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Q.

Three gold coins of weight 780gm, 840gm and 960gm are cut into small pieces, all of which have the equal weight. Each piece must be heavy as possible. If one such piece is shared by two persons, then how many persons are needed to give all the pieces of gold coins?

 A.

86

 B.

70

 C.

43

 D.

35

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Solution:
Option(A) is correct

$\text{HCF}(780, 840, 960) = 60$

Thus, total number of pieces:
$\begin{align*}
\Rightarrow \dfrac{780}{60} + \dfrac{840}{60} +\dfrac{960}{60} &= 13+14+16\\
&= 43
\end{align*}$

Total number of person required $= 43×2 =$ 86.


(5) Comment(s)


Priya
 ()

Please, can anyone solve this?

I have my elitmus exam tomorrow

how many solutions to this equation:

$\sqrt{x+7}=x \times \sqrt{x+7}$

according to me, the answer will be 3


Anamika
 ()

$x=1$ is the solution, what's there to solve here?


Andy
 ()

becoz each coin is shared by two persons..total number of coins are 43..so total number of people are 43*2=86



Deepali
 ()

Why is 43 multiplied by 2 at the end?


Balu
 ()

for one person it is 43 pieces so for 2 persons it is 43*2