Question

Ex 3.4.5.6.7 de la mijloc

Answer 1

3)
(3xa+5)x5-3=292
(3xa+5)x5=292+3
(3xa+5)x5=295
3xa+5=295:5
3xa+5=59
3xa=59-5
3a=54
a=54:3
a=18

(b-5)x2=30
b-5=30:2
b-5=15
b=15+5
b=20


(c-17)x3+15=45
(c-17)x3=45-15
(c-17)x3=30
c-17=30:3
c-17=10
c=17+10
c=27

[2x(d-30)-45]x3=165
[2x(d-30)-45]=165:3
2x(d-30)-45=55
2x(d-30)=55+45
2x(d-30)=100
d-30=100:2
d-30=50
d=50+30
d=80


4)
a= numar necunoscut
3a+147=2343
3a=2343-147
3a=2196
a=2196:3
a=732

5)
l----------l nr care se scade
l----------l---------l---------l---------l---------l tot numarul
             ....................................................
                36 diferenta

5-1=4 segmente egale
36:4=9
9x5=45 numarul la care m-am gandit


6)
4+4=8 (jumatate a patra zi)
8+4=12 a treia zi
8x2=16 (restul a doua zi)
16+4=20 a doua zi
20x2=40 (rest prima zi)
40+4=44( prima zi jumatate)
44+4=48 prima zi
44x2=88 tot drumul

verificare:
88:2+4=48 (prima zi)
88-48=40 rest prima zi
40:2+4=24 (a doua zi)
40-24=16 rest a doua zi
16:2+4=12 a treia zi
4= a patra zi



7)
1/3 = al treilea nepot

2/3=16 raman

16:2x3=24 (nr de nuci ramase dupa al doilea nepot)

24:3=8 (nr de nuci luate de al doilea nepot)

24:2x3=36 ( nr de nuci ramase dupa primul nepot)

36:3=12 (nr de nuci luate de al doilea nepot)

36:2x3=18x3=54 nr de nuci pe care le-a avut bunica

verificare:
18+12+8+16=54