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) ]]>

(17 * 35)^2 - ( 6 * 13)^2 mod 35

(17 * 35)^2 mod 35 =0

-(6 * 13)^2 mod 35 = 169 mod 35 = - 29

-29 = 35-29 = 6 ]]>

(Note that climbing 2 steps first and 1 step later is a different case from climbing 1 step first and 2 steps later) ]]>

3+3+3+1=10. So here, 4! / 3! =4

Case2:

Using only two 3s here.

3 + 3 + 2 + 2=10

So,4!/2!.2!= 6

3 + 3 + 2 + 1 + 1=10

So,5!/2!.2!= 30

3 + 3 + 1 + 1 + 1 + 1=10

So,6!/2!.4!= 15

Case3:

Using only 1 3 now

3,2,2,2,1

So,5!/3!= 20

3,2,2,1,1,1

So,6!/3!.2!= 60

3,2,1,1,1,1,1

So,7!/5!= 42

3,1,1,1,1,1,1,1

So,8!/7!= 8

Similarly,make cases when we have no 3s.

So,that comes out to be total 6 subcases are--

1,1,1,1,1,1,1,1,1,1

2,2,2,2,1,1

2,2,2,1,1,1,1

2,2,1,1,1,1,1,1

2,1,1,1,1,1,1,1,1

2,2,2,2,2

And their sum is 1+15+35+28+9+1=89

Total will be sum of all the cases=274

a. 175

b. 176

c. 180

d. 181 ]]>

now write 10 ! = 6* 4 * 5 * 6 * 7 * 8 * 9 * 10

make this expression as product of four numbers in 40's

40 * 42 * 45 * 48

so sum = 175 ]]>

.

.

.

2(27) + 1 + 1

1

=> 2 * (2 + 2 + .... + 27) + 26 * 3 + 1 = 834 ]]>

C(29, 0)^2 + C(29, 1)^2 + .... + C(29, 29)^2

Except first and last term rest all terms are divisible by 29

So, remainder will be 2. ]]>

2(a-3b/2)^2+2(5b/2-2c)=0

so a=3b/2 and 5b/2 = 2c

a = 3b/2 = 6c/5

Solve and answer is 1/3 ]]>

total cuts (8-1) + (10-1) + (12-1) = 27 = 9 + 9 + 9.

max pieces. 10 x 10 x 10 = 1000. ]]>

a) 1,1,4

b) 2,2,4

c) 3,3,4

d) 4,4,4

e) 2,3,4 ]]>

x = -1 => a - b + 4 = 4 => a = b

x = -2 => 4a - 2b + 4 = 6 => a = b = 1

a,b,c = 1,1,4 ]]>

a) 0

b) 1

c) 2

d) 3 ]]>

Would not give any negative integral solution.

Case 2 : x + 1 < 0 => x < -1

Expression becomes :

-(x + 1) = (x + 1)^2 / x

=> -x = x + 1 => x = -1/2 => No negative integral

Case 3 : x = -1 satisfies. So 1 solution only. ]]>

Sum of roots => I/(m-I) = 3x ------ (1)

Prod of roots => 1/(I-m) =2x^2 ------ (2)

(2) /(1) => x = -3/2I

Substitute in (1) => 2I^2-9I+9m=0

D >= 0 for m to be max, D=0

81=72m => m =9/8 ]]>

(1) b >= a^4

(2) a >= 4b^4

(3) a >= b^4

(4) b >= 4a^4

(5) none of these ]]>

Now x^2+y^2 >=2xy

=> 2(x^2 + y^2) >=a^2

Similarily working on 2(x^2+y^2) >=a^2

we get a^4/8 < = (x^4 + y^4)

=> a^4/8 None of these ]]>