30y = 5850

x + 23y = 4730

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7y  =  1120

=>       y  =  160

Substituting y = 160 in (i), and we get –

x + 23y  =  4730

=>     x + (23 X 160)  =  4730

=>     x  =  4730 – 3680  = 1050

So, fixed charges per month = \$ 1050, and the cost of food per day    (Ans.)

Example.4) Each one of A and B, has some money. If A gives \$ 60 to B, then B will have twice the money left with A. but if B gives \$ 30 to A, then A will have thrice as much as is left with B. How much money does each have?

Ans.)  Let A and B have \$ x and \$ y respectively.

Case.1) When A gives 60 to B : -

Then money with A = \$ (x – 60)

And, money with B = \$ (y + 60)

So,   (y + 60) = 2(x – 60)

=>    2x – y = 180  …………….(i)

Case.2)  When B gives \$ 30 to A :-

Then money with A = \$ (x + 30)

And, money with B = \$ (y – 30)

So,   (x + 30) = 3(y – 30)

=>    x + 30 = 3y – 90

=>    x – 3y = - 120  ………………(ii)

Multiplying (ii) by 2, and we get –

2x – 6y = - 240 ………………(iii)

Now we will subtract (iii) from (i), we

2x – y = 180

2x – 6y = - 240

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5 y =  420

=>           y  = 84

Now, we will substitute the value of ‘y’ in (i), and we will get –

2x – y = 180

=>    2x – 84 = 180

=>         2x = 180 + 84  = 264

=>          x = 132

Hence, A has \$ 132 money and B has \$ 84 money.   (Ans.)

Example.5) 18 pen and 32 pencils together cost \$ 642, while 32 pens and 18 pencils together cost \$ 908. Find the cost of each pen and that of each pencil.

Ans.)  Let the cost of each pen be \$ x and that of each pencil be \$ y.

As per the given condition –

18x + 32y = 642 ………………..(i)

32x + 18y = 908 …………………(ii)

Adding (i) and (ii), and we get –

18x + 32y = 642

32x + 18y = 908

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50x + 50y = 1550

=>     50 (x + y)  =  1550

=>         x + y  =  31  ……………………..(iii)

Now, we will subtract (i) from (ii), and we find –

32x + 18y = 908

18x + 32y = 642

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14x – 14y  =  266

=>       14 (x – y)  =  266

=>             x – y =  19  ……………………(iv)

Now, we will add (iii) & (iv), then we will find –

x + y = 31

x – y = 19

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2x = 50

=>       x = 25

Subtracting the value of x in (iii), and we get –

x + y  =  31

=>      25 + y = 31

=>       y = 31 – 25 = 6

So, the cost of each pen is \$ 25 and cost of each pencil \$ 6     (Ans.)