Quant Boosters  Sagar Gupta  Set 2

a < b < c < d < e
c = 7
d and e are same
a + b + c + d + e > 45
5 = a
b = 6
c = 7
18 + 2d > 45
2d > 27
d > 13.5
d = 14
e = 14
Mode =14

Q17) The number of factors of the number N = 4^6 + 6^8 is
a. 18
b. 36
c. 54
d. 72

2^12 + 2^8 * 3^8
2^8 [ 16 + 3^8 ]
2^8 [ 16 + 81 * 81 ]
2^8 [ 16 + 6561 ]
2^8 [6577]
9 * 2 = 18 factors

Q18) When a natural number, N is divided by D, the remainder is 35. When 50N is divided by D, the remainder is 11. Find D
a. 1739
b. 43
c. 47
d. Cannot be determined

n=ad +35
50n=bd +11
50n=50ad+1750
50n=bd+11
50ad+1750 = bd +11
1739= (b50a)d
d can be 1739 or 47
CBD

Q19) In a row at a bus stop, A is 9th from the right and B is 7th from the left. They both interchange their positions. If there are 20 people in the row, what will be the new position of B from the left'?
(1) 11th
(2) 12th
(3) 13th
(4) 10th

1,2,3,4,5,6,B,8,9,10,11,A,13,14,15,16,17,18,19,20
So, 12'th from left

Q20) Find the sum of all positive twodigit integers that are divisible by each of their digits

{11,22...99}
10a+b mod a=0 > b mod a=0
10a+b mod b=0 >10a mod b=0
when a=b , the above 2 conditions will be satisfied > {11,22...99}
also : {12,15,24,36,48 } will satisfy the above conditions
sum =630

Q21) If a + b + c + d + e = 8 & a^2 + b^2 + c^2 + d^2 + e^2 = 16, where a,b,c,d& e are real numbers then
maximum (a,b,c,d,e) = ?
a. 4
b. 2
c. 16/5
d. 6/5
e. None of these

Application of cauchyschwarz inequality
a+b+c+d = 8e
a^2+b^2+c^2+d^2 = 16e^2
now
(a^2+b^2+c^2+d^2)(1+1+1+1) >= (a+b+c+d)^2
or 4*(16e^2) >= (8e)^2
or 644e^2 >= 64+e^216e
or 5e^2 < = 16e
or e(5e16) < = 0
so 0 < = e < = 16/5

Q22) Let f(x) be a function such that f(x).f(y)  f(xy) = 3(x+y+2). Then f(4)=?
(1) can not be determined
(2) 7
(3) 8
(4) either 7 or 8
(5) none of these

f(0)^2  f(0) =6
f(0) =3 or 2
:
f(0) f(2)  f(0) = 3 (0+2+2)
f(0) [ f(2) 1 ] = 12
if f(0) = 3 , f(2) =5
if f(0) =2 , f(2) =5
:
f(2)^2  f(4)= 18
25 18 = f(4) = 7 ( (5)^2 = 5^2 = 25 )
Hence only 7

Q23) Number of non negative solutions for x + y = x^2  xy + y^2

x^3 + y^3 = (x + y)(x^2 xy + y^2)
=> x^3 + y^3 = (x + y)^2
Squares => Sum of cubes
0 => (0,0) ; (x,x) for all x integers
1 => (0,1) ; (1,0)
4 => None
9 => (1,2) ; (2,1)
16 => (2,2)
But (x,x) do not satisfy the original equation.
Total real solutions : Infinite
Total integral solutions : 6
Total non negative solutions : 6
Total positive integral solutions : 3

Q24) There are coins of three denominations 1 rupee, 2 rupees and 5 rupees, in a bag. The number of coins of Re1 is to the number of coins of Re 2 is 1:4 and the number of coins of Rs.2 is to the number of coins of Rs 5 is 7:2. If it is known that the total number of coins in the bag is between 200 and 250, then the total amount of money in the bag must be ?

number of Re 1 coins : x
number of Rs 2 coins : y
number of Rs 5 coins : z
x:y = 1:4
or
x:y = 7:28
y:z = 28:8
200 < x + y + z < 250
200 [35, 140, 40]
money = 35 +280 + 200 =515

Q25) The distance between A and B is 19 km. A cyclist starts from A at a constant speed towards B. A car leaves from A 15 min later in the same direction. In 10 min it catches up with the cyclist and continues towards B; after reaching B, it turns around and in 50 min after leaving, car encounters the cyclist the second time.
a) The speed of the car is?
b) The speed of the cyclist is?

The car travelled same distance in 10 min as cyclist did in 25 min
speed ratio : 5:2
lets say  5x and 2x
when car travels for 50 min, cycle has travelled for 65 min. During that time, both covered 38 kms
38 = 5x * 50/60 + 2x *65/60
x=6
speeds 12 and 30

Q26) The probability of a missile hitting a bridge is 0.5 and it takes 3 hits to destroy the bridge completely. Minimum number of missiles fired so that the probability that the bridge is completely destroyed is more than 0.99?