Problems Leading to Simple Equations

The chief difficulty in solving an algebraic problem lies in expressing correctly the condition of the problem by means of symbols.

To solve a problem on linear equations, the following steps should be adopted:

Find out from the problem what is given and what is unknown.
Represent the unknown quantity by $x$ . Other alphabets may also be used, but $x$ being used in most cases is a general convention.
According to the conditions given in the problem, write the relation between the known and the unknown. This will give rise to a simple equation.
Solve the equation to obtain the value of the unknown $x$ .

Example: If an insect creeps up a pole at the rate of $x$ centimetres per minute, how many metres will it rise in $y$ hours?

Solution:

$\because$ 1 centimetre $=\dfrac{1}{100}$ th of a metre,

$\therefore$ $x$ centimetres=\dfrac{x}{100} $th of a metre.

Hence, in 1 minute the insect creeps up $\dfrac{x}{100}$ th metres.

$\therefore$ in 60 minutes the insect creeps up $\dfrac{x \times 60}{100}$ metres.

Thus, in 1 hour, the insect creeps up $\dfrac{3x}{5}$ metres.

$\therefore$ in $y$ hours it rises $(\dfrac{3x}{5} \times y)$ metres.

Thus, the required number of metres $=\dfrac{3xy}{5}$

Example: If a man earns $x$ rupees per month, how many twenty-five paise coins will he earn in half a month?

Solution:

$\because$ 1 rupee=4 twenty-five paice coins,

$\therefore$ $x$ rupees $=4x$ twenty-five paise coins.

Clearly, the man earns $4x$ twenty-five rupee coins in a month.

$\therefore$ the number of twenty-five paise coins he earns in half a month $=\dfrac{1}{2} \: of \: 4x=2x$ .

Example: The digits of a two-digited number beginning from the left are $latex $ and $y$ . How would you represent the number?

Solution:

Say, the digits are 4 and 5.

Then the number $=4 \times 10+5 =45$

Hence, here, the number is $10x+y$

Alternate method:

Starting from the left the digits of the two-digited number are $x$ and $y$ .

$\therefore$ $x$ is in the tens’ place.

$\therefore$ place value of $x = 10 \times x=10x$

$y$ is in the units’ place.

$\therefore$ place value of $y =1 \times y =y$

$\therefore$ the number = place value of $x$ + place value of $y =10x+y$

The number can be represented as $10x+y$

Example: Find three consecutive integers such that the smallest integer plus one-fifth of the next minus half of the third integer gives 9.

Solution:

Let the smallest integer be x.

$\therefore$ the required three integers are $x, x+1, x+2$ .

Given: $x+ \dfrac{x+1}{5} - \dfrac{x+2}{2} = 9$

$or, \dfrac{10x +2(x+1) -5(x+2)}{10} = 9$

$or, 10x + 2x +2 -5x -10 = 90$

$or, 7x = 90+8 \: or, \: x= \dfrac{98}{7} = 14$

$\therefore$ the required consecutive integers are as follows:

$x, x+1, x+2$ , i.e., $14, 14+1, 14+2$ or, 14, 15 and 16.

Note:

Consecutive natural numbers , integers and whole numbers are taken as: $x, x+1, x+2, x+3,$ and so on.
Consecutive even integers, even natural numbers and even whole numbers differ by 2 and so are taken as: $x, x+2, x+4, x+6,$ and so on.
Similarly, consecutive odd integers, odd natural numbers and odd whole numbers also differ by 2 and so are taken as: $x, x+2, x+3, x+5,$ and so on.
Ideally, odd numbers are represented by $2x+1$ . This is because, for all values of $x$ , the value of $2x+1$ is odd.
Similarly, ideally, even numbers are represented by $2x$ . This is because, for all values of $x$ , the value of $2x$ is even

Example: A’s age is six times B’s age. 15 years hence, A will be three times as old as B; find their ages.

Solution:

Let B’s age $= x$ years.

$\therefore$ A’s age $=6x$ years

15 years hence:

A’s age will be $6x+15$ years.

B’s age will be $x+15$ years.

According to the given problem,

$6x+15 = 3(x+15)$

or, $x = 10$

$\therefore$ A’s age $=6x \: years =(6 \times 10) \: years = 60 \: years$ .

B’s age $= x \: years = 10 \: years$ .

Exercise:

Solve:
1. The difference of two numbers is 25. If $x$ be the smaller number, what is the greater number?
2. If a man travels $x$ hours at the rate of $y$ Kilometres per hour, how many Kilometers does he travel?
3. If a man travels at the rate of $y$ Kilometres per hour, in what time will he finish a journey of $x$ Kilometres?
4. Write down the sum of 3 consecutive numbers of which the middle one is $x$ .
5. If the digits of a two-digited number beginning from right are $x$ and $y$ , what is the number?
A man covers a distance of 25km in an hour, partly on foot at the rate of 4 Km/h and partly on motorcycle at 32 Km/h. Find the distance travelled on the motorcycle.
In 12 years, a man will be three times as old as his son; the difference of their present ages is 30 years. Find their present ages.
In a shooting competition a marksman receives 50 paise if he hits the mark and pays 20 paise if he misses it. He tried 60 shots and was paid Rs 1.30 .how many times did he hit the mark?
A worker in a factory is paid Rs 2 per hour for normal work and double the rate for overtime work. Write down an expression for his one week’s earnings in which he worked for 40 hours out of which $x$ hours was overtime. Also, find the number of hours of his normal work if he receives Rs 116 in all.
A man walks from his house to his daughter’s school at a speed of 3 Km/hr and returns at a speed of 4 Km/hr. If he takes 21 minutes for the double journey; find the distance between his house and the school.
A house and a garden cost Rs 850 and the price of the garden is $\dfrac{5}{12}$ th of the price of the house. Find the price of each.
Divide 21 into two parts such that ten times one of them may exceed nine times the other by 1.
A bag contains as many rupees in it as there are fifty paise coins. Find the number of fifty paise coins in the bag if there be Rs 30 in all.
The age of two men differ by 10 years. 15 years ago the older man was just twice as old as the younger man. Find the ages of the men.

Exercise:

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