Topic - Quant Mixed Bag

Solved ? : Yes

Source : Learningroots forum ]]>

Topic - Quant Mixed Bag

Solved ? : Yes

Source : Learningroots forum ]]>

(a) 4 : 1 : 4 : 3

(b) 1 : 2 : 2 : 1

(c) 2 : 3 : 6 : 1

(d) 5 : 2 : 7 : 3 ]]>

a) 11/12

b) 11/96

c) 29/96

d) 31/96

e) None of these ]]>

xyz=288

x^3=288×6

x^3=1728

x=12

y=6

z=4

1/24+1/24+1/32

11/96 ]]>

Original approach: Largest side is always less than the semi perimeter and greater than or equal to one third of the perimeter. Then take variables, and solve for positive solutions, remove duplicates n you are done. ]]>

a. 2

b. 3

c. 4

d. 5 ]]>

The resultant five digit number has to start with 1 and so, will be of the form 1xyx1.

1001a+110b+1001p+110q=10001+1010x+100y

1001(a+p)+110(b+q)

But as the last digit is 1, a+p has to be 11

11011+110(b+q) will be a palindrome. So, b+q should not change the central values disproportionately. This is possible when b+q=0 or b+q=11. So, 11011 or 12221. ]]>

All three together in 9 ways.

Two together in 5 ways per couple (From abcdefghi if he chooses ab, he can choose one among only defgh, similarly for the 8 other pairs).

So, two together in 9 * 5 = 45 ways.

So, no two together in 9c3 - 9 - 45 = 30 ways. ]]>

2 As and 2 Ts can be arranged in 4!/2!2!=6 ways. Only one among them AATT satisfies the condition. So, (1/6)*(11!/2!2!)=11!/4!

PS: When I say AATT... i mean all cases where 2 As will come before 2 Ts. So, the arrangement could be something like CSARAOHTPET as well. I am just talking about how As and Ts appear among themselves. As there is no binding condition, the distribution would be even. ]]>

a+b+c=132......(1)

c+d+e=119......(2)

a+b+d=129......(3)

b+d+e=121......(4)

We know the values of all possible triplets. So, 5c3=10 cases in total. So, each element occurs 6 times.

6a+6b+6c+6d+6e=1254

a+b+c+d+e=209.....(5)

From (1), (2) and (5)

c=42

From (2), (3) and (5)

d=39... putting these values in (2)

e=38

b=44

a=46 ]]>

= 436 + 87 + 17 + 3

= 543 ]]>

_ _ _ 1... as 1 has been fixed, first digit in 6 ways, 2nd in 8 ways and third in 7 ways -> 6 * 8 * 7

_ _ _ 3.. similar to the first case -> 6 * 8 * 7

_ _ _ 5.. similar to the first case -> 6 * 8 * 7

_ _ _ 7.. similar to the first case -> 6 * 8 * 7

_ _ _ 9.. first digit can be taken in 7 ways, 2nd in 8 ways and 3rd in 7 ways

6 * 8 * 7 * 4 + 7 * 8 * 7

56 * 31=1736

a. 0

b. 1

c. 2

d. More than 2 ]]>

(p/q+q/r+r/p) >= 3

So, it can never be equal to 2.

Can be done using simple substitution as well.

]]>Take a smaller case... {1,2,3,4,5} and you want the sets that have sum greater than 7. You can form 32 sets or 16 pairs of complements.

{}...{1,2,3,4,5}

{1}...{2,3,4,5}

{2}...{1,3,4,5|

{3}...{1,2,4,5}

{4}...{1,2,3,5}

{5}...{1,2,3,4}

{1,2}...{3,4,5}

{1,3}...{2,4,5}

{1,4}...{2,3,5}

{1,5}...{2,3,4}

{2,3}...{1,4,5}

{2,4}...{1,3,5}

{2,5}...{1,3,4}

{3,4}...{1,2,5}

{3.5}...{1,2,4}

{4,5}...{1,2,3}

See that the second set is the complement of the first set. The sum total of the elements of the two sets will always be 15. If the first set totals to less than or equal to 7, the second will definitely by greater than 7. The elements are such that you will not have equal values in both the sets and so, you are left with half the cases. If the sum is an odd number and the total is +/-0.5 of the average then you can go with this method. Else, normal case making and permutations i guess

(1) 12:31

(2) 59:81

(3) 345:532

(4) 23:57

(5) None of these ]]>