IPMAT 2025 Exam Solved Papers

Let us assume the number as xyzTotal outcomes = All the numbers between 100 and 400 for which the sum of digits = 10Favourable outcomes = All the numbers between 100 and 400 for which the sum of digits = 10, and the number is divisible by 4.$$P=\frac{\text { Favourable Outcomes }}{\text { Total Outcomes }}$$Now, for the total outcomes we have 3 cases as per the hundredth place digit.Case-1: When the hundredth place digit = 1 (x=1)Since the sum of digits $=10,(y+z)=9$. Thus, there are 10 possible pairs for $(y, z)=(0,9),(1,8),(2,7), \ldots \ldots,(9,0)$= > 10 possibilitiesCase-2: When the hundredth place digit = 2 (x=2)Since the sum of digits $=10,(y+z)=8$. Thus, there are 9 possible pairs for $(y, z)=(0,8),(1,7),(2,6), \ldots \ldots,(8,0)$ = > 9 possibilitiesCase-3: When the hundredth place digit = 3 (x=3)Since the sum of digits $=10,(y+z)=7$. Thus, there are 8 possible pairs for $(y, z)=(0,7),(1,6),(2,5), \ldots \ldots,(7,0)$ = > 8 possibilitiesFor 400, the sum of the digits is 4; thus, we don't need to worry about 400.Thus, the total outcomes $=10+9+8=27$For the favourable outcomes we need to check whether the number's last two digits are divisible by 4, that is, yz is divisible by 4 or not.For Case-1 = > the only possible values for $\mathrm{yz}=36$ and 72. Thus, 2 possibilities.For Case-2 = > the only possible values for $\mathrm{yz}=08,44$, and 80. Thus, 3 possibilities.For Case-3 = > the only possible values for yz = 16 and 52. Thus, 2 possibilities.So, number of favourable outcomes $=2+3+2=7$$$\Rightarrow P=\frac{\text { Favourable Outcomes }}{\text { Total Outcomes }}=\frac{7}{27}$$