M61. How many cuboids exist for which the volume is less than 100 units3 and the integer side lengths are in an arithmetic sequence?
Ans.  Let the side be xyz<=100
          The sides are in A.P. then y-x = z-y =d>=1
          Considet the cuboid of d=1,
          (x)(x+1)(x+2)
          Possible Pairs are (1,2,3),(2,3,4),(3.4.5)
          for d=2 (x)(x+2)(x+4) = (1,3,5),(2,4,6),(3,5,7)
          for d=3 (x)(x+3)(x+6) = (1,4,7),(2,5,8)
          for d=4 (x)(x+4)(x+8) = (1,5,9)
          for d=5 (x)(x+5)(x+10) =(1,6,11)
          for d=6 (x)(x+6)(x+12) =(1,7,13)
          Thus total cuboids are =11

M62. The number formed by writing any digit 6 times is always divisible by 1001 , 7 , 13 or all of these
Ans.  LCM of 1001=7x11x13 Thus LCM is 1001 digit 111111 is divisible by all of these thus any 6 digit no. is diviisible by
        all of them thus their multiple
M63. Find the exact value of the following infinite series:
1/2! + 2/3! + 3/4! + 4/5! + ...
Ans.  (2-1)/2! +(3-1)/3!+(4-1)/4!+...
         = 1 -1/2!+1/2! -1/3!+1/3!+.... = 1 ans

M64. How many digits are used to make a book containing 500 pages?
Ans.  1 digit no. 9
         2 digit no. 90x2=180
         3 digit no. from 100 to 500 = 501 no. and 501x3=1503 thus total digit

M65. The sum of the digits of a two digit number is 5. If we put the digits of the numbers in reverse order, the new number s 41 less than the original number. Find 33% of the number?
Ans.  10x+y is the no. then 10x+y=10y+x+41 Then 9(x-y)=41 then x-y = 41/9 and x+y=5 then 2x = 45+41/9 =86/9 => x = 43/9 and y = 2/9 then
          430/9+2/9 = 432/9 and 432x33/900 = 432x11/300 => 144x11/100 = 15.84 Ans.

M66. In a particular class, each student has blonde hair or brown eyes; 1/4 of students with blonde hair have brown eyes and 1/3 of the students with brown eyes have blonde hair. What fraction of the class in total have brown eyes?
Ans.  Total BE = x and Total BH =y then y/4 = x/3 then the total brown eyes = x/(x+y) = 1/( 1+y/x) = 1/( 1+4/3) = 3/7

M67. Sushil had to do a multiplication. Instead of taking 35 as one of the multipliers, he took 53. As a result, the product went up by 540. What is the new product?
Ans.  53x-35x = 540  thus x = 540/18 => 30 thus new product is 53x30 = 1590 Ans

M69. How many zeroes will be there at the end of 1 x 2 x 3 x . . . . . . x 1000 ?
Ans   It is 1000! thus the exponent of 5 and 2 is
         200+40+8 =248 for 5 power and 2 's exponent is 500+250+125+62+31+15+7+3+1 = 994 Here 248 power will give 248 zeros

M72.How many zeroes will be there at the end of 1^1 x 2^2 x 3^3x . . . . . . x 100^100 ?
Ans  Digit 5 is at 5,10,15,20,25,...Power of 5 is ( 5+ 10+15+20+25+30+...+100) = 5x105 =525 but the number 25 multiples are 25 + 50 +75+100 = 250
        Thus total is 775 'os are available ate the end.

M73. When travelling by aircraft, passengers have a maximum allowable weight for their luggage. They are then charged ?10 for every kilogram overweight. If a passenger carrying 40 kg of luggage is charged ?50, how much would a passenger carrying 80 kg be charged?
Ans. Let free weight be x and extra is y then 10y = 50 where x+y =40 then y = 5 or x = 35
        Thus 45x10 = 450 Ans

M74. Mr. and Mrs. Roberts have two daughters and three sons. At Easter time every member of the family buys one chocolate Easter egg for each other member. How many Easter eggs will be bought in total?
Ans.  2 D and 3 S. Then the Members i family is 7 then 6^7

M75. What is the least number which has no remainder when divided by any number from 1 to 10?
Ans . That is a number divisible is LCM of (1,2,3,4,5,6,7,8,9,10)