How to Avoid Silly Mistakes in Maths(With Examples & Ways to Correct)

Maths is most people’s favourite subject in school, so what can possibly go wrong with it? It’s supposed to be simple and easy. But even for people who were good at it at school, it can be difficult to remember many of the basic things. In fact, it’s only once you’re out of school that you realize how much maths you use every day in your job. So it’s important to brush up on those skills. But this isn’t an easy thing for some people. And for others, just a few silly mistakes can be enough.

Let’s learn how to avoid silly mistakes in maths with measures to correct them.

How to Avoid Silly Mistakes in Maths

We bring you a list of some of the common silly mistakes made by students in Maths and how to avoid them.

1. Unit of Ratio

Look at the question “find the ratio of 35km to 7km”.

When solving questions on measurements, students are taught to include units of measurement with the final answer. For example, if they are finding the area of a figure with dimensions in cm, then its area will be expressed in $cm^{2}$ (e.g., 64 $cm^{2}$), or finding the volume of a solid object with dimensions given in m, then its volume will be expressed in m3 (e.g., 1500 $m^{3}$), and so on.

But in the case of ratio, as in the above-mentioned case, the ratio of 35km to 7km = 5 : 1 $\left(\frac{35}{7} = \frac{5}{1} = 5 : 1\right)$.

Always remember that the unit of ratio is none, i.e., there is no unit for a ratio.

Ratio comprises of two or more similar types of quantities so the units cancel out each other and thus there is no unit for a ratio.

Try These:

1. What will be the ratio of 70 km to 700 ml?
2. If the speeds of the scooter and bicycle are in a ratio 1 : 5 and the speed of the scooter is 50 km/h, then what will be the speed of bicycle?
3. If $x : y = 2 : 7, then \frac{1}{x} : \frac{1}{y} = ?$
4. We cannot find the ratio of the areas of a triangle and a square. (True/False)
5. If a : b = 3 : 1, then (a – b) : (a + b) = ?

2. Fail to Convert Units

It is the type of one of the most common mistakes that students make while solving mensuration problems. 

Suppose students are asked to find the area of a rectangle with dimensions 1.5 m and 80 cm.

To solve such a type of problem, you always start with the formula – Area = Length × Breadth (in this case).

And then plug-in the values: Area = 1.5 × 80 = 120 $m^{2}$, which is wrong as students fail to realize that the measurement of two sides is in different units.

Whenever you are solving such types of problems, always make sure that the units associated with all the measurements involved are the same.

In our case, length = 1.5 m and width  = 80 cm.

Either convert 1.5 m to cm or 80 cm to m.

80 cm in m is 0.8 m

So, the correct answer is 1.5 × 0.8 = 1.2 $m^{2}$

Try These:

1. What will be the width of a rectangle whose length is 125 cm and area 1 $m^{2}$?
2. If the length and width of a rectangle are 1.75 m and 60 cm then find the ratio of its width to length.
3. Two rectangles have dimensions of 1.2 m, 55 cm, and 135 cm, 0.8 m. Which one has a greater area?
4. The speed of car A is 60 km/h and that of car B is 15 m/s. Which car is moving with greater speed?
5. The area of a parallelogram is given by Area = Base × Altitude. If the area of a parallelogram is 9600 $cm^{2}$ and its base is 1.2 m, then what is the measurement of its altitude?

3. $-x^{2}$ and $\left(-x\right)^{2}$, the same in magnitude

Many students evaluate $-x^{2}$ and $\left(-x\right)^{2}$ to $x^{2}$, which is absolutely wrong. $-x^{2}$ and $\left(-x\right)^{2}$ mean different things. 

While $-x^{2}$ means (-) multiplied with $x^{2}$, whereas, $\left(-x\right)^{2}$ is square of -x, i.e., -x multiplied with -x.

You might be knowing that the square of any number (whether it’s positive or negative) is always positive, therefore, $x^{2}$ will always be positive. Also, when -1 is multiplied by a positive number, the result is always negative, thus, $-x^{2}$ is a negative value.

Now, coming to $\left(-x\right)^{2}$. Since $\left(-x\right)^{2}$ is a square of -x, therefore, $\left(-x\right)^{2}$ will always have a positive value, as mentioned above the square of any number is positive.

Also, $\left(-x\right)^{2}$ = $\left(-x\right)$ × $\left(-x\right)$, which is multiplying two negative values, therefore, its result will always be positive. (Product of two negative values is always positive).

Next time, always remember that $-2^{2} = -4$ and $\left(-2\right)^{2} = 4$.

Try These:

1. $-\left(-2^{2}\right)$ and $\left(-2\right)^{2}$ have the same value. (True/False)
2. $-x^{2}$ is always less than $\left(-x\right)^{2}$. (True/False)
3. $-x^{2} + \left(-x\right)^{2} = 0$ (True/False)
4. Evaluate $-\left(-\left(-2\right)^{3}\right)$
5. Evaluate $\left(\frac{1}{3}\right)^{3} – \left(-\frac{1}{3}\right)^{3}$

4. Prime numbers between 2 and 23 are 2, 3, 5, 7, 11, 13, 17, 19 and 23

If someone asks you how many prime numbers are there between 2 and 23, then what would be the answer? Is it 7 or 9.

