New math tricks for quantitative aptitude

New Math Tricks is for competitive math. It tries to help with the quantitative aptitude question and easy solution by using math tricks. All the pages are created with the shortcut formula and aptitude question for competitive exams.

Tuesday, 30 January 2018

Cylinder mensuration tricks with solved math

Cylinder mensuration is also a valuable math chapter for competitive exams. Here we are going to discuss Cylinder properties, Surface area and volume measurement formula and shortcuts tricks. Look at the below images, it is Cylinder and its properties.
The last mensuration tutorial was on Cube and cuboid mensuration tricks. If you have read that's then you can easily adapt to this tutorial. Because, if We make a cylinder into plane figures then we will get two circles and a rectangular figure. These will help you to find the surface area easily. Generally, in question, two types surface area are asking. They are
1.Total surface area of cylinder
2. Curve surface area of the cylinder.

As you have calculated Lateral and total surface area of Cube and Cuboid. Hope, here no introduction need about the Curve and Total Surface area of Cylinder. Let's move to Formulas.

Cylinder Surface area formula

Cylinder Curve Surface area

The formula for the curve surface area without the yellow portion in above image. As you look at the left-hand side of the cylinder a rectangular picture also laid. The area of the rectangular portion is the Curve Surface area of the cylinder. If we know the height of the cylinder and also the radius of top or bottom circle then easily calculation can be done. Because the area of the rectangular is equal to (Circumference X Height).look at below.

Total surface Area

: You have seen the above images, there are a square or rectangular and two circles. Just add all area of three properties. Look at below how the shortcut formula made in a single formula.
Thanks for being here today, hope you have enjoyed the Cylinder mensuration Shortcut formula. For more mensuration chapters, visit the mensuration tricks section.

Time and Works shortcut formulas


Thursday, 25 January 2018

Cube and Cuboid mensuration tricks for volume, area-Lateral and total

New Math Tricks is covering all mensuration topics and tricks. Today we are going to discuss the properties, shortcuts formula, aptitude question and tricky solution of cube and cuboid math chapter. The tutorial has made from basic level to exam level for your confidence and performance well in the upcoming competitive exam. Before discussing the shortcut formula and aptitude solution, we should have the basic concept of Cuboid and cube. So look at below basic properties and short tricks for the Cube and cuboid mensuration tricks.

Cube and Cuboid shortcut formula

Cube and cuboid are made with height, length and base and six sides (face). And the shape of a cube or cuboid is 3 dimensional. We have covered triangle, circle and quadrilateral which are 2-dimensional shapes. Look at the below image which is a cuboid form.

As you have looked the above image where we can see six faces(side). Also, it has vertices. The vertices are the longest line of a cube and cuboid. in competitive math, generally ask volume, area and vertices length. Now we are moving to shortcuts formula for Cube and Cuboid.

Cube mensuration shortcut formulas

A cube has 12 equal edges and six faces. As all edges are equal so height, width and base are equal. Therefore, the area of each face also equal. If we have to find the total surface area, we will add areas of all (six) faces. And each face can be calculated as edge^2. So, the shortcut formula for the total surface area of cube became (6 X one face"s area) or (6*Edge^2). When we have to calculate the lateral surface area of a cube then we multiply one faced area by 4. getting confusion with this math tricks? Look at below image to be clear the shortcut formula.

Lateral and Total Surface Area

And the volume of a cube is edge^3 Square unit.

Cuboid shortcuts formula for area and volume

Every cube is cuboid but all cuboid are not a cube.  So differently, you should know shortcuts area and volume formulas of the cuboid. The difference between cube and cuboid is on edges length. The different length (maximum three variants) may be in cuboid where equal in a cube. So length, base and height may be different. That's why in the formula of the cuboid, becomes different from the cube. 
Cuboid-total-lateral-surface-area-volume-math-tricks-total-surface-lateral surface
So, friend, these are the basic shortcut formulas for the cube and cuboid to calculate volume area, total surface area and lateral surface area of cube and cuboid. In the next tutorial, we will practice some mensuration problem related to the cube and cuboid. Thanks for reading and learning to days tutorial on cube and cuboid.


