Transport for new south wales Interview Questions, Process, and Tips

390

Let cost for apple be a Cost for banana be b and Orange be c

So by first value expression becomes 17a + 13b + 9c =130 ———-1 therefore if you further solve a = (130 – 13b – 9c)/17 ———- 2 the second expression becomes: 13c + 7a + 10b = 100 ———- 3 If you put value of a in second expression it becomes: 13c + 7[(130 – 3b – 9c)/17] +10b = 100

Further if you solve you get value of b:

b = 10 – 2c ———-4

put value of b from 4 in 1

17a + 13 [10 – 2c] + 9c = 130

Further if you solve you find value of a

a = c ———-5

Put 5 in 3

13c + 7c + 10b = 100

further solve you get: c = 1 ———-6

from 5 and 6

a = c = 1 ———-7

Substitute value of c in expression 4

b = 10 – 2c b = 10 -2 * 1 b = 8 ———-8

therefore a + b + c = 10

FOR 15 = RS.20
FOR 20 = RS.18
FOR 25 = RS.16
FOR 30 = RS.14
FOR 35 = RS.12
FOR 40 = RS.10

40358

1!=1
2!=2×1=2
3!=3x2x1=6
4!=4x3x2x1=24
5!=5x4x3x2x1=120
6!=6x5x4x3x2x1=720
The answer is 120.

find out what is the problem and taking the bad out always works.

a = 2
d = 5 – 2 = 3
Then 12th term is :
This in arithmetic progression
L = a + (12 – 1) d
L = 2 + 11 * 3
L = 2 + 33 = 35

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

U is the answer as
T is sitting opposite to R and W is sitting opposite to U

(b) 16.66%

Earlier for ₹x we could purchase y gm of sugar.
Now we pay ₹1.2x for y gm of sugar

(As there was an increase in price so, x + 20%x = 1.2x)

At current rates for ₹x you can purchase y/1.2 gm of sugar

So the reduced consumption is y-(y/1.2)

Percentage change = (reduced consumption/ original consumption ) *100

That is (0.2/1.2) *100 = 16.66% (approx)

B

From the first statement
Speed=distance/time
=>240/24=10m/s
so speed=10m/s
From the second statement
distance=length of the platform+length of the train=650+240=890m
Time=distance/speed
here, the speed of the train is already calculated
=>890/10=89s
So the time taken by the train to cross the platform is 89 seconds

125

length of the base=3 times height
b=3h;
h=b/3;
so, 1/2bh=24
1/2b(b/3)=24
(b^2)/6=24
b^2=144
b=12