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

Table of Contents

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?
Famous Math Competitions for Kids

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

how to avoid silly mistakes in maths

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

how to avoid silly mistakes in maths

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
Famous Mathematicians to Inspire Your Child

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.

how to avoid silly mistakes in maths

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.

how to avoid silly mistakes in maths

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:
how to avoid silly mistakes in maths
  1. The base of a right-angle triangle is always greater than the other sides
  2. 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. $7 – \frac{1}{x} \le \frac{2}{x}$
Is your child struggling with Maths?
frustrated-kid
We can help!
Country
  • Afghanistan 93
  • Albania 355
  • Algeria 213
  • American Samoa 1-684
  • Andorra 376
  • Angola 244
  • Anguilla 1-264
  • Antarctica 672
  • Antigua & Barbuda 1-268
  • Argentina 54
  • Armenia 374
  • Aruba 297
  • Australia 61
  • Austria 43
  • Azerbaijan 994
  • Bahamas 1-242
  • Bahrain 973
  • Bangladesh 880
  • Barbados 1-246
  • Belarus 375
  • Belgium 32
  • Belize 501
  • Benin 229
  • Bermuda 1-441
  • Bhutan 975
  • Bolivia 591
  • Bosnia 387
  • Botswana 267
  • Bouvet Island 47
  • Brazil 55
  • British Indian Ocean Territory 246
  • British Virgin Islands 1-284
  • Brunei 673
  • Bulgaria 359
  • Burkina Faso 226
  • Burundi 257
  • Cambodia 855
  • Cameroon 237
  • Canada 1
  • Cape Verde 238
  • Caribbean Netherlands 599
  • Cayman Islands 1-345
  • Central African Republic 236
  • Chad 235
  • Chile 56
  • China 86
  • Christmas Island 61
  • Cocos (Keeling) Islands 61
  • Colombia 57
  • Comoros 269
  • Congo - Brazzaville 242
  • Congo - Kinshasa 243
  • Cook Islands 682
  • Costa Rica 506
  • Croatia 385
  • Cuba 53
  • Cyprus 357
  • Czech Republic 420
  • Denmark 45
  • Djibouti 253
  • Dominica 1-767
  • Ecuador 593
  • Egypt 20
  • El Salvador 503
  • Equatorial Guinea 240
  • Eritrea 291
  • Estonia 372
  • Ethiopia 251
  • Falkland Islands 500
  • Faroe Islands 298
  • Fiji 679
  • Finland 358
  • France 33
  • French Guiana 594
  • French Polynesia 689
  • French Southern Territories 262
  • Gabon 241
  • Gambia 220
  • Georgia 995
  • Germany 49
  • Ghana 233
  • Gibraltar 350
  • Greece 30
  • Greenland 299
  • Grenada 1-473
  • Guadeloupe 590
  • Guam 1-671
  • Guatemala 502
  • Guernsey 44
  • Guinea 224
  • Guinea-Bissau 245
  • Guyana 592
  • Haiti 509
  • Heard & McDonald Islands 672
  • Honduras 504
  • Hong Kong 852
  • Hungary 36
  • Iceland 354
  • India 91
  • Indonesia 62
  • Iran 98
  • Iraq 964
  • Ireland 353
  • Isle of Man 44
  • Israel 972
  • Italy 39
  • Jamaica 1-876
  • Japan 81
  • Jersey 44
  • Jordan 962
  • Kazakhstan 7
  • Kenya 254
  • Kiribati 686
  • Kuwait 965
  • Kyrgyzstan 996
  • Laos 856
  • Latvia 371
  • Lebanon 961
  • Lesotho 266
  • Liberia 231
  • Libya 218
  • Liechtenstein 423
  • Lithuania 370
  • Luxembourg 352
  • Macau 853
  • Macedonia 389
  • Madagascar 261
  • Malawi 265
  • Malaysia 60
  • Maldives 960
  • Mali 223
  • Malta 356
  • Marshall Islands 692
  • Martinique 596
  • Mauritania 222
  • Mauritius 230
  • Mayotte 262
  • Mexico 52
  • Micronesia 691
  • Moldova 373
  • Monaco 377
  • Mongolia 976
  • Montenegro 382
  • Montserrat 1-664
  • Morocco 212
  • Mozambique 258
  • Myanmar 95
  • Namibia 264
  • Nauru 674
  • Nepal 977
  • Netherlands 31
  • New Caledonia 687
  • New Zealand 64
  • Nicaragua 505
  • Niger 227
  • Nigeria 234
  • Niue 683
  • Norfolk Island 672
  • North Korea 850
  • Northern Mariana Islands 1-670
  • Norway 47
  • Oman 968
  • Pakistan 92
  • Palau 680
  • Palestine 970
  • Panama 507
  • Papua New Guinea 675
  • Paraguay 595
  • Peru 51
  • Philippines 63
  • Pitcairn Islands 870
  • Poland 48
  • Portugal 351
  • Puerto Rico 1
  • Qatar 974
  • Romania 40
  • Russia 7
  • Rwanda 250
  • Réunion 262
  • Samoa 685
  • San Marino 378
  • Saudi Arabia 966
  • Senegal 221
  • Serbia 381 p
  • Seychelles 248
  • Sierra Leone 232
  • Singapore 65
  • Slovakia 421
  • Slovenia 386
  • Solomon Islands 677
  • Somalia 252
  • South Africa 27
  • South Georgia & South Sandwich Islands 500
  • South Korea 82
  • South Sudan 211
  • Spain 34
  • Sri Lanka 94
  • Sudan 249
  • Suriname 597
  • Svalbard & Jan Mayen 47
  • Swaziland 268
  • Sweden 46
  • Switzerland 41
  • Syria 963
  • Sao Tome and Principe 239
  • Taiwan 886
  • Tajikistan 992
  • Tanzania 255
  • Thailand 66
  • Timor-Leste 670
  • Togo 228
  • Tokelau 690
  • Tonga 676
  • Trinidad & Tobago 1-868
  • Tunisia 216
  • Turkey 90
  • Turkmenistan 993
  • Turks & Caicos Islands 1-649
  • Tuvalu 688
  • U.S. Outlying Islands
  • U.S. Virgin Islands 1-340
  • UK 44
  • US 1
  • Uganda 256
  • Ukraine 380
  • United Arab Emirates 971
  • Uruguay 598
  • Uzbekistan 998
  • Vanuatu 678
  • Vatican City 39-06
  • Venezuela 58
  • Vietnam 84
  • Wallis & Futuna 681
  • Western Sahara 212
  • Yemen 967
  • Zambia 260
  • Zimbabwe 263
Age Of Your Child
  • Less Than 6 Years
  • 6 To 10 Years
  • 11 To 16 Years
  • Greater Than 16 Years

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 $x$:
    • $\vert{7x – 3}\vert \ge 18$
    • $\vert{-5x + 8}\vert \le -2$
    • $\vert{2x – 9}\vert \ge \vert{-3x + 8}\vert$
Famous Scientists to Inspire Your Child

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?
Types of Coordinate Systems

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

What causes silly mistakes in exams?

The mistakes made by students during exams are mainly due to anxiety. Whether a mistake is caused by test anxiety, haste, or carelessness, dumb mistakes result in test grades that don’t live up to your actual knowledge.

Why do students get scared of maths?

People who experience feelings of stress when faced with math-related situations may be experiencing what is called “math anxiety.”

What makes a student successful in math?

The best strategy for a student to be successful in math is to have a good grasp of basic mathematical concepts. Along with this her/his abilities to relate mathematical concepts within and across content areas and real-life situations.

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.

Recommended Reading:

Leave a Comment