Percents/ Rule of Three / Problem

In the figure bellow it is represented the outline of a rectangular room with an area of 48 square meters. We note ABCD this rectangle.

It is known that the width is three quarters (¾ ) of the length of the room.

Inside the room is a fireplace (a stove), represented in the figure by the square MNPD with the one meter side.

In the room it is installed hardwood floor except shaded area MNPD where is the stove.
a) find the length of the room;
b) knowing that losses represent 10% of the floor area that will be covered with parquet flooring, show that it is enough to buy 51.7 square meter of needed parquet flooring,
c) the parquet flooring is sold in packs each containing 2,5 square meters of parquet flooring. The price of each box with parquet flooring is 43 $. Find the minimum amount of money required to purchase needed parquet flooring.

Find a natural number by the Euclid's division

Find the natural number that simultaneously satisfy the conditions:

a) divided by 4 the remainig is 3

b) divided by 10, the remaning is 1;

c) divided by 12, the remaining is 3;

d) the sum of quotients of the three divisions from points a), b) and c) is higher by 16 than one third of the number.

 Solution:

In this issue we use the Euclidean division.

This theorem says the dividend (D) is equal to the product of the divisor (Q) and the quotient (C) added the remaining (R)

D = Q  ∙ C + R

If we denote the number  with the letter D and apply the above theorem we can write conditions:

a) D = 4 ∙ C1 + 3

b) D = 10 ∙ C2 + 1 ∙

c) D = 12 ∙ C3 + 3

For the last condition we write: C1 + C2 + C3 = D / 3 + 16

      a)       D = 4 ∙ C1 + 3

               D - 3 = 4 ∙ C1

               C1 = (D-3) / 4

       b)     D = 10 ∙ C2 + 1 ∙

                D - 1 = 10 ∙ C2

                C2 = (D-1) / 10

        c)     D = 12 ∙ C3 + 3

                D - 3 = 12 ∙ C3

                C3 = (D-3) / 12

Replacing expressions obtained for C1, C2 and C3 to the condition d) we obtain:

                    (D-3) / 4 + (D - 1) / 10 + (D - 3) / 12= D / 3 + 16

The common denominator of these fractions:

4 = 2∙2

10 = 2 ∙ 5

12 = 2∙2 ∙ 3

 3 = 3∙1

The common denominator = 2∙2 ∙ 3 ​​∙ 5 = 60

Multiply each fraction reach 60: first by 15, the second by 6, third by 5, fourth by 20 and the free term (whole number) by 60:

15 (D-3) + 6 (D-1) + 5 (D-3) =20 D  + 16 ∙ 60

15 D - 45 +6 D -6 +5 D – 15 = 20 D + 960

26 D - 20 D = 960 + 45 + 6 + 15

  6 D = 1026

      D = 171

Solution: The number required is 171

Find a two digit number

Find number ab, where a is the tens' digit and b is the units’ digit, knowing that a⁴ + a² = 5 · b.

Solution:

The problem requires us to find a two digit number ab, where a is the tens digit and b is the units’ digit.

Figures may have values ​​from 1 to 9 (for a=0 result 0⁴ + 0² = 5 · b => b=0, and the number ab=0 is not a two-digit number).

From a⁴ + a² = 5 · b. We note that  5b is a multiple of 5.

a⁴ + a² = a² (a² +1) must be multiple of 5.

a²   can’t be 5 (a is a natural number)

a² +1 = 5; 

a² = 5-1 = 4; 

a = 2

4*5 = 5 *b

b = 4 

Other case, for the next multiple of 5

a² +1=10

a²=9

a=3

a²(a²+1) = 9*10 = 90 = 5*b =>b = 90/5 = 18 >9, impossible, b is a digit, b< 9

           

The only number ab that is solution to the problem is 24.

Find the last digit of a number

Whether natural numbers x, y, z, u satisfy the relation:

7x + 5y - 2z -2u = 0, then find the last digit of the number A when

A = (11x + 9y) *(z + u - x).

