official quant thread for cat08

[linksuresh]

Originally Posted by getintoiimb

6)a watch dealer pays 10% on a wathc that costs Rs 250 abroad.for ho0w much should he mark it if he desires to make a profit of 20% after giving a discount of 25% to the buyer?
ans : Rs 440

could someone solve the above q... i cudnt comprehend it.. hence unable to solve....

others q seem quite alright...

[nikunj14_83]

Hi all,
Please help me out to solve these TSD problems....

1) A nughty bird is sitting on a car & it see's a car coming towards it at a distance of 12 Km.Th bird moves on from the bonnet of first car to the second car till the two car's crash(Assume that both car's crash). Speed of both car's is 60 Kmph & speed of bird is 120 Kmph.
a)Total distance covered by bird when it reaces the bonnet of second car for the second time?....Ans is 11.55Km
b) Therotically how many times bird reaches the second car?...Ans is Infinite times


2) Train A leaves Patna at 7:00 PM & Train B leaves Tata at 4:00 AM. Distance between Patna & Tata is 800 Km.Speed of Train A is 60 Kmph & that of Train B is 90 Kmph

a) At what distance frm Tata two trains meet?...Ans is 156Km
b) What is the time when the two trains meet?...Ans is 5:44 AM

P.S.---I have solved question 2 but I want to know how to solve it using the concept of Relative speed....
Both questions are LOD(2) from Arun Sharma

[vivs_d]

Originally Posted by linksuresh

could someone solve the above q... i cudnt comprehend it.. hence unable to solve....

others q seem quite alright...

its an easy one.
the price at which at dealer got the watch will be
250 + 0.1X250 = 275Rs.

let the price at which he has to sell it is P,
so the equation will be

275 + 0.2X275 = P - 0.25XP

on solving, we get P = 440 Rs.

[deep@k]

Originally Posted by Hi2All

Four persons can cross a bridge in 3,7,13,17 minutes.Only 2 can cross at a time.Find the minimum time taken by the 4 to cross the bridge.

23 minute is the minimum time...........

[deep@k]

4)a dealer buys eggs at Rs 36 per gross.he sells the eggs at a profit of 12(1/2)% on the CP.what is the SP per egg?
how much is a gross?i.e like one dozen is 12 how much is a gross?

1 gross = 144
cost of one egg = 36/144 = 0.25
CP = 25 paise
SP = 25(1+25/200) = 28 paise

ans : 28 paise

5)a manufacturer makes a profit of 15% by seliing a colour tv for Rs 5750.if the cost of manufacturing increases by 30% and the price paid by the retailer is inceased by 20%,find the profit percent made by the manufacturer?

SP = 5750
CP = x x(1+15/100) = 5750
x= 5000
after rate increased CP = 6500
SP = 6900
profit = 400/6500*100 = 6(2/13)%

ans : 6(2/13)%


7)a shopkeeper buys an article for Rs 400 and marks it for sale at a price that gives him 80% profit on his cost.he however gives a 15% discount on the MP to his customer.calc the actual % profit made by the shopkeeper
LET MP = x
400 = (1-0.15)x
x=470
CP = 470/1.8 = 261
profit = (400-261)/261*100 = 53%

ans : 53%

[dashing]

Originally Posted by nikunj14_83

Hi all,
Please help me out to solve these TSD problems....

1) A nughty bird is sitting on a car & it see's a car coming towards it at a distance of 12 Km.Th bird moves on from the bonnet of first car to the second car till the two car's crash(Assume that both car's crash). Speed of both car's is 60 Kmph & speed of bird is 120 Kmph.
a)Total distance covered by bird when it reaces the bonnet of second car for the second time?....Ans is 11.55Km
b) Therotically how many times bird reaches the second car?...Ans is Infinite times

For the first trip from first car to second car

distance travelled by bird = (12/180) * 120 = 8
Distance between cars after first trip = 4

distance travelled by bird in second trip = 8/3
Distance between cars after second trip = 4 - 8/3 = 4/3

distance travelled by bird in third trip = 8/9

total distance = 8 + 8/3 + 8/9 = 11.55

[man_on_mission]

Originally Posted by linksuresh

could someone solve the above q... i cudnt comprehend it.. hence unable to solve....

others q seem quite alright...

suresh here. dealer pays 10% to buy a watch which cost Rs250 abrod means ...

