Main, we have to find the number of integers in each

(I) the number of integers in set T that are multiples of 4..
Quick way... There are \(\frac{100}{4}=25\)multiples. But take out 4 and 8, so 2 of them
Total 25-2=23..
A=4*23=92

(II) the number of integers in set T that are multiples of 5..
Quick way... There are \(\frac{100}{5}=20\)multiples. But take out 5 and 10, so 2 of them
Total 20-2=18..
B=5*18=90

A>B

A

Explanation::

To find out the multiples of a given number from a set = (Last number of the set i.e a multiple of the given number - First number of the set i.e a multiple of the same given number)/(Multiple of the given number)+ 1

Option A : multiple of 4 from 11 to 100, therefore : \({\frac{(100 - 12)}{4}}+ 1 = 23\) i.e. no. of integer that are multiples of 4 from 11 to 100 is 23)

Therefore 4 times the number of integers in set = \(4 * 23 = 92\)

Option B: multiple of 5 from 11 to 100, therefore :\({\frac{(100 - 15)}{5}}+ 1 = 18\) i.e. no. of integer that are multiples of 4 from 11 to 100 is 18)

Therefore 5 times the number of integers in set = \(5 * 18 = 90\)

Hence Option A is the answer

Use the arithmetic series formula

the first multiple of 4 from 11 to 100 is 12, and 100 is the last multiple

100=12+4(n-1)
88=4(n-1)
22= n-1
23=n
23*4=92

15 is the first multiple of 5 and 100 is the last multiple
100= 15+5(n-1)
85=5(n-1)
17=n-1
18=n
18*5= 90

92>90, A is greater

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