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How to calculate the sum of differences in mathematics
There are 1,131+134 = 265 people, including twice the total number of people in Class B and Class C and Class A and Class D. ..

The number of students in Class B and Class C is less than that in Class A and Class D 1 ... that is, the number of Class A and Class B is-1= the number of Class B and Class C.

So 265- 1=264=3 times the number of people in Class B and Class C.

Number of categories b and c:

( 13 1+ 134- 1)÷3

=264÷3

=88 people

Number of Class Two of Party A and Party D:

88+ 1=89 people

Class 4 * * * has: 88+89= 177 people.

2, this question is relatively simple, first ask how much each bottle is refunded, and then ask how much a * * * is refunded?

3, 30+32+22=84 kg

84 kg = (A+B+C) ×2

So A+B+C = 84 ÷ 2 = 42kg.

Answer: 42-30 = 12kg.

B: 42-22 = 20kg。

C: 30-20= 10 kg.

4. Sum and difference problem

Sum =875, difference: one is more than two by 250, two is more than three 125, and one is more than two by 250+ 125=375.

Large number (first place) = (sum+difference) ÷3

=(875+250+375)÷3

= 1500÷3

=500 yuan

Second place: 500-250=250 (yuan)

Third place: 250- 125= 125 (yuan).

5, and double the problem.

Sum (of dividend and divisor) =43- quotient-remainder

=43-3-4

In order to make the dividend exactly three times the divisor and subtract the remainder 4,

Sum =43-3-4-4=32, multiple =3.

Decimal (divisor) = and ÷ (multiple+1)

=32÷(3+ 1)

=8

Dividend =8×3+4=28

6. The problem of differences.

Poor: The ice is 24x 2 = 48 = 48 yuan. After the jade gives ice to 30 yuan, the ice is 48+30 = 78 yuan, and the ice is three times that of jade.

Therefore, the existing money of Emerald =78÷(3- 1)=39 (yuan).

Yuyuan Garden is rich =39+30=69 (yuan)

Ice money =69+48= 1 17 (yuan).

7. It's also relatively simple. Move the decimal point to the left by 1 bit, which means reducing it by 10 times. (poor times).

8. Problems of different times.

It is the same. After A gives 30 yuan to B, A is 30× 2 less than B = 60 yuan. At this time, A is 1/5 of B, that is, B is five times that of A. ..

A How much money is there first?

A (decimal) = difference present (multiple-1)

=30×2÷(5- 1)

=60÷4

= 15 (yuan)

The original value of Party A and Party B is 15+30=45 yuan.