3. How many three digits no. can be formed also including the condition that the no. can have at least two same digits ?

Explanation:

Total number of 3 digit numbers = 9×10×10 = 900

Total number of numbers in which no digit repeats = 9×9×8 = 648

So the total number of numbers in which at least one digit repeats = 900 – 648 = 252

4. If a, b and c are forming increasing terms of G.P., r is the common ratio then find the minimum value of (c-b), given that (log a+log b+log c)/log 6 = 6.Note that r can be any real no.

a) 36

b) 24

c) 18

d) 12

Answer: d

Explanation:

a,b,c are in G.P. so let the first term of G.P. = arar, and common ratio = r.

Therefore, a = arar, b = aa, c = arar

Given, loga+logb+logclog6=6loga+logb+logclog6=6

⇒logabclog6=6⇒logabclog6=6

⇒log6abc=6⇒abc=66⇒log6abc=6⇒abc=66

put the value of a,b,c in gp format

⇒ar×a×ar=66⇒ar×a×ar=66

⇒a3=66⇒a=36⇒a3=66⇒a=36

Now a = 36r36r, b = 36, c = 36r.

We have to find the minimum value of c – b = 36r – 36.

r can be any number. So for r < 0, we get c – b negative.

When r = 1, c – b = 0

But none of the options are not representing it.

From the given options, r = 4/3, then c = 48. So option d satisfies this.

5. A natural number has exactly 10 divisors including 1 and itself.how many distint prime factors this natural number will have?

a. 1 or 2

b. 1 or 3

c. 1 or 2 or 3

d. 2 or 3

Answer: a

Explanation:

Number of factors of a number NN is (p+1).(q+1).(r+1)… where N=ap×bq×cr…N=ap×bq×cr….

Given, (p+1).(q+1).(r+1).. = 10.

From the above equation, p = 1, q = 4 or p = 9 satisfies.

So the number N is in the following two formats. a1×b4a1×b4 or a9a9

So it has either 1 or 2 prime factors.

6. How many values of c in x^2 – 5x + c, result in rational roots which are integers?

Explanation:

By the quadratic formula, the roots of x2−5x+c=0x2−5x+c=0 are −(−5)±−52−4(1)(c)‾‾‾‾‾‾‾‾‾‾‾‾‾√2(1)−(−5)±−52−4(1)(c)2(1) = 5±25−4c‾‾‾‾‾‾‾√25±25−4c2

To get rational roots, 25−4c25−4c should be square of an odd number. Why? because 5 + odd only divided by 2 perfectly.

Now let 25 – 4c = 1, then c = 6

If 25 – 4c = 9, then c = 4

If 25 – 4c = 25, then c = 0 and so on…

So infinite values are possible.