1. Not as complicated as it seems
The value C will take is '1' since max XYZ is 999 and max AB is 99.
X has to take a value '9' only then we can carry over '1'.
Looking at XYZ - AB = BGA there are two important things:
Z - B = A and A > Y so, X - 1 = B. Also 1Y - A = G
This means B = '8' since we know X is '9'.
Now looking at XYZ + AB = CDEF
Since we know X is '9' we know C is '1' and D can only be '0'.
So now '9'YZ + A'8' = '1''0'EF and
'9'YZ - A'8' = '8'GA. Digits remaining 2,3,4,5,6,7
'1'Y - A = G and '1'Z - '8' = A
Simplest way is trial and error
Simple assumption is Z = 5 which gives A = 15 - 8 = 7
'9'Y'5' - 78 = '8'G'7' so 945 - 78 = 867
945 + 78 = 1023
2. Not very straightforward:
Initially the the car is travelling for 2 hours and 10 minutes delay due to flat tire. the rest of the distance is covered with the speed of 30 miles/hour, but they are late by 1/2 an hour.
Total distance = d1 + d2, v2 = speed with which d2 is travelled if the car is not punctured.
Equation 1: 2 + (10 minutes delay) + d2/30 = t0 + ( 1/2 an hour late)
2 + 1/6 + d2/30 = t0 + 1/2 -----------> 1
When the car is not punctured the car would have travelled with the same speed.
Equation 2: 2 + d2/V2 = t0 + ( 15 minutes delay )
now v2 = d1/2
2 + 2d2/d1 = t0 + 1/4 -------------> 2
Sub 2 from 1 : 1/6 + d2/30 - 2d2/d1 = 1/4
5d1 + 120d2 = 2d1d2
Let d2 = 5 :
5d1 = 120*5
d1 = 120
Therefore the total distance will be 125 miles ( this is one of the answers )
we can substitute from 3 to 59 for d2 and can find many solutions for this problem.
3. Distance covered by Deepak and Suresh = 4 miles
If Deepak and Suresh have seen the bus after 20 minutes, means the bus takes 6 minutes to cover 4 miles. So speed of the bus is 40miles/hr
4. Speed at which two trains travel towards each other = 60+60 = 120 miles/hour.
Total length to be covered = (1/6)+(1/6) = 2/6 mile.
Time to taken to cover 2/6 mile at speed of 120 miles/hour is.....
(2/6)/120 = 1/360 hour = 10 seconds.
5. Let the actual speed of train be X miles/hour.
Let the initial distance covered by train is d1.
After fault its speed will be reduced to 3X/5 miles/hour.
So the distance travalled is 3X/5 * 2 = 6X/5 = d2 ----> (1)
So total distance = X + 6X/5 = 11X/5
Now if the fault had occurred at d1 + 50
3X/5 * 5/4 = d2 - 50 -----> (2)
(1) - (2) = 6X/5 - 3X/4 = 50
9X/20 = 50 or X = 1000/9
So the distance d2 = 133.33 miles
So d1 = 11*1000/(9*5) - 133.33 = 111.11
So total distance is 250 + 125/3 = 244.44 miles
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