Let’s check the prime numbers till 23. These are 2, 3, 5, 7, 11, 13, 17, 19, and 23, i.e., 9 in number. But we’ll not include 2 and 23 in our count, as we need to count only the prime numbers between 2 and 23.

Correct: Prime numbers between 2 and 23 are 3, 5, 7, 11, 13, 17, and 19.

Note: Between means excluding the first and last numbers.

Try These:

1. How many natural numbers are there between 1 and 150?
2. How many odd numbers are between 1 and 90?
3. The number of even numbers and odd numbers between 1 and 80 are the same. (True/False)
4. How many numbers between 5 and 75 are divisible by 5?
5. If the number of odd numbers between 11 and x (where x is an odd number) is 50, then the value of x is?

5. $\sin^{2}x$, $\sin x^{2}$ and $\left(\sin x\right)^{2}$ Have Same Meaning

While solving trigonometric problems, students interchangeably write $\sin^{2}x$ and $\sin x^{2}$. Is $\sin^{2}x$ and $\sin x^{2}$ the same? The answer is no.

$\sin^{2}x$ is the square of $\sin x$, i.e.,  $\sin x$ × $\sin x$, whereas $\sin x^{2}$ is a sine of square of x.

For example $\sin^{2}30^{0}$ = square of $\sin 30^{0}$ = $\left(\sin 30^{0}\right)^{2}$= $\left(\frac{1}{2}\right)^{2}$ = $\frac{1}{4}$ and $\sin\left(30^{0}\right)^{2}$ = $\sin 900^{0}$.  

Now, coming to $\left(\sin x\right)^{2}$. $\left(\sin x\right)^{2}$ is square of $\sin x$, i.e., $\sin x$ × $\sin x$ = $\sin^{2}x$.  

Thus, we see that $\sin^{2}x$ and $\left(\sin x\right)^{2}$ are the same, whereas $\sin^{2}x$ and $\sin x^{2}$ or $\sin x^{2}$ and $\left(\sin x\right)^{2}$ are different.

Try These:

1. For $x = 30^{0}$, evaluate $\sin^{2}x – \left(\sin x\right)^{2}$
2. For any value of x, $\sin^{2}x \lt \left(\sin x\right)^{2}$ (True/False/Both)
3. For $x = 30^{0}$, evaluate $\sin^{4}x – \left(\sin x\right)^{4}$.

6. Figure with Smaller Area has Smaller Perimeter

It’s not always that a figure having a smaller area will have a smaller perimeter

Consider the following figures

In the above figure,

Perimeter of square ABCD = 4 × 5 = 20 cm and area of square ABCD = 5² = 25 cm²

And, perimeter of square EFGH = 4 × 2 = 8 cm and area of square EFGH = 2² = 4 cm²

Clearly, you can see that ar(ABCD) > ar(EFGH) and (Perimeter of ABCD) > (Perimeter of EFGH)

Now, look at these two figures below

Perimeter of ABCD = AB + BC + CD + DA = 4 + 5 + 4 + 5 = 18 cm and area of ABCD = 4 × 5 = 20 cm²

And, perimeter of ABCDEF = AB + BC + CD + DE + EF = 4 + 2 + 2 + 3 + 2 + 5 = 18 cm  and area of ABCDEF = 20 – 2 × 3 = 14 cm²

Here, perimeters are the same but the area of ABCD > area of ABCDEF  

Try These:

1. The length of the rectangle is twice its width. If the perimeter of a square is the same as that of a rectangle, then which one has a greater area?
2. The perimeter (circumference) of a circle is 44 cm. If the perimeter of a square is the same as that of a circle, then the area of the square and that of the circle are the same. (True/False). If false, which one has the greater area?
3. The area of a circle is always greater than the area of a square with the same perimeter. (True/False).
4. If the perimeter of a square becomes half, then its area also reduces by half. (True/False)
5. Which of the following is true?
   • If the perimeter of a figure increases, it occupies more space.
   • If the perimeter of a figure increases, it occupies less space.
   • If the perimeter of a figure decreases, it occupies more space.
   • If the perimeter of a figure decreases, it occupies less space.
   • All of these

7. Base of a Right Triangle is the Side Lying on the Plane

While calculating the area of a triangle using the formula Area = $\frac{1}{2}$ × Base × Height, many students fail to identify the base of a triangle.

In the right triangle above, the base is BC and the height is AB, and hence area of triangle ABC = $\frac{1}{2}$ × BC × AB

Now, look at the triangle below.

Which of the three sides in the above triangle is the base?

