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Ans: 60 kph
Suppose Person meets the train everyday at 3 PM at Station A.
His speed is 12kph.
So normally he reaches 5 km before the meeting point (pt B) at (5/12 hr = 25 min before) 2:35PM.
But if he is late by 30 min, then he will reach that point (pt B) by 3:05 PM.
Train is traveling at its normal speed so it covers the distance of 5 Km in 5 min starting from Station A and reaches the meeting point (pt B) at 3:05 PM.
So speed of the train is 5KM/5min = 60 kph.
60%
Total=60reems
Utilized=40reems
Un utilized= 60-40=20
Percentage of remainder 20/60=.3333
0.33333*100=33.333%
pages # figures
1000 1000 4
from 999 to 100 (999-99)*3 2700
from 99 to 10 (99-9)*2 180
from 9 to 1 (9-0)*1 9
total sum= 2893
75 degrees
30h – 11/2 m = 240 – 165 = 75 ..
13
b
9936
Birds fly on water.
38 years
Let Rajan’s present age be x years. Then, his age at the time of marriage = (x – 8) years.
x = 65(x−8)
⇒5x=6x−48
⇒x=48 years
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = (x – 18) = 30 years
∴ Rajan’s sister’s present age = (30 + 8) years = 38 years
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
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