*Wednesday, August 19th, 2020*

We have learned about simultaneous equations in two and three unknowns. When we have two independent equation in two unknowns or three independent equations in 3 unknowns, we can solve for the variables. These type of equations are called “deterministic equations”. The variables have a unique value in these equations. However, if we have only one equation in three unknowns, such equations are called “Indeterminate equations”. The variables here do not have unique values but take more than one value – in general, an infinite number of values.

If we impose certain other conditions on these variables, then such indeterminate equations also can yield unique values for the variables. We take such conditions also into account while solving such equations.

Consider 3x + y =10. This, being one equation in two unknowns, is indeterminate. Suppose we impose a condition that both x and y are positive integers. With this condition the possibilities are reduced to finite number:

- x=1, y=7;
- x=2, y=4;
- x=3, y=1

If we further impose the condition that x is greater than y, then there is a unique solution x=3, y=1. So, even though we have one equation, because of additional conditions, it may have finite or sometimes even a unique solution. The conditions that we have, could be explicitly mentioned as above or could be in built into the problem as we see in the following examples.

**Examples:**

**Example 1:** Amit purchased a certain number of pencils and a certain number of pens spending Rs. 10 on the whole. If each pencil costs him Rs.3 and each pen Rs. 1, then how many pencils and pens could he have bought?

**Sol:** Solving this problem is basically solving the equation 3x+y= 10 where x and y respectively denote the number of pencils and pens purchased and hence from the context we know that x and y should be both positive integers and hence this equation has exactly the three solutions as above.

Hence we see that when certain special conditions are imposed on the variables, the indeterminate equations also can yield a finite number of solutions and sometimes even a unique solution. The conditions that we normally come across are

- Minimum values of the variables
- Maximum values of the variables
- Variables being positive integers
- Limits on the difference in the values of variables, etc.

**Example 2:** Suppose we take a date and a month, multiply the date with 12 (which is the number of months in a year) and the month with 31 (which is the number of days in a month) and add up the two products. Suppose we are given that the sum is 555 and asked to find out the date and month.

**Sol: **If we denote the date as D and month as M, we have 12D +31M=555. Here we have only one equation with two unknowns. This is an indeterminate equation. However, we have the following additional information.

- D is a positive integer less than or equal to 31
- M is a positive integer less than or equal to 12

To solve this equation and in fact to solve any equation in two variables, we shall

– first, divide the entire equation by the least coefficient

– get all the fractional values on to one side, say left and all the whole values on to the other

– replace the whole of right-hand side by k, where k is an integer

The following is the sequence of steps:

12D + 31M =555

D + 31M/12 = 555/12

D + 2M + 7M/12 =46 + 3/12

(7M – 3)/12 = k

M = 12k+3/7

We now try to find the value of k. As M has to be positive integer the value of k =5, which gives M =9. Further other values of k are not feasible. Using this we now find D which is 23. Thus, the date is 23^{rd} September. Here we get a unique solution for the equation.

**Example 3:** Nakul bought two varieties of pens, the first variety costing Rs.12 each and the second variety costing Rs.17 each, spending Rs.157 in total. In how many different combinations he could have purchased the pens?

**Sol:** Let x be the number of pens of first variety while y is the number of second variety then 12x + 17y = 157

Proceeding as above,

x + y + 5y/12 = 13 + 1/12

k= (5y-1)/12

y= (12k+1)/5

As y is a positive integer the values of k can be k =2, 7, 12 ….

When k is 2, y is 5 and this is the only possible value of y [for, if we consider the next possible value of k i.e., k=7 then y is 17 which means the amount spent on second variety is 17 *17 which is much more than the total amount spent]

As there is a unique value possible for y, it means that he can buy these pens in exactly one way.