Cost price of watch fro dealer = Rs250 + 10% of 250 = Rs 275...after this its fairly simple..

Let Marked Price be M
so, M*(3/4) = 275(6/5)
solving, M = 440...

so marked price = Rs 440

[man_on_mission]

Originally Posted by nikunj14_83

Hi all,
Please help me out to solve these TSD problems....



2) Train A leaves Patna at 7:00 PM & Train B leaves Tata at 4:00 AM. Distance between Patna & Tata is 800 Km.Speed of Train A is 60 Kmph & that of Train B is 90 Kmph

a) At what distance frm Tata two trains meet?...Ans is 156Km
b) What is the time when the two trains meet?...Ans is 5:44 AM

P.S.---I have solved question 2 but I want to know how to solve it using the concept of Relative speed....
Both questions are LOD(2) from Arun Sharma

Train A leaves @ 7pm
Train B leaves @ 4 am

As A leaves before B so in 9hrs(7pm - 4am) , a will cover 540 km....so now distance between A and B is 260 km(800-540).

Now relative speed of A & B = 60+90 = 150 kmph..distance between the two = 260 km
so time taken to cover 260km = 260/150 = 1hr 44 mins (26/15 hr).

So A & B will meet each other in 1hr 44mins after B leaves Tata.
Distance covered by B in 1hr44min = (26/15)*90 = 156 km.

a)so two train will meet @ 156km from Tata
b)time taken = (4am + 1hr 44min) = 5.44 am

[KIMSTER]

Till 4:00 am : Train A alone travels for 9hrs..so 9*60=540km covered...
4:00 am :800-540=260km left
Relative speed-->90+60=150
Time= distnce/speed = (260/150)*60minutes = 104mins..1hr and 44mins.
So trains meet at 5:44PM..

Distance from Tata : (time train b runs*its speed) = (104/60)* 90 = 156km.

2) Train A leaves Patna at 7:00 PM & Train B leaves Tata at 4:00 AM. Distance between Patna & Tata is 800 Km.Speed of Train A is 60 Kmph & that of Train B is 90 Kmph


a) At what distance frm Tata two trains meet?...Ans is 156Km
b) What is the time when the two trains meet?...Ans is 5:44 AM

P.S.---I have solved question 2 but I want to know how to solve it using the concept of Relative speed....
Both questions are LOD(2) from Arun Sharma[/quote]


MBA 2008-2010 : IIFT Delhi.

[man_on_mission]

Originally Posted by nikunj14_83

Hi all,
Please help me out to solve these TSD problems....

1) A nughty bird is sitting on a car & it see's a car coming towards it at a distance of 12 Km.Th bird moves on from the bonnet of first car to the second car till the two car's crash(Assume that both car's crash). Speed of both car's is 60 Kmph & speed of bird is 120 Kmph.
a)Total distance covered by bird when it reaces the bonnet of second car for the second time?....Ans is 11.55Km
b) Therotically how many times bird reaches the second car?...Ans is Infinite times

Both questions are LOD(2) from Arun Sharma

Distance between cars = 12km
Speeds of cars = 60kmph
Speed of bird = 120 kmph
Speed of bird relative to car = 180kmph(120+60, as they are moving in opposite direction)
Relative speed of cars = 120kmph(60+60)

Now,

when bird moves for bonnet 1 to bonnet 2 for 1st time:

Distnace Betwee cars = 12 km
So,Time taken by bird two move = 12/180
Distance covered by bird = (12/180)*120 = 8km
Distnace covered by cars = (12/180)*120 = 8km
Distance left between cars = 12-8 = 4km

when bird moves for bonnet 2 to bonnet 1 for 1st time:

Distnace Betwee cars = 4 km
So,Time taken by bird two move = 4/180
Distance covered by bird = (4/180)*120 = 8/3km
Distnace covered by cars = (4/180)*120 = 8/3km
Distance left between cars = 4-(8/3) = 4/3 km

when bird moves for bonnet 1 to bonnet 2 for 2nd time:

Distnace Betwee cars = 4/3 km
So,Time taken by bird two move = (4/3)/180
Distance covered by bird = [(4/3)/180]*120 = 8/9 km

So total distance covered by bird in whole journey = 8 + 8/3 + 8/9 = 11.55 km

Also as the distance remianing between two cars(theoritically) will keep reducing by ver very small amount for more number of trips....so theoritically bird can make infinite trips..