Question

Дама сдавала в багаж диван, чемодан, саквояж,картину, корзину, картонку и маленькую собачонку.Диван весил столько же, сколько чемодан и саквояж месте взятые и столько же, сколько картина, корзина и картонка вместе, при этом каждый из предметов - картина, корзина, картонка - по отдельности весили больше, чем собачонка. Когда загружали багаж, дама заявила, что собака не той породы. При проверке оказалось, что если к собаке на весы добавить саквояж, то вместе они перевешивают диван и то же самое происходит если к собаке на весы добавить чемодан. Докажите, что претензия дамы была справедлива. Пожалуйста, помогите!!! Очень нужно, заранее спасибо! Если можно, то с решением

Answer 1

При сдаче багажа:
диван(Д)=чемодан(Ч)+саквояж(С)=картина(К₁)+корзина(К₂)+картонка(К₃)
К₁>маленькая собачонка(МС)
К₂>MC
K₃>MC
При получении багажа:
{Ч+С=Д
{МС+Ч>Д
{МС+С>Д
МС+Ч+МС+С>2Д
2МС+Ч+С>2Д  
2МС+Д>2Д
2МС>Д
   По условию: Д=3К, значит, получается, что 2МС>3K, но также, по условию, МС<K₁; MC<K₂; MC<K₃, но, даже, если бы было, что МС=К, то, приравняв вес К  и МС к х: , получим неравенство:
     2х>3x
       2>3
     Данное неравенство неверно, значит, претензии дамы справедливы.

Answer 2

Problem Analysis

The problem states that a lady checked in a sofa, suitcase, satchel, painting, basket, cardboard, and a small dog as luggage. The weight of the sofa is equal to the combined weight of the suitcase and satchel, which is also equal to the weight of the painting, basket, and cardboard combined. Additionally, each individual item (painting, basket, and cardboard) weighs more than the small dog. During the baggage check, the lady claimed that the dog was not the correct breed. Upon inspection, it was found that if the suitcase or satchel were added to the weight of the dog, they would outweigh the sofa. The same is true if the suitcase or satchel were added to the weight of the dog. The task is to prove that the lady's claim was valid.

Solution

To prove the lady's claim, we need to compare the weights of the sofa and the combined weight of the suitcase, satchel, and dog. We also need to compare the weights of the sofa and the combined weight of the suitcase, satchel, and dog. If the combined weight of the suitcase, satchel, and dog is greater than the weight of the sofa, then the lady's claim is valid.

Let's assign variables to the weights of the items: - Weight of the sofa = S - Weight of the suitcase = C - Weight of the satchel = S - Weight of the painting = P - Weight of the basket = B - Weight of the cardboard = C - Weight of the dog = D

According to the problem statement: 1. S = C + S 2. S = P + B + C 3. P > D 4. B > D

We also know that: 5. C + D > S 6. S + D > C

To prove the lady's claim, we need to show that C + D > S.

Let's substitute the values from equations 1 and 2 into equation 5: (C + S) + D > S C + S + D > S

Since S = C + S, we can simplify the equation to: C + C + D > S 2C + D > S

Now, let's substitute the values from equations 1 and 3 into equation 6: (C + S) + D > C C + S + D > C

Since S = C + S, we can simplify the equation to: C + C + D > C 2C + D > C

From equations 2 and 4, we know that P + B + C > D. Let's substitute this into equations 5 and 6: (C + S) + (P + B + C) > S C + S + P + B + C > S

Since S = C + S, we can simplify the equation to: C + C + P + B + C > S 3C + P + B > S

(C + S) + (P + B + C) > C C + S + P + B + C > C

Since S = C + S, we can simplify the equation to: C + C + P + B + C > C 3C + P + B > C

Now, let's compare the two inequalities: 2C + D > S 3C + P + B > S

Since we know that 2C + D > S and 3C + P + B > S, we can conclude that the lady's claim is valid.

Therefore, the lady's claim that the dog was not the correct breed was valid.

Note: The solution provided above is a mathematical proof based on the given information. The actual weights of the items and the dog's breed are not provided, so we cannot determine the specific values. The proof shows that if the weights satisfy the given conditions, then the lady's claim is valid.