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Hi Mohit and Arsh  Both of you have given correct answers. Now solve this next one. 3. Hour hand of a clock is exactly at minute mark while minute hand is 6 minutes ahead of it. After some time it was observed that hour hand is again exactly at a minute mark while minute hand is exactly 7 minutes ahead of it. What is the time at the initial incidence? Kamal Lohia |
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Hi kamal sir, My method to solve the question would be wrng and so be my ans. I started with the difference between the two hands of the clock at a particular hour and then finally ended with the distance of 36 degree (6 mins) between them. IF initial distance between them be 30 degree (i.e. time in clock be 1 hr) then the minute hand will have to cover a distance of 30+36 degrees. Only in this case will the two hands be at an exactly minute mark. So initial time will be 1:12. Kindly provide the right method to approach the question. |
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4. Find the length of a side of regular hexagon inscribed in a unit square. Kamal Lohia |
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Hi kamal sir, My solution for the prob is, Let side of the regular hexagon be x. Then AP=(1-x)/2 Since its a regular polygon its one vertex will be at the mid point of the side of the square by symmetry. Therefore, 1/4+((1-x)/2)^2=x^2 x= (sqrt(28)-2)/6 Please correct me if i m wrng sir, Thanks,  |
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Hi Mohit What you have constructed is an equilateral hexagon. But think, is it regular also? Before that reconsider, what is a regular polygon? Kamal Lohia |
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Good evening, Sir Referring to the figure angle EFG=120 angle EFH=60 EH=1/2 => EH/EF= sin 60 => side of hexagon= 1/(sqrt(3))  |
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hello sir, if we anlyze dis problem then we are getting four 30-60-90 triangles on four peripherals so as we know 1:2:rt3 ratios we will end up wid 1/rt3 as the side of the hexagon but still am in dilemma as then i guess square would end it up as rectangle as one side would be unity and other would be 2/rt3.....pls throw some light |
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5. From first 100 natural numbers, 10 numbers are removed randomly and then a number is selected from the remaining numbers. Find the probability that selected number is divisible by 5. Kamal Lohia |
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hello sir, here in this question, am not gettng single unique answer as my answer depends on the numbers removed so thereby cases are generated as follows- when all 10 nums removed are multiples of 5 then prob=10/90=1/9(as in 100 we have total 20 multiples of 5) like this when all 10 nums removed are not multiple of 5 then prob=20/90= 2/9...these are the two cases i had mentioned ...there are other 9 cases more....kindly clarify!!! |
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hello sir, isnt this case is equivalent to just choosing a random number from first 100 natural numbers, so the probability should be 1/20. please let me know if my method is wrong. |
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6. What is the smallest number with exactly three odd divisors and nine even divisors? Kamal Lohia |
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| Kamal sir, am not able to get the previous probability ques? plz guide |
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since, requirement is of smallest no. let us take 3 odd divisors as- 1,3,5 and even divisors as - 2 4 6, which when multiplied by the odd divisors will give 9 even divisors as 2*1, 2*3, 2*5, 4*1, 4*3, 4*5, 6*1, 6*3, 6*5 so we have 3 odd divisors 1, 3, 5 and 9 even divisors as 2,6,10,4,12,20,6,18,30 here we see, 30 is the highest no. but 4,12,18 and 20 are not its multiple..so the required number should be a multiple of 30..next highest number is 60...which serves are purpose |
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i think i have made a mistake..60 has 4 odd divisors 1 3 5 15  |
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hello sir, My ans before was incorrect. as the ans is 72. Since it has 3 odd factors, hence the no should contain an odd square number. The smallest of such no is. 9=3*3. Total no of factors is =12 so the remaining factors should be even and should contain 2^3. Hence the no is= 3^2*2^3=72 Thanks,  |
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Hello Sir, The below is my approach. the total number of factors will be 9 + 3 = 12 let the smallest number be A and can be written in the form of (2^a)*(3^b) and (a>b) to obtain smallest number so the total number of factors will be (a+1)(b+1) so 12 needs to be written in the above form which will give (4,3) & (6,2) & (12,1). the groups(6,2) & (12,1) wont give 3 odd divisors as b will be either 1 or 0 so taking (3,4) then a= 3, b = 2 so the smallest number is (2^3)*(3^2) gives 72 Please let me know if the approach is wrong. |
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hello sir, ans would be 72 as we know no. of factors needed is 12 so start with 2 and 3,let say 2^a *3^b is the required format of the number's factors,on listing all the combination of a and b as (a+1)*(b+1)=12,we will come to a conclusion ,the required a and b are 3 and 2 resp.thereby 2^3*3^2=72 regards, Arsh Arora
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6.for exactly 3 odd divisor N=a^2 and a will be odd smallest odd num =3 N=3^2 =9 divisors of 9 =1,3,9 for exactly 9 divisors 9 can be written as product of num in two ways 1*9,3*3 to find the smallest num we choose 3*3 N=a^2* b^2 for smallest value we take a=2 ,b=3 but in this case we get 3 odd factors also (to find odd factors we remove 2^2 even factors = 9-3) so instead we take N=a^3 * b^2 even fac= total factors - odd factors 12-3=9 Ans is 9 and 72 am i correct? |
