Zeros at the end means multiple of 10, i.e. (2 × 5).

•  Now, 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60 = 1 × 5 × (5 × 2) × (5 × 3) × (5 × 22) × (5 × 5) × (5 × 3 × 2) × (5 × 7) × (5 × 23) × (5 × 9) × (5 × 5 × 2) × (5 × 11) × (5 × 22 × 3)

•  Here, number of 2s = 10

•  And number of 5s = 14

•  The lesser of the two will determine the number of zeros.

•  Hence, there will be 10 zeros at the end in the given expression.

•  XYZ is a three-digit number.

•  So, we can write this as: 100X + 10Y + Z (as X is at hundreds place and Y is at tens place)

•  So, XYZ + YZX + ZXY = (100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y) = 111 (X + Y + Z)

•  Hence, it is divisible by (X + Y + Z).

•  111 is divisible by 3, as well as 37.

•  So, the expression XYZ + YZX + ZXY is not divisible by 9.

After tearing away the page, we get a sum of 195 from remaining page numbers. Since, numbers are in sequence, we use arithmetic progression.

Sum of consecutive numbers from 1 to n = n(n+1)/2 = 195 or n(n+1)= 390 with one page torn.

If we add the digits on torn page, we will get a number more than 390.

For n to be an integer, the minimum possible value of n(n+1) over 390 is received when we put n = 20

So, n(n+1) = 20 × 21 = 420

So, sum of 20 pages = n(n+1)/2 = (20 × 21)/2 = 210

Hence, sum of the two numbers on the torn page = 210 – 195 = 15

Only option b gives 15 as sum.

Hence, 7,8 is the answer.

 •  G is 3rd from V towards the right, so position of G = 18 – 3 = 15th from right end.

•  G is 11th from A towards the right, so position of G = 16 + 11 = 27th from left end.

•  Hence, total number of boys in the row = 27 + 15 – 1 = 41 (we subtract one because G has been counted twice)

•  Let x = 1.272727….

Then 100x = 127.272727….

So, 100x – x = (127.272727….) – (1.272727….) = 126

Or x = 126/99 = 14/11