- Janatha Fish Meal & Oil Products General Aptitude Interview Questions
- Janatha Fish Meal & Oil Products Trainee Interview Questions
- Janatha Fish Meal & Oil Products Personal Questions round Interview Questions
- Janatha Fish Meal & Oil Products HR Interview Questions
- Janatha Fish Meal & Oil Products Lead Interview Questions
Statements :
Some men are educated. Educated persons prefer small families.
Conclusions :
I. All small families are educated.
II. Some men prefer small families.
126
cos if one of the factor is given for hcf and lcm and multiply hcf and lcm and them divide it by the given factor
so 18*3780/540=18*7=126
Explanation:
Let E1 be the event of drawing a red card.
Let E2 be the event of drawing a king .
P(E1 ∩ E2) = P(E1) . P(E2)
(As E1 and E2 are independent)
= 1/2 * 1/13 = 1/26
24 days
Let Suvarna, Tara, Uma and Vibha be S,T,U,V respectively
initially in the beginning each persons share be
V = x U = y T = z
S = w = (x+y+z+32) Reason: She has to double others share, so she should have each and everyone’s share and still should be left out with 32
after 1st Round of game
S loses and is out with 32 and doubles the others share
V = 2x U = 2y T = 2z
After 2nd Round of game
T loses and is out with 32 and doubles the others share
V = 4x U = 4y
This means T had 2z = 2x + 2y + 32
After 3rd round of game
U looses and is out with 32 and doubles others share
V = 8x
This means U initially has 4y = 4x + 32
In the end V = 8x = 32
Solving this we get x = 4, y = 12, z = 32 and w = 80
There fore Suvarna had highest share in the beginning
151
Second
(p*r*t)/100 = I ………………..(p*3*10)/100 = 840 …..p = 84000/30 = 2800
16 chapatis* 6 Rs= 96
5 plates of rice * 45 Rs = 225
7 plates of mixed vegetable * 70 Rs = 490
6 ice-cream * ??
The amount that Alok paid the cashier was Rs.961
96+225+490 = 811
961-811= 150
150/6 = 25
The cost of each ice-cream cup is 25 .Rs
((15 X m) + 23)/(m+1)= 16
solving , we get number of matches (m) as 7 (excluding last
match)
so he shud hit 39 runs in last innings to ake average to 18
a
11
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
Distance =speed ×time
Speed =90kmph
90km take 60min time to reach
In 10min the distance covered =?
D=90×10/60
900/60
D=15