IF you DON’T use “and” to write numbers above hundred (e.g. 101 as ‘hundred and one’), you can go on till 999. Not just ‘A’, you won’t find ‘B’ or ‘C’ either.

Just ONE…! Because, once you have placed the first penny, the box is no more ’empty’, hence anything you put subsequently, you will not be putting into an ’empty’ box.

Send that PEACOCK and the egg to me first, so that I can examine it and decide, since it needs to be confirmed whether the egg is of the peacock itself, because I never knew that a PEACOCK could lay eggs…!!

You need to plug the leak urgently.

You had filled 25 gallons but could use only 10, which let you reach just 300 miles in 5 hours, at 60 miles per hour. That means you lost 15 gallons in those 5 hours or 3 gallons per hour. Put differently, you lost 60% of your money you spent on petrol over those five hours. At around £1.2 per litre, that’s £18 that went down the drain in 5 hours or £3.6 per hour as against £2.4 you could utilize per hour.

Jacob is 23, the children are aged 2, 3 and 5.

Deduction:

Let the ages of Jacob and children be A, B, C and D. Also, let the first and second digits of Jacob’s age be X and Y.
So, we have:
A = 10X + Y  (1)
B = X             (2)
C = Y             (3)
D = X + Y      (4)
Since the sum of the ages is 33, we have
A + B + C + D = 33 => (10X + Y) + X + Y + (X + Y) = 33
=> 12X + 3Y = 33   (5)
Dividing by 3, we get
4X + Y = 11     (6)

The above equation states 11 as the sum of a multiple of 4 and another number.
The only multiples of 4 that are less than 11 are 4 and 8.

i.e, 4X = 4 OR 4X = 8 => X = 1 OR X = 2.

X= 1 is impossible because, in that case, to satisfy (6), Y must be 7, which would mean that a 17-yr old Jacob has an 8 Yr-old child, which is impossible.

Hence: X = 2, Y = 3.

Answer: £10

Derivation:

1. By reverse calculation:

The final amount you had (which you spent on Thursday) = £2.50
This amount was left after you spent HALF of you had on sweets. As you spent HALF of what you had, what remained (£2.50) should also be HALF of what yo you had. Thus, before spending on sweets, you must have had £5.
Now, this amount was what was left after spending one-third of what you had, on CD. Since you spent one-third, what remained (£5) must have been two-third of what you had before spending on CD. Thus, the amount you had before getting the CD = £7.50.
Finally, this amount was what was left after spending a quarter of what you started with. Since £7.50 was the balance after spending one quarter, it means £7.50 was three-quarters of what you started with. Hence the initial amount (the pocket money you got) must be £10.

Put differently:
Let the starting amount be X.   (1)
You spent one quarter thereof on cloths. Balance = 3X/4  (2)
Then you spent one-third of this on CD. Balance: 3X/4 x 2/3 = X/2    (3)
Thereafter, you spent half of this amount on sweets. Balance = X/2 x 1/2 = X/4.   (4)
The amount remained was £2.50. Since this is one-fourth of what you originally had as per (4) above,  X = 2.5 x 4 = 10.

Since there are TWO White stickers and only ONE Black sticker, there are only TWO possible ways to stick them on the two candidates: Either BOTH White OR ONE white and ONE black.

If one of the candidates had a BLACK sticker on, the other candidate who can see the sticker can immediately deduce that (s)he has a WHITE sticker on (because there is only one black sticker which is already on the other candidate). But this did not happen, which means that both of them must be having a WHITE sticker.

The letters ‘a’, ‘x’ and ‘z’ does not appear even once in the paragraph. All other letters appear at least once.

Next number: 6
Following numbers, in that order, completes the series: 9, 7, 8, 8.

(Hint: Number of letters in names of months)

9, 11, 13 (9 * 11 * 13 = 1287)

Deduction:

1. We know that 10*10*10 = 1000, fairly close to the target value of 1287. Thus, the required numbers should be somewhere close to 10.
2. ‘The consecutive odd numbers’ CAN’T include a number which is multiple of 5, because an odd-number multiple of 5 will ALWAYS have 5 as the last digit, wheres we require 7.
3. The only possible sequences of three consecutive odd numbers without involving ‘5’ is such that the numbers must end as 7, 9 and 1 OR 9, 1 and 3. ([1,3,5], [3,5,7] and [5,7,9] can’t be considered as all of them contain ‘5’ as a member.)
4. The ‘answer triplet’ CAN’T be such that they end in [7,9.1]  because the product of such a set will have last digit as 3, whereas the target is 7. So, from (3) above, the numbers must be ending 9, 1, 3.

From the above, we conclude that the only possible combination is 9, 11, 13.