I(N)G
D(IN)G and
S(ENDI)N
Considering option 3,
2 *(14)2 - 25*14 = 3*14.
Alternatively :
Let the age of the lad be x then,
2x2 -25x = 42.
2x2 -25x-42 = 0.
Either solve the quadratic equation or put the options to get the value.Option 3 is correct.
There are 70 red vehicles thus rest 90 are green vehicles,
out of which 18 are green trucks then we get,
Green cars = 90 - 18 = 72.
Given, Number of cars = 120.
Then, number of red cars,
= 120 - 72
= 48 Red cars.
X+24*y = 24*70 ----(1)
For the 2nd condition, we have,
X+60*Y = 60*30----(2) Now, On solving equation (1) and (2), we get
X = 1600 and
Y = 10 /3
Third Condition,
X+96*Y = 96 *N -----(3) [N = Number of Cows required]
Putting the values of X and Y in equation (3), We get
N = 20.
Mohan is older than Prabir.
Prabir < Mohan
Suresh is younger than Prabir means Prabir is older than Suresh.
Suresh < Prabir < Mohan
Mihir is older than Suresh but younger than Prabir.
Suresh < Mihir < Prabir < Mohan
Suresh is the youngest.
Let the present age of his son be X years.
So, present age of his father = X+24 (as he is 24 years older than his son)
After 2 years, father's age become (X+24+2) years and Son's age will be (X+2).
Now, according to question,
X+24+2 = 2*(X +2)
Or, X = 22 years.
Son's present age = 22 years.
Total days of rainfall = 13.
11 Mornings enjoyed and 12 Afternoons enjoyed, so it means there was rain on 6 mornings and 7 afternoons.
So, extra days without rain are 5 days (11-6 or 12-7).
Total no of day they live,
= No. of days of rain + Days without rain,
Or, 13 + 5 = 18days.
M scores more run than N.(M > N)M scores less than P. (P >M >N) Q scores more than N but less than M. So,Sequence of scoring run is P >M>Q>N. So, N is the lowest scorer.
Sequence of seating from right to left is:
Harish, Manish, Satish and Girish.
So, Girish sits farthest to right.
The alphabet starting with C, so
1st Hour = C
2nd Hour = D
3rd Hour = E, and So on. Then we get,
16th Hour = R.
16 hour is represented by letter R.