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.