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After much hesitation, now I am starting the question series today which is an auspiscious day. It will contain questions of Quantitative Aptitude and sometimes Logical reasoning as well. I hope it'll be equally beneficial for both CAT-2012 as well as CAT-2013 students. I'll try to post one question (set) at a time and I'll try to make it a regular (i.e. hopefully daily) affair. You are required to post your answer alongwith solution so that different approaches may be obtained and everybody can enhance their learning. Please try not to post any of your doubts in this thread. You can create another post for that. - Kamal Lohia A person started at some time from a point P and another man started after sometime and catch the first one at 12 noon. If second person had started 1 hour later, then he would have caught the first one at 4 p.m. 1. Find the ratio of speed of the two persons. |
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Hi Kamal sir, Let a be the speed of first person and b be the speed of second person. Relative speed= b-a Ans 1) according to it, the distance covered by the first man with speed a in 1 hr, will be covered with relative speed b-a in 3 hrs. a/b-a=3/1 hence, a/b=3/4 Ans 2) Now we have a/b=3/4. In this case a travels for 2 hrs. he will cover 6x distance. Suppose they start at 12 noon, by 2 pm a would have covered 6x distance. after this with relative speed of 4x-3x=x, b will take 6 hrs to catch up with a. Hence, 8 pm. Tell me if i m wrng. Thanks n Regards, |
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hello sirji, 1) 3:4..approach let speed of frst one is a and speed of second one is b,we can directly draw the linear diagrams and see that distance travelled by first one is 4a which the other one covers in 3 hrs as 1 hr he is late thereby 4a/b=3=a/b=3:4 2) 8...as taking clue frm the frst part 4a extra distance is travelled by one when second is one hr late so again 4a would be travelled by one,thereby time taken=8a/b=8*3/4=6 but this 6 do not include second one's wait when frst is running therefore 6+2=8hrs.. |
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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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Hi Sir, By My method, 1. Let first person with speed a start at x while second with speed b start at y So, (12-x)a = (12-y)b --(1) also, (16-x)a = (16-y)b (2nd condn 4pm is taken as 16) --(2) substracting (2) from (1)we get 4a = 3b =>a/b = 3/4 2. Let they meet at time 'k' (k-x)3 = (k-y-2)4 --(3) also (2 ) can be written as (16-x)3 = (16-y)4 -- (4) (3) - (4) gives k=20 i.e. 8pm |
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Good evening, Sir 1. Approach let s1 and s2 be the speeds of first and second persons respectively. case1: s1*t=s2*T (distances are same!) case2: s1*(t+4)=s2*(T+3) => s1/s2= 3/4 2. Approach s1*t=s2*(t-2) s1/s2=3/4 => t=8 |
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hi dishendra for the first question can you give explanation how can you equate the distances |
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Hi Mohit Your thought process is excellent and answer is also correct. Well done!! Kamal Lohia |
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Hi Dishendra and tgdel372 tg Both of you have given correct answers with proper explanations. Good job! Kamal Lohia |
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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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Hi sir, In that case the ans should be 2-sqrt(2). The hexagon should be formed taking two opposite sides of the square. Hope i m correct this time. Thanks, |
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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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Hi sir, I got ur point. The ans will be 1/sqrt(3). Hope this time i m not wrng. Thanks, |
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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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Hi Parvesh You are exactly right. Actually that keen insight is to be developed to solve problems. Good work. Kamal Lohia |
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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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hi sir, The smallest no would be 60. coz the total no of factors 12. out of which one is 9 is even and 3 odd. correct me if i m wrng. Also please explain the last question sir. Thanks, |
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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? |
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