Question 48

A, B, C, D and E are five employees working in a company. In two successive years, each of them got hikes in his salary as follows:A : p% and (p+1)%,B : (p+2)% and (p-1)%,C : (p+3)% and (p-2)%,D : (p+4)% and (p-3)%,E : (p+5)% and (p-4)%.If all of them have the same salary at the end of two years, who got the least hike in his salary?

Solution

Let the initial salary of A,B,C,D,E be $$a,b,c,d,e$$ respectively and let the final salary of everyone be $$x$$

Now, $$a*(1+ \frac{p}{100})*(1+ \frac{p+1}{100}) = x$$

$$\Rightarrow a = \frac{x}{(1+ \frac{p}{100})*(1+ \frac{p+1}{100})}$$

$$\Rightarrow a = \frac{x*100*100}{(100+p)*(100+p+1)}$$

$$\Rightarrow a = \frac{x*100*100}{(p+100)*(p+101)}$$

Similarly, $$b= \frac{x}{(1+ \frac{p+2}{100})*(1+ \frac{p-1}{100})}$$

$$\Rightarrow b = \frac{x*100*100}{(p+102)*(p+99)}$$

Similarly, $$c= \frac{x}{(1+ \frac{p+3}{100})*(1+ \frac{p-2}{100})}$$

$$\Rightarrow c = \frac{x*100*100}{(p+103)*(p+98)}$$

Similarly, $$d= \frac{x}{(1+ \frac{p+4}{100})*(1+ \frac{p-3}{100})}$$

$$\Rightarrow d = \frac{x*100*100}{(p+104)*(p+97)}$$

Similarly, $$e= \frac{x}{(1+ \frac{p+5}{100})*(1+ \frac{p-4}{100})}$$

$$\Rightarrow e = \frac{x*100*100}{(p+105)*(p+96)}$$

The numerators of the fractions are same, therefore the one with the smallest value of denominator will have the greatest value. If we compare the denominators, we can find out the fraction with the highest value. The person with the highest initial salary got the least raise, as we know that the final salary of all the candidates is same.

Thus, denominator of a, $$a_{den}= (p+100)*(p+101)= p^{2}+201p+100*101$$

Similarly, $$b_{den}= (p+102)*(p+99)= p^{2}+201p+102*99$$

$$c_{den}= (p+103)*(p+98)= p^{2}+201p+103*98$$

$$d_{den}= (p+104)*(p+97)= p^{2}+201p+104*97$$

$$e_{den}= (p+105)*(p+96)= p^{2}+201p+105*96$$

We see that we need to compare only the last terms of the denominators as the other terms are same.

Thus, last term of a, $$a_{lt}= 100*101= 100.5^{2}-0.5^{2}$$

last term of b, $$b_{lt}= 102*99= 100.5^{2}-1.5^{2}$$

last term of c, $$c_{lt}= 103*98= 100.5^{2}-2.5^{2}$$

last term of d, $$d_{lt}= 104*97= 100.5^{2}-3.5^{2}$$

last term of e, $$d_{lt}= 105*96= 100.5^{2}-4.5^{2}$$

Thus, we can see that since denominator of $$e$$ is the smallest, therefore E has the highest initial salary.

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