“Simplify” process skill to your rescue!
INTRODUCTIONHow many times have you experienced a situation when just a cursory glance of the question scares you because you see those weird looking expressions?
How many times has it happened that you reach a point in the solution and you just cannot go further because you are stuck?
To successfully combat such situations, you need to master “simplify” process skill.
What is simplify process skill?
As we discussed in this article, “simplify” process skill is defined as follows:
Let’s take our first example and learn how to apply “simplify” process skill.
Solve this 700-level Algebra question from
the Official Guide Advanced book.
EX 1If n > 4, what is the value of the integer n?
(1) \(\frac{n!}{(n-3)!}=\frac{3!n!}{4!(n-4)!}\)
(2) \(\frac{n!}{3!(n-3)!}+\frac{n!}{4!(n-4)!}=\frac{(n+1)!}{4!(n-3)!}\)
Now review the solution of this question. If you prefer a video solution, you may view it
here.
SIMPLIFY PROCESS SKILL AT PLAYBefore getting started with analyzing the given statements, let’s see what’s given to us in the question.
• We are given that n is an integer, which is greater than 4.
• And, we are asked to find the value of n.
This question looks quite simple, until we look at the expressions given in the statements. That is where simplification process skill can be applied to convert the complex looking expression/equation into a simpler form.
Let’s look at the solution to understand how it’s done.
While analyzing statement 1, we notice that all the terms in the equation are factorials. By applying the conceptual knowledge of factorial, that n! is the product of all integers from 1 to n, let’s simplify this equation:
• What is the common factor between the numerator and the denominator of \(\frac{n!}{(n – 3)!}\)?
o It is (n – 3)!
By applying the definition of factorial, we can write n! as n × (n – 1) × (n – 2) × (n – 3)!
o Thus, we can cancel out this common factor.
• Therefore, \(\frac{n!}{(n – 3)!} = n × (n – 1) × (n – 2)\)
Similarly, the right-hand side of the equation can be simplified by cancelling out the common factor in both numerator and denominator.
• The common factor between 3! and 4! is 3!
• And, the common factor between n! and (n – 4)! is n × (n – 1) × (n - 2) × (n - 3)
• Therefore,\(\frac{ (3! × n!)}{(4! × (n – 4)!)}=\frac{(n × (n – 1) × (n – 2) × (n - 3))}{4}\)
Equating both we get, n × (n – 1) × (n – 2) = \(\frac{n × (n – 1) × (n – 2) × (n – 3)}{4}\)
Now that we have a much-simplified form, let’s see whether the above equation can be simplified further or not.
• The answer is yes!
o We can simplify this equation by cancelling out the common terms on both sides of the equation
The common terms are n × (n -1) × (n – 2)
Cancelling out, we get,
• \(1 =\frac{(n – 3)}{4}\)
• Implies, n – 3 = 4
This is the simplest form of the given equation, which cannot be simplified any further. And we can directly get the value of n from this simplified equation.
Similarly, statement 2 can be simplified by applying the same process as that on statement 1.
• Take the common factors out from the respective numerators and denominators on both sides of the equation and cancel them out.
Now, some of you might think that the information, that n > 4, given in the question is redundant.
• But it’s not.
o We need to make sure that the denominators are not zeroes and
o We cannot cancel out the common factors from the numerator and denominator, unless we know that they are not equal to 0.
• Thus, for (n – 4)! and (n -3)! to be not equal to 0, the value of n must be ≥ 4
Essentially, in this question,
• We have applied the simplification process skill to reduce a complex looking equation into its simplest form by cancelling out the common factors and arrived at the answer very easily.
• In terms of conceptual knowledge, we applied the definition of factorial and the fact that the denominator of a fraction cannot be 0.
Let’s now look at another example:
First solve this question. It is from
OG Quantitative Review.
EX 2The function f is defined for each positive three-digit integer n by \(f(n)= 2^x3^y5^z\) , where x, y and z are the hundreds, tens, and units’ digits of n, respectively. If m and v are three-digit positive integers such that f(m) = 9 x f(v), then m - v = ?
Now review the solution of this question. If you prefer a video solution, you may view it
here.
SIMPLIFY PROCESS SKILL AT PLAYLet’s see what’s the role of the simplify process skill in solving this question
• As you can see, in this solution, on a high level we have used the equation f(m) = 9 x f(v), given in the question, to establish relationship between the respective digits of m and v by applying a property of exponents.
Now, most of you might think what is the role of simplify here? Let me explain that now:
• We have assumed that m = abc, and n = def.
• And then we have established a relationship between {a and d}, {b and e}, {c and f}.
Now, tell me, using the above information can we find the value of m – v directly?
