Distance covered by B to meet A=Total distance – Distance covered by A hrs
[using : distance = speed x time]
Putting value of from equation (1),
hrs
Therefore, time at which both A and B will meet is = 7 a.m. + 3 hrs =10 am
8925
That person’s struggle and dream to achieve something motivates the person.
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
Time =9 seconds
Speed =(60×518) m/sec=(503)m/sec
∴ Speed =DistanceTime
Length of the train (Distance) = (Speed × Time) = (503×9) m=150 m.
72 kmph
time (dist covrd by husband) (dist covrd by wife)
0.5hr 20 miles 0 miles
1 hr 40 miles 25 miles
1.5 hr 60 miles 50 miles
2 hrs 80 miles 75 miles
2.5 hrs 100 miles 100 miles
Therefore wife catches up after 2.5 hrs..
Mixture is 1.2 L of A + 2.4L of B which is 3.6L total. 1.2L / 3.6L = 1/3 (0.33)
South-West
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”
A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m
1. 1g
2. 3g with 1g counter
3. 3g
4. 3g plus 1g
5. 3g plus 1 g plus 1g weighed medicine
6. 9g with 3g counter
7. 9g with 1g of counter and 1g weighed medicine
8. 9g with 1g counter
9. 9g
10. 9g plus 1g
11. 9g plus 3g with 1g on counter
12. 9g plus 3g
13. All 3 weights on one side.
G.c.d=num1 *num2/lcm
Num1=20
Gcd=5
Lcm=60
5=(20*num2)/60
Num2= 5*60/20
Num2=15
Let the total number of matches to played in the tournament be ‘x’.
Given that A county cricket team has won 10 matches and lost 4.
That means total number of matches played = 14.
So,
= > 70% of x = 14
= > (70/100) * x = 14
= > 70x = 1400
= > x = 20.
So, the number of matches to be played = 20.
They have a record of exactly 75% wins = 20 * 75/100
= > 15.
Therefore, the number of matches should the team win = 15 – 10 = 5.
Hope this helps!
1/4E(E decreses by 4 times)
If discount is 25%
Profit 25%
If discount10%
Profit?
Suppose 100rs is price
25% discount make it 75
25%profit that means 75×100/125=60
Cp is 60
10 % discount make sp 90 profit is 30%
In 30 mins- Thief – 20km, Owner – 0km
In 1hr – Thief – 40km, Owner – 25km
In 1.5hrs – Thief – 60km, Owner – 50km
In 2hrs – Thief – 80km, Owner – 75km
In 2.5hrs – Thief – 100km, Owner – 100km
Answer: In 2.5 hours the owner would have overtaken the Theif
3 minutes