Topic - Quant Mixed Bag

Solved ? - Yes

Source - Compilation of posts from Sagar Gupta - 99.2 Percentile in CAT 2015 (Quant) ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - Compilation of posts from Sagar Gupta - 99.2 Percentile in CAT 2015 (Quant) ]]>

221a + 212b + 122c = 2536

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 ]]>

a. 92

b. 93

c. 97

d. 94

e. CBD ]]>

b) 90n - 125 + 93 = 90n - 32/n-1 = 32 + 58n/n-1 => n=59 -> new avg age => 91 : not possible

c)90n - 125 + 97 = 90n - 28/n-1 = 28 + 62n/n-1 => n=32 -> new avg age = 92 : not possible

d) 90n -125 + 94 = 90n -31 / n-1 = 31 + 59n/n-1 => n=60 -> new avg age = 91 : not possible

only 109 is possible : which comes from option A ]]>

252 + 10 * 210 + 120 * ( 40) + 45 * (80) + 10(210 -10-120 ) + ( 192 - 160 ) = 11584 ]]>

y=0 => |x| +|z| = 4 : 16

y=+/-1 =>|x| + |z| =2 : 8

y=+/-2 => |x| + |z| = 0 : 1

16 + 2 [ 8 + 1 ] = 34 ]]>

-y^2 * z^2 y^2+z^2 = 5

min when y=z = root ( 5/2 )

-y^2 z^2 = -25/4 = OA ]]>

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 ]]>

cp = a

sp = y

y = a ( 1 + r/100 ) => a = y / ( 1 + r/100 )

cp = b

sp = y

y = b ( 1 - r/100 ) => b = y / ( 1 - r/100 )

cp = ( 2y / 1 - r^2 / 10000 )

sp = 2y

2y ( 1 - r^2 / 10000 ) = 2y ( 1 + x / 100 )

( 1 - r^2 / 10000 ) = 1 - x/100

r^2 / 10000 = x/100

x = r^2 / 100 = loss % ]]>

take out 17 common

( 35 * 36 *...50 * 3 * 52 * 53 *....67 ) mod 17

1 * 2 *....16 * 3 * -1 * -2 * ... -16 mod 17

16! * 16 ! * 3 mod 17

16 * 16 * 3 = 1 * 3 mod 17 = 3

17 * 3 = 51

35 x 36 x 37 .......... x 67 mod 289

take out 17 common

( 35 * 36 *...50 * 3 * 52 * 53 *....67 ) mod 17

1 * 2 *....16 * 3 * -1 * -2 * ... -16 mod 17

16! * 16 ! * 3 mod 17

16 * 16 * 3 = 1 * 3 mod 17 = 3

17 * 3 = 51 ]]>

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 ]]>

2000 = 16 * 125

1776^1816! mod 16 = 0

Euler of 125 is 100.

1816! mod 100 = 0.

1776^1816! mod 125 => 1716^0 mod 125 =1

16a = 125b + 1

Remainder = 1376 ]]>

when all equal : 1 way

when 2 equal : 2a + c = 18 -> a = 0 to a = 9 , except a = 6 :

total 9 ways 3c2 * 9 = 27

Req cases = (190 - 1 - 3 * 9) / 3! + 1 + 9 = 37 ]]>

a) 9 cm

b) 16 cm

c) 4 cm

d) none of these ]]>

4a = 15b/2 = 17c/2 = k

we know that : area = inradius * semi perimeter

semi perimeter =

( a + (8a/15) + (8a/17) ) /2 = 511a/510

inradius = to find

4a = r * 511a / 510

r = 510 * 4 / 511 = 2040 / 511 , which is less than 4

Hence none of these ]]>