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required number =H.C.F of (73-25),(97-73) & (97-25)
=H.C.F of 48 , 24 and 72 = 4 (c)
answer can be 1,2,3,4 can’t be determined exact number
just explaining a case:
there are 10 people in the party.
name of people no. of people with they made handshake list of those people(this can vary but showing the possibility)
1 1 9
2 2 8, 9
3 3 7, 8, 9
4 4 6, 7, 8, 9
5 5 6, 7, 8, 9. 10
6 6 4, 5, 7, 8, 9, 10
7 7 3, 4, 5, 6, 8, 9, 10
8 8 2, 3, 4, 5, 6, 7, 9, 10
9 9 1, 2, 3, 4, 5, 6, 7, 8, 10
10 4 5, 6, 7, 8, 9
jack got 9 different answer so jack can be either 4th number or 10th number and jack’s wife know jack very well so she can’t have handshake with jack so if 4th is jack then she can’t be handshake with 6,7,8,9, in this case she can be 1,2,3, 5, 10 and now depending upon which no is jack’s wife she can have hand shake with- 1- 4 people, and if jack is number 10 then she can’t be 5,6,7,8,9 so again depending upon her number she can handshake with people in range of 1-4
12 %
Yes, it is possible by using a HDMI cable
both a and b is correct
Statements :
Most teachers are boys. Some boys are students.
Conclusions :
I. Some students are boys.
II. Some teachers are students.
In 3500 g rod contains 74% of silver = 3500 * 74/100 = 2590 g
Then 3500 g + 500 g of rod contains 84% of silver
Let x be the silver contained in 500 g of silver
(2590/3500 * 100) + (x/500 * 100) = 84
74 + x/5 = 84
(370 + x) /5 = 84
370 + x = 420
x = 50
Then the percentage of silver contained in 500 g of rod = 50/500 *100 =10%
7770-7077=693
9
10, 25, 45, 54, 60, 75, 80
54
198.20
3 groups
A is travelling at 50kmph, B is travelling at 40
kmph……..according to the formula
time taken to meet = distance between them
———————-
relative speed of two vehicles
so, time taken to meet= 15/(50-40)=15/10=3/2hrs
Father’s sister
I agree with @TUMWINE PETER
R>P>Q>T
V>S>P
so the lightest is T.
length of train = 150 m
5
307,311,313,317,319
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds