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The statement is of the type: if A (it flows) then B (have viscosity).

Let us assume the present age of the person be P years.
3 years hence the age of person = P + 3
3 years ago the age of person = P - 3
According to question,
Take my age three years hence, multiply it by 3 and then subtract 3 times my age three years ago,
= 3(P + 3) - 3(P - 3)
= 3P + 9 - 3P + 9
= 18 years
Hence, age of the person = 18 years

On the basis of above given figure , we can see that
City A is connected by 2 roads, B by 4 roads, C by 3 roads, D by 3 roads and E by 2 roads.
For a city to be starting city for such a route, it has to be connected by odd number of roads.
Hence the required answer is 2, i.e. C and D.

Draw a table with Person, Sex, Vehicle and Destinations.
P is travelling in Honda City and is going to Hyderabad.
S is travelling by Ford Ikon.
T, a male is travelling with only Z.
R is not travelling with Q and W. T, a male is travelling with only Z and they are not going to Chennai.
S is the sister of P and is travelling by Ford Ikon. V and R are travelling together.
W is not going to Chennai.
Based on the above information, the classification of the group is as follow in the table.

If the day before yesterday was Saturday, then today is Monday.
Thus, tomorrow will be Tuesday and day after tomorrow will be Wednesday.

From the answer of previous question,
Total number of ways from A to B is 90.
Now we need to calculate total number of ways from B to C that is 13 (on the same logic)
Hence, total number of paths from A to C is 90 x 13 = 1170.

Here , number of vertical steps ( v ) = 3
Number of horizontal steps ( h ) = 5
Then in this case total number of ways is given by ^h+vC_h = ^h+vC_v = ⁸C₃ = 6 x 7 x ( 8/6 ) = 7 x 8 = 56.
Hence , 56 distinct routes can a car reach point B from point A .

This is the only square in the group.

As per the given diagram , we can see that
Let the toll charged at junctions A, B, C and D be a, b, c and d respectively.
Then 1^st we will list down all the routes and corresponding cost of travel.
Here in this case all 5 routes have the same toll charge hence 14 + a = 7 + b + c = 13 + d = 9 + a + b = 10 + c + d
After solving we will get a = 1, b = 5, c = 3 and d = 2

The series is an alternate series, having
Odd position series :-
First term = 2
Third term = ( First term × 3 ) - 1
? Third term = ( 2 × 3 ) - 1 = 5 ;
Fifth term = ( Third term × 3 ) - 1
? Fifth term = ( 5 × 3 ) - 1 = 14 ;
Seventh term = ( Fifth term × 3 ) - 1
?Seventh term = ( 14 × 3 ) - 1 = 41 ;
Even position series :-
Second term = 3
Fourth term = ( Second term × 3 ) - 1
? Fourth term = ( 3 × 3 ) - 1 = 8 ;
Sixth term = ( Fourth term × 3 ) - 1
? Sixth term = ( 8 × 3 ) - 1 = 23 ;
Eighth term = ( Sixth term × 3 ) - 1
? Eighth term = ( 23 × 3 ) - 1 = 69; 69 is wrong. It should be 68.