Simple Equations  Nikhil Goyal  Learn Quest

Let us begin our discussion with a question.
What is an expression and what is an equation?
An expression is of the form  ‘3X + 4’ or ‘x^2+ 5x + 7’Here we are not interested in the dictionary meaning of an algebraic expression. These are forms as we
understand then and as you would recall it.Now, can we find a unique value of x in these cases? No, we can't.
In order to have a value, the expression will need to be assigned a value, which is done through an equation.
So, if we say
3x + 5 = 17 or
x^2 + 5x + 7 = 7
we can find the values of x in either cases.So an equation is the process of ‘equating’ or making the expression equal something so that we can final its value. Let us spend some time understanding the equation.
3x + 5 = 17
(1) Here ‘x’ could have taken various values depending upon what it was being equated to. Therefore it is called the ‘variable’.
(2) The term qualifying the variable is called the coefficient. Here coefficient of x = 3
(3) The highest power of a variable in an expression is called the DEGREE of the expression. If the Degree is 1, we call it a linear equation. Eg  2x +3 = 0  Here the highest power of variable is 1
If the highest power of variable is 2, it it is called a quadratic equation.
If its 3, it is called a cubic equation and so on.(4) The number of variables in an equation can be 1, 2 etc.
Together, the degree and the no. of variables are expressed as, ‘first degree one variable, ‘first degree – two variable’, ‘second degree – one variable’ equation etc.(5) In general, the no. of equations required to find a unique combination of values that will simultaneously satisfy them is equal to the no. of variables. This unique combination of values is called the ‘solution’.
Now, an equation or a group of equation may have no solution, an unique solution or even Infinite solutions.
Let us spend some time understanding how to solve linear equation. We will look at 3 types of equations in two unknown and three equations in 3 unknowns.
Case(a) One equation and one unknown
Ex: 3x + 7 = 31We can easily solve this equation
3x = 24
So X = 8Case(b) Two equations in two unknown
Ex: 2x + 3y = 12 (1)
3x + 4y = 17(2)#Method : Elimination method
I believe every one knows this method
First We make Coefficient of one variable equal and then cancel out it
And solve for the remaining variable
In above example
Suppose I am making coefficient of x equal
(1) X 3 => 6x + 9y = 36
(2) X 2 => 6x + 8y = 34
Now subtracting above equations
You will get Y = 2
Similarly we can find the value of x
I.e X = 3
Therefore with two independent simple equations in two variables we are able to get a unique value of the
variables.#Note: To find a unique value of the variables in an equation we need as many independent simple equations as the number of variables present.
#No solution
If we get the below two equations
3x + 4y = 8 (i)
6x + 8y = 15 (ii)
and if we multiply the first equation by 2, we have 6x + 8y = 16;
but the second equation says 6x + 8y = 15.
Naturally, both can not be simultaneously true and therefore, we have no solution.So we can generalise this thing
If a1x + b1y = c1 and a2x + b2y = c2 are two equations then if a1/ a2 = b1/b2 ≠ c1/c2 then we have no Solution#Infinite solution
If 3x + 4y = 8 (i)
For a single equation in two variables (for example above )
Then there would be a unique value of y corresponding to each value of x that satisfies the equation and we have infinite such combinations of x and y.Right ?
I.e one equation with 2 variables will have infinite number of Solutions
Also, Had we been given
3x + 4y = 8 (i)
and 6x + 8y = 16 (ii)
Basic both equations are same
Which would have again meant infinite solutions.Right ?
So we can generalise this thing
if a1x + b1y = c1
a2x + b2y = c2
Then if a1/a2 = b1/b2 = c1/c2 then we have infinite number of SolutionsClear ?
The cost of 3 chairs and 4 tables is 2500.
The cost of 4 chairs and 3 tables is 2400.
find the cost of 6 chairs and 6 tablesIf you represent the statements in the form of simple equations we get
3c + 4t = 2500
4c + 3t = 2400By Visual inspection we can say that by adding the equations and dividing by ‘7’
we get the combined cost of 1c + 1t => 4900/7 => 700
Now since we require cost of 6c + 6t, multiply it by 6
I.e 700 * 6 = 4200For the above problem some students would find the cost of a chair and a table independently and then will find the required answer. But in such cases there is no need to find cost of chair and table individually. So it’s highly important to read the problem carefully and then solve it accordingly, in a way which helps you to even save time.
#Tip: Speed of calculations is also very important along with conceptual knowledge, to gain an competitive edge in entrance exam. 🙂
Clear ?
