Topic - Quant Mixed Bag

Solved ? - Yes

Source - Compilation of posts by Shubham Ranka (Shubham aced CAT with 99.65 and is an alum of IIM C and IIT Gandhinagar) ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - Compilation of posts by Shubham Ranka (Shubham aced CAT with 99.65 and is an alum of IIM C and IIT Gandhinagar) ]]>

N^2 = (4019.5^2-.5^2) (4019.5^2-1.5^2) + 1. This is slightly lesser than 4019.5^2

Now, observe that N^2 must end with either 1 or 9.

And, N ~ 4019.5^2 = 16156380.25

so, the integer ending with either 1/9 just before 16156380.25 = 16156379. Cross check !!

Method-2:

N^2 = 1+ a(a+1)(a+2)(a+3) == > N= (a+1)(a+2) -1

a. 60 kg

b. 62 kg

c. 64 kg

d. Cannot be determined ]]>

GIven, a+b = 110 and d+e = 121

Adding all the 10 eqns will give: a+b+c+d+e = 289

== > c= 58

Now, next min combination is 112 (a+c)== > b = 56, a = 54

Now, max = 121 -- > 59 + 62 or 60 + 61 (does not satisfy other combinations)

== > 62 Kg ]]>

a) –3 only

b) an integer only

c) any real number

d) None of these ]]>

-1/1 = -1/1 ==== > any real number (C) ]]>

2x^2 -5x + 3 - t = 0

x = 5 +- rt(1+8t))/4

== > f(t) = 3[5 +- rt(1+8t))/4] + 2 = [23 +- 3rt(1+8t)]/4

x^2 = 1/16[26+8t+-10rt(1+8t)]

== > f(x^2) = f(1/8[13+4t+-5rt(1+8t)]) = [23 +- 3rt(p)]/4

where, p = 1+8*1/8[13+4t+-5rt(1+8t)] = 14+4t+-5rt(1+8t)

Converting to x :

1+8t = 16x^2 -40x + 25 = (4x-5)^2

== > p = 14+4(2x^2 -5x + 3) +- 5(4x-5) = 8x^2 -20x +26 +- (20x-25) = 8x^2 +1 or 8x^2 -40x +51

f(x^2) = 23+-3rt(8x^2 +1 or 8x^2 -40x +51) / 4

Since, f(0) = 5

f(x^2) = [23 - 3rt(8x^2+1)]/4 satisfies ]]>

a. 22

b. 24

c. 44

d. 48 ]]>

It makes an angle of 180 degrees 11 times in one circle. 00:30:xx, 1:35:xx, 4:50:xx, 6:00, 7:05:xx, ... 11:25:xx

== > 22 times.

0 degrees -- > 00:00: 1:05, ..10:50:xx. == > 2*11 = 22 times

== > 22 + 22 = 44 times (C)

]]>a. Monday

b. Wednesday

c. Tuesday

d. Friday ]]>

100 years -- > 5 odd days -- > Friday

200 years -- > 3 odd days -- > Wed

300 odd days -- > 1 odd day -- > Mon

400 odd days -- > 0 odd day -- > Sunday

== > (C) ]]>

I got n+2 in rs 12 = > Price per dozen = 144/n+2

144/n+2 = 144/n - 1

== > n = 16 ]]>

^2 + 2 divides 2 + 2001n == > It will also divide 2+2001n - (n^2+2) = 2001n-n^2

n^2 + 2 divides 2001n-n^2 ----(1)

Case-1:

If n is even = 2k:

4k^2 + 2 divides (4002k - 4k^2)

2k^2 + 1 divides 2001k-2k^2 = k(2001-2k)

== > 2k^2 + 1 divided 2001-2k -(A)

Again, Since 2k^2 + 1 divides 2001k-2k^2, it will also divide 2001k-2k^2 + (2k^2 +1)= 2001k+1

=== > 2k^1 + 1 divides (2001-2k) and (2001k+1)

Trying to remove k:

2k^1 + 1 will also divides 2001(2001-2k) + 2(2001k+1) = 2001^2 + 2 = 4004003 = 19*83*2539

By hit and trial, k = 0,3 == > n = 6 (positive integer)

Case-2:

If n is odd:

n^2 + 2 dvides 2001n-n^2 = n(2001-n)

== > n^2 + 2 divides 2001-n -------(1)

Also, since n^2 + 2 dvides 2001n-n^2, it also divides: 2001n-n^2 + (n^2+2) = 2001n+2 --(2)

Trying to remove n:

2001*(1) + (2):

n^2 + 2 divides 2001^2 + 2 = 19*83*2539

Using hit and trial:

== > n = 9, 2001

Hence n = 6,9,2001

And Kanti takes 3y no. of days to complete the task

== > 3xy/x+y = 11(13/16) = 189/16

== > 16xy = 63x+63y ---(1)

== > xy should be a multiple of 63 = 7*9 == > either of x or y should be a multiple of 7

1/3rd work -- Ram --x days

2/3rd work--Kanti --2y days

== > x+2y = 25 ----(2)

Now, either of x or y should be 7/21 == > x = odd == > 7

x = 7 == > y =9 == > satisfies 1

Kranti can complete the work in 3y = 27 days, working alone. ]]>

(a) 5

(b) 10

(c) 11

(d) 12 ]]>

a and b are distinct and a,b > = 1 and a+b > =3 == > k < =17

k will be a factor of 50 < =27 == > 10,5,1

Max. of k = 10 ]]>

x = D _ _ _ _ _ = 5! = 120

= _ D _ _ _ _ = 3c1 (Sice A/B can not come in the first place now) * 4! = 72

= _ _ D _ _ _ = 3c2 * 3! * 2! = 36

= _ _ _ D _ _ = 3! * 2! = 12

== > x = 120 + 72 + 36 + 12 = 240

== > Total no. of ways = 720 = 480 = 240 ]]>

a) 40

b) 42

c) 44

d) 48 ]]>

F = 6s

F+8/S+8 = 10/3

3(6S+8 ) = 10(S+8 )

S = 7

F = 42

== > Current age is 44 ]]>

a) 9 cm

b) 10 cm

c) 11 cm

d) 12 cm ]]>