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31
cos
434-403=31
465-434=31
RABBIT
6.25
Let the distance between each pole be x m. Then, distance up to 12th pole = 11x m
∴ Speed = 11x22m/s
∴ Time to cover total distance up to 20th pole
= 19x×2411x
= 41.45 s
42m
Ans-44
4x+8=6Y
5X=5Y
X=Y
FROM EQ 1
2X=8
X=4
6Y+5Y=11*4=44
4.224 days
3 miutes
each pencil 100/25 = 4
each book 100/15 = 6.66
100 – 15 = 85
85-5*4 = 65
65/6.66 = 9
he can buy 9 books that is not in given options currently so
answer is NONE
Let the number of males be given the name M.
Let the number of females be given the name F.
If 15 females are absent, then M will be twice that of
present females.
This means that M = 2 * (F – 15)
M = 2 * F – 30.
or 2 * F – M = 30.
Now if in addition to the 15 females being absent, we also
have 45 males being absent,
then this gives the equation,
(F – 15) = 5 * (M – 45)
which simplifies to
F – 15 = 5 * M – 225
5 * M – F = 210
Pulling the equations together, we get
5 * M – F = 210
-M + 2 * F = 30
Multiply the first equation by 2, and keep the second
equation as is.
10 * M – 2 * F = 420
– M + 2 * F = 30
Add the equations.
9 * M = 450
M = 50
Verify answer.
Calculate F
from – M + 2 * F = 30
-50 + 2 * F = 30
2 * F = 30 + 50
F = 40.
If 15 females are absent, then number of males will be twice
that of females.
40 – 15 = 25.
50 = 2 * 25. Confirmed.
If also 45 males were absent, then female strength would be
5 times that of males.
Female strength is 25 due to the 15 females being absent.
50 – 45 = 5.
25 = 5 * 5. Confirmed.
3500*10/100
3500*11.5/100 = 402.5 per year
402.5×3 = R1207.5