If q is positive two-digit integer and p is the units digit of q, is 10<q<50 ?
1. p is more than the tens digit of q.
2. 4p-1 = q
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adthedaddy
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IMO: B
1. Insufficient: since the number can be 12 or 89 or 56; Since the units digit is larger than the Units digit.
2. Sufficient: 4p-1= q; given
but we know that q= 10x+p
hence equating two, we have; 4p-1=10x+p; p= {10x+1}/3;
hence q= {40x+1}/3 and X can take values from 1 till 9
Hence substituting 1 we have 41 which is not divisible by 3
Substituting 2 we have 81; then q= 27 and we cannot substitute 3 inplace of x as it leads to a 3 digit number.
Hence, B is the sufficient.
1. Insufficient: since the number can be 12 or 89 or 56; Since the units digit is larger than the Units digit.
2. Sufficient: 4p-1= q; given
but we know that q= 10x+p
hence equating two, we have; 4p-1=10x+p; p= {10x+1}/3;
hence q= {40x+1}/3 and X can take values from 1 till 9
Hence substituting 1 we have 41 which is not divisible by 3
Substituting 2 we have 81; then q= 27 and we cannot substitute 3 inplace of x as it leads to a 3 digit number.
Hence, B is the sufficient.
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Matt@VeritasPrep
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S1: Not sufficient, we could have 47 or 89 or many other things.
S2: q + 1 = 4p, so (q + 1) is a multiple of 4. Since p is a single digit integer and q is a two-digit integer, we could have p = 3, 4, 5, ..., 9, and hence q = 11, 15, ..., 35. So whatever q is, it must be between 11 and 35, inclusive; SUFFICIENT.
S2: q + 1 = 4p, so (q + 1) is a multiple of 4. Since p is a single digit integer and q is a two-digit integer, we could have p = 3, 4, 5, ..., 9, and hence q = 11, 15, ..., 35. So whatever q is, it must be between 11 and 35, inclusive; SUFFICIENT.
