Aptitude Discussion

Q. |
A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible? |

✖ A. |
3 |

✖ B. |
4 |

✖ C. |
5 |

✔ D. |
6 |

**Solution:**

Option(**D**) is correct

Let the no. of students in front row be $x$.

So, the no. of students in next rows be $x–3,x−6,x–9$ ... so on

If $n$ i.e. no. of rows be then no. of students $(n=3)$

$x+(x–3)+(x–6)=630$

$3x=639$

$x = 213$

So possible,

Similarly for $n = 4$

$x+(x–3)+(x–6)+(x−9)=630$

$4x–18=630$

$\Rightarrow x=162$

If $n = 5$

$(4x–18)+(x−12)=630$

$5x–30=630$

$x=120$

Again possible.

If $n = 6$

$(5x−30)+(x−15)=630$

$6x−45=630$

$6x=675$

$x ≠$ Integer

Hence **n ≠ 6**

**Edit:** Thank you **Priyanshi** for providing an alternate solution in the comments.

**Pooja**

*()
*

The other method could be as follows,

We know that given series is A.P. with common difference $l=-3$.

Now, sum of A.P.

$S_n =\dfrac{n}{2} (2a+(n-1) l)$

put values,

$630 =\dfrac{n}{2} [2x+(n-1) \times -3]$

$\dfrac{630 \times 2}{n}=[2x-3(n-1)]$

$\dfrac{1260}{n}+3(n-1)=2x$

$\Rightarrow x=\dfrac{1}{2} \times\left( \dfrac{1260}{n}+3(n-1)\right)$

Now we only need to check for what value of $n$, $x$ is not an integer.

Upon checking from the options, we see that $n=6$ is the right choice.

give easier methods if possible.......