Quant Boosters  Ravi Handa  Set 2

Consider the side as a and hypotenuse as (L  a)
Calculate third side and find out area.
You will find out that it depends on your ability to maximize a * a * (L2a)
Its sum is constant at L.
=> Product will be maximum when all the three terms are equal.
=> Hypotenuse = Twice of side
=> Angle between them is 60 degrees

Q27) The positive integer N and N^2 both end in the same sequence of four digits 'abcd' when written in base 10 where digit 'a' is not zero. find three digit number abc.
a) 984
b) 937,
c) 987
d) 321

N^2  N will end in 000
=> N(N1) will be divisible by 625
=> abc5 or abc0 is divisible by 625
=> From the options only 9375 is divisible by 625
=> So, 937 is the answer

Q28) If p, q and r are in A.P & x, y and z are in G.P, Then x^(qr) * y^(rp) * z^(pq) = ?

Using options (if available) and assuming values works best in such cases.
Take, p = 1, q = 2, r = 3 and x = 1, y = 2, z = 4
=> x^(qr) * y^(rp) * z^(pq) = 1^(1) * 2^2 * 4^(1) = 1To solve it properly, take the AP values as (ad), a, (a+d) and the GP values as b/r, b, br
=> x^(qr) * y^(rp) * z^(pq)
= (b/r)^(d) * b^(2d) * (br)^(d)
= b^(d + 2d  d) * r^(d + 0  d)
= b^0 * r^0
= 1

Q29) LCM of 2 numbers is 315. Which of the following cannot be the sum of two numbers, given that their HCF is prime number > 3
a) 322
b) 98
c) 320
d) 126

First of all, find out the factors of 315
315 = 3^2 * 5 * 7
=> HCF can be 5 or 7
Case 1: HCF is 5
=> The two numbers are 5a and 5b, where a & b are coprime to each other
=> 5 * a * b = 315
=> a * b = 63
=> a & b are (1,63) or (7,9)
=> Sum of the numbers = 5a + 5b = 5 * 64 or 5 * 16 = 320 or 80.
=> This eliminates option C.
Case 2: HCF is 7
=> The two numbers are 7a and 7b, where a & b are coprime to each other
=> 7 * a * b = 315
=> a * b = 45
=> a & b are (1,45) or (5,9)
=> Sum of the numbers = 7a + 7b = 7 * 46 or 7 * 14 = 322 or 98
=> This eliminates options A & B
=> Answer is Option D

Q30) Find Remainder [47^17/19]

Remainder [47^17 / 19]
= Remainder [9^17/19]
= Remainder [9 * 81^8/19]
= Remainder [9 * 5^8/19]
= Remainder [9 * 25^4/19]
= Remainder [9 * 6^4/19]
= Remainder [9 * 36^2/19]
= Remainder [9 * (2)^2/19]
= Remainder [9 * 4/19]
= Remainder [36/19] = 17

@handakafunda Sir is answer 20c1018c8 is correct for this question??