Topic - Quant Mixed Bag

Solved ? - Yes

Source - Elite's Grid Prep Forum ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - Elite's Grid Prep Forum ]]>

a) 100 - 1/2^99

b) 101 - 1/2^99

c) 100 + 1/2^99

d) 100 + 1/2^100 ]]>

now if this is true then we can represent other numbers also in this form

3/2 = (2 - 1/2^1) = 3/2 ( n = 1)

3/2 + 5/4 = (3 - 1/2^2) = 11/4 ( n = 2) and so on.

so OA = 100 - 1/2^99 ]]>

so in one day A and B can complete 221 * 9/100 of the work.

so when they work in alternate way they will do this work in 2 days

so in 22 days they will do 221 * 99/100 unit work so remaining work = 1/100

and days =89/4

so remaining days = 1/4

so in 1/4 days A wil do 1/100 work

so in 1 day A will do 1/25 th work

so work done by B = 9/100 - 1/25 = 5/100 = 1/20

so ratio = 4 : 5

so B's sharing will be 5/9 * 1800 = 1000 rs ]]>

a) S88 = S188

b) S66 = S160

c) S100 = S160

d) S120 = S142

[note:digits after S is in subscript] ]]>

so subscript should yield the same remainder when divided by 6. use this in options and only S100 = S160 satisfies. (Both 100 and 160 yield remainder as 4 when divided by 6) ]]>

a) 3f(x)

b) f(x)/(3 + 2f(x))

c) f(x)/(3 - 2f(x))

d) f(x)/(2 - 3f(x)) ]]>

f(3x)=1/(1+3x)

now Question is simple and date is not that much . if data is lengthy then we can apply Options approach

take x=1

so f(1)=1/2

and f(3)=1/4

now f(3)=f(1)/2

so f(1)=2f(3)

a) out of picture

b) f(1)/(3 + 2f(1) = 1/2/(3 + 1)) = 1/6 not equal to f(3)

c) f(1)/3 - 2f(1) = 1/2/((3 - 1)) = 1/4 so OA = C ]]>

a) 104

b) 112

c) 120

d) 128 ]]>

Each person left with exactly one of their own shoes means either they will get rite one or left one so derangement of 5

so D(5) = 44

so 2 * 44 = 88

now 4 of them will get right then obviously 5th one will get right

now 3 of them correct then

so 5c3 * 2 = 20

and 2 of them correct = 2 * 5c3 = 20

so 88 + 40 = 128

Quicker method:

Every friends has two options (left or right) so ans must be multiple of 2^5 and only 128 satisfies.

a) 16

b) 20

c) 24

d) Cannot be determined ]]>

and in second case they lost 1/3rd job

12 men and 12 woman can do (1/3) + (1/4) = 7/12 work in 1 day

so They can finish the work in 12/7 days

so 1 woman and 1 man can finish the work in 12 * 12/7 = 144/7 or 20.57 days ]]>

a) 4

b) 6

c) 8

d) 10 ]]>

so S(2) = 3 * S(1)

S(1) is sum of first term means first term itself.

S(2) = 3 and S(1) = 1

So sum of 2 terms = 3

Sum of first term = 1

So 2nd term = 3 - 1 = 2

First term = 1

Second term = 2

So d = 1

Means third term = 3

So S(3) = 1 + 2 + 3 = 6

S(3)/S(1) = 6 ]]>

a) 31! − 1

b) 15 × 31

c) 30! − 1

d) 31! + (15 × 31)

e) 30! + (15 × 31) – 1 ]]>

so here in this case 31! - 1 ]]>

it is of the form 3k + 2

to make it multiple of 3 (A got 2x and B got x = > Total 3x) we should subtract a number of the form 3k + 2.

23 is present here in form of 3k + 2

Another approach:

A has even number of cookies

even = even + even + even

even = odd + odd + even

three even is not here so he will choose two odd and one even

so 34 + 19 + 21 or 34 + 19 + 23 or 34 + 19 + 25 or 34 + 21 + 23 or 34 + 23 + 25

so 74 or 76 or 78 or 78 or 82

now B has 37 or 38 or 39 or 41

37 = 18 + 19 (19 is already used so not possible)

38 is not pssible

39 = 18 + 21 (this is possible combination)

41= 23 + 18

so 23 almond cookies are there

Understand Basic - average speed = total distance/total time

car 1__________________12d _______________ car2

let distance between them = 12d

speed of fly is double so when they both will cover 4d distance fly will cover 8d distance

now speed is triple so when both cars will travel d then fly will travel 3d and meets ann again so

so distance travelled by fly in forward direction is 8d and in backward direction = 3d

so avg speed = (8d + 3d)/(8d/20 + 3d/30)

11/(8/20 + 3/30)

11 * 60/30 = 22 km/hr = 110/18 m/s

time taken by ann and anne to collide = 50/(100/18) = 9 seconds

fly will cover= 9 * 110/18 = 55 m

Method 2 :

Suppose that at a given instant the fly is at Ann and the two cars are 12d apart. Then, while each of the cars travels 4d, the fly travels 8d and meets Anne. Then the fly turns around, and while each of the cars travels d, the fly travels 3d and meets Ann again. So, in this process described, each car travels a total of 5d while the

fly travels 11d.

So the fly will travel 11/5 times the distance traveled by each bumper car = 11/5 * 50/2 = 55 meters

therefore father reached to his son 12 minutes before than he use to be at 3:18

His son walked for 48 minutes. (from 2:30 to 3:18)

so distance travelled by son in 48 minutes is equal to distance travelled by father in 12 minutes

hence ratio of their speed is 1 : 4 and thus 24 km/h. ]]>

x^4 + 8x^2 + 16 = 4x^2 - 12x + 9

a) 0

b) 1

c) 2

d) 3

e) 4 ]]>

(x^2 + 4)^2 = (2x - 3)^2

x^2 + 4 = ± (2x - 3)

x^2 - 2x + 7 = 0 or x^2 + 2x + 1 = 0

First equation has no real roots and second has equal roots so only 1 different root ]]>

a) 60740

b) 64720

c) 62826

d) 61741 ]]>