Largest two-digit number

 What is the largest two-digit number that is divisible by 5 and whose digits differ by 3?

 * Using reasoned guess and check is the easiest way to solve this problem

Let's start with the largest 2-digit number divisible by 5  - 95. The difference between the digits in 95 is 4, this doesn't fit our second criteria. 

Moving on to the next largest number divisible by 5 - 90. The difference between these 2 digits is 9 which again does not fit our criteria. 

Now let's try 85. The difference between the digits is 3! 

85 is the largest two-digit number that is both divisible by 5 and its digits differ by 3.

Because we started with 95 (the absolute largest 2-digit number divisible by 5) and worked our way down, we can ensure that 85 is indeed the LARGEST number that fits these criteria. 

Answer: 85

How many different choices does Peter have?

 Peter is picking a number for his jersey. He wants to have a 2-digit number with an odd digit in the tens-place and an even digit in the ones-place. How many different choices does Peter have? 

* This problem is a permutation and combination.

Peter wants an odd digit in the tens place. The numbers eligible are: 1, 3, 5, 7, 9 (5 digits)

He also wants an even digit in the ones-place. The numbers eligible are: 0, 2, 4, 6, 8 (5 digits)

To start off let us pretend that Peter chooses 1 as his tens place digit. What are all of the different possibilities for his jersey number? He can have 10, 12, 14, 16, or 18. This gives peter 5 different options that all have 1 in the tens place. 

Now let us pretend Peter chooses 3 as his tens digit. Once again he has 5 different options for his jersey number  - 30, 32, 34, 36, and 38. 

Note that for every single tens place digit that Peter chooses, he will have 5 different options for the ones digit. 

Since there are 5 different options for the tens digit and each of those 5 options has 5 possibilities for the ones digit. Peter has 5x5 = 25 options total. See the image below for more clarification on why we are multiplying 5x5. 

 Answer = Peter has 25 different choices

How tall is the tree?

Anne is standing next to a tree that casts a shadow of 21 feet long. Anne is 4 feet tall and her shadow is 3 feet long. How tall is the tree?

* This problem involves ratio and proportions

The height of something and its shadow will be directly proportional, so we can use a ratio to find out how tall the tree is. 

Let us define the height of the tree as h.

In this case, we know that the ratio between Anne's height and Anne's shadow is 4:3

Similarly, we can say that the ratio between the tree's height and the tree's shadow is h:21

As mentioned earlier, all height to shadow relationships will be directly proportional. 

We multiplied Anne's shadow length by 7 to get the shadow length of the tree. Since the ratios are proportional this means we can also multiply Anne's height by 7 to get the height of the tree. 

Anne's height is 4 feet so 4x7=28. The tree's height is 28 feet tall. 

Ans: The tree is 28 feet tall

What is the difference between the second number and the first number?

 The sum of two numbers is 76. One-third of the first number plus half of the second number is 34. What is the difference between the second number and the first number?

Let us define the first number as x and the second number as y. We can write the following two equations

x + y = 76 .......... (a). This equation is derived from the statement "the sum of the 2 numbers is 76"

1/3 x + 1/2 y = 34 ............ (b). This equation is derived from the statement " 1/3 of the first number plus 1/2 the second number is 34. 

Now let's multiply (b) by 6

2x + 3y = 204 ............ (c)

Now multiply (a) by 2

2x + 2y = 152 ............ (d)

Subtract d from c to find out what the second number (y) is. 

y = 204 - 152 = 52

Now plug y into equation (a) to solve for the first number (x). 

x + 52 = 76

Subtracting 52 from both sides we get:

x = 76-52 = 24

Now we know that the first number is 24 and the second number is 52, but we are asked for the difference between the two numbers. 

52 - 24 = 28

Ans: 28

What is the maximum possible age?

 Mark, Mike, Joe, Jim and Dan are brothers. All of their ages are different. The sum of their ages is 95. Dan is 15 years old and is the youngest. Mike is the second oldest. What is the maximum possible age for Mike?

The sum of the 5 brothers' ages is 95 years. 

We know that Dan is 15 years old, so we can subtract 15 from 95. 

 95 - 15 = 80 

We know that the sum of the ages of the remaining 4 brothers is 80 years. 