Solution:

The problem gives us an equality between the numbers x, y, z and u.

7x+5y – 2z – 2u = 0 (the hypothesis)

If you look at this equality you can see that the relation could be processed moving two terms on the right side of the equality.

-2z and -2u moved to the right side with the changed sign become:    2z and 2u.

7x + 5y = 2z + 2u. We can remove the common factor 2.

7x + 5y = 2 (z + u).

From this form of the last relationship we note the following: 

to the right we have a bracket multiplied by 2, which means that this product is an even number.

If on the right side of the equality it is an even number, then we also have an even number on the left of equal sign.

So, 7x + 5y is an even number.

Here the sum is even.

How do we get an even number in the sum? if we add two even numbers or two odd numbers!

How 7 and 5 are odd, it follows that x and y are either both even or both odd (in other words: they have the same parity).

We conclude that the amount x + y is even (if we add two even numbers the result is an even number, or if we add two odd numbers the result is an even number also).

The problem asks us the last digit of the number A

A = (11x + 9y) *(z + u - x).

.

Number A is a product of two brackets. We see that z is assembled with u and this amount (z + u) is in the hypothesis multiplied by 2:    7x + 5y = 2 (z + u).

.

To be able to use the hypothesis equality we multiply by 2 the number A and we replace 2(z + u) by 7x+5y. And we get:

2 A = (11x + 9 y) * 2 * (z + u - x) = (11x + 9 y) (2 (x + u) - 2 x) = (11x + 9 y) (7 x + 5 y - 2 x) = (11x + 9 y) (5x + 5y) = (11x + 9 y) * 5 (x + y)

We have demonstrated that x and y have the same parity, and x + y is an even number, x + y is divisible by 2.

For parenthesis (11x + 9y) we note:

If x and y are both even 11x + 9y is an even number, so it is divisible by 2.

If x and y are both odd 11x is odd, 9y is also odd and the sum of two odd numbers is always an even number, so 11x + 9y is divisible by 2.

It follows that number 2A is divisible by the number 2 from the bracket 11x + 9y, once it is divisible by 2 by the even number x + y and it is also divisible by 5.

Number A will be divisible by 2 and 5.

Therefore, number A is divisible by 2 and 5, so it is divisible by 10.

From the criterion of divisibility by 10 (any number that has the last digit “0” is divisible by 10) the last digit of the number A is „0”.

Angles of triangle and the direct proportionality

Angles A, B and C of triangle ABC are directly proportional to numbers 3, 7 and 8, respectively.  Calculate the size of the smallest angle of triangle.

Solution:

We know that the directly proportional sizes form a string of equal ratios.

So:

(size of angle A)/3 = (size of angle B)/7 = (size of angle C)/8 = k (the constant of proportionality); 

                                                                                                                                      (relationship 1)

Note: (angle A)/3 = (angle A)-out-of-3 

Because the angles A, B and C are the angles of the triangle, these three angles always add to 180⁰.

(size of angle A) + (size of angle B) + (size of angle C) = 180 ⁰ ;   (the relationship 2)

Also, we know: from a string of equal ratios we can obtain a new ratio with the sum of the numerators-out-of-the sum of the denominators. Using the above relationship in 1 and 2 we obtain

(angle A)/3 = (angle B)/7 = (angle C)/8 = (angle + angle B + angle C)/(3 + 7 + 8) = 180 ⁰ /18 = 10 ⁰ = k (the constant of proportionality).

Now, we calculate the measures of the angles by matching each report with k.

(angle A)/3 = 10 ⁰; the measure of the angle A = 3*10 = 30 ⁰

(angle B)/7 = 10 ⁰; the measure of the angle B = 7*10 ⁰ = 70 ⁰

(angle C)/8 = 10 ⁰; the measure of angle C = 8*10 = 80 ⁰

The smallest angle in the triangle is ∡A =30 ⁰