In a triangle shown above many times, students take BC as the base of $\triangle$ ABC and proceed to calculate the area which is wrong.

In the case of the above triangle, the base is either AB or CA, and hence the area of $\triangle$ ABC = $\frac{1}{2}$ × AB × CA

The base of a triangle is always the one including the right angle ($90^{0}$ angle).

Try These:

1. Other than hypotenuse, any of the two remaining sides of a right-angled triangle can be considered it’s base. (True/False)
2. If a right-angle triangle is rotated, its base changes (True/False)
3. Identify the base in each of these triangles:

4. The base of a right-angle triangle is always greater than the other sides
5. Area of right triangle = $\frac{1}{2}$ × (Product of two sides other than hypotenuse)

8. $\frac{1}{x + y} = \frac{1}{x} + \frac{1}{y}$

$\frac{1}{x + y}$ is not equal to $\frac{1}{x} + \frac{1}{y}$

In fact, you cannot separate or split the denominator part of a fraction. Although you can separate or split its numerator part.

$\frac{x + y}{z} = \frac{x}{z} + \frac{y}{z}$

But $\frac{1}{x + y}$ is not equal to $\frac{1}{x} + \frac{1}{y}$

Let’s consider an example:

$\frac{1}{5} = 0.2$ 

$\frac{1}{5}$ can be written as $\frac{1}{2 + 3}$

Now, $\frac{1}{2} + \frac{1}{3}$ = 0.5 + 0.333… = 0.833…

Try These:

1. Given $a = 5$ and $b = 10$, find
   • $\frac{1}{a}$
   • $\frac{1}{b}$
   • $\frac{1}{a}$ + $\frac{1}{b}$
   • $\frac{1}{a + b}$
2. If a and b are two numbers, such that a > b, then $\frac{1}{a + b}$ is always greater than $\frac{1}{a – b}$
3. $\frac{1}{a + b}$ is always smaller than $\frac{1}{a}$ + $\frac{1}{b}$
4. Given a = 0.5 and b = 1.5, find
   • $\frac{1}{a}$
   • $\frac{1}{a}$
   • $\frac{1}{a}$ + $\frac{1}{b}$
   • $\frac{1}{a + b}$

9. Forget to Invert the Inequality Sign While Multiplying by a Negative Number

Generally while solving inequalities of the type $-x \lt 3$, many students multiply both sides of the inequality by -1 to remove the negative sign with the variable and proceed as:

$-x \lt 3$ => $-x × \left(-1\right) \lt 3 × \left(-1\right)$ => $x \lt -3$, which is wrong

Always remember to invert the inequality sign while multiplying a negative number on both sides of the inequality

Correct way is $-x \lt 3$ => $-x × \left(-1\right) \gt 3 × \left(-1\right)$ => $x \gt -3$

While multiplying an inequality by a negative number do the following changes

• < to >
• > to <
• <= to >=
• >= to <=

Try These:

Solve the following inequalities:

1. $2x – 3 \lt -5$
2. $-2x + 3 \gt 7$
3. $\frac{x + 2}{-3x + 5} \gt 1$
4. $\frac{x}{3} – \frac{x}{2} \lt -3$
5. $2\left(\frac{x}{6} – \frac{x}{3}\right) \gt -5$

10. Forget to Invert the Inequality Sign While Taking Reciprocals

How will you solve an inequality of the form $\frac{1}{x + 7} \gt \frac{2}{3}$

Students generally proceed as  $\frac{1}{x + 7} \gt \frac{2}{3}$ => $x + 7 \gt \frac{3}{2}$ => $x \gt \frac{3}{2} – 7$ => $x > -\frac{11}{2}$ which is wrong

The correct way of solving such types of inequalities is to invert the inequality sign while taking the reciprocal of both sides

$\frac{1}{x + 7} \gt \frac{2}{3}$  => $x + 7 \lt \frac{3}{2}$ => $x \lt \frac{3}{2} – 7$ => $x \lt -\frac{11}{2}$

Try These:

Solve the following inequalities:

1. $\frac{2}{3x – 2} \le \frac{5}{2}$
2. $\frac{1}{3} \lt \frac{5}{x – 7}$
3. $-\frac{1}{2x – 9} \le 3$
4. $-\left(\frac{1}{2x}\right) \gt \frac{5}{2}$
5. $7 – \frac{1}{x} \le \frac{2}{x}$

11. If $a^{2} = 9x^{2}$, then $a  = 9x$

Let’s see if $a = 9x$, then how much is $a^{2}$?

