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Mathematics application problems in the fourth grade of primary school
1.

Every hour114.

B completed every hour 1/20.

In a cycle, Party A and Party B can each type 1 hour, one ***2 hours, to complete:

1/ 14+ 1/20= 17/ 140

14017 = 8 is greater than 4.

That is, after 8 cycles, the remaining 4/ 140= 1/35.

It still needs:1/35 ÷114 = 0.4 hours to complete a.

1 * * * Requirements: 8×2+0.4= 16.4 hours.

2.

When my brother is twice as old as my sister, my father is 34 years old.

Compared with this year, the age of the three people has changed as much.

Imagine this:

Counting the age that my dad changed for my brother, that is, my dad is still 34 years old.

In this way, my sister changed 1 and my brother changed 2.

My brother is twice as old as my sister.

Sister this year: (64-34)÷( 1+2)= 10 years old.

So, when my sister was 9 years old, it was 1 years ago.

The sum of their ages is: 64-3=6 1 year.

The total age of father and brother is 6 1-9=52 years old.

Elder brother's age: 52÷(3+ 1)= 13.

Brother this year: 13+ 1= 14 years old.

Dad: 64- 10- 14=40 years old.

A: This year, my father is 40 years old, my brother 14 years old and my sister 10 years old.

3.

Similar to the above question. .

The difference between Party A and Party B is 2 1- 17=4 years old.

When A is 18 years old, C is three times that of D.

At this time B: 18-4= 14 years old.

Compared with this year, the age changes of the four people are the same.

If the ages of both parties are given to Party C,

A is still 18 years old and b is still 14 years old.

Then d becomes 1, c becomes 3, and the age of c is three times that of d.

This year: (64-18-14) ÷ (3+1) = 8 years old.

4.

Think of it as a problem that chickens and rabbits are in the same cage. .

If this 14 day is done by B, it can be completed:

1/ 12× 14=7/6

Exceeding: 7/6- 1= 1/6.

Every day, A does less than B:

1/ 12- 1/20= 1/30

A did: 1/6÷ 1/30=5 days.