Monday, 25 December 2017

Mensuration problems and aptitude related to the Square

In the mensuration topic quadrilateral properties and formula, we have briefly discussed the various shape, properties and shortcut formula. Square is one of the important figures of the quadrilateral. Today we are going to solve some mensuration problems related to the square. As you have learnt that the square is four equal-armed and right angular figure. Before going to solve square mensuration, please read the opening tutorial of quadrilateral properties and mensuration tricks if you need.

Square mensuration and math tricks

Mensuration Problem 1: Find the area and perimeter of a 20 metres armed square. 
Solution: Arm length of the square is 20m. So the area will be (Arm * Arm) square unit. So area= 20*20 Sq metre =400 Sq metre.
The perimeter is the outer length of the figure. The perimeter of the square is (4*Arm). So, perimeter = 4*20 = 80 metre.
mensuration problem 2: The diagonal of a square is 10√2. Now find the area of the square.
Solution: Just look at below how we find the area of the square. 
Mensuration problem 3:  20 meters and 40 meters armed two square land marge into a rectangular land, if the base of the rectangular lad is 50 meters then find the width of the land.
Solution: To solve the mensuration problem, we have to calculate the sum of the area of both square and that is the area of rectangular land.  Look at the below how we solved it. 
 Problem 4: A house owner wants to cover a courtyard with marvels. The base and the width of the courtyard are 10 and 8 meters respectively. The area of each marvel is 1/4 sq metre. Then how many marvels should have to buy to cover the courtyard?
Solution: We can solve this mensuration problem easily without pen and paper. First of all, we will calculate the area of the courtyard. As the courtyard is rectangular, so the area can be calculated multiplying by base and width. Area= 10*8 sq metre
=80 sq metre. Now we have to divide the area of the courtyard by the area of a marvel tile. 80/(1/4)=80*4
=320 tile needs to cover the courtyard.


Friday, 22 December 2017

Triangle math tricks for quantitative aptitudes

Classification and properties of the triangle, we have already discussed on the opening tutorial of this topic. Today we are going to resume triangle mensuration with all the shortcut formula and math tricks. Triangle math tricks are very important to the point of competitive math. We strongly recommend you to visit Triangle properties and classification for the clear concept. 
Classification and properties of the triangle.

Formula for area and perimeter of triangle

Perimeter formula for Triangle: The perimeter is the outer length of any figure. So, it easy to find the perimeter of any triangle when we know all the length of all arms. Just we have to add all arm's length. When we got equilateral triangle then we have to multiply an arm by 3. The formula became 3*an arm's length. When we will not get all arm's length then we have to follow the type of triangles or other formulae and then have to apply required formula or concept to find out the perimeter of a triangle. We hope, at the end of this tutorial, you will get it very well. 

Triangle math tricks for the area


We know the above formula for calculating the area of the triangle. Where S represents the half of perimeter of the triangle and a,b and C are the lengths of three arms. This formula can be used to any triangles where we know the perimeter or the all arm's length of a triangle. But sometimes we can use some tricky shortcut formula to find the area of a triangle on some special triangles. Even we can simplify to find the area of a triangle by using some concept of triangles or merging with the quadrilateral figures. 

Area formula for right angle triangle When we know a triangle is right angular and have to calculate it's area, then use the below formula. 


How made the formula. This is a simple fact. Look at below, First one(pic 1) is a right angular triangle and the second one (pic2) is an imaginary figure of rectangular. As you know that the hypotenuse is the longest arm of a triangle as well as rectangular. If you imagine hypotenuse as a diagonal of a rectangular, the figure divides into two equal triangles. As the area formula of rectangular base*width. As the triangle is the half of the rectangular that we say 1/2 of the area of the rectangular.  Also, note that all squares are rectangular. 


Above formula, you can apply on any kind of triangle to calculate the area.Because you may have proved in geometry that when a triangle made with same base and height then the triangle will be half of the rectangular in point of the area.
In the next tutorial on the triangle, we will apply all formulas to a triangle mensuration problems.


Saturday, 16 December 2017

Solved quantitative aptitude question using math tricks #3

We are preparing for quantitative aptitudes with the question-answer series for our upcoming competitive maths. Here, we take most important math problems which have already asked in previous years competitive exam or a relevant question to perform well in the exam. Look at below, we have solved some exam level arithmetic problems using math tricks. This is the 3rd tutorial on solved quantitative aptitudes. If you hadn't viewed previous two tutorials, then visit by clicking below.

