Three sprinters A,Band C has to sprint from points P to Q and back again ( starting in that order).The time interval between between the starting times of the three sprinters A,B and C was 5 sec each .Thus C started 10 seconds after A while B started 5 seconds after A .The three sprinters passed a certain point R , which is somewhere between P and Q , simultaneously (none of them having reached point Q yet ) Having reached Q and reversed the direction , the third sprinter met the second one 9 m short of Q and met the first sprinter 15 m short of Q .Find the speed of the first sprinter if the distance between PQ is equal to 55m
a 4m/s b 3 m/s c 2 m/s d 1 ms e none of these
Let the time taken by A,B and C to reach the point R be t, t-5 and t-10 respectively.
distance is same v(A)/v(C) = (t-10)/t v(B)/v(C) = (t-5)/(t-10) ---->1 After meeting at R, till the next meeting A and C cover x and (x+30) respectively. x+30=>x+15+15(I hope u get this)
After meeting at R, till the next meeting B and C cover (x+6) and (x+24) respectively. x+6 => x+15-9 and x+24 =>x+15+9
here after meeting at R all meet at same time.so apply distance speed ratio v(A)/v(C) = x/(x+30) v(B)/v(C) =(x+6)/(x+24) -------->2
now equate va/vb from 1 and 2, u will get 2 equations..solve variables to get t = 20 and x = 30. so PR=55-30-15=10 A covered PR in 20 secs speed=10/20=1/2m/s so speed is 1/2m/s
in a survey among students at all the iims.it was found that 48% preferred coffee,54% liked tea and 64% smoked.of the total 28% liked coffee and tea,32% smoke and drank tea and 30% smoked and drank coffee.only 6% did none of these.if the total number of student is 2000 then
1.the two items havin the ratio 1:2 are a.tea only and tea and smoking only b.coffee and smoking only and tea only c.coffee and tea but not smoking and smoking but not coffee and tea
2.the no. of persons who like coffee and smoking only and the no. who like coffee only bear a ratio a.1:2 b.3:2 c.5:1 d.2:1
Mensuration LOD2, (Q43) Circles are drawn with 4 vertices as the centre and radius equal to the side of a square .If the sqaure is formed be joining the mid-pointsof another square of sides 2(6)^.5. Find the area common to all the 4 circles???
in a survey among students at all the iims.it was found that 48% preferred coffee,54% liked tea and 64% smoked.of the total 28% liked coffee and tea,32% smoke and drank tea and 30% smoked and drank coffee.only 6% did none of these.if the total number of student is 2000 then
1.the two items havin the ratio 1:2 are a.tea only and tea and smoking only b.coffee and smoking only and tea only c.coffee and tea but not smoking and smoking but not coffee and tea
2.the no. of persons who like coffee and smoking only and the no. who like coffee only bear a ratio a.1:2 b.3:2 c.5:1 d.2:1
yours 2 one's answer is correct please give the solution and 1st one answer is 'b'
bro if u r telling the answr 4m arun sharma..Then it mite b wrng becoz i have double chckd it..Nd if 2nd one is rite then da 1st cnt b wrng..There many wrng answers in arun sharma.. C i cnt xplain it here since its a venn diagram question but tis is a very basic one..I solvd it with the help of variables..Trying using variable method
This one's from PnC : - I am not able to understand the solution - Can someone please explain: The no of circles that can be drawn out of 10 points of which 7 r collinear. ? Answers :1)130 2)85 3)45 4)cannot be determined
All the numbers that can be formed by using one or more of the digits 3, 4, 5, 6, 7 without repetition, are arranged in increasing order. The rank of 76354 is Please elaborate the process of finding out the rank for the above.......
total numbers=5+20+60+80+120=285 now 76354 is 5th from last.. So it's rank should be=281
All the numbers that can be formed by using one or more of the digits 3, 4, 5, 6, 7 without repetition, are arranged in increasing order. The rank of 76354 is Please elaborate the process of finding out the rank for the above.......
Simple 😃 Since there are 5 different digits, the total number of combinations would be 120, with the last rank going to 76543.