At a certain place, there are some rabbits and some peacocks. The total number of heads is 78 and the total number of legs is 240. Find the number of peacocks and rabbits
Let the no. of peacocks be P.
Let the no. of rabbits be R.
From first statement,
P + R = 78  (1)
From second statement
2P + 4R = 240  (2)
Solving equations (1) and (2) we get
number of peacocks = 36
Number of rabbits = 42Alternate solution: If all the animals were peacocks then we would have total 156 legs. However total legs is (240 – 156) = 84 more which can be achieved by replacing 42 peacocks with rabbits (as each rabbit will contribute 2 extra legs). Hence number of rabbits is 42.
Clear ?
Fifteen years hence a man would be twice as old as he was fifteen years ago. Find the present age of the man.
Let the present age of the man be ‘x’.
Fifteen years ago his age was x – 15.#Note : Meaning of the word ‘Hence’ is after
Fifteen years hence his age would be x + 15
From the statement in the problem
x + 15 = 2(x – 15)
Solve the above equation
and we get the present age of the man
i.e. ‘x’ as 45 years.Alternate solution: If 15 years ago the man was x years old, after 15 years he is 2x years old. Thus x must
be equal to 30 years (15 years ago) as this is the time gap between the two ages. Hence presently he should be 45 years old.Clear ?
A tells B “ When I was 3/5th your present age you were 5/7th of my present age”. If the present age of A is 28 years, what is the present age of B?
A B
Present 28 X
Past (3/5)X (5/7)28Suppose the age of B is X
So when A was 3/5 of X
That time B was 5/7 of 28 =>20Now ,
28  (3/5)X = X  20
=> 48 = (8/5)X
=> X = 30 years
I.e B is 30 years old
Only the gap between present age and past age will be same
Take example
Suppose you are 17 years old now
And your brother is 5 years old
2 years before
You would be 15 years old and your brother was 3 years old
Means if you subtracted your present age and Past age ,you will get 2
Same for brotherSo in such cases of Age always we are sure about the gap
So we solve for gap onlyThe sum of the digits of a two digit number is 9. The fraction formed by taking 9 less than the number as numerator and 9 more than the number as denominator is 3/4 . Find the number.
Let the number be XY
So value of the number is 10X + Y
Now according to the given condition
(10X + Y  9)/ (10X + Y + 9) => 3/4
On solving
10x + Y => 63
Hence the number is 63
There is no need to solve for individual values of x and y.Note1: Now, alternate method will take more time.
Let the number be ‘xy’
From first statement x + y = 9
The various possibilities are 18, 27, 36, 45, 54, 63, 72, 81
Using second condition to sort out the answer you need to check out for all above options which takes more time.Note2: Hence even if you knows a shortcut method its mandatory that you should know the conventional method which is a better option in cases like this.
In a fraction, if 1 is added to both the numerator and the denominator the fraction becomes 3/4. If 1 is subtracted from both the numerator and the denominator, the fraction becomes 5/7. Find the fraction
Now let the original fraction be x/y
Now using first statement we get
(X+1)/(Y+1) = 3/4
4x + 4 = 3y + 3
=> 4x – 3y = –1  (1)
Using the second statement ,we get
(X1)/(Y1) = 5/7
=> 7x 7 = 5y  5
=> 7x  5y = 2 (2)
On solving you will get x= 11 and y = 15
I.e. Fraction = 11/15Approach2
You can solve by assuming fraction
Fraction 2/3 can't be possible
(Although it satisfies conditions first but not second )
Luckily if you directly move on to second fraction
And try to satisfy according to this you will get your answer
To satisfy second condition
Fraction has to be 6/8
But which is same as 3/4 so it won't satisfy first condition
Now 5/7 can be written as 10/14
Now to satisfy second condition
Taking 10/14 as 11/15
It will satisfy both the statements
So we got the answer
See you can try in this way but again we are not sure everyone we will get the answer.The sum of the digits of a two digit number is 9. If the digits are reversed then the number formed exceeds the original number by 27. Find the two digit number. (a question on these lines came in CAT 2006)
Let the two digit number be xy
Since the sum of digits is 9
=> x + y = 9 (1)the original number is xy
i.e.10x + y and when the digits are reversed it becomes 10y + x and it is said that second number exceeds first number by 27
i.e.
(10y + x) – (10x + y) = 27
=> 9y – 9x = 27
=> y – x = 3  (2)Solving equations (1) and (2) we get the value of
x and y as x = 3, y = 6
∴ The required two digit number is 36.Second method:
Let's see how this problem can be solved in less time,
Sum of digits is 9.
