[implex]

Originally Posted by selebratinglife

Theres an obvious flaw in the above explanation..Following it up..since 481 =37*13..The equation should be divisible by 37 as well as 13..Now if we take b and take n^2 out of the equation..16(m/n)^2-15 which will not satisfy for 37 as well as 13..So the minimum possible value is 481...Even this seems a bit inadequate to me though!!!

answer is correct !
the proof requires fermat's theorem and is a bit involved

[Singhania_Niket]

Originally Posted by prade

2^-6 mod 167 ;
let 2x mod 167 = 1
2x = 167y+1
x = (167y+1)/64
=>y = 1
x = 84

=> 84^6 mod 167 solve this to get
=> 107 mod 167

Hi can you please elaborate on the approach....


x=84????
84^6 mod 167 => 107??????? plz help....got stuck.... how do you solve that???

[Hi2All]

There are 10 coins. 6 coins showing head. And 4 showing tail. Each coin was randomly flipped (not tossed) seven times successively.after flipping the coins are 5 heads 4 tails one is hided the hided coin will have what.

I am unable to understand this question.. Help me to solve this..

[srikar2097]

Originally Posted by Hi2All

There are 10 coins. 6 coins showing head. And 4 showing tail. Each coin was randomly flipped (not tossed) seven times successively.after flipping the coins are 5 heads 4 tails one is hided the hided coin will have what.

I am unable to understand this question.. Help me to solve this..

The word 'randomly' in the question is a bit ambiguous. But any way let me try..

Given that; 1H ----> flip 7 times ---> 1T
1T ----> flip 7times --->1H

Initially, (6H+4T), but after the exercise we have (5H+4T). So one possible case for this to happen is that the 4T were not flipped at all and only 1H was chosen for flipping which results in a 1T.

or we have a case where all the 4T are flipped. We now have 10H. But we need 5H. So we flip 5H to give 5T. Again we get the hidden coin as 1T.

So I would go for the Hidden coin to be a T.

[Topgear]

Here is my Question..

Train X started from point A at 9:00 am with a speed 72km/hr towards station Y. 2 hrs after train Y starts from point B towards X at a speed of 90 km/hr. They cross each other at 1:30 pm.

But owing to signal problems at 12:00 noon the speeds of both the trains is reduced by same quantity such that they now cross each other at 4:30 pm.

Calculate the new speed of train that started from point A.

Please do post the procedure. Answer is 18kmph. How ? please puys.. gimme the procure.. i hangout always at time distance.. :lookroun:

[getintoiimb]

poys pls help me with these qns..pls explain with steps

1)the short and the long hands of a clock are 4 cm and 6 cm long.what will be the sum of distances travelled by their trips in 4 days?

ans : 3818.24 cm

2)find the num of bricks each measuring 25 cm x 12.5 cm x 7.5 cm,required to construct a wall 12 m long,5 m high and 0.25 m thick, while the sand and cement mixture occupies 5% of the total volume of the wall?

ans : 6080

3)a road that is 7 m wide surrounds a circular path whose circumference is 352 m.what will be the area of the road?

ans : 2619 sq cm

4)in a swimming pool measuring 90 m by 40 m,150 m take a dip.if the average displacement of water by a man is 8 cubic metres what will be rise in water level?

ans : 33.33 cm

5)how many metres of cloth 5 m wide will be required to make a conical tent the radius of whose base is 7 m and height 24 m?

ans : 110 m

6)a conical tent is to accomodate 10 persons.each person must have 6 sq m space to sit and 30 cm^3 of air to breadth.what will be the height of the cone?

ans : 80 m

7)a closed wooden box measures externally 10 cm long,8 cm broad and 6 cm high.thickness of wood is 0.5 cm.find the volume of wood used?

ans : 165 cubic cm

8 )3 cubes each of volume of 216 m^3 are joined end to end.find the surface area of the resulting figure?

ans : 504 m^2

9)the largest cone is formed at the base of a cube side measuring 7 cm.find the raion of volume of cone to cube?

ans : 11:42

10)a cube whose edge is 20 cm long has circle of each of its faces painted black.what is the total area of the unpainted surface of the cube if the circles are of the largest area possible?

ans : 514.28 cm^2

11)the areas of 3 adjacent faces of a cuboid are x,y,z.if the volume is V then V^2 will be?

ans : xyz

12)a wire is looped in the form of a circle of radius 28 cm.it is bent again into a square form.what will be the lenght of the diagonal of the largest square possible thus?

ans : (88 )/2 sqrt 2

Thanks!

