Iex Interview Questions, Process, and Tips

Answer is 124 because in this series is cube root sequence 2 cube root is 8 ,3 cube root is 24,4 cube root is 64, 5 cube root is 125 but there is 124 which is wrong ,6 cube root is 216 ,7 cube root is 343 .

( b ) ANGEME

125

(N * 1.1) * 0.9 = 7920
N=8000

Marks in thrice in english as in science
so if science marks is x
then english marks = 3x
lets maths marks = y
ratio of english and maths marks is
3x/y = 3/5
=> y = 5x

and total marks = 162
3x + x + 5x = 162
x = 18 —-answer

Employee Performance Standards. Employee performance measurements can determine an employee’s compensation, employment status or opportunities for advancement.

189

390

LNTKCHMF

Monday =y cheques
Tuesday =3y cheques
wednesday=4000 cheques
totaling to 16000 cheques
therefore y+3y+4000=16000 cheques
4y+4000=16000 cheques therefore 4y=16000-4000 cheques= 4y=12000cheques
therefore y=12000/4=3000 cheques
monday=3000 cheques
tuesday=3×3000=9000cheques

c

X×x-x=272
X^2-x-272=0
(X-16) or (x-17)
X=16 or x=17

17^2-17=272
Answer Is 17

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

Let the number of males be given the name M.
Let the number of females be given the name F.
If 15 females are absent, then M will be twice that of
present females.
This means that M = 2 * (F – 15)
M = 2 * F – 30.
or 2 * F – M = 30.
Now if in addition to the 15 females being absent, we also
have 45 males being absent,
then this gives the equation,
(F – 15) = 5 * (M – 45)
which simplifies to
F – 15 = 5 * M – 225
5 * M – F = 210
Pulling the equations together, we get
5 * M – F = 210
-M + 2 * F = 30
Multiply the first equation by 2, and keep the second
equation as is.
10 * M – 2 * F = 420
– M + 2 * F = 30
Add the equations.
9 * M = 450
M = 50
Verify answer.
Calculate F
from – M + 2 * F = 30
-50 + 2 * F = 30
2 * F = 30 + 50
F = 40.
If 15 females are absent, then number of males will be twice
that of females.
40 – 15 = 25.
50 = 2 * 25. Confirmed.
If also 45 males were absent, then female strength would be
5 times that of males.
Female strength is 25 due to the 15 females being absent.
50 – 45 = 5.
25 = 5 * 5. Confirmed.

First write equations from info:
(A) (Mon + Tue + Wed)/3 = 111 Rearrange as ——–> Tue + Wed = 111 – Mon
(B) (Tue + Wed + Thu)/3 =102 Rearrange as ——–> Tue + Wed = 102 – Thu
(C) Thu = 0.8(Mon)

Substitute equation C into B:
(B) Tue + Wed = 102 – 0.8(Mon)

At this point I changed the values for clearer algebra:
Mon = x
Tue + Wed = y

Re-write equations A & B with new values:
(A) y = 111 – x
(B) y = 102 – 0.8x

Solve simultaneous equations:

111 – x = 102 – 0.8x
111 – 102 = x – 0.8x (Re-arraged)
9 = 0.2x
x = 45

Thus, Mon = 45C

Thu = 0.8(45)
Thu = 36C

So the answer is it was 36C on Thursday