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Let x be fathers present age and y be son present age.
5 yrs ago, the age of father and son be x-5 & y-5.
Then,
x-5+y-5=40
x+y-10=40
x+y=50
y=50-x ———–> (1)
ratio between father and son in present age
x:y=4:1
x/y=4/1
x=4y
Apply eq (1) ,
x=4(50-x)
x=200-4x
x+4x=200
5x=200
x=200/5
x=40,,
=> The present age of father is 40.
1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.
B
let (xy) be original number.
then, 10y + x +45 = 10x + y; eq 1
and x + y = 13; eq 2
solve ab equation,
then xy = 94;
Its very simple..
consider the fraction of s in the mixture = 1/3
So if we add one more R the the fraction wil be = 1/4
Automaticaly S becomes 25% of the mixture
90 inches
Slab = 6(x) where x is each step
Slab = 5(x + 3)
6(x) = 5(x + 3)
6(x) = 5(x) + 15
x = 15 inches
So 6(x) = 6(15) = 90
Check: 5( x + 3) = 5(18) = 90
Slab size is 90 inches
13km
Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:
1. Weigh ABC versus DEF.
Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.
2. Weigh G versus H.
Scenario a.i: If these (2) balance, the oddball is I.
Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.
Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.
2. Weigh A versus B.
Scenario b.i: If these (2) balance, the oddball is C.
Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.
answer is maximum of 2.
Example 1:
Assign, A=20, B=10, C=5, D=5(Because C is equal to D as
given), E=1.
A/B = 20/10 = 2. So A/B = 2
A/C = 20/5 = 4. So A/C = 4
A/E = 20/1 = 20. So A/E = 20
Therefore “A/E is Greatest”
Example 2:
Assign, A=100, B=50, C=20, D=20(Because C is equal to D as
given), E=10.
A/B = 100/50 = 2. So A/B = 2
A/C = 100/20 = 5. So A/C = 5
A/E = 100/10 = 10. So A/E = 10
Therefore “A/E is Greatest”