Please enter the following details
Free Daily Targets & Mocks
Study Material & Formulas PDFs
1 on 1 Mentorship
Question 63
For natural numbers x, y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is
Correct Answer: 34
Solution
It is given, y(x + z) = 19
y cannot be 19.
If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.
Therefore, y = 1 and x + z = 19
It is given, z(x + y) = 51
z can take values 3 and 17
Case 1:
If z = 3, y = 1 and x = 16
xyz = 3*1*16 = 48
Case 2:
If z = 17, y = 1 and x = 2
xyz = 17*1*2 = 34
Minimum value xyz can take is 34.
Video Solution
Create a FREE account and get:
- All Quant CAT complete Formulas and shortcuts PDF
- 35+ CAT previous year papers with video solutions PDF
- 5000+ Topic-wise Previous year CAT Solved Questions for Free
CAT Quant Questions | CAT Quantitative Ability
CAT Logarithms QuestionsCAT Time Speed Distance QuestionsCAT Inequalities QuestionsCAT Functions, Graphs and Statistics QuestionsCAT Progressions and Series Questions
CAT DILR Questions | LRDI Questions For CAT
CAT Games and Tournamnents QuestionsCAT Venn Diagrams QuestionsCAT Coins and Weights QuestionsCAT Data Interpretation QuestionsCAT Quant Based DI Questions
