# GRE Algebra | Applications

**Algebra **and its application cover a large chunk of the quantitative reasoning part of the GRE test paper. This is due to the fact that algebra has a good number of topics under its head. The topics which form a part of algebra as follows:

- Expanding expressions
- Basic equations
- Systems of equations
- Quadratic equations
- Equations with exponents
- Equations with fractions
- Equations with square roots
- Equations with absolute values
- The coordinate plane
- Equations of lines
- Graphs of quadratics
- Simplifying algebraic expressions

The application of algebra is in word problems. The most important and preliminary step of solving any word problem is in understanding the textual description and translating it into an equivalent numerical equation.

The word problems mainly can be divided into given topics:

- Average and Mixture Word Problems
- Distance, Rate, Time Word Problems
- Work Word Problems
- Word Problems involving simultaneous equations and inequalities.

A few examples to show how to translate word problems into equations:

- The fourth power of
**y**is added to cubes of**x**,y^4+ x^3

- Smita salary is
**x**and is increased by**10%**every year,x = x + x/10

Some useful tricks and shortcuts which will help while solving problems in algebra:

**FOIL Method:**

In binomial problems like the one given below it is at times difficult to know which term to multiply with which term. F.O.I.L can be used in this case:

(x^4 + y^3)(x^5 + y^2)

F:multiply thefirsttermO:multiply theoutertermI:multiply theinnertermL:multiply thelastterm

**Examples:**

**Ques-1:** Given two equations. If x and y satisfy the system of the equation shown, what is the value of x-y ?

7x + 3y = 12 3x + 7y = 6

**Options:**

(a)2/3(b)3/2(c)1(d)4(e)6

**Explanation:**

Solving the above two equations,

7x + 3y = 12 ------ (1) 3x + 7y = 6 --------(2) Equation (1) X 3 Equation (2) X 7 Solving we get, x = 231/70 y = 3/20 x-y = 210/140 = 3/2

So, option (b) is correct.

**Ques-2:** A mixture of 160 gallons of wine and water contains 25% water. How much water must be added to the mixture in order to increase the percentage of water to 40% of the new mixture ?

**Options:**

(a)40 gals(b)50 gals(c)80 gals(d)33 gals

**Explanation:**

Initially in the 160 galloons mixture, 25% water was present which makes ratio of wine to water as 120:40 or 3:1.

Now we need to add **W** gallons more water so that the percentage of water in the mixture is 40%.

That means the water would become **40 + W **, while the wine remains at 120.

120 gallons of wine corresponds to 60% of the mixture.

Let **M** be the total mixture:

60% of M = 120 M = 200 gallons W = 40% of 200 = 80 gallons

Initially, 40 gallons of water was added.

Hence more 80 – 40 = **40 gallons** of water has to be added.

So, option (a) is correct.

**Ques-3:** The average age of a group of men is increased by 6 when a man ages 26 years is replaced by a new person of age 56 years. How many men are there in the group ?

**Options:**

(a)3(b)4(c)5(d)6

**Explanation:**

Let the number of men be **X** and the total age of all men be **T**.

When a person aged 26 years is replaced by a person aged 56 years, the total age of the group goes up by 30 years.

This leads to an increase in the average age by 6 years.

Hence,

T/(X+30) - T/X = 6Solving for X, we getX = 5

The number of men in the group is 5.

So, option (c) is correct.

**Ques-4:** Nishu and Archana can do a piece of work in 10 days and Nishu alone can do it in 12 days . In how many days can Archana do it alone ?

**Option:**

(a)3 60 days(b)3 30 days(c)3 50 days(d)3 45 days

**Explanation:**

Nishu and Archana can do a piece of work in **10 days**.**In 1 day**, Nishu and Archana can do **(1/10)th** part of the work.

Nishu alone can do the work in **12 days**.

**In 1 day**, Nishu can do **(1/12)th** part of the work.

So in 1 day, Archana can do,

1/10 - 1/12 = (1/60) th part of the work

So, Archana can do the work alone in 60 days.

So, option (a) is correct.