Appliance center Interview Questions, Process, and Tips

Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.

If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.

Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.

Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).

Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.

answer : 36 –Explaination:
X= Chickens
Y = Qty per day

X*Y*9= 12 *.9 *Y*30

x=36

5

D

( A ) 124

X-340 = X x 4 x 8/ 100

100X – 34000 = 32X
68X = 34000
X = 34000/68
X= 500.

(b) 16.66%

Earlier for ₹x we could purchase y gm of sugar.
Now we pay ₹1.2x for y gm of sugar

(As there was an increase in price so, x + 20%x = 1.2x)

At current rates for ₹x you can purchase y/1.2 gm of sugar

So the reduced consumption is y-(y/1.2)

Percentage change = (reduced consumption/ original consumption ) *100

That is (0.2/1.2) *100 = 16.66% (approx)

It is About 35 Billion

Answer: 3121 gold coins
Let total no of coins be M
Let the disbursement D to each son:

D1 = 1 + (M – 1)/5 = (M + 4)/5
D2 = 1 + ( M – D1 -1)/5 = (D1) * 4/5
D3= (D2) * 4/5
D4= (D3) * 4/5
D5= (D4) * 4/5
Total disbursements to sons=
= ∑D= (M+4)*1/5[ 1+4/5+(4/5)(4/5)+ (4/5)(4/5)(4/5)+(4/5)(4/5)(4/5)(4/5) ]
= (2101/3125)*(M+4)
Thus balance left for daughters =M-{(2101/3125)*(M+4)}
=(1024M-8404)/3125
This balance should be a positive integer ( assuming M and all disbursements are full coins )
Thus 1024M-8404 should be a multiple of 3125….so….
1024M – 8404 = N*3125 where N is an integer
Using Python code:
n=int(input(“Enter num n: “))
X=int()
a=int()
a=0
X=’ ‘
for a in range(0,n+1):
a=a+1
X= (3125*a + 8404)/1024
if (3125*a + 8404)% 1024== 0:
print(X,a)
Enter num n: 10000
3121.0 1020
6246.0 2044
9371.0 3068
12496.0 4092
15621.0 5116
18746.0 6140
21871.0 7164
24996.0 8188
28121.0 9212
We get minimum value of N = 1021 and M = 3121 gold coins

40
40
30
60
– 10 cows
+ 20 days
if 40 cows can eat for forty days
then 20 cows can eat for 80 days

let a = 9 * 9 * 9;
let b = 12 * 12 * 12;
let c = 15 * 15 * 15;

sum = a + b + c;

edge = Math.cbrt(sum);

console.log(edge);