All Operations in the Order of Operations: Identify the greater value

Let's solve exercise A:

$(4\times3)+(5\times12)=12+60=72$

Let's solve exercise B:

$(10\times3)+(7\times5)=30+35=65$

Let's solve exercise C:

$(7\times7)+(7\times3)=49+21=70$

Let's solve exercise D:

$(25\times2)+(15\times2)=50+30=80$

Let's solve exercise A:

$8\times8=64$

Let's solve exercise B:

$10\times6=60$

Let's solve exercise C:

$12\times5=60$

Let's solve exercise D:

$15\times4=60$

We solve the exercise:

$24:6:2=$

According to the rules, we solve the exercise from left to right:

$24:6=4$

$4:2=2$

Now we solve the exercise:

$24:(6:2)=$

According to the rules, first we solve the exercise inside the parentheses and then we divide:

$6:2=3$

$24:3=8$

Therefore:

8 > 3

First, we solve the exercise$14+16:4\times2$According to the order of operations, multiplication and division come before addition and subtraction, therefore we first divide 16 by 4 and then multiply the result by 2:

$16:4=4$

$4\times2=8$

Then, we solve the exercise:
$14+8=22$

Now, we solve the exercise:$14+16:(4\times2)=$

According to the order of operations, the exercise within parentheses is first, so we first multiply 4 by 2 and then proceed to the division operation.

$4\times2=8$

$16:8=2$

In the last step we add:$14+2=16$

Since 22 is greater than 16, the corresponding result is >

First, we solve the exercise:

$4+7-3+1-5 =$

According to the rules of the order of operations, we solve the exercise from left to right:

$4+7=11$

$11-3=8$

$8+1=9$

$9-5=4$

Now we solve the exercise:

$4+(7-3+1-5) =$

According to the rules of the order of operations, we first solve the parentheses and then we will join:

$(7-3+1-5)=$

$(4+1-5)=$

$(5-5)=0$

$4+0=4$

Therefore:

$4=4$

A:

$100-3\times20=100-(3\times20)=100-60=40$

B:

$10:5\times20=2\times20=40$

C:

$50-4\times3=50-(4\times3)=50-12=38$

D':

$20:4+10=(20:4)+10=5+10=15$

A:

$250-50:5=250-(50:5)=250-10=240$

B:

$100+30=130$

C:

$20\times4-7=(20\times4)-7=80-7=73$

D:

$10+4:2=10+(4:2)=10+2=12$

A:

$9\times7:3+3=(9\times7:3)+3=(63:3)+3=21+3=24$

B:

$20+4\times3=20+(4\times3)=20+12=32$

C:

$15\times4:6=60:6=10$

D:

$12-3+20=9+20=29$

Let's solve exercise A:

$(4\times11)+10=44+10=54$

Let's solve exercise B:

$(9\times3)+(9\times3)=27+27=54$

Let's solve exercise C:

$9+(8\times5)+7=9+40+7=49+7=56$

Let's solve exercise D:

$5+(5\times10)=5+50=55$