$a = 9x$ => $a^{2} = \left(9x\right)^{2}$ => $a^{2} = \left(9x\right)^{2}$ => $a^{2} = 9^{2} × x^{2} = 81x^{2}$

Clearly, you can see that $a = 9x$, if $a^{2} = 81x^{2}$

$a^{2} = 81x^{2}$ can be written as $a^{2} = \left(9\right)^{2}x^{2} = \left(9x\right)^{2}$ => $a = 3x$

Now, if you take square of both sides of $a = 3x$, you’ll get

$a^{2} = \left(3x\right)^{2}$ => $a^{2} = 9x^{2}$

Try These:

1. If $b^{2} = \frac{1}{64}y^{2}$, then find the value of $b$
2. If $a^{2} = \frac{25}{49}x^{2}$, then find the value of $a$
3. If $a^{2} = \frac{4}{9x^{2}}$, then find the value of $a$
4. Factorize $16x^{2} – 25y^{2}$
5. Factorize $25x^{2} – 40xy + 16y^{2}$

12. Solving Absolute Value Equations

What will be the solution of $\vert{x – 2}\vert= 5$?

Most of the students write $\vert{x – 2}\vert = 5 => x – 2 = 5$ => $x = 5 + 2 => x = 7$, which is incomplete

Whenever you solve an absolute value equation, always remember to consider both positive and negative values.

$\vert{x – 2}\vert = 5$ => $\pm \left( x – 2 \right) = 5$ => $+\left(x – 2\right) = 5$ or $-\left(x – 2\right) = 5$ => $x – 2 = 5$ or $-x + 2 = 5$ => $x = 7$ or $x = -3$ 

Try These:

Find the value of x in each of the following:

1. $\vert{x – 7}\vert = 7$
2. $\vert{-x + 6}\vert = 8$
3. $-\vert{2x – 5}\vert = 13$
4. $-\left(-\vert{-x + 8}\vert\right) = 10$
5. $\vert{\frac{1}{2}x + 9}\vert = 11$

13. Solving Absolute Value Inequalities

$\vert{2x + 3}\vert \lt 6$ => $\left(2x + 3\right) \lt 6$ or $-\left(2x + 3\right) \lt 6$ => $2x \lt 6 – 3$ or $-2x \lt 6 + 3$ => $2x \lt 3$ or $-2x \lt 9$ => $x \lt \frac{3}{2}$ or $x \gt -\frac{9}{2}$

=> $-\frac{9}{2} \lt x \lt \frac{3}{2}$

Easy way to solve Absolute Value Inequalities

For Less Than Inequality: Just sandwich the expression between negative and positive values of right-hand side

$\vert{2x + 3}\vert \lt 6$

=> $-6 \lt \left(2x + 3\right) \lt 6$ => $-6 -3 \lt 2x < 6 – 3$ => $-9 \lt 2x \lt 3$ => $-\frac{9}{2} \lt x \lt \frac{3}{2}$

For Greater Than Inequality: Make two inequalities – one with less than and negative value and the other with greater than and positive value

$\vert{2x – 3}\vert \gt 5$ => $2x – 3 \lt -5$ or $2x – 3 \gt 5$ => $2x \lt -2$ or $2x \gt 8$ => $x \lt -1$ or $x \gt 4$

Try These:

1. $\vert{x}\vert \gt \vert{-x}\vert$ (True/False)
2. $\vert{x}\vert > -\vert{x}\vert$ (True/False)
3. Solve for $x$:
   • $\vert{7x – 3}\vert \ge 18$
   • $\vert{-5x + 8}\vert \le -2$
   • $\vert{2x – 9}\vert \ge \vert{-3x + 8}\vert$

14. $\sin\left(x + y\right) = \sin x + \sin y$

Students use the associative property of multiplication over addition or subtraction while solving expressions of the form $a\left(b + c\right) = ab + ac$

While solving an expression of the form $\sin\left(x + y\right)$, they try to use the same property and write $\sin\left(x + y\right)  = \sin x + \sin y$, which is incorrect. The reason here in the case of $a\left(b + c\right)$, $a$ and $\left(b + c\right)$ are two different values/entities, whereas in the case of $\sin\left(x + y\right)$, $sin$ and $\left(x + y\right)$ are not two different values. $\sin\left(x + y\right)$ is one value. ($x + y$ is an argument of $sin$).

Similarly $sin\left(x – y\right)$ is not equal to $\sin x – \sin y$ and the same rule follows with other trigonometric ratios viz, $cos$, $tan$, $sec$, $cosec$ or $cot$.

Try These:

1. For $A = 30^{0}$ and $B = 60^{0}$, find
   1. $sin A$
   2. $sin B$
   3. $sin A + sin B$
   4. $sin \left(A + B\right)$
2. For $A = 60^{0}$ and $B = 30^{0}$, find
   1. $cos A$
   2. $cos B$
   3. $cos A – cos B$
   4. $cos \left(A – B\right)$
3. For $A = 0^{0}$ and $B = 45^{0}$, find
   1. $tan A$
   2. $tan B$
   3. $tan \left(A + B\right)$

15. $\left(x – a\right)\left(x – b\right) \lt 0$ => $x – a \lt 0$ and $x – b \lt 0$

$\left(x – a\right)\left(x – b\right) \lt 0$ means $\left(x – a\right) \left(x – b\right)$ is negative and you might be knowing that the product of two values is negative only when one is positive and the other is negative.

