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4%
The answer is A)
y1 = 62 Rs/kg
y2 = 72 Rs/kg
y = 64.5 Rs/kg
y2 – y1 = 10 Rs/kg
The distance between the y and y1 is
y – y1 = 64.5 – 62 = 2.5
x1 = (y – y1)/(y2 – y1) = 2.5/10 = 0.25
x2 = 1 – x1 = 1 – 0.25 = 0.75
The target price is calculated by the lever method.
x1 * y2 + x2 * y1 = 0.25 * 72 + 0.75 * 62.5 = 64.5
The ratio is of y1 to y2 is
0.75 : 0.25
Divide by both by 0.25
3 : 1
no combinations is there because a number divisible by 36
also divisible by its divisors anu.sum of 4 digits=20,whis
is not multiple of 3.so it is cannot divisible by 3
Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125
When none of the digits are repeated:
The hundred’s place can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 except the one which has already been used at the thousand’s place, so it can be filled in 5 ways.
Similarly tens’ place can be filled in 4 ways: only those 4 numbers which have not been use either at hundred’s or thousand’s place.
Unit’s place can be filled in only 3 ways. So, total number of nos. Possible =4×5×4×3= 240
38 years
Let Rajan’s present age be x years. Then, his age at the time of marriage = (x – 8) years.
x = 65(x−8)
⇒5x=6x−48
⇒x=48 years
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = (x – 18) = 30 years
∴ Rajan’s sister’s present age = (30 + 8) years = 38 years
Day 1: 30-3ft = 27+2ft = 29
Day 8: 23 jumps 3ft = 20 slips 2ft = 22
Day 16: 13 jumps 3ft = 10 slips 2ft = 12
Day 24: 3 jumps 3ft = Out?
Greatest number of 5 digits=99, 999
Smallest number of 5 digits=10, 000
And their sum=99, 999+10, 000=109, 999
A. 14th