Burgess rawson Interview Questions, Process, and Tips

In the example first we have to find the area of the field

we have given the following values ,

cost of fencing per meter = Rs 1.50

Diameter of circular field = 28

We have to find the area , a = ?

We know that,

Area = 2 π r

= 2 x 22/7 x 14

= 2 x 22 x 2

= 44 x 2

= 88 sq. m

So, area of circle is 88 sq.m

Now just multiply it with 1.50

cost of fencing = 88 x 1.50

= Rs. 132

So, the cost of fencing the circular field is Rs. 132

10

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

Answer: 1 / 3

Lets say each brand had 16 laptops
So Sold : A = 3 , B= 1 and C = 12 , total sold: 16 ,
Since each brand has 16 , total = 48
so 16/ 48 = 1/3 is the sold fraction

D = 180m
S = 42 – 6 = 36 * 5/18 = 10
T = D/S = 180/10
T = 18s

5c2*4c1/9c3 +5c3/9c3=25/42

419

Days taken by A to complete whole work= 140 days
Days taken by B to complete whole work= 105 days

A’s 1 day work= 1/140
B’s 1 day work= 1/105
On adding these together we get: 1/60 (1 day work by working together)

So, together they will be able to complete it in 60 days

13. One group of each = 2+3+5 = 10 and 1 remainder for each = 1+1+1 = 3. 10+3 = 13

Hockey

5
307,311,313,317,319

Gain=(profit/c.p)*100

Gain =selling price-cost price =900-675
=225.
Gain=(225/675)*100.
=33.3%

9 sec=4 times
x sec=12 times
x=(9*12)/4=27 sec

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​

Soldier have to move 0.75 miles to West and then 0.375 miles to South to reach the camp.
How:-
Firstly soldier moving 1 mile to East from camp and then 1/2 miles= 0.50 miles to North .

Then it is moving 1/4 miles =0.25 miles to west
And then 1/8 miles =0.125 miles to South

Now we just have to count the difference and minus the value of camp to East with the value of North to west = 1mile – 0.25miles(1/4) = 0.75 miles

Same case with north and south = 0.50 miles(1/2) – 0.125 miles(1/8) = 0.375 miles

Hence proves to return camp the soldier have to move 0.75 miles to west and 0.375 miles to south

d

c