A and B run a km race. If A gives B a start of 100 m, A wins by 20 sec and if A gives B a start of 5

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15 Jun 2017 Race

A and B run a km race. If A gives B a start of 100 m, A wins by 20 sec and if A gives B a start of 50 sec, B wins by 30 m. The time taken by A to run a km is

OPtion

1) 188.846 sec
2) 190.846 sec
3) 192.846 sec
4) 196.846 sec
5) 198.846 sec
6) 234.94 sec
7) 236.94 sec
8) 238.94 sec
9) 240.94 sec
10) None of these

Solution

Answer: 196.846 sec
Let time taken by A to run 1 KM be a and by B be to run 1KM be b
If A gives B a start of 100 m, A wins by 20 sec

Distance travelled by A = 1000

Distance travelled by B = (1000-100)=900

Time taken by B to run 900 m - Time taken by A to run 1000 m = 20 sec
900b/1000 - a = 20 sec
a = (900b/1000) - 20


If A gives B a start of 50 sec, B wins by 30 m

Distance covered by B in b seconds - Distance covered by A in b-50 seconds = 30m
1000 - 1000*(b-50)/a = 30
=> 970a = 1000b - 50000

970*(900b/1000) - 970*20 = 1000b - 50000
873b - 970*20 = 1000b - 50000
=> 127b = 30600
=> b = 240.94

so a = (900b/1000) - 20 = 900*240.94/1000 -20 = 196.846 sec

Option 4)

Correct Option: 4 Best Solution (4)

G.S.Kantharao   7 years AGO

Let A runs 1000mts in = p seconds
B runs 1000mts in =q seconds.
Step 1: If A gives B a start of 100mts.
That means A ran 1000mts, B ran 900mts
Time difference = 9q/10 -p= 20sec
9q-10p=200-------Eqn.1
If A gives B a head of 50 sec
Then the distance covered by B in q sec.- the distance covered A in (q-50)sec=30mts
That is 1000- 1000(q-50)/p= 30
1000(q-50)/p=1000-30=970
10q-500=9.7p
10q- 9.7p=500
100q-97p=5000--––--Eqn.2
By the method of elimination
We can get 127p=25000
Then the time taken by A to complete the race = p sec = 25000/127 = 196.85sec
Hence the right option is 4.😇😇😇👌👌👌

Like? Yes (7) | No (1)  

Nitish Sinha   7 years AGO

Let, Time taken by A=a & B=b
If gives a start of 100m A wins by 20s
Distance travelled by A=1000m then
B=1000-100=900m
Time taken by B to run 900m- Time taken by A to run 1000m=20s
(900*b/1000)-a=20...(1)
If A gives B a start of 50s, B wins by 30m (b-50)
=> 1000- (1000/a)* (b-50)=30
=> 1000b- 50000=970a...(2)
eq1*eq2, we get:
(90*97/100)- 97a=20*97
=>873b-970a=1940...(3)
Now, eq2-eq3;
=>1000b-873b=50000-19400
=>127b=30600
=>b=240.94
Substituting the value of b in eq1;
(90/100)*240.94-a=20
=>a=196.84
ans: option4

Like? Yes (3) | No (3)  

ganesh chemanchula   7 years AGO

A gives B 100 meter start and wins the race by 20 sec=(1000/900-20b)=X
if A gives a 50 sec start then, wins by 30 m=(970/950-50b)=Y
X=Y
by solving the value of b is 4.1503
let speed of A be x..the (1000/900-20(4.1503))=(x/4.1503)
then the value of x will be 5.0799
time taken by A to complete the 1000 meter race is 1000/5.0799
by solving the ans is 196.8 sec approx
so option 4.

Like? Yes (2) | No (3)  

Sampad Mukherjee   7 years AGO

Aptitude on Race
Time taken by A to run 1000 m = a
Time taken by B to run 1000 m = b
(900/1000)b - a = 20 --------- Equation 1
1000 - (1000/a)×(b - 50) = 30 –---------- Equation 2
Solving equation 1 and 2
Answer is 196.85

Like? Yes (4) | No (8)