Planet Honda (India) Interview Questions, Process, and Tips

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NabasishGogoi

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0L -> 1 way
1L -> 3 ways
2L -> 7 ways
3L -> 4 ways
4L -> 1 way
total 16 ways

740

Lets call the 5 litre jug as jug A and 3 litre jug as jug B. Now, follow the steps:

Fill jug A completely. Now it contains 5 litres.
Slowly pour the water from jug A to jug B until jug B is completely filled. Now, jug A contains 2 litres and jug B contains 3 litres.
Throw away the water in jug B so that it is completely empty. Now, jug A contains 2 litres and jug B is empty.
Transfer the water from jug A to jug B. Now, jug A is empty and jug B contains 2 litres.
Fill jug A completely. Now, jug A contains 5 litres and jug B contains 2 litres.
Transfer water from jug A to jug B until jug B is completely filled. Now, jug A contains 4 litres and jug B contains 3 litres.

Now you have 4 litres of water in jug A.

Greatest number of 5 digits=99, 999
Smallest number of 5 digits=10, 000
And their sum=99, 999+10, 000=109, 999

ans: 40
6/3*12/3*15/3=40

let the speed limit be x.
speed by first rider S1 and 2nd rider be S2.
S1=x+10;
ans S2=x+2*10; and given S2=35.
solving we get x=15mph.

let the total no of breads be x.
1st man 2nd man
x- (x/2) – 1/2 – 1/2((x-1)/2) – 1/2 ….. so on.
ans is 31.
first ate : 15.5 + .5 = 16 remaining 15
second ate : 7.5+0.5 = 8 remaining 7.
third ate : 3.5 +0.5 = 4 remaining 3.
fourth ate : 1.5 + 0.5 = 2 remaining 1
fifth ate : 0.5 + 0.5 = 1 remaining 0

let x be lenghth of candles.

thicker one—- in 1 hr 1/6th will go.
thinner one— in 1 hr 1/4th will go.

let he lit for n hrs.

so in n hrs thicker one goes n/6.
and thinner one goes n/4.

so remainini length are x-n/6 and x-n/4.so x-n/6=2(x-n/4)
implie n=3x. if x=1 n=3hrs.

(x-2)^3==x^3-6x^2+12x-8
(x-2)^3==(2^(2/3)+2^(1/3))^3
therefore,by solving
Ans: x^3-6x^2+x== -8-5*2^(2/3)-5*2^(1/3)

( c ) 15 litres

8 cubes

2years

Iron and steel