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Mamma: "Let me think, Tommy. Well, our three ages add up to exactly seventy years."
Tommy: "That's a lot, isn't it? And how old are you, Papa?"
Papa: "Just six times as old as you, my son."
Tommy: "Shall I ever be half as old as you, Papa?"
Papa: "Yes, Tommy; and when that happens our three ages will add up to exactly twice as much as today."
Tommy: "And supposing I was born before you, Papa; and supposing Mamma had forgot all about it, and hadn't been at home when I came; and supposing..."
Mamma: "Supposing, Tommy, we talk about bed. Come along, darling. You'll have a headache."
Now, if Tommy had been some years older he might have calculated the exact ages of his parents from the information they had given him. Can you find out the exact age of Mamma?
[Ref: ZPYU] Mamma's Age. Amusements In Mathematics by Henry Ernest Dudeney (1917).
Hint: The answer involves years and months.
Answer:
The age of Mamma must have been 29 years 2 months.
That of Papa, 35 years; and that of the child, Tommy, 5 years 10 months. Added together, these make seventy years. The father is six times the age of the son, and, after 23 years 4 months have elapsed, their united ages will amount to 140 years, and Tommy will be just half the age of his father.
The answer above is taken from the original book, here is another version of the answer:
If we call Tommy T, Mamma M and Papa P we can see that:
"our three ages add up to exactly seventy years" leads to
T + M + P = 70 (1)
"Just six times as old as you" leads to
P = 6 x T (2)
In an unknown number of years (Y) "Shall I ever be half as old as you" leads to:
P + Y = 2 x (T + Y) (3)
and "our three ages will add up to exactly twice as much as today" leads to:
(T + Y) + (M + Y) + (P + Y) = 140
which can be written as
T + M + P + 3Y = 140 (4)
We can see from (4) and (1) that
3Y = 70
so
Y = 70 ÷ 3 (5)
Using (2) and (5) in (3) we have
P + Y = 2 x (T + Y)
6 x T + 70÷3 = 2 x (T + 70÷3)
4 x T = 70÷3
T = 70÷12 (6)
We can now use (6) in (2) to see that:
P = 6 x T
P = 6 x 70÷12
P = 70÷2
And using the values for T and P in (1) we have:
T + M + P = 70
70÷12 + M + 70÷2 = 70
Multiply throughout by 12 to give:
70 + 12 x M + 420 = 840
12 x M = 840  420  70
12 x M = 350
M = 350÷12
So: Tommy = 70÷12 = 5.83333 = 5 years 10 months.
Papa = 70÷2 = 35 = 35 years.
Mamma = 350÷12 = 29.1666 = 29 years 2 months.
You find yourself playing a game with your friend.
It is played with a deck of only 16 cards, divided into 4 suits:
Red, Blue, Orange and Green.
There are four cards in each suit:
Ace, King, Queen and Jack.
Ace outranks King, which outranks Queen, which outranks Jack  except for the Green Jack, which outranks every other card.
If two cards have the same face value, then Red outranks Blue, which outranks Orange, which outranks Green, again except for the Green Jack, which outranks everything.
Here's how the game is played: you are dealt one card face up, and your friend is dealt one card face down. Your friend then makes some true statements, and you have to work out who has the higher card, you or your friend. It's that simple!
Round 1:
You are dealt the Green Ace and your friend makes three statements:
1. My card is higher than any Queen.
2. Knowing this, if my card is more likely to beat yours, then my card is Blue. Otherwise it isn't.
3. Given all of the information you now know, if your card is more likely to beat mine, then my card is a King. Otherwise it isn't.
By #1: your friend's card is higher than any Queen, so your friend can only have one of these cards: Red Ace, Blue Ace, Orange Ace, Red King, Blue King, Orange King, Green King, Green Jack.
By #2: 4 of these 8 cards could beat your card, so your friend's card is not more likely to beat yours, so your friend's card is not blue. Leaving Red Ace, Orange Ace, Red King, Orange King, Green King, Green Jack.
By #3: of the remaining 6 cards, your card can beat 3, so your card is not more likely to beat your friend's, so your friend's card is not a King. Leaving Red Ace, Orange Ace, Green Jack  all of which beat your card.
Puzzle 19
If four kittens can catch four mice in four minutes...
...how many mice can twelve kittens catch in twelve minutes?
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