Quant Boosters by VP  Set 1

Use Descartes rule
And see the sign changes in F(X)
And F(x)
If Number of sign changes in F(X) =k
number of Positive roots can be k , k2 ,k4 and so on
Similarly
Number of sign Changes in F(x) = k
number of negative roots = k , k2 , k4 and so on

Q18) Find the largest 5 digit number which when divided by 8 leaves a remainder of 5 and when divided by 7 leaves a remainder of 2.

7k +2 => 8p +5
=> 7k = 8p + 3
For p = 4
It satisfiesI.e least number which satisfies the condition is
8 * 4 +5 = 37
And number is of the form
56k + 37
Now to find largest 5 digit number
99999/56
=> 1785.xx
So k = 1785
56 * 1785 + 37
=> 99997

Q19) From the first 25 natural numbers, how many arithmetic progressions of 6 terms can be formed such that common difference of the AP is a factor of the 6th term.

Let the first term of the AP be a and the common difference be d.
The sixth term of the series will be a+5d
Given that d should be a factor of a+5d
=> a+5d is divisible by d
=> a should be divisible by d
So the required cases are
d = 1, a = 1,2,3.......20
d= 2 , a = 2,4,6.......14
d = 3, a = 3, 6,9
d= 4, a = 4
So the required number of AP’s are 20 + 7 + 3 + 1 = 31

Q20) 101010101010..... (94 digits) / 375. Find the remainder.

375 => 125 * 3
Use CRT
With 3
Sum of digits
47 mod 3 = 2
Now ,
With 125
You have to check only last three digits
As 10^3 mod 125 = 0
So last three digits would be 010
So 125k+10
3b +2 = 125k +10
=> For k = 2
It satisfies
So 260

Q21) Four Prime Numbers are written in ascending order. The product at first three is 7429 and the product at last three is 12673. Find the sum of first and last Number?

7429 = 17 x 19 x 23
12673 = 19 x 23 x 29
∴ four prime numbers are 17 x 19 x 23 x 29.
∴ The sum is = 17 + 29 = 46.

Q22) Hypoteneus of a Right angle triangle is 26. What could be its perimeter?
(a) 51
(b) 56
(c) 45
(d) 60

We know 5  12  13 is a Pythagorean triplet
∴ 10 – 24 – 26.
Will also be a Pythagorean triplet. ∴ Perimeter is 60. Hence (d).

Q23) P, 2P, q + P, 2P – 2q – 6 is an A.P. then what is its 99th term?
(a)–99
(b) –98
(c) 99
(d) 100

A.P. So differences will be equal
2P – P = q + P – 2P = 2P – 2q – 6 – q – P
P = q – P = P – 3q – 6
∴ q = 2P and 2P – 6 = 4q
On solving, we get P = – 1 and q = – 2
∴ A.P. is –1, –2, –3………….
∴ T99 = a + 98d = – 1 + 98 x –1 = – 99

Q24) What is the range of x satisfying x^2 + 2x – 15 > 0 and x^23x4 < 0
(a) 5 < x < 4
(b) 1 < x < 4
(c) 5 < x < 3
(d) 3 < x < 4

x^2 + 2x – 15 > 0 or (x + 5)(x  3) > 0
∴ x <  5 or x > 3
x^ 2 – 3x – 4 < 0
(x – 4) (x + 1) < 0
⇒ 1 < x < 4
∴ common solution is 3 < x < 4

Q25) There are 100 people in a room. Some or all of them shake hands. Probability that there will be two people in the room who have shaken hands with the same number of people?

No of people = 100
Possible no of handshakes = 0 to 99 => 100So it seems all 100 ppl can have different no of handshakes, but notice 0 and 99 both won't b possible. If out of 100 ppl, there is one person who shakes hand with 0 ppl(none) then can there be any person among them who will shake hand with 99 ppl(all except himself? Not possible since he won't b shaking hand with himself and that guy who didn't shake hand)
Hence it means 0 and 99 both won't coexist, instead any one of those two will come
Hence now we got 100 people and 99 different no of handshakes. So even if 99 people got those 99 different number, 100th person will get no of handshakes same as one of those 99, and hence there will be atlst two people with same no of handshakes no matter what. Hence it's a certain event and probability is 1.

Q26) If x + (1/x) = 1, find x^127 + (1/x)^127

(x + 1/x) = 1
x^2 + 1 = x
x^2 + 1  x = 0
Now try to observe
Where have you seen such terms
a^3 + b^3 = (a+b) (a^2 + b^2  ab)
So here we have to multiply by
(x+1) on both sides
(x+1) (x^2 x +1 ) = 0
=> (x^3 +1) = 0
x^3 = 1
Now x^127 => (x^3)^42 * x
=> (1)^42 * x
=> x
So x + (1/x) which is equal to 1

Q27) A natural number equals 75 times the average of its digits. How many three digit numbers satisfy this condition