(a) 38km/hr (b) 40km/hr (c) 45 km/hr (d) 35 km/hr
A man travels from Agra to Gwalior at an average speed of 30km/hr and returns at an average speed of 60km per hour. What is his average speed?
@padmanabhan1989 said:A man travels from Agra to Gwalior at an average speed of 30km/hr and returns at an average speed of 60km per hour. What is his average speed? (a) 38km/hr (b) 40km/hr (c) 45 km/hr (d) 35 km/hr
40?
@padmanabhan1989 said:A man travels from Agra to Gwalior at an average speed of 30km/hr and returns at an average speed of 60km per hour. What is his average speed? (a) 38km/hr (b) 40km/hr (c) 45 km/hr (d) 35 km/hr
40 km/hr
@deedeedudu said:a)6
Incentre lies on the circle hence distance is equal to the radius
approach , bhai .....
@bhatkushal said:Can you please specify the reason for this..........I mean whu will iy lie on the circle..........
Draw a line joining point R to the centre intersecting the circle and PQ at 2 points say A & B.
Now by alternate segment theorem prove that PA & QA r angle bisectors of triangle PQR meeting at A
Hence A is the incentre which lies on the circle
Now by alternate segment theorem prove that PA & QA r angle bisectors of triangle PQR meeting at A
Hence A is the incentre which lies on the circle
@padmanabhan1989 said:A man travels from Agra to Gwalior at an average speed of 30km/hr and returns at an average speed of 60km per hour. What is his average speed? (a) 38km/hr (b) 40km/hr (c) 45 km/hr (d) 35 km/hr
(b) 40km/hr
Total distance be 120 kms (say)
Agra to Gwalior 120/30=4hrs
Return journey=120/60=2hrs
Total distance = 240
Total time 6 hrs
Average speed = 240/6=40 km/hr
Total distance be 120 kms (say)
Agra to Gwalior 120/30=4hrs
Return journey=120/60=2hrs
Total distance = 240
Total time 6 hrs
Average speed = 240/6=40 km/hr
@padmanabhan1989 said:A man travels from Agra to Gwalior at an average speed of 30km/hr and returns at an average speed of 60km per hour. What is his average speed? (a) 38km/hr (b) 40km/hr (c) 45 km/hr (d) 35 km/hr
40
@padmanabhan1989 said:A man travels from Agra to Gwalior at an average speed of 30km/hr and returns at an average speed of 60km per hour. What is his average speed? (a) 38km/hr (b) 40km/hr (c) 45 km/hr (d) 35 km/hr
40== HM of the 30 60
@padmanabhan1989 said:A man travels from Agra to Gwalior at an average speed of 30km/hr and returns at an average speed of 60km per hour. What is his average speed? (a) 38km/hr (b) 40km/hr (c) 45 km/hr (d) 35 km/hr
40??
s=2d/(d/30+d/60)
s=2*60/3=40 km/h
@padmanabhan1989 said:A man travels from Agra to Gwalior at an average speed of 30km/hr and returns at an average speed of 60km per hour. What is his average speed? (a) 38km/hr (b) 40km/hr (c) 45 km/hr (d) 35 km/hr
2*30*60/(30+60)=40kmph
@padmanabhan1989 said:A man travels from Agra to Gwalior at an average speed of 30km/hr and returns at an average speed of 60km per hour. What is his average speed? (a) 38km/hr (b) 40km/hr (c) 45 km/hr (d) 35 km/hr
b) 40 km/hr
@jain4444 said:It takes six days for 3 women and 2 men, working together, to complete a work. 3 men would do the same work 5 days sooner than 9 women. How many times does the output of a man exceeds that of a woman?
