Special equations constitute an important part of several MBA aptitude entrances including CAT, XAT, SNAP, IIFT, NMAT and others. In a typical special equations question the number of variables is more that the number of equations provided. Let us have a detailed look at special equations now.
Special Equations: More number of variables than number of equations
If we abide by the law of mathematics we need to have same number of equations as the number of variables to be able to solve any set of equations. So strictly speaking we should not be able to find the proper solution in general case. But in case of special equations the variables have some possible set of values (generally natural numbers or whole numbers) and thereby restricting the number of possible solutions.
We’ll discuss problems related to this in this article.
Two variables and one equation
Let us start with an equation 2x + 3y = 12.
Now if I ask you to find out the solution then obviously your answer would be not possible to solve. So let us introduce a constraint that x and y are natural numbers.
Now the set of possible values are given below,
Now let us discuss a case where we will get more than one solution possible.
Take the equation 5x + 7y = 197. Now here the values of (x, y) can be (38, 1); (31, 6) and so on. Now since we have more than one possible solution so this is not possible to solve. But since x and y are natural numbers so the number of solutions will be a finite number.
Now the question that arises here is many such solutions will be there and what will they be.
To understand this let us bring back the first equation we considered.
Now here the value of y increases by 1 and the value of x decreases by 3/2 in subsequent cases.
From this we can say that if the value of y increases by 2 then the value of x decreases by 3. Now here 2 and 3 are the respective co-efficient of x and y.
And we will build on this particular pattern and find out the pattern for the equation 5x + 7y = 197.
Find out the maximum or minimum value of any of the variables. We can easily see that the least value of y is 1 and the respective value of x is 38.
So first possible solution is (38, 1)
Decrease the value of x by the co-efficient of y. Here the co-efficient of y is 7 so next possible value of x is 38 – 7 = 31. Now for value of y increases (current value is 1) by the co-efficient of x (which is 5) so we get 1 + 5 = 6. So the next solution is (31, 6).
We need to keep on doing this till we reach the minimum permissible value of x.
So we have (31 – 7, 6 + 5) which is (24, 11); then we have (17, 16); (10, 21) and (3, 26).
So here we have 6 possible solutions.
Least possible value of x is 3 and highest is 38. So the answer is 38 – 3/7 + 1 = 6.
Here 7 is the co-efficient of y.
One equation and three variables
Let us start with a problem to understand the approach to solve this kind of question.
In how many ways can you pay 17 Rs. using only 1 rupee, 2 rupee and 5 rupee coins such that you need not to pay coin of each denomination?
Let us assume the number of 1 rupee, 2 rupee and 5 rupee coins are A, B and C respectively.
So we can say A + 2B + 5C = 17
Now, C can be 0, 1, 2 or 3.
When C is 0 then we have A + 2B = 17
Now 2B can only be an even number (including 0) so we can say 2B can be 0, 2 …,16 or we have 9 possibilities here.
Now if C is 1 then A + 2B = 12 so 2B can be 0, 2 …,12 or we have 7 possible ways here.
Similarly if C is 2 then A + 2B = 7 and 2B can be 0, 2, 4 or 6 so we have 4 possibilities here.
Now if C is 3 then A + 2B = 2 so 2B can be 0 or 2 so we have 2 possibilities here.
So here total possible ways = 9 + 7 + 4 + 2 = 22
Now again one question that should arise here – what should be the approach if the amount to be paid is a larger number?
I am assuming the amount to be 117 instead of 17.
Then the equation we have is A + 2B + 5C = 117.
Now here also values of C can be 0, 1, 2 …23.
Let us take first C = 0, so A + 2B = 115 so B can have 58 possible values.
Then for C = 1, A + 2B = 110 so B has 56 possible values.
For C = 2 we have A + 2B = 105 and we have 53 possible values and,
for C = 3 we will have A + 2B = 100 and there are 51 possible solutions.
So if we notice the series goes like 58, 56, 53, 51 and then we will have 48, 46, 43, 41 and so on. The possible number of solution decrease by 2 and 3 alternatively.
Find out first solution which is 58 then you find the second one which is 56, next the third one which is 53 and at last fourth one is 51 so you have two arithmetic progressions.
58 + 53 + 48 + 43 + … 3 (taking the 1st, 3rd, 5th and so on) = 61 x 12/2 = 366
56 + 51 + 46 + 41 + … 1 (taking 2nd, 4th, 6th and so on) = 57 x 12/2 = 342
Total possible ways = 366 + 342 = 708
The article has been written by Subhankar Dhar sir, the expert faculty at Career Anna, with the track record of 99+ percentiles in CAT and other MBA Exams.
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