[srikar2097]

Originally Posted by Topgear

Here is my Question..

Train X started from point A at 9:00 am with a speed 72km/hr towards station Y. 2 hrs after train Y starts from point B towards X at a speed of 90 km/hr. They cross each other at 1:30 pm.

But owing to signal problems at 12:00 noon the speeds of both the trains is reduced by same quantity such that they now cross each other at 4:30 pm.

Calculate the new speed of train that started from point A.

Please do post the procedure. Answer is 18kmph. How ? please puys.. gimme the procure.. i hangout always at time distance.. :lookroun:

Easy! There are 2 cond. here. Before signal and after.
Initially
Train Y travels in 2 hrs d1=90*2=180km.
train X & Y meet after 4.5hrs. (from 9am)
=>dist. covered by Y, d2=90*4.5 = 405km
and dist.covered by X, d3=72*4.5 = 324km.
=>total dist. between stations AB=d1 + d2 + d3 = 909km

Later
Both trains travel till 12:00 at their usual speeds. So each train covers-
train Y covers d4=90*5 = 450km
and train X covers d5 =72*3=216km
=> rem. dist. to cover = 909 - (450+216) = 243km

Let the speed for both trains be reduced by a kmph
=> Y covers d6 in 4.5 i.e. d6=(90-x) * 4.5
and lllly for X, d7 = (72-x)*4.5
but d6+d7 = 243
=> 243 = 4.5(162-2x)
on simplifying x = 54.

Thus the new speed of X = 72-54 = 18kmph.

[Topgear]

Originally Posted by srikar2097

Easy! There are 2 cond. here. Before signal and after.
Initially
Train Y travels in 2 hrs d1=90*2=180km.
train X & Y meet after 4.5hrs. (from 9am)
=>dist. covered by Y, d2=90*4.5 = 405km
and dist.covered by X, d3=72*4.5 = 324km.
=>total dist. between stations AB=d1 + d2 + d3 = 909km

Later
Both trains travel till 12:00 at their usual speeds. So each train covers-
train Y covers d4=90*5 = 450km
and train X covers d5 =72*3=216km
=> rem. dist. to cover = 909 - (450+216) = 243km

Let the speed for both trains be reduced by a kmph
=> Y covers d6 in 4.5 i.e. d6=(90-x) * 4.5
and lllly for X, d7 = (72-x)*4.5
but d6+d7 = 243
=> 243 = 4.5(162-2x)
on simplifying x = 54.

Thus the new speed of X = 72-54 = 18kmph.

Thanks.. What i did is that Took that X started at 9 am and took Y to start at 11am(2hrs after)

How do u differentiate between what i did and what u did with the language mentioned here.. why cant we take that Y starts 2 hrs after X..

[srikar2097]

Originally Posted by Topgear

Thanks.. What i did is that Took that X started at 9 am and took Y to start at 11am(2hrs after)

How do u differentiate between what i did and what u did with the language mentioned here.. why cant we take that Y starts 2 hrs after X..

It clearly states in the problem that X starts 2 hrs after Y and X starts at 9:00 => Y starts at 7:00. Seems clear to me. Where do you find the confusion?

[Topgear]

Originally Posted by srikar2097

It clearly states in the problem that X starts 2 hrs after Y and X starts at 9:00 => Y starts at 7:00. Seems clear to me. Where do you find the confusion?

Yeah .. misinterpreted d language.. ryte... .. thanks...