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cube root of 5 is 125 so 128 is nearer to 125 and hence 5+5+5=15 and the nearest answer amongst options is 17 so 17 is correct answer:) yipppy……..
15%
option D
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
A = (B*H)/2
H = 2B
formula becomes A = (B*2*B)/2
this can be rewritten as A = (2*B^2)/2
the 2 in the numerator and denominator cancel out and you get:
A = B^2
150 miles
21
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
24 days
10
40
40
30
60
– 10 cows
+ 20 days
if 40 cows can eat for forty days
then 20 cows can eat for 80 days
time is inversely proportional to speed.
So if speed taken is 4/5 of usual speed, time taken will be 5/4 of usual time
and the difference between time is 10 min
So,
(5/4)t -t=10 min
usual time,t=40 min
late time=40+10=50 min
121
180*2/3 = 60degree
General formula for simple interest is SI=PTR/100
Then P= SI x 100 / TR
Here SI= rs.24, T= 2 years, R = 10%
P= 24 x 100 / 2 x 10 = 120 Rs
Answer : Rs. 120