Algebra
Aptitude

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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

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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.


(2) Comment(s)


Pooja
 ()

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


Priyanshi
 ()

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.