A four-digit number 4ab5 is divisible by 55. Then the value of b - a is
(a) 0
(b) 1
(c) 4
(d) 5
Solution:
55 is a product of 11 × 5
Hence for 4ab5 to be divisible by 55 it has to be divisible by 11 and 5.
For 4ab5 to be divisible by 11 the rule is
(4 + b) - (a + 5) = 0 or 11
b - a - 1 = 0 or 11
b - a = 1 or 12
Since 12 is not an alternative in the choices of answers given.
Therefore the value of b - a = 1
Answer is (b)
✦ Try This: A four digit number 9ab2 is divisible by 77 then the value of b - a is (a) 0, (b) 1, (c) 4, (d) 5
77 is a product of 11 × 7
Hence for 9ab2 to be divisible by 77 it has to be divisible by 11 and 7.
For 9ab2 to be divisible by 11 the rule is
(9 + b) - (a + 2) = 0 or 11
b - a + 7 = 0 or 11
b - a = -7 or 4
Since -7 is not an alternative in the choices of answers given.
Therefore the value of b - a = 4
Answer is (c)
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 16
NCERT Exemplar Class 8 Maths Chapter 13 Problem 8
A four-digit number 4ab5 is divisible by 55. Then the value of b - a is (a) 0, (b) 1, (c) 4, (d) 5
Summary:
If a four digit number 9ab2 is divisible by 77 then the value of b - a is 1
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