RRB NTPC Prelims Model Paper 1 with solutions

Concept:The sum is found using permutations and place values.Explanation:The digits available are $0,2,4,6,8$.Numbers greater than $10,000$ must be five-digit numbers.Total five-digit permutations with no repetition and no leading zero are $5!-4! = 120-24 = 96$.First, find the sum of all permutations of these five digits, allowing leading zero.Each digit appears in each place $5!/5 = 24$ times.Sum of digits is $0+2+4+6+8 = 20$.So the place-value sum is $24 \times 20 \times (10000+1000+100+10+1)$.$=480 \times 11111 = 5333280$.Now subtract permutations with leading zero.These are four-digit numbers formed by $2,4,6,8$.Each digit appears in each place $4!/4 = 6$ times.Their sum is $6 \times 20 \times 1111 = 133320$.Therefore, required sum $=5333280-133320=5199960$.Answer:The required sum is $5199960$, so the correct option is A.