  1. Solved quantitative aptitudes part-1
  2. Solved quantitative aptitudes part-2

Quantitative question 1: In two blends mixed tea, the ratios of Darjeeling and Assam tea are 4:7 and 2:5. The ratio in which these two blends should be mixed to get the ratio of Darjeeling and Assam tea in new mixture as 6:13 is-
Solution:  For the quick solution of this problem, we will use the Allegation method.
Problem 2:   Two trains 180 metres and 120 metres in length are running towards each other on parallel rails, first one at the rate 65 km per hours and second at 55 km per hour. In how many seconds will they can be clear each other from the moment they meet?

Solution: Here, we have to calculate the relative speed. As per the question both trains are not running in the same direction. So, the relative speed of both trains will be the sum of their speed. Relative speed =65+55=120 km per hours. And they have to cover the length of both trains. The sum of both trains is 180+120 =300 metre. 
Problem 3: A furniture seller allows 4% discount on his marked price. If the cost price of an article is Rs. 8 000  and he has to make a profit of 20%, then how many rupees have to mark as the price of the furniture?
Problem 4: There are 480 coins in half rupees, quarter rupees and 10 paise coins and their values are proportional to 5:3:1. The numbers of coins in each case are 
Solution: The ratio of value = 5:3:1 and ratio of numbers =10:12:10 or 5:6:5.
Therefore, numbers of 50 paise coins = (5/16) *480=5*30=150.
Number of 25 paise coins =(6/16)*480 =6*30=180.
As ration of 10 paise coins and 50 paise coins are equal so it will be same. Therefore, the numbers of 50, 25 and 10 paise coin will be 150,180 and 150 respectively. 


Friday, 15 December 2017

Solved quantitative aptitude problems for competitive exam #2

 This is the second part of solved quantitative aptitude problems for competitive exams. We are providing some important quantitative aptitudes for quicker math solution using maths tricks. Let's looks some previous years math problems and easy solution. 

Problem1: A boat travels 24 km in upstream in 6 hours and 20 km downstream in 4 hours. Then the speed of the boat in still water and the speed of the current are-
Solve: We knew the travelled distance of upstream and downstream.We need the speed of upstream and downstream of the boat to execute the answer by shortcut formula.
Upstream rate =24/6 =4 kmph.
Downstream =20/4 =5 kmph. 
solved-quantitative-aptitude-problems-for competitive exam-by-math-tricks
Problem 2: The single discount equivalent discount to two successive discounts of 40% and 10% is-
Answer:  Look at below math tricks, how solved it quickly. 

Problem 3: A single discount equivalent to consecutive discount of 20%, 20% and 10% is-
Quick solve: 
Problem 4: Average of three numbers is 40. The first number is twice the second and the second one is thrice the third number. The difference between the largest and the smallest number is-
Solution: here 1st no>2nd no>3rd no. If we assume 1st no as x, then 2nd and 3rd number will be fractional numbers. To avoid that we assume the lowest number (3rd no) as x. let's look at the quick solution. 
Problem 5: The average score of a cricketer has 60 in ten innings. How many runs to be scored in the eleventh innings to raise the average score to 65?  
Solution: The average score of 10 innings is 60, means the batsman scored (60*10) =600 runs. As it will be (65*11) =715 after eleventh innings. So the batsman must be the score in eleventh innings (715-60) =115 runs.
Problem 6: In an election, 80% of the voter has voted and three candidates contested. The first candidate got 30% votes and second got 46% votes. If the total number of registered voter were 50,000, then find the number of votes got by the 3rd candidate. 
Solution: Look at the question 
  • All voters are not voted, only 80% voters are voted and fifty thousand of voter are registered. 
  • The 3rd candidates got votes 100-(30+46) =24%. 

Read more: 

Tuesday, 12 December 2017

Quantitative aptitude problems and solutions - Math Tricks

This is the first page of solved quantitative aptitude problems. Here, we will provide most important competitive math question and easy solution using math tricks. Earlier, we have provided you chapter-wise concept and shortcut math formula.  On that tutorials, we have taken some easy math problems to apply shortcuts formula. But now, we are going to apply shortcuts formula in exam level math problems. All the quantitative aptitude will from various chapters. If you need chapter wise math concept, shortcut formula and tricks then click below and read your desire chapter.

