# Math is Fast #2

How many of these can you solve quickly (and without using pen and paper)

- A teacher gave her students 3 pieces of rambutan each and had 18 left. If she wants to give her students 5 each she would need 32 more? How many students are there in the class?
- Pedro has a total of thirty 5 and 10 pesos coins worth a total of P245. How many 10 pesos coins are there?
- A train travelling at a speed 45km faster than a tourist bus covered 33 kms at the same time the bus travelled a distance of 18km. What is the average speed of the train?
- One leg of a right triangle with a hypothenuse of length 85 mm measures 84 mm. Find the length of the other leg.
- Find a right triangle given one leg is 13.
- What is 2019 x 20202020 – 2020 x 20192019?

Here are the solutions

- A teacher gave her students 3 pieces of rambutan each and had 18 left. If she wants to give her students 5 each she would need 32 more? How many students are there in the class?

Conventional solution: Algebra

Let s = number of students

Total number of rambutan = 3s + 18 or 5s – 32

3s + 18 = 5s – 32

5s – 3s = 18 + 32

2s = 50; s = 25

MATH-Inic Alternative Solution: Reasoning

The teacher has 18 left but needs an additional 32 to give 2 more rambutan per student. So number of students is (18 +32)/2 = 25

2) Pedro has a total of thirty 5 and 10 pesos coins worth a total of P245. How many 10 pesos coins are there?

Conventional solution: Algebra

Let t = number of 10-peso coins

30 – t = number of 5 peso coins

5(30-t) + 10t = 245

150 – 5 t + 10t = 245

(10-5)t = 245 – 150

5t = 95

t = 19 – 10 peso coins

Check 19 * P10 + 11 * P5 = P190 + P55 = P245

MATH-Inic Alternative solution: Reasoning

If all 30 coins were P 5, the total would be P150. The excess of P245-150 or P95 comes from the P10 coins each contributing P5 more. So P95/5 = 19 ten peso coins.

3) A train travelling at a speed 45km faster than a tourist bus covered 33 kms at the same time the bus travelled a distance of 18km. What is the average speed of the train?

Conventional solution:

Let x = the speed of the train

x – 45 = speed of the bus

Time travelled by the bus = time travelled by the train

18/( x – 45) = 33/x

18x = 33x – 1485

-15x = – 1485

x = 99 kph

MATH-Inic Solution: Reasoning

The difference in the distance travelled by the train and the bus is 33- 18 = 15 kilometers. This is exactly 1/3 of the difference of their speed. So the speed the of train is 3 x 33 = 99kph.

4) One leg of a right triangle with a hypothenuse of length 85 mm measures 84 mm. Find the length of the other leg.

Conventional solution: long multiplication

^{ }a^{2} = c^{2} – b^{2} ; Pythagorean theorem

a^{2} = 85^{2} – 84^{2}

a^{2} = 7225 – 7056

a^{2} = 169

a = 13mm

MATH-Inic Alternative solution: factoring

a^{2} = c^{2} – b^{2} ; Pythagorean theorem

a^{2} = (c + b) ( c – b)

a^{2} = (85 + 84)(85-1)

a^{2} = (169)(1) = 169

a = 13mm

5) Find a right triangle given one leg is 13.

Conventional solution: Trial and error: There is one equation and two unknowns, so there may be multiple solutions. Find two numbers whose squares differ by **a ^{2}**or a

^{2}= c

^{2}– b

^{2}. We must find two numbers whose square differ by 13

^{2}or 169.

MATH-Inic Alternative Solution: Factoring

Since a^{2} = c^{2} – b^{2} = (c + b)( c – b)

If a^{2 }is expressed as a product of two factors, one can be considered (c + b) and the other (c – b). Adding the two, we get ( c + b) + (c – b) = 2c or twice the hypothenuse. By solving for their difference, we get ( c + b) – ( c – b)= 2b or twice the unknown leg.

The simplest way to factor 13^{2 }or 169 is (169 x 1) so the hypotenuse is (169 + 1)/2 = 85 and the other leg is (169 – 1)/ 2 = 84.

But since (c-b) = 1, c and b are consecutive counting numbers. So the problem becomes finding two consecutive numbers adding up to 169.

6) What is 2019 x 20202020 – 2020 x 20192019?

Conventional solution: Long Multiplication/Using calculator

2019 x 20202020 – 2020 x 20192019

= 40,787,878,380– 40,787,878,380 = 0

MATH-Inic Alternative Solution: Factoring

2019 x 20202020 – 2020 x 20192019

= 2019 x (2020 x 10001) – 2020 x (2019 x10001) = 0