a) who started with small amount of money?
b) who started with greatest amount of money?
c) what amount did B have?
answer 10
if initially the price was Re 1.the new price nw bcoms to b
0.7..
so ,if the price z 0.7 ,the increment required is 0.3
so % in crement= 0.3/0.7*100=42.87…
Traffic Jams
Public Speaking
What to Wear to a Party
Noises on an Airplane
Bees
Let the numbers be x and x + 2.
Then, (x + 2)2 – x2 = 84
⇒ 4x + 4 = 84
⇒ 4x = 80
⇒ x = 20.
∴ The required sum
= x + (x + 2)
= 2x + 2
= 42
562.
True
let’s say the Cost Price is 1000x. (CP)
The selling price is also the same as the Cost price. So, here SP=CP (But he sells 950gm instead of 1000gm)
Instead of 1000gm, he is selling 950 grams at the CP. So he sells 050 gms @1000x price. So his net profit is 50gm.
Now 1000gm is 1000x Rs
So, 50gms is 50x rs. [Apllied Unitary Method]
So his profit percentage is:
Profit Percentage Formula: {(Profit/CP)*100%}
So, Here profit is 50x;
CP is 1000x;
so putting the value in the formula we get, Profit Percentage is: (50x/1000x)*100% =(5000x/1000x)%=5%.
5
307,311,313,317,319
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
Length of train = 125Mtrs
Speed of man = 5Km/Hr
=(5*1000)/(1*3600)
=1.4M/s
Time taken for train to pass = 10s
Spd = D/T
Total Distance (Man distance+ Train Length)
#Distance covered by Man in 10s
D=Spd*T
= (1.4M/s*10s)
= 14Mtrs
#Train length = 125Mtrs
Total D = 14Mtrs + 125Mtrs
= 139Mtrs
Time taken = 10s
Sp = D/t
=139M/10s
=13.9M/s
B
0
The answer cannot be determined as there is a particular formula where the consecutive numbers start.
brother
status after 3 games
A-40
B-40
c-80
d-16
status after 2 games
A-20
B-20
C-40
D-96
status after 1 game
A-10
B-10
C-108
D-48
status after 0 games
A-5
B-93
C-54
D-24
so answers are
a)A
b)B
c)93