Moving in decreasing order, the next numbers would be:
please solve this question.... give complete solution !!!
two bodies moving along a circle in same direction meets after every 49 minutes. had they moved at the same speeds in opposite direction they would have meet at every 7 minutes. If, moving in opposite direction the bodies are at the distance of 40 mts from each other along the arc at t=0 then at t=24 seconds their distance will be 26 mts ( the bodies donot meed during these 24 secs ). find the speed of the faster body in mts/min ????
in eqn 2 ... u have assumed speed of both the bodies as v and u . but in question its a separate condition where speed can be anythng ..... moreover the speed is equal ..."had they moved at the same speeds in opposite direction they would have meet at every 7 minutes."... pls explain ...
in eqn 2 ... u have assumed speed of both the bodies as v and u . but in question its a separate condition where speed can be anythng ..... moreover the speed is equal ..."had they moved at the same speeds in opposite direction they would have meet at every 7 minutes."... pls explain ...
umm i missed that part of question :shocked:
if thats the case then the eqs would be
v+u=14/24 =>7/12 so 49=D/v-u---------->1
also 7=D/u+v-------->2
since u=v =>7=D/2v
u know u+v=7/12, here u=v, so 2v=7/12
2v=(7/12) m/sec, but in eq 2 it is m/min.. 7x60 x (7/12) = D =>D=49x5=245m
so its one and the same...
PS:IF even the below condition is different then i guess there would be no solution for this. =>( If, moving in opposite direction the bodies are at the distance of 40 mts from each other along the arc at t=0 then at t=24 seconds their distance will be 26 mts)
A watch losses 2/3% time during the 1 st week and 1/3% time during the next week .If on a Sunday noon , it showed the right , what time will it show at noon on the Sunday after the next ??
a 11:26:24 am b 10:52:18 am c 10:52:48 am d 11:36:24 am
Can anyone please answer these questions (with solution method, if possible):
1) What is the highest power of 3 in 58!-38!
2) How many number are in a set S such that:
a) S has elements of first 1000 natural numbers b) There is no prime number in S c) If any two number of S are selected randomly then they are always co-prime...
3) Last one is :
A number 83p796161q is completely divisible by 11. Given that q>p>0, find the sum of remainders if N is divided by (p+q) and p successively.
Can anyone please answer these questions (with solution method, if possible):
1) What is the highest power of 3 in 58!-38!
2) How many number are in a set S such that:
a) S has elements of first 1000 natural numbers b) There is no prime number in S c) If any two number of S are selected randomly then they are always co-prime...
3) Last one is :
A number 83p796161q is completely divisible by 11. Given that q>p>0, find the sum of remainders if N is divided by (p+q) and p successively.
Thnx in anticipation !!
1. Since 38! can be taken out common from 58! - 38!, we only need to find out the highest power of 3 in 38!, which can be found out as follows:
3^1 = 12 3^2 = 4 3^3 = 1 Total = 17
Thus, highest power would be 3^17.
3. We have two sets of added digits:
19+p and 22+q (first set is sum of odd digits, next of even digits)
For the number to be divisible by 11, the difference between both the numbers should be zero or a multiple of 11. Since q>p, the option of zero difference doesn't exist. Thus, the only combination exists, p = 1 and q = 9, due to which the two numbers become 20 and 31, a difference of 11.
Thus, the number becomes 8317961619, which is divisible by 11.
Now, we need to divide it by p+q = 1+9 = 10, which would give a remainder of 9. Then again, we need to divide the quotient by p = 1, which would anyway give a remainder of 0. Thus, the sum of remainders is 9.
PS: Have to think a bit for the 2nd one 😃 Would be better if others reply.
1. Since 38! can be taken out common from 58! - 38!, we only need to find out the highest power of 3 in 38!, which can be found out as follows:
3^1 = 12 3^2 = 4 3^3 = 1 Total = 17
Thus, highest power would be 3^17.
3. We have two sets of added digits:
19+p and 22+q (first set is sum of odd digits, next of even digits)
For the number to be divisible by 11, the difference between both the numbers should be zero or a multiple of 11. Since q>p, the option of zero difference doesn't exist. Thus, the only combination exists, p = 1 and q = 9, due to which the two numbers become 20 and 31, a difference of 11.
Thus, the number becomes 8317961619, which is divisible by 11.
Now, we need to divide it by p+q = 1+9 = 10, which would give a remainder of 9. Then again, we need to divide the quotient by p = 1, which would anyway give a remainder of 0. Thus, the sum of remainders is 9.
PS: Have to think a bit for the 2nd one 😃 Would be better if others reply.
Can you please tell me as to why only the part that is common is being checked for the highest power and not what is inside the bracket...
Can you please tell me as to why only the part that is common is being checked for the highest power and not what is inside the bracket...
Both your answers are correct !! :)
When we take 38! common, we get the other factor as (39*40*41*.....58 - 1), which anyway would be a smaller figure than 38!. So to get the highest power of 3, we need to consider 38!.
