# Quant Boosters - Sagar Gupta - Set 6

• 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 cauchy-schwarz inequality
a+b+c+d = 8-e
a^2+b^2+c^2+d^2 = 16-e^2
now
(a^2+b^2+c^2+d^2)(1+1+1+1) >= (a+b+c+d)^2
or 4*(16-e^2) >= (8-e)^2
or 64-4e^2 >= 64+e^2-16e
or 5e^2 < = 16e
or e(5e-16) < = 0
so 0 < = e < = 16/5

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

no. 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

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

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

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?

P [ destroying ] = 1 - p [ not destroying ]
it needs 3 hits to destroy
so when will it not able to destroy ??
when it will either hit 0 or 1 or 2 missiles
p [ 0 hit ] = nc0 * ( 1/2 )^0 ( 1/2 )^n
p [ 1 hit ] = nc1 * ( 1/2 )^1 ( 1/2)^n-1
p [ 2hits ] = nc2 * ( 1/2)^2 ( 1/2)^n-2
sum of these 3 will give p [ not destroying ]
(1/2)^n + n * 1/2 * (1/2)^n-1 + n(n-1)/2 * 1/4 * (1/2)^n-2
Put n=14
(1/2)^14 + 7 * (1/2)^13 + (91/4) * (1/2)^12 = 0.0064
1 - 0.0064 = 0.9935 which is just more than 99%
Hence 14

Each day, I choose at random between my brown trousers, my grey trousers and my expensive but fashionable designer jeans. Also in my wardrobe, I have a black silk tie, a rather smart brown and fawn polka-dot tie, my regimental tie, and an elegant powder-blue cravat which I was given for Christmas. With my brown or grey trousers, I choose ties (including the cravat) at random, except of course that I don’t wear the cravat with the brown trousers or the polka-dot tie with the grey trousers. With the jeans, the choice depends on whether it is Sunday or one of the six weekdays: on weekdays, half the time I wear a cream-coloured sweat-shirt with E = mc2 on the front and no tie; otherwise, and on Sundays (when naturally I always wear a tie), I just pick at random from my four ties. This morning, I received through the post a compromising photograph of myself. I often receive such photographs and they are equally likely to have been taken on any day of the week. However, in this particular photograph, I am wearing my black silk tie. Find, on the basis of the complete information, the probability that the photograph was taken on Sunday ?

p(sunday) = 1/7
p(weekday) = 6/7
On Sunday :
p(brown trousers) = p(grey trousers) = p(jeans) = 1/3
with brown trousers...p[black silk tie ] = 1/3
with grey trousers ...p[black silk tie ] = 1/3
with jeans ...p[black silk tie ] = 1/4
On weekday :
p(brown trousers) = p(grey trousers) = p(jeans) = 1/3
with brown trousers...p[black silk tie ] = 1/3
with grey trousers ...p[black silk tie ] = 1/3
with jeans ...p[black silk tie ] = 1/2 * 1/4 = 1/8
total : (1/7) [ 1/9 + 1/9 + 1/12 ] + (6/7) [ 1/9 + 1/9 + 1/24 ] = 11/68

Books and More sells books, music CDs and film DVDs. In December 2009, they earned 40% profit in music CDs and 25% profit in books. Music CDs contributed 35% towards their total sales in rupees. At the same time total sales in rupees from books is 50% more than that of music CDs.

(Q1) If Books and More have earned 20% profit overall, then in film DVDs they made
(1) 15.2% profit
(2) 10.0% profit
(3) 10.0% loss
(4) 16.3% loss
(5) 23.4% loss

(Q2) If Books and More made 50% loss in film DVDs, then overall they made
(1) 12.3% profit
(2) 8. 7% profit
(3) 0.4% loss
(4) 6.25% loss
(5) 20% loss

cp books = x
sp books =1.25x
cp music = y
sp music = 1.40y
cp film = z'
sp film = z
1.25x = 1.5 * 1.4 y
1.40 y / (1.40y + 1.25x + z) = .35
y = 25x/42
z = 25x/84
2z = y
total cp = x+y+z'
total sp = 1.25x + 1.40y +z

First Part : 1.25x + 1.40y +z = 1.2 ( x+y+k )
1.25x + 1.40y + z = 1.2x + 1.2y + 1.2k
0.05x + 0.20y + z = 1.2k
0.05(84z/25) + 0.20(2z) + z = 1.2k
1.568z = 1.2z'
z = 0.765 z'
so loss of 23.4 %

Second Part 2z = z'
cp = x + y + 2z = x + 25x/21 = 46x/21 = 2.19 x
sp = 1.2x +1.40y + z = 1.2x + 1.4 *25x/42 + 25x/84 = 2.33x
0.14x / 2.19x ~ 8% profit

Two brothers, each aged between 10 and 90, "combined" their ages by writing them down one after the other to create a four digit number, and discovered this number to be the square of an integer. Nine years later they repeated this process (combining their ages in the same order) and found that the combination was again a square of another integer. What was the sum of their original age?

xyab = m^2
(xy+9)(ab+9) = n^2
b+10a + 100y+1000x = m^2
9 + b+10a + 900 + 100y + 1000x = n^2
n^2 - m^2 = 909
(m+n) ( m-n) = 9*101
n = 55 : 3025 :
m = 46 : 2116 : original sum = 37

If x = 3 - 2 * root (2) then find x^6 + (1/x^6)

x = 3 - 2 root 2
1/x = 3 + 2 root 2
x^6 + (1/x)^6 => 2 [ 3^6 + 6C2 * 3^4 * ( 2 root 2 )^2 + 6C4 * 3^2 * ( 2 root 2 )^4 + ( 2 root 2 )^6 ]
2 [ 729 + 15 * 81 * 8 + 15 * 9 * 64 + 512 ] = 39202

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