Hence, $\left(x – a\right)\left(x – b\right) \lt 0$ means either of these two:

$\left(x – a\right)$ is positive and $\left(x – b\right)$ is negative => $\left(x – a\right) \gt 0$ and $\left(x – b\right) \lt 0$

OR

$\left(x – a\right)$ is negative and $\left(x – b\right)$ is positive => $\left(x – a\right) \lt 0$ and $\left(x – b\right) \gt 0$

Now, coming to $\left(x – a\right) \lt 0$ and $\left(x – b\right) \lt 0$

$\left(x – a\right) \lt 0$ => $\left(x – a\right)$ is negative and $\left(x – b\right) \lt 0$ => (x – b) is negative

And the product of two negative values is positive, therefore, $\left(x – a\right)\left(x – b\right) \gt 0$

Thus, $\left(x – a\right) \lt 0$ and $\left(x – b\right) \lt 0$ => $\left(x – a\right)\left(x – b\right) \gt 0$ and NOT $\left(x – a\right)\left(x – b\right) \lt 0$.

Try These:

Solve the following for x:

1. $\left(x + 5\right)\left(x – 5\right) \ge 0$
2. $-\left(x + 7\right)\left(x – 2\right) \lt 0$
3. $\left(-2x + 3\right)\left(3x + 5\right) \ge 0$
4. $\left(\frac{x – 7}{2}\right)\left(\frac{x + 2}{5}\right) \lt 0$

16. $\frac{\left(2x + 7\right)\left(3x – 4\right) + 4x}{\left(3x – 4\right)\left(2x + 11\right)}$ = $\frac{\left(2x + 7\right) + 4x}{2x + 11}$

A student generally mistakenly cancel out common-looking expressions in the numerator and denominator to simplify

$\frac{\left(2x + 7\right)\left(3x – 4\right) + 4x}{\left(3x – 4\right)\left(2x + 11\right)}$ = $\frac{\left(2x + 7\right) + 4x}{2x + 11}$ which is wrong

In $\frac{\left(2x + 7\right)\left(3x – 4\right) + 4x}{\left(3x – 4\right)\left(2x + 11\right)}$, you cannot take $\left(3x – 4\right)$ as common.

The reason is $4x$ is also part of the numerator and $\left(3x – 4\right)$ is not common with $4x$.

17. Rounding too Early in Your Responses

Many students lose unnecessary marks just because of minor differences between their responses and the correct answer. This is due to the rounding of decimal places early on in a solution. 

Let’s consider the following example

(1.28 × 2.57) + (0.13 × 4.87)

= 3.2896 + 0.6331 = 3.29 + 0.63 (After rounding off to 2 decimal places) = 3.92 

The correct way is 3.2896 + 0.6331 (Carrying the decimals to the last step) = 3.9227 

The rule is simple; never round any values until you reach your final answer. If you are hell-bent on doing so, however, make sure you take enough decimal places (4 to 5 would be ideal) so your answer is still correct.

Try These:

Simplify the following and round off the answer to the nearest of 2 decimals

1. 4.96 × (5.89 – 2.78) × 1.5
2. (3.96 × 2.97) – (1.08 × 0.045)
3. $\left(\frac{3}{5} + \frac{7}{3}\right) × \left(\frac{1}{2} + \frac{5}{6}\right) + 0.25$
4. 7.28 ÷ (1.25 × 0.001)
5. (15.09 – 12.98) ÷ (1.01 × 1.25)

18. $\left(x – a\right)\left(x – b\right) = 0$ => $x – a = 0$ & $x – b = 0$, then $\left(x – a\right)\left(x – b\right) = c$ => $\left(x – a\right) = c$ & $\left(x – b\right) = c$= c

The equations of the type $\left(x – 2\right)\left(x – 5\right) = 0$ means

$\left(x – 2\right)\left(x – 5\right) = 0 × 0$ (Since, 0 can be written as 0 × 0)

=> $x – 2 = 0$ or $x – 5 = 0$ => $x = 2$ or $x = 5$

Many students use the same steps to solve the equations of the type $\left(x – 2\right)\left(x – 5\right) = 10$, following the steps

$\left(x – 2\right)\left(x – 5\right) = 10$ => $x – 2 = 10$ or $x – 5 = 10$ => $x = 12$ or $x = 15$, which is incorrect

Here, 10 on the right-hand side is written as 10 × 10 (=100)

The proper way of solving these types of equations is splitting 10 properly and then solving:

$\left(x – 2\right)\left(x – 5\right) = 10$ => $\left(x – 2\right)\left(x – 5\right) = 5 × 2$ => $x – 2 = 5$ and $x – 5 = 2$ (Since, 8 × 5 = 40)

=> $x = 7$ and $x = 7$ => $x = 7$ 

Try These:

1. $\left(x – 2\right)\left(x + 5\right) = 120$
2. $\left(x + 2\right)\left(x + 5\right) = 70$
3. $\left(x – 3\right)\left(x + 2\right) = 50$

19. 4 Points Lying on a Line Segment Divide it into 4 Parts

The statement n points lying on a line segment divided into n parts is wrong. In fact, n points lie on a line segment divided into (n + 1) parts.

