UPSC CDS 2 2021 Math Paper

Let all the men do the same amount of work in one day and one day work of each men = 1 unit; we get
Amount of work done by (3n−1) men in (2n+1) days
=x=(3n−1)(2n+1) units of work.
Similarly, amount of work done by (3n+1) men in (4n−3) days
=y=(3n+1)(4n−3) units of work.
According to the question,
‌

x

y

=‌

6

11

⇒‌

(3n−1)(2n+1)

(3n+1)(4n−3)

=‌

6

11

⇒‌

6n2+3n−2n−1

12n2−9n+4n−3

=‌

6

11

⇒‌‌‌

6n2+n−1

12n2−5n−3

=‌

6

11

⇒‌‌66n2+11n−11=72n2−30n−18
⇒‌‌0=6n2−41n−7
⇒‌‌6n2−42n+n−7=0
⇒6n(n−7)+1(n−7)=0
⇒‌‌(n−7)(6n+1)=0
⇒‌‌n=7,n≠−‌

1

6

[ ∵ number of men can't be negative]
Hence, the value of n is 7 .