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Profit one with 3 & 2 nd 2.4

G.c.d=num1 *num2/lcm
Num1=20
Gcd=5
Lcm=60
5=(20*num2)/60
Num2= 5*60/20
Num2=15

Given:

In a group of 15 students,

7 have studied Latin,

8 have studied Greek,

3 have not studied either.

To find:

The number of students who studied both Latin and Greek.

Solution:

In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.

Therefore,

n(A∪B) = 15 – 3

n(A∪B) = 12

7 have studied Latin,

n(A) = 7

8 have studied Greek,

n(B) = 8

n(A∩B) is the number of students who studied both Latin and Greek.

n(A∩B) = n(A) + n(B) – n(A∪B)

n(A∩B) = 7 + 8 – 12

n(A∩B) = 15 – 12

n(A∩B) = 3

The number of students who studied both Latin and Greek is 3

Final answer:

3 of them studied both Latin and Greek.

Thus, the correct answer .3

P+C+M=80
P+C=70
Therefore M=10
Given , P+M=90
If M=10
P=80

East

Assuming the youngest has X years, so we have
X+(X+3)+(X+6)+(X+9)+(X+12)=50
5X+30=50
5X=20
X=4
The youngest has 4 years

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.

Each PAIR of stations means a PAIR of tickets (A to B and B to A)

2(old stations)(new stations) + 2(new stations)(new stations – 1) = 46

(N * X) + (X * (X – 1)) = 23

factoring ___ X (N + X – 1) = 23

23 is a prime number with only two factors, 1 and 23

so N = 23 and X = 1

Pen

here the 30% of 100 is 30
and 10% of the 30 = 3
so 20% = 6%
so ans = 6%