Problems 1: If the cost price of an article is 80% of its selling price, then profit percentage will be-

Solution:- Let's selling price is 100 units. And cost price will be 80 unit as per the question. And gain =100-80=20. So, apply to the executing of the percentage of profit formula. Because we know gain and cost price. 

Math Problems 2:- A milkman bought 70 litres of milk for rupees 630 and added costless 5 litres water. If he sells it 9 rupees per litres, then find his profit percentage. 

Solution:- We will use above formula again. To use that formula, we have to find total profit he made.  

Profit= (Selling price - Cost price)
  (70+5)*9 -630=  675-630 = 45 Rupees.
Now apply the above formula: 

Problem 3: A train starts running from a place at 1 P.M at the rate of 18 km per hours. Another train starts from the same place at 3 P.M in the same direction and over-takes the first train at 9 P.M. The speed of the second train in km/hour is -
Problem 4: Water and sugar in two vessels A and B are in the ratio5:3 and 5:4 respectively. In what ratio, the liquids in both the vessels are mixed to obtain a new mixture in vessel C in the ratio 7:5?
Solution: We can solve this problem using arithmetic operation. But the easiest method is the allegation. We use it for the quicker solution. 


Sunday, 10 December 2017

Average shortcut formula for competitive exam- New Math Tricks

Average math tricks one of the most important aptitude chapters in competitive math. So we should learn the average math tricks very carefully. Even you should need average maths trick to solve other math chapters. If you have the clear concept of average shortcuts formulas then you will get both advantages. Therefore, we are going to discuss the arithmetic chapter average. Before discussing the math tricks of average, let's clear the concept.
What is average?: The average is the mean value of two or more numbers. Suppose you and your two friends had gone for fishing and caught 15 fishes in total. Now you are going to distribute fish among you equally. As you are distributing equally, obviously distributing value is the mean value. Now you are getting five fish each. How it calculated? Simply you added yours caught fishes and divide by three (observation numbers) because you are three friends. In this way tried to find out the mean value. It is the average and the condition is easy. When the condition becomes more complicated, then some average math tricks help you to solve problems quickly and effectively.

Average quantitative aptitudes and math tricks

Aptitude question 1: Find out the average (or mean value) of 1 to 100 numbers?
Solve: You can solve this problem by two methods. First, you can add all the numbers like 1+2+3+4+.......+98+99+100 and then divide by 100 because there are 100 numbers between 1 to 100. Or you can apply a math trick which is (1+100) / 2.  This is the advantage of average math tricks. But the condition should be increase or decrease in sequence order. Otherwise, this formula will not apply. Look at another aptitude question on average.
Find the mean value or average of 2, 4, 6, 8, ........96, 98 and 100?
Here you are looking that the number is increasing 2 by its previous number. Using the mentioned math tricks you can find out the mean value or average of the series. Let’s look at below how we find out the result. (2+100) / 2 =102/2 =51. 
In above we have just calculated the average of some numbers. By using this formula you can also calculate the sums of numbers.Just you have to multiply with the observation numbers. Look at below all shortcut formulas in an image. 



Tuesday, 31 October 2017

Time and works shortcut formula and aptitude solution as math tricks 1.

In competitive arithmetic, time and works math chapter is the most asked chapter than other arithmetic chapters. Take any question paper, whether it group D level examination or upper-level examination, always several questions are asked from this section. Whenever you are preparing for competitive exam and Arithmetic is listed as a part of the syllabus, then you have to give effort in this chapter. Because one or two number score makes a significant difference from your competitor. So why not we should give time to know the section accurately? Another thing, we want to inform you that the Time and Works shortcut formula also helps you to solve some math problems of pipe and cistern chapter. Here we are going to discuss the shortcut formula of time and works, though we will give you some aptitude question and solution with traditional methods which you had used in school level education. That is not as valuable as like the shortcut formula in competitive examination. Because of competitive exams, always take in short time.
The basic concept of time and works:-Whenever we face a question of the time and works math chapter, you may see Worker or machine works for a specific work. And work is completed by the worker or machine by taking time. So worker, times and works these three thing is essential to solve any aptitude question from this chapter. Actually, these three things have a relationship, and we have to drive it on an equation, then the math problem will be easy to evaluate the correct answer.
Time and works aptitude tricks
1st shortcut formula of time and works:-When in a question, declare two-person or machine and their separate workability to complete work and ask you how much time they will take to finish the job if they work together. Then use the below formula to get the answer quickly.
In the above formula, you are looking that 'man1' and 'Man2' are the time value of two persons. Be careful this when you use this method.

