If division of 3 is allowed then I might do 5 divisions in one go & do in less iterations. I would assume max 2 division as you have to hands, and compare the weight.
Given y/x = 1/3, x+2y =10
3y=x
Then substitute x=3y in x+2y=10
3y+2y=10
5y=10
y=2
Then substitute y=2 in x=3y
x=3*2
x=6
: x=6, y=2
10.66 kmph
cost price(c.p) of one fruit = 24/16
(c.p) = Rs 1.50
selling price(s.p) of one fruit = 18/8
(s.p) =Rs 2.25
profit for one fruit = 2.25 – 1.50 = 0.75
profit
profit % = ———– * 100
cost price
0.75
= ——- * 100
1.50
= 50%
the answer is E
2.5 sec
let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9
so X<=3 and any of the digit in the number is <=9
and also given that 3 pairs sum is 11...
so make trial and error..
if X=1...any of the no is 10 which is wrong trial....
if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296
total no of 4 wheelers: 40
total no of 2 wheelers: 18
Given:
In a group of 15 students,
7 have studied Latin,
8 have studied Greek,
3 have not studied either.
To find:
The number of students who studied both Latin and Greek.
Solution:
In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.
Therefore,
n(A∪B) = 15 – 3
n(A∪B) = 12
7 have studied Latin,
n(A) = 7
8 have studied Greek,
n(B) = 8
n(A∩B) is the number of students who studied both Latin and Greek.
n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = 7 + 8 – 12
n(A∩B) = 15 – 12
n(A∩B) = 3
The number of students who studied both Latin and Greek is 3
Final answer:
3 of them studied both Latin and Greek.
Thus, the correct answer .3
The Bear was white in color as it was a polar bear
It started moving from the north pole towards the south 1
mile and again after moving 1 mile east and one mile north
he reached his original position
12
Statements :
All mangoes are golden in colour. No golden coloured things are cheap.
Conclusions :
I. All mangoes are cheap.
II. Golden coloured mangoes are not cheap.
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds
Speed = 13+4 km/hr
Distance = 68 km/hr
Time=distance/speed
68/17=4hrs