We can write this solution briefly by focusing only on the remainders 12x + 17y =157proceeding as discussed in the above example. Dividing the equation by 12 we have Rem(5y/12) =1, y can be 5, 17, 29 etc. i.e., values can be obtained by adding (or subtracting) 12 successively and from (A) when y=5 then the x value is 6 and the remaining values of x can be obtained by (subtracting or adding) 17 to 6 successively i.e., (x, y) could be (6,5), (-11, 17), (-28, 29) etc…

As x>0 and y>0 only (6,5) is the acceptable solution.

Now let’s do some advanced questions:

**Question 1 :** A student purchases gel pens, ball point pens and pencils by spending a total of Rs.28. Each gel pen, ball point pen and pencil cost Rs.15, Rs.5, and Rs.3 respectively. In how many ways can he purchase them if he buys at least one of each item?

**Sol:** Let the number of gel pens, ball point pens, and pencils the student purchased be x, y, and z respectively

Total amount spent is 15x + 5y +3z = 28

As x>0, therefore the value of x could be only 1 (making x=2 will make the total 30 however, the total amount spent is 28 which is less than 30)

Therefore, y=2 and z=1.

Hence only one combination is possible. Ans: (1)

**Question 2:** On a particular day, a salesman sold three types of toys. Each toy of the 3 varieties costs Rs.100, Rs.50, and Rs.25 respectively. If the total sale on that day was Rs.300 and the salesman sold at least one toy of each variety, find the maximum number of toys he could have sold.

**Sol: **Let the number of toys of each variety sold be x, y and z respectively.

The total cost is 100x + 50y + 25z = 300

He sold at least one of each variety

Therefore the amount received by selling one of each toy is

100+50+25=175

Amount left with him:300-175=125

For the number of toys sold to be the maximum he has to sell toys of third variety for the remaining amount 125,

For Rs.125 he can sell 125/25

= 5 toys of this variety.

Therefore the maximum number of toys he can sell:

=1+1+6 =8 toys

**Question 3:** Ram told Krishna “12 times the date of my birth added to 31 times the month of my birth is 376”. On which date way Ram born?

**Sol:** Let the day on which Ram was born to be D and month be M. Then we have 12D +31M =376

Divide the equation by the least coefficient 12, and collect all fractions on left and all integers on the right. Now denoting the combination of all integers on right by ‘k, we get,

7M/12 -4/12 = k

M =(12k+4)/7

By a few hit and trials, we observe for k=2, we are getting M=4. Now next possible of k=7 gives M =16 but M can’t be more than 12 as denotes month, therefore, M has to be 4.

Using this in 12D + 31M =376, we get D=21.

Hence Ram was born on 21^{st} April.

**Question 4:** Shefali purchased some pastries and cookies, each cookie costing Rs.3 and each pastry costing Rs.8. In how many different combinations can she buy the items if she spends a total amount of Rs.68?

**Sol: **Suppose Shefali purchased ‘P’ pastries and ‘C’cookies

Now 8P+3C=68

Dividing by the least coefficient and collecting all fraction on left and all integeters on right (and denoting the number of combinations of integers on right by k) we get

2P/3 -2/3=k

P = (3k-2)/2

K=0,2,4,6…

When k=0 P=1 C=20

when k=2 P=4 C=12

when k=4 P=7 C=4

when k=6 P=10

P=10 not possible as pastries should now cost her 80 while the total amount she has to spend is only 68.

Hence she can buy in 3 different combinations.

**Alternate solution:**

2P/3 -2/3 = k. Now multiplying the equation with 2 ( a number which makes the coefficient of P, 1 more than the denominator) we get P=3K +1. Using this in 8P + 3C=68, we get 3C =60-24K i.e., C=20-8k. Clearly K≤2 i.e, K =0,1,2 giving that there are 3 different ways in which Shefali can purchase her items.

Ans(3).

**Question 5:** The average weight of a certain group of ‘n’ men is 75kg. Three men whose weights are 80kg, 76kg, and 74 kg join the group and one man whose weight is between 90kg and 100kg leaves the group. The average of the group has now come down by 2kg. If the number of men initially is a perfect square, then the weight of the man who left would be?