Consider a line segment AB

A ______________________________________________________ B

Let’s take 4 points on it say C, D, E, and F

A _________C___________D___________E__________F_________ B

How many line segments do you have now?

There are 5 line segments, viz., AC, CD, DE, EF, and FB

So, n points lying on a line segment are divided into (n + 1) parts.

Try These:

1. The number of points required to divide the circumference of a circle into n equal parts is
   1. n
   2. (n + 1)
   3. (n – 1)
   4. Any of the above
2. If you want to divide a straight line into 10 equal parts, how many points will you locate on it?
3. The circumference of a circle needs to be divided into 6 equal parts, how many points will you locate on it?

20. Omitting Negative Sign (-) When Taking Square Root of a Number

What is the square root of 16?

Many of the students write $\sqrt{16}$ as 4.

It is not incorrect but is incomplete. It’s true that $\sqrt{16}$ is 4 (since 4 × 4 = 16), but $\sqrt{16}$ is also -4 (since -4 × (-4) = 16).

Therefore, $\sqrt{16}$ is 4 as well as -4 and you should always specify both values by writing $\sqrt{16}$ as $\pm{4}$ ($\sqrt{16}$ = $\pm{4}$).

21. $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ or $\sqrt{a – b} = \sqrt{a} – \sqrt{b}$

Another common mistake students make is in simplifying expressions like 

$\sqrt{12 + 4} = \sqrt{12} + \sqrt{4}$, which is again incorrect

Here, LHS is  $\sqrt{12 + 4} = \sqrt{16} = \pm \sqrt{4}$

And, RHS is $\sqrt{12} + \sqrt{4} = \pm 2 \sqrt{3} + \left(\pm2\right) = \pm 2 \sqrt{3} \pm 2 = \left(2 \sqrt{3} + 2 \right) $ or $ \left(-2\sqrt{3} + 2 \right)$ or $\left(2\sqrt{3} – 2\right)$ or  $\left(-2\sqrt{3} – 2 \right)$.

Same applies for $\sqrt{12 – 4}$. It’s not equal to $\sqrt{12} – \sqrt{4}$. Here,

LHS is $\sqrt{12 – 4} = \sqrt{8} = \pm 2\sqrt{2} $

RHS is $\sqrt{12} – \sqrt{4} = \pm 2\sqrt{3} – \left(\pm 2\right) = \pm 2\sqrt{3} \pm 2$ $=\left(2\sqrt{3} + 2\right)$ or $=\left(2\sqrt{3} – 2\right)$ or $=\left(-2\sqrt{3} + 2\right)$ or $=\left(-2\sqrt{3} – 2\right)$

Try These:

1. For $a = 24$ and $b = 12$, find
   • $\sqrt{a}$
   • $\sqrt{b}$
   • $\sqrt{a + b}$
   • $\sqrt{a – b}$
   • $\sqrt{a} + \sqrt{b}$
   • $\sqrt{a} – \sqrt{b}$
   • Which one of the above is the greatest and which one is the least?
2. For any two positive integers, $\sqrt{a + b} \gt \sqrt{a} + \sqrt{b}$ (True/False)

22. Canceling $x$ on Both Sides While Solving Equations of the Type $x^{2} – 5x = 20x$

Many students solve equations of type $x^{2} – 5x = 20x$ as shown below:

$x^{2} – 5x = 20x$ => $x\left(x – 5\right) = 20x$ => $\left(x – 5\right) = 20$ (Canceling $x$ on both sides)

=> $x = 20 + 5$ => $x = $25

Here $x = 25$ is not a wrong answer, but there exists one more solution for it and it’s 0.

Let’s check:

Put $x = 25$ in equation $x^{2} – 5x = 20x$

LHS = $x^{2} – 5x  = 25^{2} – 5 × 25 = 625 – 125 = 500$

RHS = 20 × 25 = 500

Put $x = 0$ in equation $x^{2} – 5x = 20x$

LHS = $x^{2} – 5x  = 0^{2} – 5 × 0 = 0 – 0 = 0$

RHS = 20 × 0 = 0

Let’s see where you went wrong and what’s the correct way of solving such types of equations.

