Arey Hai..Srry.. It shld be 1144.. 
Wht have i Dne..
Wht have i Dne..@Amrofa said:Eighty five children went to an amusement park where they could ride on the merry-go-round, roller coaster, and Ferris wheel. It was known that 20 of them took all three rides, and 55 of them took at least two of the three rides. Each ride cost Re 1, and the total receipt of the amusement park was Rs 145.How many children did not try any of the rides?(1) 5 (2) 10 (3) 15 (4) 2015. How many children took exactly one ride?(1) 5 (2) 10 (3) 15 (4) 20
15,15?
total cost of 3 rides = 3*20 = 60
no. of children took exactly two rides = 55 - 20 = 35
cost of 2 rides = 35*2 = 70
cost of exactly 1 ride = 145 - 70 - 60 = 15
no. of children took exactly 1 ride = 15
no. of children took no ride = 85 - 35 - 20 - 15 = 15
@Logrhythm said:N = 2^4*3^4*2^2*5^2 = 2^6*3^4*5^2terms with one 5 -> 7*5*2 = 70terms with two 5's -> 7*5 = 35total power of 5 -> 70+35 = 105...hence, option 2...
the number of factors having two 5s will be 35. these 5s will be like 5^2 in each factor. so when the product is taken there will be 35x2 = 70 5s ... plus the terms with one 5 =70. so total = 70+70 = 140 powers.. dnt u think?
sorry my bad...i wrote 5 instead of 15...ya 15,15!!
@CSK23 said:Q: In a base n, 32 * 45 = 2133. Express the number (424) of base 10 in base n.
(32)n*(45)n = (2133)n
(3n+2)(4n+5) = 2n^3 + n^2 + 3n + 3
=> n = -1,-1/2 and 7
so only 7 holds..
hence, 1144
@Amrofa
15 , 15
3 rides = 20
2 rides = 55 -20 = 35
1 ride = 85 - 55 = 30 (incl. no ride)
no ride + 1 ride = 145 - (60 + 70) = 15 rs
since no ride = 0rs , so 1ride ppl =15 and no ride =15.
@TootaHuaDil said:Copying mistake..Really sorry...here is it....@LogrhythmWhat is the product of all factors of the number N = 6^4 x 10^2, which are divisible by 5?1. 2^210 × 3^102 × 5^1402. 2^210 × 3^140 × 5^1053. 2^140 × 3^210 × 5^1024. 2^140 × 3^102 × 5^2105. 2^102 × 3^210 × 5^140
5 ?
techgeek..dude dont post single one three times!!!
How many natural number
Q : If x^2 + y^2 + z^2 = 4 and k = xy + yz + zx then
1. k>= -2
2. k
3. k >= -4
4. k
@techgeek2050 said:the number of factors having two 5s will be 35. these 5s will be like 5^2 in each factor. so when the product is taken there will be 35x2 = 70 5s ... plus the terms with one 5 =70. so total = 70+70 = 140 powers.. dnt u think?
i din't get u...
see i'll do the complete sum again..
N = 2^6*3^4*5^2
now remember that we need to calculate the prod of those factors which are div by 5...
1) to determine the power of 5...
take one 5 out -> 5(2^6*3^4*5) -> 7*5*2 = 70
now take both the 5's out -> 5^2(2^6*3^4) = 35
this makes the power of 5 as 70+35 = 105..
2) to determine the power of 3..
take one 3 out -> 3(2^6*3^3*5) = 7*4*2 = 56
take two 3's out -> 3^2(2^6*3^2*5) = 7*3*2 = 42
take three 3's out -> 3^3(2^6*3*5) = 7*2*2 = 28
take all four 3's out -> 3^4(2^6*5) = 7*2 = 14
hence, total power of 3 -> 56+42+28+14 = 140..
3) to determine the power of 2...
take one 2 out -> 2(2^5*3^4*5) = 6*5*2 = 60
take two 2's out -> 2^2(2^4*3^4*5) = 5*5*2 = 50
take three 2's out -> 2^3(2^3*3^4*5) = 4*5*2 = 40
take four 2's out -> 2^4(2^2*3^4*5) = 3*5*2 = 30
take five 2's out -> 2^5(2*3^4*5) = 2*5*2 = 20
take all the 2's out -> 2^6(3^4*5) = 5*2 = 10
hence, the total power of 2 -> 60+50+40+30+20+10 = 210...
PS - I reduced the power of 5 by one in case of 2 and 3 coz we need to find the factors which are divisible by 5...
Therefore, our answer = 2^210*3^140*5^105
----------------------------------------
In my previous explanation I found out only the power of 5 as all the options had different powers of 5...
-----------------------------------------
This question can actually be done in 1 minute....
look at N -> 2^6*3^4*5^2
here, power of 5
the same patter should be in the answer....which was available only in option 2...
I hope samaj aya hoga...isse jayada type ni hoga bhai.. 😛
@CSK23
1: since its square..its safe to approach a value of 2 for one variable & 0 for others which brings us to option 1
@Logrhythm said:i din't get u...see i'll do the complete sum again..N = 2^6*3^4*5^2now remember that we need to calculate the prod of those factors which are div by 5...1) to determine the power of 5...take one 5 out -> 5(2^6*3^4*5) -> 7*5*2 = 70now take both the 5's out -> 5^2(2^6*3^4) = 35this makes the power of 5 as 70+35 = 105..2) to determine the power of 3..take one 3 out -> 3(2^6*3^3*5) = 7*4*2 = 56take two 3's out -> 3^2(2^6*3^2*5) = 7*3*2 = 42take three 3's out -> 3^3(2^6*3*5) = 7*2*2 = 28take all four 3's out -> 3^4(2^6*5) = 7*2 = 14hence, total power of 3 -> 56+42+28+14 = 140..3) to determine the power of 2...take one 2 out -> 2(2^5*3^4*5) = 6*5*2 = 60take two 2's out -> 2^2(2^4*3^4*5) = 5*5*2 = 50take three 2's out -> 2^3(2^3*3^4*5) = 4*5*2 = 40take four 2's out -> 2^4(2^2*3^4*5) = 3*5*2 = 30take five 2's out -> 2^5(2*3^4*5) = 2*5*2 = 20take all the 2's out -> 2^6(3^4*5) = 5*2 = 10hence, the total power of 2 -> 60+50+40+30+20+10 = 210...PS - I reduced the power of 5 by one in case of 2 and 3 coz we need to find the factors which are divisible by 5...Therefore, our answer = 2^210*3^140*5^105----------------------------------------In my previous explanation I found out only the power of 5 as all the options had different powers of 5...-----------------------------------------This question can actually be done in 1 minute....look at N -> 2^6*3^4*5^2here, power of 5 the same patter should be in the answer....which was available only in option 2...I hope samaj aya hoga...isse jayada type ni hoga bhai..
mahn! d explanation ws superb.. but mai short me bhi samajh jata.. 
what i meant to say was dis
taking 5^2 out we get
5^2(2^6*3^4) .... the number of factors of this form will be 35. when we take the product of these 35 factors , the power of 5 will be added and since each factor will have 2 powers of 5, the total should be 35x2 = 70.. i hope m clear dis time.
@CSK23 said:Q : If x^2 + y^2 + z^2 = 4 and k = xy + yz + zx then1. k>= -22. k3. k >= -44. k
(x + y + z)^2 >= 0 so x^2 + y^2 + z^2 + 2xy +2yz + 2zx >=0
so 4 + 2k >= 0
so k>= -2