Because we are working with the sum of the brothers' ages, in order to maximize someone's age (we are maximizing Mike's age in this case) we have to minimize the rest of the brothers' ages. 

Since Mike has two younger brothers after Dan, we will minimize their ages. 

Joe and Jim cannot be the same ages as each other or as Dan, which means that the smallest possible ages they can be are 16 and 17 years old, so we will assign those ages to Joe and Jim. 

Jim = 16; Joe = 17

We can now subtract the total ages of Joe and Jim so that the remaining number is the sum of the age of Mark and Mike. 

80 - (16 + 17) = 47

Now that Mike is the younger of the two siblings remaining and we still want him to be as old as possible, we will make Mike and Mark as close in age as possible. Specifically, they will be one year apart. 

As Mark gets older, Mike will have to get younger in order to maintain that the sum of their ages is still 47. This means that in order for Mike's age to be maximized, he must be exactly one year younger than Mark. 

We can define Mike's age as x and Mark's age as (x + 1)

x + (x + 1) = 47

2x + 1 = 47

2x = 46

x = 23

Answer: Mike is 23 years old

What is the value of C?

If digits A,B, and C are added, the sum is the two-digit number AB as shown at the right. What is the value of C?

Given A+B+C = AB
AB means 10th digit is A and B is the units digits, AB is 10A+B

i.e; A+B+C = 10A+B
A+C = 10A
C = 9A

As C is a single digit number, A can be only 1.
If A = 1, C = 9.

Hence C is 9.

What is the smallest number of children the class could have?

If a kindergarten teacher places her children 4 on each bench, there will be 3 children who will not have a place. However if 5 children are placed on each bench, there will be 2 empty places. What is the smallest number of children the class could have?

Let N be the number of children in the class.

If 4 children are placed on each bench, there will be 3 children left out.
=> N-3 is divisible by 4

If 5 children are placed on each bench, there will be 2 empty seats left out.
i.e. 3 children in one bench.
=> N-3 divisible by 5.

Hence N-3 should be divisible by 4 and 5 and smallest number divisible by 4 and 5 is 20.
=> N-3 = 20
=>N = 23

Smallest number of children the class could have is 23.

What is the perimeter of Figure B?

The small boxes in Figures A and B at the right are congruent squares. The perimeter of Figure A is 48 inches. What is the perimeter of Figure B? (The perimeter of a figure is the distance around it.) 

Let S be the side of the square in inches.

Perimeter of Figure A = 5S+5S+1S+5S = 16S = 48 inches

One side of the square S = 48/3 = 16 inches

Perimeter of Figure B = 3S+3S+3S+4S+3S+4S = 20S = 20 *3 = 60

Perimeter of Figure B is 60 inches.

How much does one loaf of bread cost?

One loaf of bread and six rolls cost $1.80. At the same prices, two loaves of bread and four rolls cost $2.40. How much does one loaf of bread cost?

Let B represent the cost of one loaf of bread and R represent the cost of one roll.

1B+6R=$1.80 = 180 cents ------(1)
2B+4R=$2.40 = 240 cents ------(2)

(1) * 2 gives
2B+12R=360 ----(3)

(3)-(2) gives
12R-4R=360-240
8R=120
R=15

B=180-6R=180-6*15=180-90=90

One loaf of bread costs 90 cents.

Which is greater, D or N, and by how much?

D is the sum of the odd numbers from 1 through 99 inclusive, and N is the sum of the even numbers from 2 through 98 inclusive:
           D = 1+3+5+ ........+99 and N = 2+4+6+ ...... +98
Which is greater, D or N, and by how much?

MetroEye: Refer Progression03 for formulae.

Number of terms in odd number series = \(\dfrac{t_n+1}{2}\)= \(\dfrac{99+1}{2}\) = \(\dfrac{100}{2}\)  = 50

Number of terms in even number series = \(\dfrac{t_n}{2} = \dfrac{98}{2}\)  = 49

Sum, D = \( n^2\) = 50*50

Sum, N = n(n+1) = 49*50

=> D is greater than N.

Difference between D and N = D-N = 50*50-49*59 = 50(50-49) = 50

Hence D is greater than N by 50.