I guess no solution
4x +3 = 4X + 18
4X - 4X = 18 - 3
0 = 15
Call (F) the age of the father and (J) the age of Julio
The F & J are related in this way: F=4J
Now you have a restriction in the form of inequality: The sum of both ages has to be greater or equal than 55.
Algebraically that is: F + J ≥ 55
You can substitute F with 4J to find the solution for J:
4J + J ≥ 55
5J ≥ 55
Now divide both sides by 5
5J/5 ≥ 55/5
J ≥ 11
That Imposes a lower boundary for the value of J of 11, meaning that the youngest age of Julio can be 11
The correct answer is option a
but i think there s a mistake .
is the value w represent length
130x5=650
650-104=546
546-137=409
409-154=255
255-131=124
x=124
<span>Let the total hour of exam is 3hrs ie 180 minutes.( since answer required in minutes).
Let as assume maths questions be in x minutes.
Let the assume rest of the question is in y minutes.
Given that out of 320 qns 40 is maths qns.
therefore, maths qn is =40 and rest of qns is 280.
Also given the time of maths qns is twice of rest. They related in the following equation. ie x:y=2:1 ; x/y =2/1
ie x=2y
therefore 2x+y= 180 min ; ie 2(2y)+y= 180 ; y=36
therefore resolving for x = 2*36 = 72
time spent for maths qn is 2*x = 2* 72 =144min.
rest of maths question is 36 min.</span>
