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by VJesus12 » Mon Mar 19, 2018 4:16 am
$$\frac{1}{3}+\frac{1}{2}−\ \frac{5}{6}+\frac{1}{5}+\frac{1}{4}−\ \frac{9}{20}=\ ?$$ $$(A)\ \ 0$$ $$(B)\ \ \ \frac{2}{15}$$ $$(C)\ \ \ \frac{2}{5}$$ $$(D)\ \frac{9}{20}$$ $$(E)\ \ \frac{5}{6}$$ The OA is A.

What is the best way to solve this PS question? May any expert give me some help? Thanks in advanced.
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by GMATGuruNY » Mon Mar 19, 2018 4:35 am
VJesus12 wrote:$$\frac{1}{3}+\frac{1}{2}−\ \frac{5}{6}+\frac{1}{5}+\frac{1}{4}−\ \frac{9}{20}=\ ?$$ $$(A)\ \ 0$$ $$(B)\ \ \ \frac{2}{15}$$ $$(C)\ \ \ \frac{2}{5}$$ $$(D)\ \frac{9}{20}$$ $$(E)\ \ \frac{5}{6}$$ The OA is A.

What is the best way to solve this PS question? May any expert give me some help? Thanks in advanced.
$$\frac{1}{3}+\frac{1}{2}−\ \frac{5}{6}+\frac{1}{5}+\frac{1}{4}−\ \frac{9}{20}=\ x$$

$$60(\frac{1}{3}+\frac{1}{2}−\ \frac{5}{6}+\frac{1}{5}+\frac{1}{4}−\ \frac{9}{20})=\ 60x$$

$$20+30-50+12+15-27 = 60x$$

$$0 = 60x$$

$$0 = x$$

The correct answer is A.
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by Vincen » Mon Mar 19, 2018 4:40 am
Hello Vjesus12.

I would solve it as follows:

$$\frac{1}{3}+\frac{1}{2}-\ \frac{5}{6}+\frac{1}{5}+\frac{1}{4}-\ \frac{9}{20}=\left(\frac{1}{3}+\frac{1}{2}-\ \frac{5}{6}\right)+\left(\frac{1}{5}+\frac{1}{4}-\ \frac{9}{20}\right)$$ $$\left(\frac{5}{6}-\ \frac{5}{6}\right)+\left(\frac{9}{20}-\ \frac{9}{20}\right)=0+0=0.$$ Hence, the correct answer is the option is [spoiler]A=0[/spoiler].

I hope it helps you.
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by Jeff@TargetTestPrep » Tue Mar 20, 2018 4:04 pm
VJesus12 wrote:$$\frac{1}{3}+\frac{1}{2}−\ \frac{5}{6}+\frac{1}{5}+\frac{1}{4}−\ \frac{9}{20}=\ ?$$ $$(A)\ \ 0$$ $$(B)\ \ \ \frac{2}{15}$$ $$(C)\ \ \ \frac{2}{5}$$ $$(D)\ \frac{9}{20}$$ $$(E)\ \ \frac{5}{6}$$ The OA is A.
We can find the sum the first 3 terms and the sum of the last 3 terms first, and then add the two sums together:

1/3 + 1/2 - 5/6 = 2/6 + 3/6 - 5/6 = 0

and

1/5 + 1/4 - 9/20 = 4/20 + 5/20 - 9/20 = 0

Thus the sum of all 6 terms is 0 + 0 = 0.

Answer: A

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by VJesus12 » Tue Mar 27, 2018 3:00 am
I'm very thankful for all your answer. <i class="em em---1"></i><i class="em em-grinning"></i>
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