2nd shortcut formula of time and works:- When in question declare three persons and their separate workability to complete the whole works. And ask you, if they work the entire job altogether then how long time they will take to finish works? As this kind of question, use the below formula.  
Please note here X. Y & Z is time value for three people or machine.
 3rd shortcut formula of time and works:-When three pairs of workers can( A&B, B&C  and C&A) do a piece of work in X, Y and Z days respectively. If they work altogether then how long time they will take to finish the job? As this kind of condition uses the shortcut formula. 
in the next tutorial, we will apply these time and works math tricks to some aptitude question.  


Saturday, 23 September 2017

Quadrilateral properties, shortcut formulas and aptitude question

Quantitative aptitude related to quadrilateral properties is one of the important mensuration chapters. Because every competitive examiner asks one or several questions from this chapter in quantitive aptitude section. It is same important as Triangle, Circle and many other mensuration chapters. Here we are going to inform you the quadrilateral properties, shortcut formula, aptitude question and easy solution by math tricks.
What is quadrilateral? The quadrilateral is a figure which has made by four arms and having four angles. There are many types of quadrilateral figures. So, we have to get the clear idea of each quadrilateral figure and shortcut formula to solve any aptitude question related to the quadrilateral. Here we are going to discuss the properties and shortcut formulas one by one. 
Quadrilateral properties and mensuration formulas.
Mensuration formula of the square:- Square one of the important part of the quadrilateral. It is mainly four equal-arms and Angeles figure where all angels are the right angels.

Formula for the perimeter, area and diagonal of square

Perimeter formula for the square is straightforward. The perimeter is the outer length of the figure. So you can calculate the perimeter of a square by adding the length of all arms or multiply an arm by four because of all sides is equal.
Formula for the area of Square:- 
 As we know that all arms are equal to the square. So, we can calculate as Arm^2 or Arm*Arm.
Formulas-related to diagonal of the square:- A diagonal divides a square into two equal triangles. We can find the area of a square by knowing the length of a diagonal. So examiner asks questions by giving the length of diagonal. Here we included triangle in the square because we will use the Pethegoius formula.

Find area, perimeter and arm length by the diagonal of a square.

Above, we describe all formulas to find out the area, perimeter and arm length by knowing the diagonal length. Examiner always asks perimeter, area and side length by giving the length of diagonal of a square. Here we used the Pythagorean theorem to find the hypotenuse of a triangle. As you know that the diagonal divides a square into two equal triangles. So the diagonal is the hypotenuse of a triangle. 
Properties of Rectangular:- 
The rectangular is a particular shape of the quadrilateral.The properties of rectangular are shown below.

  • Opposite arms are equal and parallel.
  • All angels are the same and right angle.
  • Rectangular has two equal diagonals and bisect each other equally.
  • All square are Rectangular.

Mensuration shortcut formulas of Rectangular:-
The perimeter of a rectangular is the sums of all outer lines. As opposite arms are equal, so we can calculate the perimeter of rectangular by the formula 2(length+base).
Area of a rectangular =(lenght*Base)
Diagonal of a rectangular:- To calculate diagonal of rectangular you can use the Pethegoius formula. Because diagonal of rectangular divide it into two equal triangles. And diagonal become the common hypotenuse of two triangles. As per the Petthegoius formula hypotenuse is 

To use this formula in rectangular length will be length and hight will be the base of the rectangular.
In the next tutorial, we will apply all formulas which are maintained on this topic. After then we will give you few more kind of quadrilateral properties, shortcut formula and some solved aptitude problems.

Please comment below if you got any difficulty to get understand and also don't forgot to share this post if you enjoyed learning about the quadrilateral properties and shortcut formulae.

Copyright © New math tricks for quantitative aptitude | ---------- Terms of Use-Privacy Policy II II About Us-Contact Us Design by ronangelo >