$x^{2} – 5x = 20x$ => $x^{2} – 5x – 20x  = 0$ (Bring $20x$ to left hand side)

$=> x^{2} – 25x = 0 => x\left(x – 25\right) = 0$

$=> x = 0$ or $\left(x – 25\right) = 0$

$=> x = 0$ or $x = 25$

Hence, both 0 and 25 are solutions for equation $x^{2} – 5x = 20x$

Try These:

Solve for $x$:

1. $x^{2} + 3x = 12x$
2. $x^{3} + 5x^{2} + 6x = 10x^{2}$
3. $x^{3} – 2x^{2} + x = 3x^{2} – 5x$

23. $\left(x^{2}\right)^{3} = x^{5}$

Another common mistake students make while solving problems involving exponents and powers is $\left(x^{a}\right)^{b}$. While solving such problems, they add the two powers and get $x^{a + b}$ as an answer which is wrong.

To understand this, let’s consider an example of $\left(2^{2}\right)^{3}$

If you write $\left(2^{2}\right)^{3}$ as $2^{\left(2+3\right)} = 25$, it’s absolutely wrong.

$\left(2^{2}\right)^{3}$ is $4^{3} = 4 × 4 × 4 = 64$, whereas $2^{5}$ is 2 × 2 × 2 × 2 × 2 = 32

And $2^{\left(2×3\right)} = 2^{6}$ = 2 × 2 × 2 × 2 × 2 × 2 = 64

Always remember, $x^{a} × x^{b} = x^{a+b}$ and $\left(x^{a}\right)^{b} = x^{ab}$

Try These:

1. Simplify the following
   1. $\left(x^{2}\right)^{-1}$ 
   2. $\left(a^{-3}\right)^{-2}$
   3. $\left(2^{\frac{1}{5}+\frac{2}{5}}\right)^{\frac{5}{3}}$
2. $\left(2^{2}\right)^{-2} = \left(2^{-2}\right)^{2}$ (True/False)
3. $\left(\frac{1}{3^{3}}\right)^{-4} = \left(3^{4}\right)^{3}$ (True/False)

24. $x^{2} + x^{3} = x^{5}$

Also, $x^{a} + x^{b} = x^{a+b}$ is wrong. In fact, you cannot add two exponents directly. To add two exponents, the first step is to simplify them.

For example, $2^{2} + 2^{3} \ne 2^{5}$

Here, LHS =  $2^{2} + 2^{3} = 4 + 8 = 12$

And, RHS = $2^{5} = 32$

Try These:

1. Evaluate $\left(2^{3} + 2^{5}\right) × 2^{2}$
2. Which of the following is correct for $2^{8}$?
   1. $2^{6} + 2^{2}$
   2. $2^{6} × 2^{2}$

25. $\left(\sin x\right)^{-1} = \sin ^{-1}\left(x\right)$

$\left(\sin x\right)^{-1}$ is actually the reciprocal of $\sin x$ i.e., $\frac{1}{\sin x}$, whereas $\sin^{-1}\left(x\right)$ is inverse trigonometric ratio

An inverse trigonometric ratio actually represents the measure of an angle.

If $y = \sin\left(x\right)$, then here 

$x$ is an angle (in degrees or radians)

$y$ is a ratio (of two numbers)

$y = \sin\left(x\right) => x = \sin^{-1}y$, here again

$x$ is an angle (in degrees or radians)

$y$ is a ratio (of two numbers)

26. $\frac{5}{3}x = \frac{5}{3x}$

$\frac{5}{3}x \ne \frac{5}{3x}$

$\frac{5}{3}x = \frac{5x}{3}$, where numerator is $5x$ and denominator is $3$.

And, in $\frac{5}{3x}$ numerator is $5$ and denominator is $3x$.

Try These:

1. Evaluate $\frac{2}{5}x + \frac{1}{5x} = \frac{3}{5x}$. Is it equal to
   1. $\frac{3}{5x}$
   2. $\frac{3x}{5}$
2. Evaluate $\frac{2x}{5} + \frac{2}{5x} + \frac{2}{5}x$

27. $\frac{\frac{a}{b}}{\frac{c}{d}}$ = $\frac{ac}{bd}$

$\frac{\frac{a}{b}}{\frac{c}{d}} \ne \frac{ac}{bd}$

$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$ (Recall dividing $a$ by $b$ means multiplying $a$ by reciprocal of $b$)

= $\frac{ad}{bc}$ 

Try These:

1. Evaluate $\frac{2}{3} \div \frac{5}{7}$ 
2. Is $\frac{\frac{2}{3}}{\frac{3}{5}} = \frac{\frac{3}{2}}{\frac{5}{3}}$?
3. Evaluate $\frac{\frac{7}{9}}{\frac{3}{5}}$

28. $\log\left(x + y\right) = \log x + \log y $

$\log\left(x + y\right) \ne \log x + \log y$

Students use the associative property of multiplication over addition or subtraction while solving expressions of the form $a\left(b + c\right) = ab + ac$

While solving an expression of the form $\log\left(x + y\right)$, they try to use the same property and write $\log\left(x + y\right)  = \log x + \log y $, which is incorrect. The reason here in the case of $a\left(b + c\right)$, $a$ and $\left(b + c\right)$ are two different values/entities, whereas, in the case of $\log\left(x + y\right)$, $\log$ and $\left(x + y\right)$ are not two different values. $\log\left(x + y\right)$ is one value. ($x + y$ is an argument of log).

