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Word Problem Solver

Word problems in math test students on their ability to decipher and use mathematical information when it is presented verbally. Word problems often appear complicated and lengthy, and contain a lot of information which makes it confusing. However, to solve a problem we only need the relevant points, and the first step to solving word problems successfully is to identify and separate the required information.

Math Word Problems

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Like any other type of math problem, word problems get easier with practice. An easy way to understand the relation between the given parameters is to learn the vocabulary that is used to denote the different operations. There are stock phrases and words which repeat themselves in math word problems, so once students get familiar with these, their work is half done. Making formulas to solve the problem also becomes simpler.

Solved Example

Question: The sum of the square of 3 consecutive numbers is 110. Find the numbers.
Solution:
Let the smallest integer be m, then 3 consecutive integers are m, m + 1, m + 2.

From the statement :

$m^2 + (m + 1)^2 + (m + 2)^2 = 110$

$m^2 + m^2 + 1 + 2m + m^2 + 4 + 4m = 110$

$3m^2 + 6m + 5 = 110$

$3m^2 + 6m - 105 = 0$

Solve for m:

$3m^2 + 6m - 105 = 0$

$3m^2 + 21m - 15m - 105 = 0$ (By factorization)

$3m(m + 7) - 15(m + 7) = 0$

$(3m - 15)(m + 7) = 0$

$3m - 15 = 0$   or  $m + 7 = 0$

=> $m = 5$   or  $m = -7$

Therefore the three consecutive integers are 5, 6, 7  or  -7, -6, -5.
 

Algebra Word Problems

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There are plenty of algebraic solved examples online which can help students when they get stuck or are unsure about the solution. The entire solution is included so that students can learn and practice form the beginning to the end.

Solved Examples

Question 1: Sujen leaves 15420 dollars behind. According to her wish, the money is to be divided between her daughter and son in the ratio 2 : 3. Find the sum received by her daughter.
Solution:
We know that, if a quantity x is divided in the ratio m : n then two parts are $\frac{mx}{m+n}$ and $\frac{nx}{m+n}$.

The sum received by the daughter = $\frac{2 * 15420}{2 + 3}$

= 6168

$\therefore$ The sum received by her daughter is $\$$6168.
 

Question 2: The age of a father is twice the square of the age of his daughter. Four years hence, the age of the father will be one year more than the three times the age of the daughter. Find the present age of his daughter.
Solution:

Let the present age of the daughter be $x$ years.

From the statement:


The present age of the father is 2$x^2$ years.

4 years hence:

The age of the daughter will be $(x + 4)$ and that of the father, (2$x^2$ +

$4$)years.

From the question:


$2x^2 + 4 = 1 + 3(x + 4)$

$2x^2 + 4 = 1 + 3x + 12$

$2x^2 + 4 = 3x + 13$

2$x^2 - 3x - 9 = 0$

2$x^2 - 6x + 3x - 9 = 0$

$2x(x - 3) + 3(x - 3) = 0$

$(2x + 3)(x - 3) = 0$

=> $x$ = $\frac{-3}{2}$  or  $x$ = 3

But age cannot be negative. So $x$ $\neq$ $\frac{-3}{2}$. Therefore, $x$ = 3.

$\therefore$ The present age of the daughter is 3 years.