IPMAT 2025 Exam Solved Papers

Let the total work done be 1 unit.Now, let us assume the total number of days taken by Arpita is A, then her efficiency will be $=\frac{1}{A}$And the total number of days taken by Nikita is N, then her efficiency will be $=\frac{1}{N}$It is given that they complete the work together in 12 days. Thus $=\frac{1}{A}+\frac{1}{N}=\frac{1}{12} \rightarrow 1$Now, it is given that Arpita works initially to complete 40% of the job, which means she does 0.4 units of work.For 1 unit, she was taking A days; for 0.4 units of work, she will take 0.4 A days.Then, Nikita completes the remaining job alone. Thus, Nikita will do 0.6 units of work, which will take 0.6 N days. It is given that in this way, they take 24 days to complete the job.$$\begin{aligned} & \Rightarrow 0.4 A+0.6 N=24 \rightarrow 2 \\ & \Rightarrow 2 A+3 N=120 \\ & \Rightarrow A=\frac{(120-3 N)}{2} \rightarrow 3 \end{aligned}$$From eq. 1, we can get $= > \frac{(A+N)}{A N}=\frac{1}{12}$$$\Rightarrow 12 A+12 N=A N$$Substituting the value of A from eq. 3 -$$\begin{aligned} & \Rightarrow 12\left(\frac{120-3 N}{2}\right)+12 N=\left(\frac{120-3 N}{2}\right) N \\ & \Rightarrow 1440-36 N+24 N=120 N-3 N^{2} \quad \text { (Dividing the entire equation by 3) } \\ & \Rightarrow 480-4 N=40 N-N^{2} \\ & \Rightarrow N^{2}-44 N+480=0 \\ & \Rightarrow N^{2}-24 N-20 N+480=0 \quad \text { (Splitting the middle term) } \\ & \Rightarrow(N-24)(N-20)=0 \\ & \Rightarrow N=20 \text { or } N=24 \end{aligned}$$Thus, Nikita alone takes to complete that work is either 20 or 24 days.