SSC CGL Tier 1 Math Short Tricks & Formulas
The level of questions asked in Math section in Tier-1 level of SSC CGL exams is fairly basic. Usually nothing beyond secondary school-level math is asked. However, the questions might be tricky or calculative. Anyone can solve these questions if enough time is provided but the shortage of time, especially in tier-1 examination makes you feel as if you were participating in a 100-meter dash. You can save yourself from this torture if you are able to remember and apply some formulas/short tricks correctly. So today we will give you some of the most commonly used shortcut methods and formulas to help you score good marks in Math section of SSC CGL Tier-1.
A word of caution!
Before you start memorizing the tricks and formulas,
Make sure you should have a thorough understanding of the concept. These short tricks are only helpful to those students who have totally mastered a concept. Those who learn the shortcut before fully understanding the concept will likely use the shortcut incorrectly or fail to use it when the problem is presented differently. Additionally, many shortcuts do not apply to all situations. If your concepts are clear you might even be able to develop your own shortcut methods as you practice that topic.
Never try to memorize the formula/tricks without practicing it. The best way to memorize them is to use them by practicing examples.
Try to solve each question using different methods.
Following are some shortcut methods which you can use to save time
1. Assuming a particular value for the unknown variable.
Many of the questions don’t require you to solve all complex equations. One can avoid spending time in solving complex equations by simply picking up a number and substituting it for the unknown variable.
In a class of x students, one-third of the students did not participate in the quiz and while one-fourth of those who participated failed in the quiz. What percent of the total number of students in the class passed the quiz?
In this case, we should choose a number that is the nearest multiple of both 4 and 3. Let us take x=12 i.e. the lowest common multiple. It means 4 students did not participate in the quiz, and the rest i.e. 8 students participated. One fourth i.e. 2 failed the quiz. So, the number of students who passed the quiz=6 i.e. 50% of the total students.
2. Option elimination technique.
This option involves going through all the options before proceeding to solve the given question.
Sometimes the questions are asked in such a way so as to see if one can get the right answer efficiently. Approximation technique is very helpful to eliminate wrong options.
Q. A box contains 80 white and 100 black shoes. If 10% of the white shoes and 20% of the black shoes are defective, what percentage of the total number of shoes (including black and white are defective?
Judging by the question we can say that our option will be somewhere between 10 and 20, so option A and D will be eliminated. The unweighted average of 10% and 20%, comes out to be 15% but as the number of black shoes is more than the number of white shoes, this average will be somewhat higher than 15, so 15.5 will be our answer.
3. By looking at the last Digit of each option,
Sometimes we can also determine correct answer or eliminate wrong answers by determining the last digit of the answer.
Q. What will be the resultant of 1x2x3……x9
A. 362770 B. 362842 C. 372824 D. 362885
As the option includes multiplication of 2 and 5, our answer must include a zero, so our option B, C, and D will be eliminated.
4. Hit and Trail method.
This technique involves back solving by using different answer choices. This method can not only save our time but is also useful when one does not know how else to solve a given question. In this method, we start with an option that lies in/near the middle of the two extreme options provided. Even if the chosen option does not balance the equation, we will be able to determine whether the number will be higher or lower and choose our next alternative accordingly.
What is the smallest positive integer x for which x2 + 10x is more than 50?
Sol. Starting with the middle value is x=4, we get x2 + 10x = 56, which satisfies the given condition, so option D and E will be eliminated. Now trying option B, i.e. x=3 we get x2 + 10x = 39 which is less than 50, so option A and B will also be eliminated, and we get C as our correct answer choice.
5. Shortcut method to calculate square of any number between 11 and 99.
Step 1. Find the nearest multiple of ten.
For example: To calculate square of 18.
Then add and subtract the difference between that multiple and the number to obtain two different numbers.
20-18=2=D, So the numbers are 16 and 20.
Step 3. Multiply those numbers and add the square of D to the resultant.
16×20=320 + 4= 324
Besides the above-mentioned techniques, following are some important formulas will be helpful in preparing for Tier-1 of SSC CGL.
(a + b)2= a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(a + b)3 = a3 + b3 + 3ab(a + b)
a3 + b3 = (a+b)3 −3ab(a + b)
(a − b)3 = a3 − b3 − 3ab(a − b)
a3 − b3 = (a−b)3 + 3ab(a − b)
a2 − b2 = (a+b)(a − b)
a3 − b3 = (a−b)(a2 + ab + b2)
a3 + b3 = (a+b)(a2 − ab + b2)
am .an = am+n
am/ an = am-n
(am)n = amn
a0 = 1
Avg. of n numbers in arithmetic progression = (first number + last number)/2
For eg. Average of 2,4,6,8 = (2+8)/2= 5
Sum of all the first n natural numbers =n(n+1)/2
For eg. Sum of first four numbers = 10 (1+2+3+4)
Sum of first n odd numbers = n2
Sum, of first n even numbers = n (n + 1)
Sum of Squares of first n natural numbers = [n(n+1)(2n+1)] /6
Sum of cubes of first n natural numbers = [n(n+1)/2]2
When the distance is constant,
If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr, then the average speed = 2xy/(x+y)
When the time is constant,
If a person travels at a speed of x km/hr for one half of the time and at the speed of y km/hr for rest half of the time, then the average speed = (x+y)/2
If the price of a commodity increases by x% , then the reduction in consumption so as not to increase the expenditure= [x/(1+x) * 100]%
If the price of a commodity decreases by x% , then the increase consumption so as not to decrease the expenditure= [x/(1-x) * 100]%
If the value of a number is first increased by x% and later decreased x%, then the net change is always a decrease of x2/100
If the value is increased by x%, followed by a decrease of y%, then net change =[x-y-(xy/100)]
If the value is increased by x%, followed by an increase of y%, then net change =[x+y+(xy/100)]
If a shopkeeper cheats by using false weights, then
Gain % = [ (Actual weight- claimed weight)/claimed weight] x 100
If two trains have started from two opposite points towards each other at the same time.
Then the ratio of their speeds =
St1/St2 = (time taken by second train to reach the destination after crossing the first train/ time taken by first train to reach the destination after crossing the second train)
Also check our NCERT Solutions app which we have launched recently for Class 6 – Class 12.