ISI JRF Previous Year Papers with Solution

Although with cyclicity this would be easy. Here is an alternative using binomial theorem only: $3^{999} \equiv 3 (10 - 1)^{499} \equiv 3 (10 - 1) \equiv - 3 \equiv 7 \text{ mod } 10$ and $7^{1000} \equiv (10 -3)^{1000} \equiv (-1)^{1000} (3)^{1000} \equiv (10 - 1)^{500} \equiv 1 \text { mod 10}$ Now $(3^{999} \times 7^{1000}) \text { mod }  10 \equiv \left[(3^{999} \text{ mod } 10) \times (7^{999} \text{ mod } 10)\right] \text { mod 10 } \equiv 7 \text { mod } 10 \equiv 7.$ [...]

I think the answer is option B, I am not going to describe everything but I will say this, see where we subtract cross product result with the actual result, now here is the catch say one female student has enrolled into every subject will she be appearing on the list, No because when taking the cross product with female std if and all the subjects and subtracting with set where a female has taken all subjects will cancel and we will not the get of female students who have taken all courses but only I'd of female STDs who [...]

In each iteration of the bubble sort, the ith largest element finds its correct position.For example, given the sequence 1, 2, 4, 3, in the first pass, 4 moves to its correct position [1, 2, 3, 4]. But bubble sort will do another pass to check it is sorted or not Subsequently, as there are no swaps in the second pass, the algorithm recognizes that the array is sorted and terminates.Hence, in just two passes, we achieve a sorted array [...]

B [...]

9 [...]

Given information: - Dick is thrice as old as Tom. Let Tom's age be x. - Harry is twice as old as Dick. - Dick's age is 1 year less than the average age of all three. Step 1: Express Dick's age in terms of Tom's age (x). Dick's age = 3x Step 2: Express Harry's age in terms of Dick's age. Harry's age = 2 × Dick's age = 2 × 3x = 6x Step 3: Set up an equation for Dick's age being 1 year less than the average age of all three. Dick's age = (Tom's age [...]