Apart from some basic maths tricks, we will try to make you aware some of the most important maths tricks for any competitive exam. Just focus on how the concepts work rather than the fancy name of the method. Just joking! Remember the name of the method and how it works so that it can help you and your kids too!

While we would try to explain the concepts just keep patience and go on.

## What is Line Method?

See the following image how this process work. The Fraction \displaystyle \frac{x}{y} is given value and \displaystyle \frac{x}{y+x} and \displaystyle \frac{x}{y-x} are changed value depending whether the value decreases or increases.

Notice that for both the cases numerator **x **remains same. Only denominator decreased to **(y – x)** to increase the fraction in one case. On another case denominator increased to **(y + x)** to decrease the overall fraction.

## How to apply Line Method?

- Convert the given percentage data to fraction
- Apply line method
- Convert fraction to percentage data

See Fraction to percent conversion and vice versa here.

## Application 1: Change of Base

**Question**. A is 25% taller than B. B is what percentage shorter than A?

**Ans**: Given data (**taller**) is in positive sense. So it will be in top of line.

Now, \displaystyle \frac{1}{5}=20\%

So, B is 20% shorter than A.

**Question**: Ram’s salary is \displaystyle 33\frac{1}{3}\% less than Shyam’s. Shyam’s salary is what percent over than Ram’s?

**Ans**: Given data (**less**) is in negative sense. So it will be in bottom of line.

Now, \displaystyle \frac{1}{2}=50\%

So, Shyam’s salary is 50% over than Ram’s.

**Question**: In the semester break Golu’s weight increases by 25%. By what percent should he reduce his weight so as to retain his original weight?

**Ans**: ** increases **is in positive sense

**Question**: Ram is taller than Shyam by 10%. By what percent Shyam is shorter than Ram?

**Ans**: ** taller **is in positive sense

So, Shyam is 9.09% shorter than Ram.

**Question**: A can complete a work in 40 days. B is 25% more efficient than A. In how many days B alone can complete the work?

**Ans**:

Time taken by B = 40 - 40 \times \displaystyle \frac{1}{5} = 32 days

\begin{aligned} B &=125\%A \\ \frac{B}{A} &=\frac{125}{100}=\frac{5}{4} \\ \end{aligned}`: efficiency`

Alitertime = complete work = const.×

Now, if time taken by B and A are x and y days respectively

\begin{aligned} Bx &=Ay \\ \Rightarrow 5 \times &x=4 \times 40 \\ \Rightarrow x &=\frac{4 \times 40}{5}\\ &=32\\ \end{aligned}**Similar problem**: Ram can complete a work in 30 days. Shyam is 50% more efficient than Ram. In how many days Shyam can complete the work alone?

## Application 2: ab = Constant

\begin{aligned} a \times b=\text{const.}\\ a\uparrow \times (b \downarrow?)=\text{const.}\\ a\downarrow \times (b \uparrow?) =\text{const.} \end{aligned}Line method will also be applicable where product of two terms remains constant. If one term increases another term must be decreased by some percent and vice versa.

**Question**: Height and breadth of a rectangle are 40 cm and 20 cm respectively. If the length increases by 25%, by what percent breadth must be decreased to maintain the same area?

Area = length × breadth = const.\begin{aligned} &\frac{1}{4} \\ &\downarrow \\ \frac{1}{5} &=20\% \\ \end{aligned}

**Aliter:** By basic method

Area = 40 × 20 = 800 cm^{2}

Now, (40 + 40 × 0.25) × b = 800

⇒ b = 16

Decrease in breadth = 20 – 16 = 4

**Question**: The length of diagonal of a rhombus are 17.5 and 14.5 cm respectively. If the first diagonal is increased by 10%, by what percent must the other diagonal be decreased so as to maintain the same area?

`Area of rhombus = \displaystyle \frac{1}{2}d_1 d_2 `

\begin{aligned}
&\frac{1}{10} \\
&\downarrow \\
\frac{1}{11} &=9.09\% \\
\end{aligned}
**Question**: The base of a prism increases by 20%, by what percent must the height be decreased so as to maintain the same volume?

Volume of prism = base area × height\begin{aligned} &\frac{1}{5} \\ &\downarrow \\ \frac{1}{6} &=16.67\% \\ \end{aligned}

**Question**: The price of petrol increases by 10%, by what percent must the consumption be decreased to keep the expenditure same?

price × consumption = expenditure\begin{aligned} &\frac{1}{10} \\ &\downarrow \\ \frac{1}{11} &=9.09\% \\ \end{aligned}

**Question**: The price of sugar increases from ₹20/kg to ₹30/kg. By what percent consumption must be decreased to maintain the same expenditure?

**Question**: The commission of a broker decreases from 10% to 8%. Find what percent should he increase his business so as to retain his original income?

\begin{aligned}
\frac{1}{4} &=25\% \\
&\uparrow \\
\frac{10 \%-8 \%}{10 \%} &=\frac{1}{5} \\
\end{aligned}

Increases in business = 25%.

**Question**: Walking at \displaystyle \frac{3}{4} of his original speed, a man is able to reach his destination by 10 minutes late. Find his original time.

velocity × time = distance

Late is due to the decrease in speed = 1 - \displaystyle \frac{3}{4} = \displaystyle \frac{1}{4}

\displaystyle \frac{1}{3} = 10 min

⇒ 1 = 30 min

Hence original time was 30 min.

**Aliter**

Distance = vt = \displaystyle\frac{3}{4}v(t+10)

On solving, t = 30

**Similar Problem**: Walking at \displaystyle \frac{4}{5} of his original speed, a man reaches his destination by 5 minutes late. Find his original time. [ **Ans**: 20 min]

**Question**: A train at \displaystyle \frac{1}{3} of its original speed reaches its destination by 4 hours late. Find the original timing.

Must Read other math tricks posts:

Most Important Maths Tricks for Any Exam : Total = 100%

Most important maths tricks for any exam – AB Method

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