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How many numbers are there from 300 to 650 which are completely divisible by both 5 and 7? (SSC CGL 2017)
- a)8
- b)9
- c)10
- d)12
Correct answer is option 'C'. Can you explain this answer?
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How many numbers are there from 300 to 650 which are completely divisi...
Solution:
To find the solution of this question, we need to find the common multiples of 5 and 7 between 300 and 650.
Multiples of 5: 300, 305, 310, 315, ………., 645, 650
Multiples of 7: 301, 308, 315, 322, ………., 637, 644
To find the common multiples of 5 and 7, we need to find the LCM (Least Common Multiple) of 5 and 7.
LCM of 5 and 7 = 35
So, the common multiples of 5 and 7 are:
35 x 9 = 315, 35 x 10 = 350, ………., 35 x 18 = 630
To find the count of numbers which are completely divisible by both 5 and 7, we need to count the number of terms in this series.
Number of terms = (630-315)/35 + 1 = 9
Therefore, the number of numbers from 300 to 650 which are completely divisible by both 5 and 7 is 9.
Hence, the correct answer is option (C) 10.
To find the solution of this question, we need to find the common multiples of 5 and 7 between 300 and 650.
Multiples of 5: 300, 305, 310, 315, ………., 645, 650
Multiples of 7: 301, 308, 315, 322, ………., 637, 644
To find the common multiples of 5 and 7, we need to find the LCM (Least Common Multiple) of 5 and 7.
LCM of 5 and 7 = 35
So, the common multiples of 5 and 7 are:
35 x 9 = 315, 35 x 10 = 350, ………., 35 x 18 = 630
To find the count of numbers which are completely divisible by both 5 and 7, we need to count the number of terms in this series.
Number of terms = (630-315)/35 + 1 = 9
Therefore, the number of numbers from 300 to 650 which are completely divisible by both 5 and 7 is 9.
Hence, the correct answer is option (C) 10.
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Community Answer
How many numbers are there from 300 to 650 which are completely divisi...
In the range of 300 to 650
315 to 630 are divisible by 35 (LCM of 7,5)
Tn = a+(n-1)d
630=315+(n-1)35
630-315=(n-1)35
315=(n-1)35
315/35=(n-1)
9=n-1
9+1=n
10=n Ans (i.e. no. between 300 to 650 which are divisible by both 35)
315 to 630 are divisible by 35 (LCM of 7,5)
Tn = a+(n-1)d
630=315+(n-1)35
630-315=(n-1)35
315=(n-1)35
315/35=(n-1)
9=n-1
9+1=n
10=n Ans (i.e. no. between 300 to 650 which are divisible by both 35)
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