SSC CGL Exam > SSC CGL Questions > How many natural numbers are in between 120 ...
Start Learning for Free
How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?
- a)19
- b)17
- c)15
- d)21
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
How many natural numbers are in between 120 and 519 which are exactly...
Solution:
Step 1: Find the first natural number which is divisible by 3 and 7.
The LCM of 3 and 7 is 21. Therefore, the first natural number which is divisible by both 3 and 7 is 21.
Step 2: Find the last natural number which is divisible by 3 and 7.
The last natural number which is less than or equal to 519 and divisible by both 3 and 7 is 504. This can be obtained by finding the largest multiple of 21 which is less than or equal to 519. 21 × 24 = 504.
Step 3: Find the number of natural numbers between 21 and 504 which are divisible by 3 and 7.
The required numbers are 21, 42, 63, 84, …, 504.
To find the number of terms in this sequence, we use the formula for the nth term of an arithmetic sequence:
an = a1 + (n - 1)d
where a1 is the first term, d is the common difference, and an is the nth term.
In this case, a1 = 21, d = 21, and an = 504. We need to find n.
504 = 21 + (n - 1)21
Simplifying this equation, we get:
(n - 1)21 = 483
n - 1 = 23
n = 24
Therefore, there are 24 terms in the sequence.
Step 4: Find the number of terms between 120 and 519.
The first term of the sequence which is greater than or equal to 120 is 126. The last term of the sequence which is less than or equal to 519 is 504. Therefore, we need to find the number of terms between 126 and 504.
To do this, we subtract the number of terms before 126 from the total number of terms in the sequence.
The first term of the sequence is 21, which is less than 126. Therefore, we need to find the number of terms in the sequence from 21 to 126.
To find this, we use the formula for the sum of the first n terms of an arithmetic sequence:
Sn = n/2(2a1 + (n - 1)d)
where Sn is the sum of the first n terms, a1 is the first term, d is the common difference, and n is the number of terms.
In this case, a1 = 21, d = 21, and the last term of the sequence less than or equal to 126 is 105. Therefore, we need to find the sum of the first 5 terms of the sequence:
S5 = 5/2(2 × 21 + (5 - 1)21)
S5 = 5/2(42 + 84)
S5 = 315
Therefore, there are 5 terms in the sequence from 21 to 126.
Now, we can find the number of terms between 126 and 504:
Number of terms = Total number of terms - Number of terms before 126 - Number of terms after 504
Number of terms = 24 - 5 - 0
Number of terms = 19
Therefore, there are 19 natural numbers between 120 and 519 which are
Step 1: Find the first natural number which is divisible by 3 and 7.
The LCM of 3 and 7 is 21. Therefore, the first natural number which is divisible by both 3 and 7 is 21.
Step 2: Find the last natural number which is divisible by 3 and 7.
The last natural number which is less than or equal to 519 and divisible by both 3 and 7 is 504. This can be obtained by finding the largest multiple of 21 which is less than or equal to 519. 21 × 24 = 504.
Step 3: Find the number of natural numbers between 21 and 504 which are divisible by 3 and 7.
The required numbers are 21, 42, 63, 84, …, 504.
To find the number of terms in this sequence, we use the formula for the nth term of an arithmetic sequence:
an = a1 + (n - 1)d
where a1 is the first term, d is the common difference, and an is the nth term.
In this case, a1 = 21, d = 21, and an = 504. We need to find n.
504 = 21 + (n - 1)21
Simplifying this equation, we get:
(n - 1)21 = 483
n - 1 = 23
n = 24
Therefore, there are 24 terms in the sequence.
Step 4: Find the number of terms between 120 and 519.
The first term of the sequence which is greater than or equal to 120 is 126. The last term of the sequence which is less than or equal to 519 is 504. Therefore, we need to find the number of terms between 126 and 504.
To do this, we subtract the number of terms before 126 from the total number of terms in the sequence.
The first term of the sequence is 21, which is less than 126. Therefore, we need to find the number of terms in the sequence from 21 to 126.
To find this, we use the formula for the sum of the first n terms of an arithmetic sequence:
Sn = n/2(2a1 + (n - 1)d)
where Sn is the sum of the first n terms, a1 is the first term, d is the common difference, and n is the number of terms.
In this case, a1 = 21, d = 21, and the last term of the sequence less than or equal to 126 is 105. Therefore, we need to find the sum of the first 5 terms of the sequence:
S5 = 5/2(2 × 21 + (5 - 1)21)
S5 = 5/2(42 + 84)
S5 = 315
Therefore, there are 5 terms in the sequence from 21 to 126.
Now, we can find the number of terms between 126 and 504:
Number of terms = Total number of terms - Number of terms before 126 - Number of terms after 504
Number of terms = 24 - 5 - 0
Number of terms = 19
Therefore, there are 19 natural numbers between 120 and 519 which are
Free Test
FREE
| Start Free Test |
Community Answer
How many natural numbers are in between 120 and 519 which are exactly...
Given
The numbers are between 120 and 519
they are exactly divisible by both 3 and 7
Concept used:
The number divisible by both 3 and 7 is divisible by 21
(LCM of 3 and 7)
Calculation:
The total numbers divisible by 21 in between 1 and 120 is
⇒ 120 / 21 = 5
The total numbers divisible by 21 in between 1 and 519 is
⇒ 519 / 21 = 24
So total number that is divisible by 21 between 120 and 519 is
(24 - 5) = 19
So, the total number is 19
Attention SSC CGL Students!
To make sure you are not studying endlessly, EduRev has designed SSC CGL study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in SSC CGL.
|
Explore Courses for SSC CGL exam
|
|
Similar SSC CGL Doubts
Top Courses for SSC CGLView all
How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer?
Question Description
How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer? for SSC CGL 2024 is part of SSC CGL preparation. The Question and answers have been prepared according to the SSC CGL exam syllabus. Information about How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC CGL 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer?.
How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer? for SSC CGL 2024 is part of SSC CGL preparation. The Question and answers have been prepared according to the SSC CGL exam syllabus. Information about How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC CGL 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer?.
Solutions for How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for SSC CGL.
Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free.
Here you can find the meaning of How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer?, a detailed solution for How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?a)19b)17c)15d)21Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice SSC CGL tests.
|
|
Explore Courses for SSC CGL exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup