Medium Puzzles 

Workings

Hint
Can Q1 have A as its answer?

Answer
Q1. C – the first question with B as the correct answer is Q3.
Q2. D – the answer to Question 4 is C.
Q3. B – the answer to Question 1 is C.
Q4. C – the number of questions that have D as the correct answer is one.
Q5. B – the number of questions that have B as the correct answer is two.

Reasoning
A complicated, and sometimes, confusing answer!

Can Q1 be A? No, because this would tell us that Q1 was the first question with B as the answer, which would be a contradiction (as we've just assumed Q1 was A).

Can Q1 be B? No, because this would tell us that Q4 was the first question with B as the answer, which would be a contradiction (as we've just assumed it was Q1).

Can Q1 be C? Possibly, because Q3 points back to Q1 correctly, and is logically consistent.

Can Q1 be D? No, because:

   Q2 would be B

   Q4 would be A – which means that we have three questions with D

   which means that Q3 would have to be D, and this would tell us that Q1 is A, which would be a contradiction (as we've just assumed that Q1 is D).

Therefore, Q1 is C, which means that Q3 is B.

We can ignore Q2 for a moment, as it asks us about Q4, and look at Q4 first.

Looking at Q4 (how many questions have D as the answer) it can't be D (zero), as this would contradict itself.

It can't be A (three) as we only have two other questions without an answer.

If Q4 was B, then the remaining questions (Q2 and Q5) would both have to be D, and:

   Q5 being D would mean that only one question was B, which would be a contradiction (as Q3 is already B, and we've just assumed that Q4 is also B).

So Q4 must be C, which means that Q2 is D.

Looking at Q5, it can't be A (as Q3 is B), it can't be D (as Q4 tells us that we only have one question with D as the answer – Q2). It can't be C as we don't have three questions that are B. So Q5 is B (the two questions are Q3 and Q5).

Workings

Hint
One of the words is CLOCKWISE.

Answer
FOO + TPR + INT = FOOTPRINT
SOM + EWH + ERE = SOMEWHERE
PIN + EAP + PLE = PINEAPPLE
TEC + HNI + QUE = TECHNIQUE
REF + ERE + NCE = REFERENCE
PRO + FES + SOR = PROFESSOR
BIO + GRA + PHY = BIOGRAPHY
CLO + CKW + ISE = CLOCKWISE
BAR + OME + TER = BAROMETER

Workings

Hint
Start by labelling the number as ABCDE.

Answer
24976.

Reasoning
We can start by labelling the digits as ABCDE.

We know that:

(i)  B = 2 x A

and:

     E = A + B

And using (i) we get:

     E = A + (2 x A)
(ii) E = 3 x A

If A = 1, then B = 2, and E = 3, but this isn't allowed (as there are no 1's in the puzzle).

If A = 2, then B = 4, and E = 6.

If A = 3, then B = 6, and E = 9, but this isn't allowed (as C has to be the largest digit).

So, A = 2, B = 4, E = 6, and we now have to find C and D.

We also know that:

D = (A + B + C + D + E) ÷ 4

And using (i) and (ii) we get:

D = [A + (2 x A) + C + D + (3 x A)] ÷ 4

so:

3 x D = (6 x A) + C

so:

(iii) D = [(6 x A) + C] ÷ 3

C can only be 7, 8 or 9 (as it's the largest digit, and we've already found 6) and (iii) tells us that it must be a multiple of 3, which means that C = 9. Leaving D = 7.

So the final number is: 24976.

Double-Checking
The answer is 24976.

The fourth digit is a quarter of the total of all of the digits.
   A + B + C + D + E = 2 + 4 + 9 + 7 + 6 = 28, and 28 ÷ 4 = 7.

The second digit is twice the first digit.
   4 = 2 x 2.

The third digit is the largest.
   9 is the largest digit.

The last digit is the sum of the first two digits.
   6 = 2 + 4.

Workings

Hint
These are the first letters of the words: A, P, T, I, P, U, A, R, R, W, D, C, R, B, E.

Answer
----nst = against ----nip = parsnip ----fth = twelfth ----rno = inferno ----dox = paradox ----sil = utensil ----nac = almanac ----iem = requiem ----arb = rhubarb ----ior = warrior ----bey = disobey ----van = caravan ----ipt = receipt ----dth = breadth ----nox = equinox