Latest TCS Aptitude Question SOLUTION: Of 6000 apples harvested every third apple was too small,evey 4rth apple was too green and every tenth apple was bruised.the remaining apples were perfect.how many perfect apples w

Calculating 3's there are (6000-3)/3+1=1500 4's (6000-4)/4+1=2000 and 10's 6000-10/10+1=600..nw eliminating common terms b/w 3 and 4 it occurs for every 12 so 6000-12/12+1=500..so subtracting frm no of 3's and 4's we get 2000+1500-500=3000;then 10's ll contain 3 for every 30 nos 6000-30/30+1=200 and 4's for every 20 6000-20/20+1=300 then eliminate common cases b/w 3 and 4 containing 10's it ll occur for every 60 so 6000-60/60+1 = 100 then 300+200-100=400;so nos containing 10's are 600-400=200....then ans is 3000+200=3200 then harvested apples are 6000-3200=2800..one who knows eliminating cases ll understand this:p

every 3rd apple 6000/3 = 2000
every 4th apple 6000/4 = 1500
every 10th apple 6000/10 = 600

Total defected apples = 2000+1500+600 => 4100, so 6000-4100 = 1900 (Perfect apples)

ans is b) 2700
3rd apple was too small, so: 6000 =3+ (n+1)3 = 2000 .......... (1)
4th apple was too green, so: 6000 =4+ (n+1)4 = 2000 .......... (2)
10th apple was bruised, so: 6000 =10+ (n+1)10 = 2000 .......... (3)

now some apples would come under both too small and too green, or too green and bruised, or too small and brused, or all the three conditions
so,
too small and too green: 6000/12 = 500 ......(4)
too green and bruised : 6000/40 = 150 .....(5)
too small and bruised: 6000/30 = 200 ....(6)
and under all three conditions: 6000/120= 50 ....(7)
6000 - 2000-1500-600 + 500+150+200+50 =2800 (from 1,2,3,4,5,6,7)

ans) c - 1900
as every 3rd apple was too small, so the no of too small apple is- 6000=3+(n-1)3 = 2000
as every 4th apple was too green, so the no of too green apple is- 6000=4+(n-1)4 = 1500
as every 10th apple was bruised, so the no of bruised apple is- 6000=10+(n-1)10 = 600
no of perfect apple were = 6000-(2000+1500+600) = 1900

1).n(small)=6000/3=2000
2).n(green)=1500
3).n(bruised)=600
4).n(small& green)=500
5).n(small & bruised)=200
6).n(green &bruised)=300
7).n(small & green & bruised)=100
now do 1). +2). +3). +7). - 4). -5). -6).=3200
6000-3200=2800=perfect apples

First, I found the least common multiple of 3, 4 and 10 (since every
3rd, 4th and 10th apple was "bad"). The least common multiple is 60.

Knowing that, we now know that every 60 apples will have the same
number of "bad" and "perfect" apples.

1 P
2 P
3 S
4 G
5 P
6 S
7 P
8 G
9 S
10 B
11 P
12 S G
13 P
14 P
15 S
16 G
17 P
18 S
19 P
20 G B
21 S
22 P
23 P
24 S G
25 P
26 P
27 S
28 G
29 P
30 S B
31 P
32 G
33 S
34 P
35 P
36 S G
37 P
38 P
39 S
40 G B
41 P
42 S
43 P
44 G
45 S
46 P
47 P
48 S G
49 P
50 B
51 S
52 G
53 P
54 S
55 P
56 G
57 S
58 P
59 P
60 S G B

Now after every 60 apples, the pattern will repeat itself, giving us
the same number of ?perfect? and ?bad? apples.

Going back and counting them, we know that in every 60 apples there will be:
20 apples that were too small (S)
15 apples that were too green (G)
6 apples that were too bruised (B)
28 apples that were perfect (P)

(Remember, sometimes apples were ?bad? in more than one way.)

Next, 18,000 divided by 60 is 300. So the above pattern will be counted 300 times.

Which means of the 18,000 apples, there were:
300 x 20 = 6000 apples that were too small
300 x 15 = 4500 apples that were too green
300 x 6 = 1800 apples that were too bruised

and

300 x 28 = 8400 apples that were perfect.

You can do anyways u want.start wit 10,3 or 10,4..or the three at d same tym..its upto you.bt the tng is you have to eliminate evry nos in common