QBM009

M91. A girl bought 15 pens costing ?1.84. She paid one pence more for each red pen than each blue pen. How many of each kind did she buy and at what price? M92. Consider the following series. S(n) = [1/(√1+√2) ] + [1/(√3+√2) ] + [1/(√3+√4) ]+ . . + [1/(√(n+1)+√n) ] For which values of n is S(n) rational? M93. the product of three consecutive integers, plus their mean, is always a. cube b. square c. both M94. How many zeros will be there at the end of (2!)2! x (4!)4! x (6!)6! x (8!)8! M95. What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm? M96. When I was 14 years old my father was 42 years old, which was three times my age. Now he is twice my age, how old am I? M97. What is the remainder when N is divided by 37 where N = 3636 + 36. M98. There are six simple fractions involving ninths that cannot be cancelled down. How many simple fractions with a denominator equal to 24 cannot be cancelled down? M99. What proportion of 3-digit numbers contain the digit one? M100. Matilda's father takes 20 minutes to mow the back garden lawn, but Matilda takes 30 minutes to do the same job. If they worked together, how long would it take to cut the lawn? M101. N = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + . . . . . . + 1/156, the value of N is 12/13 13/12 1/13 None M102. Convert 423 from base 5 to base 8 M103. A person starts multiplying consecutive positive integers from 11. How many numbers should he multiply before the will have result that will end with 3 zeroes? M104. How many 2-digit numbers have an odd product? (i.e the product of the 2-dgit numbers is an odd number) M105. Convert 671 from base 9 to base 3 END

QBM009

M92. Consider the following series.
S(n) = [1/(√1+√2) ] + [1/(√3+√2) ] + [1/(√3+√4) ]+ . . + [1/(√(n+1)+√n) ]
For which values of n is S(n) rational?
Ans.S(n) =  (√2-√1)  + (√3-√2)  + (√4-√3) ]+ . . + (√(n)-√n-1) ]+(√(n+1)-√n) ] = √(n+1) - 1 Thus for value of rational for n+1 =k^2 thus
       n = k^2 -1 where k varies from 1 to m
M93. the product of three consecutive integers, plus their mean, is always a. cube b. square c. both
Ans. x(x^2-1)+x is always cube.

M94. How many zeros will be there at the end of (2!)^2! x (4!)^4! x (6!)^6! x (8!)^8!
Ans. The power of 5 in 2 = 0 in 4! = 0 and 6! = 1 and 8! = 1 thus Power of 5 in the series = 6!+8! =720+40320 =41040 Ans.

M95. What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm?
Ans. HCF of ( 578 and 374) = 34 and 34 x m = 578 and 34 x n = 374 => m = 17 and n =11 so no. of tiles = 187 ans.

M96. When I was 14 years old my father was 42 years old, which was three times my age. Now he is twice my age, how old am I?
Ans. 1:3 and 14+x/(42+x) = 1/2 => 28 + 2x = 42 + x => x = 14 thus now I amd 28 yr old.

M97. What is the remainder when N is divided by 37 where N = 36^36 + 36.
Ans. Now a^p-1 -1 is divisible by p where p is a prime no. thus thus remainder is 0.

M98. There are six simple fractions involving ninths that cannot be cancelled down. How many simple fractions with a denominator equal to 24 cannot be cancelled down?
Ans. Relatively prime to 24 and less than 24  factors of 24 = 2^3x3 thus the numberator must contain 1,5,7 which is 3 only.
        with 9 is 1,2,4,5,7,8
M99. What proportion of 3-digit numbers contain the digit one?
Ans. First no. is 100 to 199 = 100 Nos after that 200 to 299 has 1 coming is 9 + 10 times=19 times and same is for other s
        Total is 100+19x9 =100+171=271 thus proportion is 271 : 1000
M100. Matilda's father takes 20 minutes to mow the back garden lawn, but Matilda takes 30 minutes to do the same job. If they worked together, how long would it take to cut the lawn?
Ans. 20x30/50 = 12 min

M101. N = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + . . . . . . + 1/156, the value of N is
12/13 13/12 1/13 None
Ans.  N = 1/2+1/6 = 4/6=2/3 and 2/3+1/12 = 9/12 =3/4 at first term it is 66 % and at term it is 75 % and 3/4+1/20 = 16/20 = 4/5 = 80 %
         Thus series appear to be approaching 12/13 state .Last digit is multiple of 13.

M103. A person starts multiplying consecutive positive integers from 11. How many numbers should he multiply before the will have result that will end with 3 zeroes?
Ans. Now n! is the number then n/5 = 3 then n= 15 Thus the number is less than 14

M104. How many 2-digit numbers have an odd product? (i.e the product of the 2-dgit numbers is an odd number)
Ans.  Product of odd digits is odd Numbers are ending in {1,3,5,7,9} is 5 ways and balance numbers are 9 thus 45 ways number can be formed.Product is also
         45x45  ways = 2025 ways.

M98

I didnt understand the question can neone please elaborate wats being asked ?

M 99

i think the answer is 17/60.... using sets we get O(1) no of digits having 1 in units place = 90 T(1) ----------------,, ------ Tenths place = 90 H(1) ----------------,,----- Hundreths place = 100 O U T U H = O + T + H -(O int T + T int H + H int O) + H int T int O = 255 no of three digit numbes = 900 therefore fractions = 255/900 = 17/60

RE OF M99

Dear frn, think once again:

If one appears for one time

If unit digit is one: then such no is possible by 90(9 x10) ways,
If tens digit is one: then such no is possible by 90(9 x10) ways,
If hundred digit is one: then such no is possible by 100(10 x10) ways,

If one appears for 2 times

Its possible for  X11 (9 nos) [X is any digit frm 1-9] 11X (10nos) 1X1(10 nos)
As a whole 29 no.s

Another no. is 111
 


TOTAL (280 + 29 +1)=310....
Ratio (310/900)=31/90..... Hp its clear......

Giv ur best to d world.
Nd d best will cm back to u.....

Regards,
Dipanjan......

M99

Dear Dipanjan, u have made a mistake .. let me show wat it is .. ur claim if units digit is one : then such no is possible by 90(9 x10) ways; so got it by multiplying no ways u can fill 10th and 100ths place. So according to u tenths place can be filled in 10 ways i.e (0,1,2,...,9) .. so in ur 10 ways u have included 1 again which contradicts ur claim of - "If one appears for one time" ... cause 1 now from ur calculation is repeating twice .. and doing the same for 100ths place u have accounted for 111. !!!! no i think u see the mistake thats happened thanks for giving ur best, helped me work on my maths basics !!!

correction in my answer for M99

O U T U H = O + T + H -(O int T + T int H + H int O) + H int T int O = 90+90+100-(9+10+10)+1 = 252 required answer = 252/900 = 7/25

M105

answer is 60701

M102

answer is 1051

THNKZ YAAR

Thnkz for correction..... Giv ur best to d world. Nd d best will cm back to u... Regards, Dipanjan.....