6(3w+2m)=1
3.w*d1=6(3w+2m)
m.d2=6(3w+2m)
so,
d2-d1=5
we get,
m= 6w
Hi....i have a qustion from set theory:-
In a group if 60% of people drink tea and 70% drink coffee, what is the maximum percentage of people drinking either tea or coffee but not both?
a.)100% b.)75% c.)30% d.)10%
@abhinayapratap said:Hi....i have a qustion from set theory:-In a group if 60% of people drink tea and 70% drink coffee, what is the maximum percentage of people drinking either tea or coffee but not both?a.)100% b.)75% c.)30% d.)10%
Let the people drinking both be x, and total is 100
-> Only Tea = 60 - x
-> Only Coffee = 70 - x
(60-x) + (70-x) + x + None = 100 => x = None + 30
Now, people drinking either tea or coffee but not both = 130 -2x
for this to be maximum, x should be minimum => None = 0
=> x = 30 => Ans = 130 - 60 = 70 ? 😞 ... Not matching any options :(
@abhinayapratap said:Hi....i have a qustion from set theory:-In a group if 60% of people drink tea and 70% drink coffee, what is the maximum percentage of people drinking either tea or coffee but not both?a.)100% b.)75% c.)30% d.)10%
getting 70. are options correct?
@abhinayapratap said:Hi....i have a qustion from set theory:-In a group if 60% of people drink tea and 70% drink coffee, what is the maximum percentage of people drinking either tea or coffee but not both?a.)100% b.)75% c.)30% d.)10%
30+40 = 70 so will go with 75 
😛
😛 @chandrakant.k said:30+40 = 70 so will go with 75
But 75 is not possible. 😃 .. Go with 70 only, I think the options are messed up OR this is some ground breaking question 😃
@maddy2807 said:getting 70. are options correct?
even i am getting 70.....thats why i posted this question......i think answer should be 70..options must be wrong
Find the number of ternary sequences of length 4 where 0 is not followed by 1 and 1 is not followed by 2.(ternary sequence of length n is a sequence having n terms and each term is either 0,1 or 2)
47
37
72
54
44
For Maximum and Minimum of values, the key point to note is:
If you allot the a value to the intersection, it will get added to all the indivudal sets but will bring down the total.
Example: In a survey it was found that 80% like tea whereas 70% like coffee. What is the maximum and minimum number of those who like both?
Ans: First thing to note is that no information is mentioned about the people who don't like either of them. So that value is flexible and can change.
n(tea) = 80
n(coffee) = 70
n(total) = 100 {This includes those who like neither.}
n(tea U coffee) = ??? {We don't know this value and it is flexible}
If we want to maximize those who like both, we have to maximize the value in the intersection. So, we have to minimize the value of the union.
n(tea INT coffee)max = 70 {It is limited by the higher of the two values}
In this case 20% of people like neither tea nor coffee.
If we want to minimize those who like both, we have to minimize the value in the intersection. So, we have to maximize the value of the union.
We know that the maximum possible value of the union ie n(tea U coffee) = 100
So, we need to figure out the surplus
n(tea) + n(coffee) = 80 + 70 = 150.
The surplus is = 150 – 100 = 50
So, the value of the intersection = value of the surplus = 50
This could have also been obtained by the formula
n(a U b) = n(a) + n(b) – n (a INT b)
If you allot the a value to the intersection, it will get added to all the indivudal sets but will bring down the total.
Example: In a survey it was found that 80% like tea whereas 70% like coffee. What is the maximum and minimum number of those who like both?
Ans: First thing to note is that no information is mentioned about the people who don't like either of them. So that value is flexible and can change.
n(tea) = 80
n(coffee) = 70
n(total) = 100 {This includes those who like neither.}
n(tea U coffee) = ??? {We don't know this value and it is flexible}
If we want to maximize those who like both, we have to maximize the value in the intersection. So, we have to minimize the value of the union.
n(tea INT coffee)max = 70 {It is limited by the higher of the two values}
In this case 20% of people like neither tea nor coffee.
If we want to minimize those who like both, we have to minimize the value in the intersection. So, we have to maximize the value of the union.
We know that the maximum possible value of the union ie n(tea U coffee) = 100
So, we need to figure out the surplus
n(tea) + n(coffee) = 80 + 70 = 150.
The surplus is = 150 – 100 = 50
So, the value of the intersection = value of the surplus = 50
This could have also been obtained by the formula
n(a U b) = n(a) + n(b) – n (a INT b)
In this case, there is no one who likes neither coffee nor tea.