Latest TCS Aptitude Question SOLUTION: In school there are some bicycles and 4 wheeler wagons.One tuesday,there are 190 wheels in the campus.how many bicycles are there?

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In school there are some bicycles and 4 wheeler wagons.One tuesday,there are 190 wheels in the campus.how many bicycles are there?

Asked In TCS trupti (11 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

sharmila yakkala  (11 years ago)

actually this question shud have options,there are many solutions.
say 2x+4y=0
consider y=0
so x,y=95,0
as multiple of 2,4 is 4, the x and y values are also reduced in the same order like
93,1
91,2
89,3
.
.
.
goes on
options shud be given

Is this solution Helpfull? Yes (16) | No (2)  

tarun  (11 years ago)

question is incomplete

Is this solution Helpfull? Yes (5) | No (1)  

juhe rawat  (11 years ago)

let no of bicycle is x and wagons is y
so 2x+4y=190
if we take x=69 and y=13 , it satisfy the question...
so no of bicycle is 69

Is this solution Helpfull? Yes (4) | No (9)  

upendraprakash  (11 years ago)

69 bicycle and 13 car

Is this solution Helpfull? Yes | No (9)  

vinoth kanna  (11 years ago)

65 bicycles

Is this solution Helpfull? Yes | No (7)  

jaya akula  (11 years ago)

can any1 explain in detail??

Is this solution Helpfull? Yes | No (1)  

parminder maan  (11 years ago)

pls xplIn sltion in detail

Is this solution Helpfull? Yes | No (1)  

Bhavana Desa Sri Madhuri  (11 years ago)

95.,AS 2 WHEELS PER EACH BICYCLE.,190/2=95.,

Is this solution Helpfull? Yes | No (8)  

Artham Vinay  (11 years ago)

prime factors of 190 are 2,5,19.
19*2=38
190-38=152
152/4=38
hence 19 bicycles n 38 4 wheel wagons :p

Is this solution Helpfull? Yes | No (1)  

varshish  (9 years ago)

If there are 20 bicycles, they would have 40 wheels, leaving 150 wheels for the wagons, but 150 is not divisible by 4.

If there are 18 bicycles, they would have 36 wheels, leaving 154 wheels for the wagons, but 154 is not divisible by 4.

19 and 15 are both possible. More information is needed to choose one of those options.

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