Please, solve these maths

[n_sun30]

1. Given that, ab = b+1 and a(b+c) = ab + c

Col A: C
Col B: a/(b + c)

2. In a set of numbers from 1 to 10. If two numbers are to be selected from this 10 numbers with replacement, then what is the probability that at least one of them will be even?

3. Col A: sqrt [1+ sqrt [a + sqrt (2)]]

Col B: sqrt [ sqrt (2) + sqrt [{sqrt (2) + 1}/sqrt (2)]]

4. If Standard daviation of r is 13 and standard daviation of t is 7

Col A: mean of r

Col B: mean of t

5. If a(n) = a(n-1) - a(n-2), a(1)= -5 and a(2)= 4, then find the sum of first 100 terms.

[Riyad Mohammad]

1. Given that, ab = b+1 and a(b+c) = ab + c;

then a(b+c) = ab + c ; or ab + ac = ab + c ; or ac = c ; or a=1.

If a =1 then ab = b+1 ; or 1*b = b + 1 ; or 1=0 ; so there is not enough information about the math. The answer is D or it is mistyped question.

[Riyad Mohammad]

a(n) = a(n-1) - a(n-2) ;
then a3 = 9 ; a4 = 5; a5 = -4 ; a6 = -9 ; a7 = -5 ; a8 = 4 ......so on. So it repeats at a7. Now sum up from a1 to a6 and the sum is zero. so up to 96th term the sum is zero. Then the sequence is being repeated from the 97 as well. a97 = -5 ; a98 = 4 ; a99 = 9 ; a100 = 5. Add them and you will get 13.

[Riyad Mohammad]

3. Col A: sqrt [1+ sqrt [a + sqrt (2)]]

Col B: sqrt [ sqrt (2) + sqrt [{sqrt (2) + 1}/sqrt (2)]]

There is not enough information about a. Is it necessary to solve the problem?

[pureen]

2. 1 - probability of both the nos. are odd ;
1-5/10 * 5/10 ;
3/4 ( probability of at least one no. is even ) .
4. The answer must be D ( as, you cant predict the mean while looking merely on the standard deviation )
5. 13

[n_sun30]

thank you all for the solutions