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To solve this problem, we can break it down into steps:
Step 1: Determine the individual rates of work for A, B, and C.
If A needs 8 days to finish the task, then their work rate is 1/8 of the task per day.
If B needs 12 days to finish the task, then their work rate is 1/12 of the task per day.
If C needs 16 days to finish the task, then their work rate is 1/16 of the task per day.
Step 2: Calculate the combined work rate of A and B.
If A works for 2 days, their contribution will be 2 * (1/8) = 1/4 of the task completed.
If B works until 25% of the job is left for C, then they will complete 75% of the task.
Step 3: Calculate the time it takes for B to complete 75% of the task.
Since B’s work rate is 1/12 of the task per day, it will take B (75%)/(1/12) = 9 days to complete 75% of the task.
Step 4: Calculate the remaining work for C.
If B completes 75% of the task, then the remaining work for C is 100% – 75% = 25% of the task.
Step 5: Calculate the time it takes for C to complete the remaining work.
Since C’s work rate is 1/16 of the task per day, it will take C (25%)/(1/16) = 4 days to complete the remaining 25% of the task.
Step 6: Calculate the total time required.
A worked for 2 days, B worked for 9 days, and C worked for 4 days, totaling 2 + 9 + 4 = 15 days.
Therefore, it will take a total of 15 days for A to work for 2 days, B to work until 25% of the job is left, and C to complete the remaining work.
9936
Complete step-by-step answer:
Investment made by A for 1 year, IA=2000
Investment made by B for 2 year,
IB=2×3000IB=6000
Investment made by C for 2 year,
IC=2×4000IC=8000
The ratio of their investment is given by
IA:IB:IC=2000:6000:8000
To simplify the ratio divide it by 1000,
IA:IB:IC=2:6:8
Now again to simplify divide the ratio by 2,
IA:IB:IC=1:3:4
The total parts of this investment =1+3+4=8
A’s share in the investment is 1 out of the 8 parts; B’s share is 3 out of 8 while C’s share is 4.
Therefore, the profit share of A’s investment of Rs. 2000=18×3200=400.
The greatest number that will divide 187, 233 and 279 leaving the same remainder in each case.
To find : The number ?
Solution :
First we find the difference between these numbers.
The required numbers are
233-187=46
279-233=46
279-187=92
Now, We find the HCF of 46 and 92.
Therefore, The required largest number is 46.
A
p = prob(1st digit not 7)*prob(2nd digit not 7)*prob(3rd
digit not 7)
=(8/9)*(9/10)*(9/10)
=0.72
Sandhya had it right except that there are 900 numbers
between 100 and 999 inclusive (not 899).
1(100-1)+ 2(100-2)+…..99(100-99)
N=100
solution: N(N-1)
100(99)= 9900
3, 8, 15, 24, 34, 48, 63
8-3 = 5
15-8 = 7
24-15 = 9
34-24 = 10 this one should be 35-24 = 11
48 – 34 = 14 based on the correction will be 48-35 = 13
63-48 = 15
the difference will create a series 5,7,9,11,13,15…..etc
789, 645, 545, 481, 440, 429, 425
b
5m/sec
The ans is -20.
Solution:
A-1
B-2
C-3 D-4 …like wise till Z-26.
ELECTRICITY-5+12+5+3+20+18+9+3+9+20+25==129
GAS-7+1+9==27
ELECTRICITY-GAS=129-27-(Minus 2)=100
so
JACK-JILL=(10+1+3+11-(10+9+12+12)-(minus(2))==(-20)