Verbal Reasoning - Arithmetic Reasoning - Discussion
Discussion Forum : Arithmetic Reasoning - Section 1 (Q.No. 2)
2.
Two bus tickets from city A to B and three tickets from city A to C cost Rs. 77 but three tickets from city A to B and two tickets from city A to C cost Rs. 73. What are the fares for cities B and C from A ?
Answer: Option
Explanation:
Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.
Then, 2x + 3y = 77 ...(i) and
3x + 2y = 73 ...(ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.
Putting y = 17 in (i), we get: x = 13.
Discussion:
47 comments Page 1 of 5.
Raviraushan Kumar said:
11 months ago
I can't understand this, Anyone, Please, explain me to get it.
Tobiyo Kageyama said:
1 year ago
Thanks all for the explanations.
Jennie said:
3 years ago
Why should we multiply the first equation and the second equation with 3and 2? please explain.
Prince thakur said:
4 years ago
Here:
Ticket from city A and b = 2 or 2x.
Ticket from city A to C= 3 or 3x.
All cost is;
2x+3y = 77 ----> equation 1
Similarly
Ticket from city A and b = 3 or 3x
Ticket from city A to C= 2 or 2x
All cost is;
3x+2y = 73 ----> equation 2.
Now,
Cross multiply 1 and 2 equation.
We get
146x +219y=231x +154y
219y-154y=231x-146x
65y=85x.
65/85=x/y.
65 and 85 divide by 5 we get.
13/17=x/y.
Here
X=13.
Y=17.
Ticket from city A and b = 2 or 2x.
Ticket from city A to C= 3 or 3x.
All cost is;
2x+3y = 77 ----> equation 1
Similarly
Ticket from city A and b = 3 or 3x
Ticket from city A to C= 2 or 2x
All cost is;
3x+2y = 73 ----> equation 2.
Now,
Cross multiply 1 and 2 equation.
We get
146x +219y=231x +154y
219y-154y=231x-146x
65y=85x.
65/85=x/y.
65 and 85 divide by 5 we get.
13/17=x/y.
Here
X=13.
Y=17.
(4)
Gajjela said:
5 years ago
@All.
According to me, the shortcut method is;
Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A. And also let Rs. z be the fare for 2tickets from A to B and 2tickets from A to C.
then; z+y=77 and z+x=73,
results y-x=5.
x=13 and y=17 only satisfies the equation.
So, the answer is an option [B].
According to me, the shortcut method is;
Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A. And also let Rs. z be the fare for 2tickets from A to B and 2tickets from A to C.
then; z+y=77 and z+x=73,
results y-x=5.
x=13 and y=17 only satisfies the equation.
So, the answer is an option [B].
(1)
Vanshika said:
5 years ago
I can't understand is why the 1st equation is multiplied by 3 and 2nd by 2.
Vanshika said:
5 years ago
How do we get to know that 1st equation will be multiplied 3 and second by 2?
Sunil gampa said:
6 years ago
Please explain clearly how to take x y values in an equation?
Chandini said:
6 years ago
2A+3B=77.....(1)
3A+2B=73.....(2)
FROM 1&2 we get A=30-B
put A value in 1 then we get 2(30-B)+3B=77.
Finally, we get B=17.
put B value either1 or 2 eqn,
we A=13
final answer 13, 17.
3A+2B=73.....(2)
FROM 1&2 we get A=30-B
put A value in 1 then we get 2(30-B)+3B=77.
Finally, we get B=17.
put B value either1 or 2 eqn,
we A=13
final answer 13, 17.
Umar said:
6 years ago
2x+3y=77-->(1)
3x+2y=73-->(2)
73(2x+3y)=77(3x+2y),
146x+219y=231x+154y,
65y=85x,
13y=17x.
3x+2y=73-->(2)
73(2x+3y)=77(3x+2y),
146x+219y=231x+154y,
65y=85x,
13y=17x.
(2)
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