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CLAT Exam  >  CLAT Questions  >  For a real number x the condition |3x - 20| +... Start Learning for Free
For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if
  • a)
    10 < x < 15
  • b)
    7 < x < 12 
  • c)
    9 < x < 14
  • d)
    6 < x < 11
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
For a real number x the condition |3x - 20| + |3x - 40| = 20necessaril...
Case 1: x ≥ 40/3
we get 3x-20 +3x-40 = 20
6x=80

Case 2
we get 3x - 20 + 40 - 3x = 20
we get 20 = 20
So we get x 
Case 3x  < 20/3
we get 20-3x+40-3x =20
40=6x
x = 20/3
but this is not possible
so we get from case 1,2 and 3

Now looking at options
we can say only option C satisfies for all x .
Hence 7<x<12.
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Most Upvoted Answer
For a real number x the condition |3x - 20| + |3x - 40| = 20necessaril...
Understanding the Equation:
To understand why option 'B' (7 < x="" />< 12)="" is="" the="" correct="" answer,="" let's="" first="" analyze="" the="" given="" equation:="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20." this="" equation="" involves="" the="" absolute="" value="" of="" expressions="" containing="" />

Solving the Equation:
1. We first need to consider the two cases for the absolute value:
a) When 3x - 20 ≥ 0, then |3x - 20| = 3x - 20
b) When 3x - 20 < 0,="" then="" |3x="" -="" 20|="-(3x" -="" 20)="20" -="" />
2. Similarly, for the second absolute value:
a) When 3x - 40 ≥ 0, then |3x - 40| = 3x - 40
b) When 3x - 40 < 0,="" then="" |3x="" -="" 40|="-(3x" -="" 40)="40" -="" />
3. Now, we substitute these values back into the original equation and simplify:
(3x - 20) + (3x - 40) = 20
6x - 60 = 20
6x = 80
x = 80/6
x = 13.33

Checking the Options:
Now we need to check which option satisfies the condition 7 < x="" />< />
- If x is less than 7 or greater than 12, the equation will not hold true.
- Therefore, the correct range for x is 7 < x="" />< 12,="" making="" option="" 'b'="" the="" right="" />
Therefore, the correct answer is option 'B' (7 < x="" />< 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" true.="" 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" />
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For a real number x the condition |3x - 20| + |3x - 40| = 20necessarily holds ifa)10 < x < 15b)7 < x < 12c)9 < x < 14d)6 < x < 11Correct answer is option 'B'. Can you explain this answer?
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