Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
65 views
in Aptitude by (101k points)
closed by
If \(\frac{{3 + \;\frac{1}{{2 + \;\frac{1}{3}}}\; \times 7}}{{0.2}} \div 1.2 + 2\sqrt {99} = \;\sqrt a + \;\sqrt b + 5\) than find the value of a/b.
1. 1.2223
2. 2.2424
3. 3.5252
4. 2.1515

1 Answer

0 votes
by (108k points)
selected by
 
Best answer
Correct Answer - Option 1 : 1.2223

Concept used:

Follow BODMAS rule to solve this question, as per the order given below:

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

formula (a + b)2 = a2 + 2ab + b2

Calculations:

 

\(\frac{{3 + \;\frac{1}{{2 + \;\frac{1}{3}}}\; \times 7}}{{0.2}} \div 1.2 + 2\sqrt {99} = \;\sqrt a + \;\sqrt b + 5\)

 ⇒ \(\frac{{3 + \frac{3}{7}\; \times 7}}{{0.2}} \div 1.2 + 2\sqrt {99} = \;\sqrt a + \;\sqrt b + 5\)

⇒ 20 + \(2\sqrt {99} = \;\sqrt a + \;\sqrt b \)

(99 = 11 × 9 and 11 + 9) so we can write by above formula is:

\(\sqrt {11} + \;\sqrt 9 = \;\sqrt a + \;\sqrt b \)

hence a = 11 and b = 9

∴ a/b = 11/9 = 1.2223

 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...