• No, right?
o Because we know m – v = abc – def, but we don’t have any expression where we can use the established relations between the respective digits of m and v.
For that we need to simplify the expression, m – v = abc – def, by applying the place value concept.
• The three-digit number, abc, can be written as 100a + 10b + c, and
• The three-digit number, def, can be written as 100d + 10e + f.
Thus, abc – def = 100a + 10b + c – 100d – 10e – f, which can be further simplified as 100(a - d) + 10 (b – e) + (c – f).
Now, in this above expression we can use the relations that we have established earlier to get to the final answer.
Therefore, in this approach,
simplify process skill helped us get to an expression where we can use the established relations between the respective digits to get to the final answer.
Let’s work on another example, where simplify process skill is required, though there are no weird expressions given to us in the question.
First solve this question.
EX 3If x and y are integers, what is the remainder when \(x^2 + y^2\) is divided by 5?
1. When x – y is divided by 5, the remainder is 1.
2. When x + y is divided by 5, the remainder is 2.
Now review the solution of this question. If you prefer a video solution, you may view it
here.
SIMPLIFY PROCESS SKILL AT PLAYNow, let’s look at how we applied simplification process skill in this question
• What can we infer from the derived equation, \(x^2 + y^2 = 25a^2 + 10a + 1 + 2xy\), in statement 1 analysis? Are we able to draw any inferences quickly from this complicated looking equation?
o No, right?
o Because of the multiple variables present, {xy, a} and their powers, it seems difficult or time taking
• So, what can we do?
o Let’s try to simplify this complicated looking expression
o How to do it?
If we observe the right-hand side of the equation \(25a^2 + 10a + 1 + 2xy\), carefully, we notice that the first two terms – {\(25a^2\) and 10a} - are multiples of 5
So, we can take 5 common from both the terms and represent the remaining part with some other variable ”c” as shown below
• \(25a^2 + 10a = 5 × (5a^2 + 2a) = 5c\), where \(c = 5a^2 + 2a\)
• Thus, we have a simplified form of the equation
o \(x^2 = y^2 = 5c + 1 + 2xy\)
Now, by applying the conceptual knowledge, we can infer that this equation is of the form:
• Dividend = divisor × quotient + remainder, where dividend = \(x^2 + y^2\), divisor is 5 and the remainder term is 1 + 2xy
From this we can easily infer that the value of the remainder will depend upon the values of x and y
Hence, we can conclude that Statement 1 is not sufficient to answer the question
Similarly, in statement 2 we have simplified the complex looking expression, \(25b^2 + 20b + 4 - 2xy\), by replacing, \(25b^2 + 20b\), with 5d and inferred that the value of the remainder depends upon the value of xy.
Further, when we combined the information from both statements, we got
\(2 (x^2 + y^2) = 5c + 5d + 5\)
Which is further simplified to get to the form \(x^2 + y^2 = 5q + r\), where r is the remainder
o This is done by taking 5 common from the three terms on the RHS of the equation and replacing it with a new variable “e”
o \(2(x^2 + y^2) = 5 × (c + d + 1) = 5e\)
The above equation then helped us to infer that \(x^2 + y^2\) is a multiple of 5
All in all, in this question,
simplify process skill is used to simplify the complex looking expression derived in the analysis of both statements and get them to the general remainder form, so that we can draw the inferences easily and get to the final answer.
In contrast to the first two examples, there are no weird or complex looking expressions in this question. So, most of you might think simplification is not necessary to solve this question, until you start manipulating the equations given in the statements to get the LHS of the equations to the form \(x^2 + y^2\).
CONCLUDING REMARKSThrough these three examples from Number Properties we understood that:
1. Simplify Process Skill is applied on complex or weird looking expressions in the question.
2. Simplify Process Skill can be applied on simple expressions, in the question, which cannot be used directly in the solution.
3. Simplify Process Skill can be applied on the expressions that are not present in the question but are derived in the solution.
Example 1 is a direct application of the definition of Simplify Process Skill. In this question,
we converted the complex equations given in the statements to their simplest forms and arrived at the answer.
Whereas, in Example 2 we do not have any weird expression given in the question, but we have an expression in which
we cannot use the established relations. Hence,
we applied Simplify Process Skill to convert the given expression into a form in which the established relations can be used to get to the final answer.
And finally, in Example 3, which looks pretty simple and straight forward, most of us might think there is no role of Simplification. But we have seen how Simplify Process Skill is applied on an equation that is derived in the solution, and
further made the process of drawing inferences from the simplified equation much easier.
In summary, we can say that
the Simplify Process Skill can be applied on the complex expressions given in the question or derived in the solution or the expressions that can be manipulated to be used in the solution.
So, start building this process skill to get closer to your Q50/Q51 score.
Happy learning!
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