**Example 1**. In an objective type test containing 25 questions, a student is to be awarded +5 marks for every correct answer, –5 for every incorrect answer and zero for not writing any answer. Mention the ways of scoring 110 marks by a student.

**Example 2**. What is the first time after 4:00 p.m. at which the minute hand and hour hand of a clock will meet up rounded to the nearest minute?

**Example 3**. The product of four positive integers *a*, *b*, *c* and *d *is 8!, and they satisfy the equations

*ab*+*a*+*b*= 524*,**bc*+*b*+*c*= 146*,*and*cd*+*c*+*d*= 104*.*

Find the value of *a *− *d*.

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “**the only way to learn mathematics is to do mathematics**”.

**Solution 1**. Total marks scored = +110

So, minimum correct responses = 110 ÷ (+5) = 22.

Case 1. Suppose the number of correct responses = 22Marks for one correct response = + 5

Marks for 22 correct respones = (22 × 5) = +110

But, marks scored = +110

Marks obtained for incorrect answer = 0

So, no incorrect response.

And, therefore, 3 questions were unattempted.

Case 2. Suppose the number of correct responses = 23Marks for one correct response = + 5

Marks from 23 correct responses = (23 × 5) = + 115

But, marks scored = + 110

Marks obtained for incorrect answers = 110 – (+115) = –5

Marks for 1 incorrect answer = –5

Number of incorrect responses = (–5) ÷ (–5) = 1.

So, 23 correct, 1 incorrect and 1 unattempted.

Case 3. Suppose the number of correct responses = 24Marks for one correct response = + 5

Marks from 24 correct responses = (24 × 5) = + 120

But, marks scored = + 110

Marks obtained for incorrect answers = +110 – (+120) = –10

Number of incorrect responses = (–10) ÷ (–5) = 2

Thus the number of questions = 24 + 2 = 26. Whereas, total number of questions is 25. So, this case is not possible.

So, the possible ways are:

**22 correct, 0 incorrect, 3 unattempted**; or**23 correct, 1 incorrect, 1unattempted.**

**Solution 2**. In this problem, we will calculate the rates of the minute and hour hands in revolutions per minute.

Our main formula will be *rate* (in revolutions per minute) × *time* (in minutes) = *revolutions*. This is identical to rate × time = amount of work done, except it is a special case.

Let us picture our desired scenario. This scenario, where the minute and hour hands coincide, will be sometime after 4:20 but before 4:25.

Let us now find the rates of the minute and hour hands. Since the minute hand of a clock makes one full revolution every 60 minutes, it makes 1/ 60 revolutions per minute. Since the hour hand of a clock makes one full revolution every 12 hours or 720 minutes, it makes 1/ 720 revolutions per minute.

At 4:00 p.m., the hour hand has a 1/ 3 of a revolution head start on the minute hand. We want the fraction of a revolution that the minute hand makes to equal the fraction of a revolution that the hour hand makes plus the head start that it has.

If we set the time that this takes as *t*, we can write the equation

Our answer is this amount of time after 4:00 p.m., which turns out to be 4:22 to the nearest minute.

**Solution 3**. A critical observation in the solution to this problem is that the three equations can be rewritten as

- 525 = 524 + 1 =
*ab*+*a*+*b*+ 1 = (*a*+ 1)(*b*+ 1)*,* - 147 = 146 + 1 =
*bc*+*b*+*c*+ 1 = (*b*+ 1)(*c*+ 1)*,*and - 105 = 104 + 1 =
*cd*+*c*+*d*+ 1 = (*c*+ 1)(*d*+ 1)*.*

Now factor the constant terms in each equation to obtain facts about the products.

- (
*a*+ 1)(*b*+ 1) = 525 = 3 · 5^{2}· 7*,* - (
*b*+ 1)(*c*+ 1) = 147 = 3 · 7^{2}*,*and - (
*c*+ 1)(*d*+ 1) = 105 = 3 · 5 · 7*.*

Since (*a *+ 1)(*b *+ 1)has a factor of 5^{2} = 25, but (*b *+ 1)(*c *+ 1)has no factor of 5, (*a *+ 1)must be divisible by 25. In a similar manner, (*d *+ 1) must be divisible by 5.

Because (*b *+ 1)(*c *+ 1)= 3 · 7^{2}, the possibilities for (*b *+ 1)and (*c *+ 1) are either (*b *+ 1)= 7 and (*c *+ 1)= 3 · 7 or are (*b *+ 1)= 3 · 7 and (*c *+ 1)= 7.

However, if (*b *+ 1)= 7, then (*a *+ 1)= 3 · 25 = 75 and *a *= 74. But we are given that *a *· *b *· *c *· *d *= 8!, and 74 does not divide 8!.

So, we cannot have (*b *+ 1)= 7. Hence we must have (*b *+ 1)= 3 · 7 and (*c *+ 1)= 7. This implies that (*a *+ 1)= 25 and (*d *+ 1)= 3 · 5 = 15. Thus *a *= 24, *b *= 20, *c *= 6, and

In conclusion, *a *− *d *= 24− 14= 10.

Such types of wonderful maths… great…. thanks sir…

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