4) a- Since a+b is the highest & c is the lowest, so (a+b)/c is the max.
5) c – Since x,y,z are consecutive integer greater than 0 & x+y+z is odd, so xz are even.Thus 4 could be value of z
6) d – As xy+x=x(y+1) is odd, so x&(y+1) are odd & y must be even.
7) a – 4^41(1+4+4^2)=4^41(21) which is divisible 7.
8) c – Since y is divisible by 3 but not 2, so y is not divisible by 3*2=6 or multiple of 6 that means y/24 can never be integer.
9) d - very simple. The L.C.M of 3,4,7 is 84.
10) b – Only 113 is not divisible by any number except 1 & itself.
Skill Development
1) c – Since p is odd then 3p is odd & (3p+1) is even. Hence (3p+2) & (3p+4) are the next two odd numbers & sum(3p+2+3p+4)=6p+6
2) c- the sum means 5+1+7+3+2+4=22, which is not divisible by 3. the nearest value divisible by 3 is 24. So '*' has to be replaced at least by 2.
3) a – 101 is not divisible with any other number.
4) d – see the rule of divisibility of 11.
5) c – Since x,y,z are consecutive integer greater than 0 & x+y+z is odd, so xz are even.Thus 4 could be value of z
6) d – As xy+x=x(y+1) is odd, so x&(y+1) are odd & y must be even.
7) a – 4^41(1+4+4^2)=4^41(21) which is divisible 7.
8) c – Since y is divisible by 3 but not 2, so y is not divisible by 3*2=6 or multiple of 6 that means y/24 can never be integer.
9) d - very simple. The L.C.M of 3,4,7 is 84.
10) b – Only 113 is not divisible by any number except 1 & itself.
Skill Development
1) c – Since p is odd then 3p is odd & (3p+1) is even. Hence (3p+2) & (3p+4) are the next two odd numbers & sum(3p+2+3p+4)=6p+6
2) c- the sum means 5+1+7+3+2+4=22, which is not divisible by 3. the nearest value divisible by 3 is 24. So '*' has to be replaced at least by 2.
3) a – 101 is not divisible with any other number.
4) d – see the rule of divisibility of 11.




