Clearly, we have :
A = B - 3 ...(i)
D + 5 = E ...(ii)
A+C = 2E ...(iii)
B + D = A+C = 2E ...(iv)
A+B + C + D + E=150 ...(v)
From (iii), (iv) and (v), we get: 5E = 150 or E = 30.
Putting E = 30 in (ii), we get: D = 25.
Putting E = 30 and D = 25 in (iv), we get: B = 35.
Putting B = 35 in (i), we get: A = 32.
Putting A = 32 and E = 30 in (iii), we get: C = 28.
Originally, let number of women = x. Then, number of men = 2x.
So, in city Y, we have : (2x - 10) = (x + 5) or x - 15.
Therefore Total number of passengers in the beginning = (x + 2x) = 3x = 45.
L.C.M. of 6, 5, 7, 10 and 12 is 420.
So, the bells will toll together after every 420 seconds i.e. 7 minutes.
Now, 7 x 8 = 56 and 7 x 9 = 63.
Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.
Let son's age be x years. Then, father's age = (3x) years.
Five years ago, father's age = (3x - 5) years and son's age = (x - 5) years.
So, 3x - 5 = 4 (x - 5) ⇔ 3x - 5 = 4x - 20 ⇔ x = 15.
Let the number of boys be x. Then, (3/4)x = 18 or x = 18 x(4/3) = 24.
If total number of students is y, then (2/3) y = 24 or y = 24 x (3/2) = 36.
Therefore Number of girls in the class = (36 - 24) = 12.
Let money with Ken = x. Then, money with Mac = x + £ 3.
Now, 3x = (x + x + £ 3) + £ 2 ⇔ x = £ 5.
Therefore Total money with Mac and Ken = 2x + £ 3 = £ 13.
Let x and y be the number of deer and peacocks in the zoo respectively. Then,
x + y = 80 ...(i) and
4x + 2y = 200 or 2x + y = 100 ...(ii)
Solving (i) and (ii), we get) x = 20, y = 60.
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