Topic - Quant Mixed Bag

Solved ? - Yes

Source - Compilation of my solutions from various CAT prep forums. ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - Compilation of my solutions from various CAT prep forums. ]]>

so product of these = 12^80 * (2 * 3^3 * 5^9)^40 = 2^200 * 3^83 * 5^360.

total product = (2^3 * 3^4 * 5^9)^(100) = 2^300 * 3^400 * 5^900

so not divisible ke liye divide the above two.

2^100 * 3^x * 5^540 so 10^100

100 trailing zeros. ]]>

122(a+b+c) + (99a + 90b) = 2536

2536 mod 9 = 7

99a + 90b mod 9 = 0

122(a+b+c) mod 9 = 7

5(a+b+c) mod 9 = 7

so : a+b+c = 14 as 70 mod 9 = 7 ]]>

x^2 - yx + (y^2 - 727 ) = 0

D = y^2 - 4(y^2 - 727)

D = 2908 - 3y^2

2907-3y^2+1 should be a perfect square

3(969-y^2)+1 should be a perfect square

clearly y=31 , which gives 2 possible x = 13 and x= 18

values can be interchanged here :

y =13 will give x=31 , x=18

y=18 will also give x=31 , x=13

same will be possible when all these values are of negative sign

so total 3 * 2 * 2 = 12 values ]]>

12 = 2^2 * 3

max power of 3 in 100! = 100/3+100/9+100/27+100/81 = 33+11+3+1 = 48

max power of 3 in 60! = 60/3 + 60/9 + 60/27 = 20 + 6 + 2 = 28

max power of 3 in 40! = 40/3 + 40/9 + 40/27 = 13 + 4 + 1 = 18

3^48 / 3^28 * 3^18 = ( 3^2 )

max power of 4 in 100! = 100/4 + 100/16 + 10/64 = 32

max power of 4 in 60! = 60/4 + 60/16 = 18

max power of 4! in 40! = 40/4 + 40/16 =12

2^32 / 2^18 * 2^12 = 2^2

so highest power is 1 ]]>

Sp * 2 = b

Sq * 2 = 500 - 5b

10Sp + 2 Sq = 500

(42, 40) satisfies ]]>

12^144 + 12^143 +....12^145-P

12^144 [ 1 - (1/12)^P ] / [ 11/12 ]

12^145 [ 1 - 1/12^P ] / 11 = Ap

Bp = A1 + A2 + .... Ap

12^145 [ 1 - 1/12^P ] / 11

12^145 / 11 [ 1 - 1/12 + 1-1/12^2+....1/12^P ]

12^145 / 11 [ 11P/11 - 1/11 + (1/12)^P/11 ]

12^145/121 [ 11P -1 + (1/12)^P ] ]]>

a0 : a1 to a24 + sara

b0 : b1 to b24+ sara

c0 : c1 to c24 + sara

d0 : d1 to d24 + sara

e0 : e1 to e24+ sara

2 friends of sara are connected to each other : say a-b

24 * 5 + 3 = 123

all five friends are mutual friends

each has 20 distinct friends and give people a0,b0,c0,d0,e0 : 20 * 5 + 5 = 105 ]]>

vinegar : 40

water : 60

20 ml vinegar added

conc of vinegar = 50%

For B

vinegar =80 + 3x/5

water = 120-3x/5

80+3x/5 = 120 - 3x/5

x = 100/3 ml was added

20 : 100/3 = 3:5

CP=50, MP=80.

3 * (SP-50) * 100/SP = 2 * D (D=discount%)

Also, SP = (1-D/100) * 80 = 80 * (100-D)/100

so 3 * {80(100-D)/100 -50 } * 100 = 2D * {80 * (100-D)}/100

150 { 75-2D} = D {200-2D}

D^2 - 250D + 5625 = 0

D= (250-200)/2 = 25% ]]>

(a) 22

(b) 26

(c) 24

(d) 25 ]]>

minute and hour hands meet every 65 5/11 minutes = 720/11 mins

so number of extra meetings: 240*11/720 = 3.xx so 3 more meetings than usual.

so 22+3=25 times ]]>

a. 100

b. 140

c. 200

d. 280 ]]>

(x-a)(x-a-5)=1 let (x-a)=t

t (t-5) = 1

t^2-5t =1

divide by t

t-1/t= 5

cube both sides.

t^3-1/t^3 -3*5 = 5^3

t^3-1/t^3 = 125+15=140 ]]>

a) 24/25

b) 125/116

c) 125/108

d) 116/125

e) 108/125 ]]>

S/6 = 2/6^2 + 12/216 + 30/6^4 + 56/6^5...

5S/6 = 2/6 + 10/6^2 + 18/6^3 + 26/6^4...

5S/36 = 2/6^2 + 10/6^3+ 18/6^4....

25S/36 = 2/6 + 8 (1/6^2 + 1/6^3....)

25S/36 = 2/6 + 8 ( 1/30) = 18/30 = 3/5

S = 108/125 ]]>