Word Problems Shortcuts

Shortcuts for LIC AAO, Bank (PO + clerk), SSC, TCS, Infosys, HCL, Wipro, Accenture, Capgemini placement papers

Numbers play a very important role in our daily life. Questions based upon numbers are known as word problems. In order to solve word problems we have to extract information correctly from the question, then form equations based upon the information provided in the question. Equations thus formed can be linear equations in one variable, linear equations in multi variables, quadratic etc. depending upon the information provided in the question. Forming equations and then solving equations to get the answer to the question is a complex and a time consuming task. We need an approach that can help us solve word problems within seconds thus saving a lot of time. Time plays a very crucial role for students appearing for placement exams in companies like TCS, Infosys, HCL, Wipro, Cognizant, Accenture etc or for government jobs like SBI PO, SBI Clerk, IBPS PO, IBPS Clerk, LIC AAO, SSC (10+2, CGL, CPO), Railways, Hotel Management, CMAT, MAT etc to name a few. Now let us consider a word problem from SSC CGL 2013 examination: Question: On the occasion of Christmas, sweets were to be distributed amongst 300 children. But on that particluar day 50 children remained absent, so each child got one extra sweet. How many sweets were distributed? 1) 1450 2) 1700 3) 1500 4) 1650 Solution: Method-1: Form equations to solve the question. Let x be the total number of sweets. According to the question, Method-2 (Shortcut): As sweets are distributed among children equally so total number of sweets should be divisible by 300 (Total number of children). Hence, total number of sweets are 1500. So, we can see that using Method-2 we have solved this question within seconds. Question: If the numerator of a certain fraction is increased by 2 and the denominator is increased by 1, then the resulting fraction is equal to 1/2. If however, the numerator is increased by 1, and the denominator is decreased by 2, then the resulting fraction is equal to 3/5. Find the original fraction. 1) 2/7 2) 3/5 3) 1/7 4) 2/5 Solution: Method-1: Form equations to solve the question. Let the fraction be x/y. According to the question, Again, according to the question, On solving the two equations formed above, we get that Therefore, the original fraction is . Method-2 (Shortcut): Let us make use of options. Option 1 is . When numerator is increased by 2 and denominator is increased by 1 then our fraction becomes Hence, 1st condition of the question is satisfied. When numerator is increased by 1 and denominator is decreased by 2 then our fraction becomes Hence, 2nd condition of the question is satisfied. As option 1 satisfies both the conditions specified in the question so it is the answer. So, we can clearly see that using Method-2 (By making use of options) we have solved this question within seconds thus saving a lot of time. Download: How to use substitution method Let us discuss some previous year questions from some other competitive exams Question: Find the HCF of 2n + 13 and n + 7, where n is a natural number. 1) 1 2) n + 6 3) n + 7 4) Depends on n Solution: As ‘n’ is a natural number. So, we can substitute any natural number in place of n. Put n = 1. So, HCF of 2n + 13 and n + 7 = HCF of 15 and 8 = 1 When n = 1, then option 2 and 3 are eliminated. So, answer is either option1 or option 4. Now, put another value of n. Say n = 2. HCF of 2n + 13 and n + 7 = HCF of 17 and 9 = 1 Option 4 is also eliminated as the answer does not depend on ‘n’. Question: If a/b = 1/3, b/c = 2, c/d = 1/2, d/e = 3 and e/f = 1/4, then what is the value of abc/def? 1) 3/8 2) 27/8 3) 1/4 4) 27/4 Solution: a/b = 1/3 = 2/6, b/c = 2 = 6/3 c/d = 1/2 = 3/6, d/e = 3 = 6/2 e/f = 1/4 = 2/8 So, a = 2, b = 6, c = 3, d = 6, e = 2, f = 8 abc/def = 3/8 Question: If a + b + c = 0 the value of is 1) 2 2) 3 3) 4 4) 5 Solution: In this question, a + b + c = 0. We can substitute any values of a, b, c such that the given condition is satisfied. Put a = 1,b = 0 so c = -1 = 2