Aptitude Discussion

Q. |
$a, b, c, d$ and $e$ are five consecutive numbers in increasing order of size. Deleting one of the five numbers from the set decreased the sum of the remaining numbers in the set by 20%. Which one of the following numbers was deleted? |

✖ A. |
a |

✖ B. |
b |

✔ C. |
c |

✖ D. |
d |

✖ E. |
e |

**Solution:**

Option(**C**) is correct

Since $a, b, c, d$ and $e$ are consecutive numbers in the increasing order,

We have $b = a + 1$, $c = a + 2$, $d = a + 3$ and $e= a + 4$.

The sum of the five numbers is:

$a + ( a + 1) + ( a + 2) + ( a + 3) + ( a + 4) = 5 a + 10$

Now, we are given that the sum decreased by 20% when one number was deleted.

Hence, the new sum should be:

$\begin{align*}

&=(5 a + 10)\left(1 - \dfrac{20}{100}\right) \\

&= (5 a + 10)\left(1 - \dfrac{1}{5}\right)\\

&= (5 a + 10)\left(\dfrac{4}{5}\right)\\

&= 4 a + 8.

\end{align*}$

Now, since $\text{New Sum = Old Sum - Dropped Number}$

We have,

$(5 a + 10) = (4 a + 8) + \text{Dropped Number}$.

Hence, the number dropped is $(5 a + 10) - (4 a + 8) = a + 2$.

Since $ c = a + 2$, the answer is **(C).**

**Geetha**

*()
*

**Apoorv Jain**

*()
*

another method:-

assume numbers as 1,2,3,4,5.

sum = 15.

now what is the 20% of 15.... ans is 3.

so we can conclude that if 3 is remove then the sum will be decrease by 20%.

hence, 3 (or the 3rd number is the answer).

can we do in another method like

20% of the total sum

$=(5a+10)*(20/100)$

then also we are getting same answer $a+2$