JEE Novice - (25)

Algebra Level 2

1 2 + 1 4 + 1 6 + 1 8 + 1 10 + 1 12 \Large \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10}+\dfrac{1}{12}

Which terms must be removed from the above summation , so that the sum of the remaining terms becomes equal to 1?

This question is a part of JEE Novices .
1 6 , 1 8 , 1 12 \dfrac{1}{6} , \dfrac{1}{8}, \dfrac{1}{12} 1 2 , 1 6 , 1 10 \dfrac{1}{2} , \dfrac{1}{6} , \dfrac{1}{10} 1 4 , 1 12 \dfrac{1}{4} , \dfrac{1}{12} 1 8 , 1 10 \dfrac{1}{8} , \dfrac{1}{10} 1 8 , 1 12 \dfrac{1}{8} , \dfrac{1}{12} 1 4 , 1 8 \dfrac{1}{4} , \dfrac{1}{8}

View wiki
Nihar Mahajan by Nihar Mahajan , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

4 solutions

Majed Musleh
May 28, 2015

t h e ∗ s u m m a t i o n ∗ e q u a l ∗ t o 49 40 the *summation* equal *to \frac{49}{40}

49 40 = 40 40 + 9 40 \frac{49}{40}=\frac{40}{40}+\frac{9}{40}

9 40 = 1 8 + 1 10 \frac{9}{40}=\frac{1}{8}+\frac{1}{10}

11 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Great job!

Ahmed Obaiedallah
Jun 10, 2015

1 6 + 1 12 = 2 12 + 1 12 = 3 12 = 1 4 \frac16+\frac{1}{12}=\frac{2}{12}+\frac{1}{12}=\frac{3}{12}=\frac14

so;

1 2 + 1 4 + 1 6 + 1 12 = 1 \frac12+\frac14+\frac16+\frac{1}{12}=1

and

1 8 + 1 10 \frac18+\frac{1}{10} will be the removed parts

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Nasim Bappi
Aug 8, 2015

First find the LCM of the denominators,that is 120.now we can write the expression as below: (60+30+20+15+12+10)/120 Now we have to find 120 by deducting numbers from numerator and its 15 & 12.that means 8 & 10.so the result is 1/8 & 1/10...

1 Helpful 1 Interesting 1 Brilliant 0 Confused

i thought it has to do with the Least Common Multiple. so i choose 1/8 & 1/0

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...