Burrito Gringo Mexican Grill Interview Questions, Process, and Tips

5 is the answer….
Remaining all are the multiples of 3 expect 5

Let ‘x’ be the number of perons in the group and let ‘y’ be
amount per head which they have to pay.
Then xy = 2400.
or y = 2400/x
Since two friends have forget the purse, x-2 persons should
share the total amount(2400).
If they share, they have to make an extra contribution of
Rs 100 to pay up the bill
That is x-2 persons should pay y+100 each
or (x-2)(y+100) = 2400
or y+100 = 2400/(x-2)
or y = 2400/(x-2) – 100
Therefore we have got two equations namely,
y= 2400/x and y = 2400/(x-2) – 100
Comparing these two, we get
2400/x = 2400/(x-2) – 100
Solving this we get
x^2 – 2x -48 = 0
or (x-8) (x+6) = 0
or x = 8 or x = -6
Since ‘x’ denote the number of persons, it should be
positive.
So, x = 8.

a)0
b)206
c)0
d)250
e)39
f)92

area is given by Area of Rectangle = Length of Rectangle*(sqrt((Diagonal of Rectangle^2)-(Length of Rectangle^2)))
so substituting we get area=4*(sqrt((25-16)))
area=4*(sqrt(9))
since area cannot be negative hence we take sqrt of 9 as 3 (as sqrt of 9 is 3 and -3 as well)
area=4*3=12m^2

5c1 / 9c1

1972

Divide 30*60 seconds by LCM of Numbers =15

Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.

A. 80 : 81 : 84

17and+-12=5-48=-43

No loss no gain.

( c ) 10

10000

one-legged =5% of 10000=500

remaining=10000-500=9500

barefooted=9500/2=4750

remaining people= 9500-4750=4750

hence required number of shoes= 4750*2+500*1=100005% of 10000 = 500 one legged

9500 / 2 = 4750 bare foot

minium no of shoes = 4750*2 + 500*1 = 10000

let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240