In fact, $\log x + \log y = \log\left(xy\right)$

Similarly $\log\left(x – y\right)$ is not equal to $\log x – \log y$ but $\log x – \log y = \log\left(\frac{x}{y}\right)$.

Try These

1. If $\log\left( 2\right) = 0.3010$, $\log\left(3\right) = 0.4771$ and $\log\left(5\right) = 0.6990$, find
   • $\log\left(1.5\right)$
   • $\log\left(0.6\right)$
   • $\log\left(1.6667\right)$
   • $\log\left(0.005\right)$
   • $\log\left(0.0003\right)$

29. $\log\left(\frac{a}{b}\right) = \frac{\log a}{\log b}$

While solving problems involving logarithms, writing $\log\left(\frac{a}{b}\right)$ as $\frac{\log(a)}{\log(b)}$ is one of the most common mistakes made by students. It is actually incorrect.

$\log\left(\frac{a}{b}\right) = \log\left(a\right) – \log\left(b\right)$.

Let’s consider the following example:

$\log\left(3\right)$ can be written as $\log\left(\frac{6}{2}\right)$.

$\log\left(3\right) = 0.4771$, $\log\left(6\right) = 0.7781$, $\log\left(2\right) = 0.3010$

$\log\left(6\right) – \log\left(2\right) = 0.7781 – 0.3010 = 0.4771$, which is equal to $\log\left(3\right)$.

Now, let’s see the value of $\frac{\log\left(6\right)}{\log\left(2\right)}$. It is $\frac{0.7781}{0.3010} = 2.5851$, which is not equal to $\log\left(3\right)$.

Remember, that $\log\left(\frac{a}{b}\right) \ne \frac{\log\left(a\right)}{\log\left(b\right)}$, but is equal to $\log\left(a\right) – \log\left(b\right)$.

Try These:

1. If $\log\left(2\right) = 0.3010$, $\log\left(3\right) = 0.4771$ and $\log\left(5\right) = 0.6990$, then find
   • $\log\left(1.5\right)$ and $\frac{\log\left(3\right)}{\log\left(2\right)}$. Are they equal in values?
   • $\log\left(0.6667\right)$ and $\frac{\log\left(2\right)}{\log\left(3\right)}$. Are they equal in values?
2. If $\log\left(2\right) = 0.3010$, $\log\left(3\right) = 0.4771$ and $\log\left(5\right) = 0.6990$, then find
   • $\log\left(0.2\right)$
   • $\log\left(0.5\right)$
   • $\log\left(0.3333\right)$
   • $\log\left(0.1667\right)$
   • $\log\left(0.0333\right)$

30. If $a + b = c$, then $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$

To check the above statement, let’s consider these values: $a = 2$, $b = 3$, and $c = 5\left(=2 + 3\right)$.

Now, $\frac{1}{a} + \frac{1}{a} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$

And, $\frac{1}{c} = \frac{1}{5}$

Obviously, $\frac{5}{6} \ne \frac{1}{5}$ 

If $a + b = c$, then $\frac{1}{a} + \frac{1}{b} \ne \frac{1}{c}$.

Try These:

1. For $a = 3$ and $b = 7$, find
   1. $\frac{1}{a} + \frac{1}{b}$
   2. $\frac{1}{a + b}$
2. For any two values $a$ and $b$, which of the following is true?
   • $\frac{1}{a} \lt \frac{1}{a + b}$
   • $\frac{1}{b} \lt \frac{1}{a + b}$
   • $\frac{1}{a} \gt \frac{1}{a + b}$
   • $\frac{1}{b} \gt \frac{1}{a + b}$
   • (a) and (b)
   • (c) and (d)
3. For two fractional numbers with the same denominators, the one with a smaller numerator is smaller. (True/False)
4. For two fractional numbers with the same numerators, the one with a smaller denominator is smaller. (True/False)

Practice Problems

1. Which of the following is a unit of ratio?
   • unit
   • $\text{unit}^{2}$
   • No unit
   • None of these
2. Are the values $-\left(5^{2} \right)$ and $\left(-5 \right)^{2}$ the same in magnitude?
3. Figure with Smaller Area has Smaller Perimeter (True/False)
4. $\frac{1}{2 + 5} = \frac{1}{2} + \frac{1}{5}$ (True/False)
5. Solve $|2x – 4| \le 3$
6. Solve $|3x = 8| \ge 4$
7. Solve the equation $x^{2} – 6x = 30x$

FAQs

Conclusion

In this article, we discussed some silly mistakes students make and how to avoid silly mistakes in maths. These are the kind of mistakes that they make when they are doing their homework or during their tests. We hope the article will help you remember these mistakes and also how to avoid them in the future.

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