So the various possibilities are 18, 27, 36, 45, 54, 63, 72, 81
From the problem,
when the digits are reversed, the second number is greater than the first number by 27,
it is clear from the second number is greater then first one and hence the original number can be
18, 27, 36, 45.The original number and the reversed number can be
18 → 81;
27 → 72;
36 → 63;
45 → 54In the above we can find out the difference and
only in case of 36 → 63,
the difference is 27.
Hence the required two digit number is 36The difference between the digits of the two digit number is 4. The sum of the number and the number formed by reversing the digits is 110. Find the two digit number.
"Difference”
I.e Difference between the numerical value irrespective of sign between two numbers
ExDifference of 4 and 2 is 2
Difference of 2 and 4 is 2
Now going back to the problem let the two digit number be xy.
using first statement we get two equations.
x – y = 4  (1)
y – x = 4  (2)
Using second statement I get 10y + y + 10y + x = 110
=> 11x + 11y = 110
=> x + y = 10 → (3)
Now take equations (1) and (3) and solve
You will get value of
of the number is ‘73’.
Now take equations (2) and (3)
and solve to get value of the number as’37’.#Note: Hence this problem has two different answers.
So you should be careful when such questions come in data sufficiency or in LR (Logical reasoning )Alternative Approach : Let the number be ‘xy’. Difference of digits is 4
this leaving different options of ‘xy’ as 15, 26, 37,48, 59, 51, 62, 73, 84, 95.
Now using the second statement that sum of
number and after reversing the digits is 110
we get eliminate many options leaving two options  37, 73
So Answer is 37 and 73Concept:
Sum of a two digit number and the number formed by reversing the digits is divisible by 11
The difference between the number and the number formed by reversing the digits is divisible by 9.Carrying forward the same concept for 2/3/4 and higher digited numbers:
#Difference between the number and the number formed by reversing the digits:
2 digited: Divisible by 9
3 digited: Divisible by 99
4 digited: Divisible by 9
5 digited: Divisible by 99
I.e.
Even number of digits: Divisible by 9
Odd number of digits: Divisible by 99
Take an example and cross checkSimilarly
#Sum of the number and the number formed by reversing the digits:
2 digited: Divisible by 11
3 digited: Not divisible by any specific number
4 digited: Divisible by 11
5 digited: Not divisible by any specific number
even number of digits: Divisible by 11
odd number of digits: Not divisible by any specific number
Take example and cross checkThe unit's digit of a 3digit number is the square of the digit in hundreds place. Which of the following can be the difference between the number and the number formed by reversing the order of the digits ?
a. 200
b. 198
c. 196
d. 194We know the difference between a 3digit number and it's reverse will always be a multiple of 99.
From the given options, only option 2 satisfies this condition
Hardly 5 second questionWhich of the following cannot be the sum of a 6 digit number and the number formed by reversing the digits?
a) 1999998
b) 357753
c) 947749
d) 906706Option (d) is not divisible by 11 and hence cannot be sum of a 6 digit number and the number formed by reversing the digits. So you can see two type of Questions which was directly based on that concept.
A person starts with X rupees and Y paisa. He spent 3.50 and was left with 2Y rupees and 2X paisa. Then what is the amount he started with ?
Y  50 => 2X
So definitely
2X < 50
As maximum value of Y can be 99
Or we can say X < 25 (1)
Now
X  3 = 2Y
I.e 2Y < 25 (from (1) )
Right ?But we are subtracting 50 from Y
So it can't be possible
We have to take a borrow from X
And during borrow ,we will get 100
So 100 + Y  50 = 2X
Y  2X = 50 (1)Similarly
(X1)  3 = 2Y
=> X  2Y = 4 (2)
On solving (1) and (2)
You will get X = 32 and Y = 14
I.e Amount = 32.14A person starts with x rupees and y paisa. He spent 8.40 and was left with 2y rupees and 2x paisa. Then what is the amount he started with?
Y  40 => 2X
So definitely
2X < 100
As maximum value of Y can be 99
Or we can say X < 50
Now
X  8 = 2Y
I.e 2Y < 50
Y < 25
Right ?But we are subtracting 40 from Y
So it can't be possible
We have to take a borrow from X
And during borrow ,we will get 100
So 100 + Y  40 = 2X
Y  2X = 60 (1)Similarly
(X1)  8= 2Y
=> X  2Y = 9(2)
On solving (1) and (2)
You will get
X = 37 and Y = 14
I.e Amount = 37.14