**Sol:** The average weight of ‘n’ is 75kgs. The total weight of the group is 75n. Let x be the weight of the man who left the group.

We have 90<x<100

Now (75n+(80+76+74)- x)/(n+3-1)

=73

75n +230 –x =73n +146

i.e. 2n+84 =x

As 90< x< 100, we have 90< 2n +84 < 100

3< n< 8. As n is a perfect square, is has to be 4.

Hence, x=2*4 +84 =92 kg.

**Question 6:** The average marks of a group of n students in a subject is 65. Three students with marks 48, 66, and 56 leave the group and one student with marks between 55 & 65 joins the group. As a result, the average of the group goes up by 3. If the number of students in the group initially is an odd perfect square then the marks of the student who joins the group would be?

**Sol:** The average of the n students is 65.

Total marks of n students = 65n

When three students whose marks are 48, 66, and 56 left, then the total marks of the remaining (n-3) students should be:

65n-(48+66+56) = 65n-170

Let the marks of the student who joined the group be x, (55<x<65)

Therefore, the average of these (n-2) students is (65n- 170-x)/(n-2)

Now, this value is given 68.

i.e. (65n- 170-x)/(n-2)=68 ⇒ 3n = x-34

and n= (x-34)/3. As 55 <x< 65 the values of x (that give integral values of n) are 58, 61, and 64 and the corresponding values of n are 8, 9 and 10.

The initial value of nn is an odd perfect square.

Therefore n=9.

The marks of the person who joins the group is 3(9) =x-34

which gives x=61.

**Question 7:** Pradyuman who is interested in numismatics, came across some old coins consisting of 1 rupee, 8-annas and 4-anna coins totaling 36 coins and their total value being 19 rupees and 8-annas. If the number of 4-annas is at the most two more than the eight-anna coins and there are at least four 4-anna coins more than one-rupee coins then:

- How many 4-anna coins does Pradyumna have with him?
- IfPradyumna lost two of 4-anna coins then of which denomination he would have equal number of coins?

Given: (16 annas make a rupee)

**Sol:** We note that

16 annas = 1 rupee

8 annas = ½ rupee

and 4 anna= ¼ rupee

Let r, e, and f denotes the number of 1 rupee, 8-anna, and 4-anna coins respectively.

r + e + f=36… (1)

r + ½e + ¼f =19½

⇒ 4r + 2e + f=78… (2)

Eliminating r from (1) and (2), we get,

3r + e = 42

⇒ e = 42 -3r

⇒ r≤15

Substituting for e in (1),

we get r + (42-3r) + f = 36

⇒ f = 2r -6

r≥4

We also have f≤ e+2 and f≥r+4

we get, 2r-6≤ 42 -3r +2 and 2r – 6≥ r+4

⇒ r≤10 and r≥10

⇒ r=10. Consequently e=12 and f=14

- Hence 4 annas coins are 14.
- On losing 2 4-annas coins, Pradyumna now has with him 12 of 4 anna coins and 12 of 8 anna coins which are equal.

In CAT Exam, if you come across a question in which the number of equations is forming is less than the number of variables and one of the options in the answer is “Cannot be determined”, so before marking this option as answer, analyze whether it is a question of special equations or not.

You can check whether any other constraints such as value are integers or natural numbers are given or not and also are the number of possible has been asked or a definite value.

After analyzing, if it is a special equation question, solve it by the above-mentioned methods.

Baisc Idea of Remainders

Cyclicity Of Remainders

Basic Application of Remainder Theorem

Remainders Advanced

Divisibility Rules for CAT Quantitative Aptitude Preparation

Dealing With Factorials

Factor Theory

Determining the second last digit and the last two digits

How to Find Number of Trailing Zeros in a Factorial or Product

Application of LCM (Lowest Common Multiple) in solving Quantitative Aptitude Problems

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