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

### 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.

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

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.

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:

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.

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