Verbal Reasoning - Arithmetic Reasoning - Discussion

Let, A to B fare as X and A to C as Y

(2x + 3y = 77)X 3 ---------------(1 line)
(3x + 2y = 73)X 2 ---------------(2 line)
now, after multiplications, we get

6x + 9y = 231---------------(1 line)
6x + 4y = 146---------------(2 line)

now, after change the signs and solve the equation, we will get

6x + 9y = 231
-6x - 4y =-146
----------------
ox + 5y = 85

OR,5y = 85

Y= 85/5
OR, Y= 17
NOW, put the value of y in any line, for Eg. in first line, as under:-

2x+3y=77------(1 line)
2x+3(17)=77
2x+51=77
OR, 2x=77-51
2x=26
x=26/2
x=13
SO,
13 and 17 is the answer

@Kamaho

We have to make either x or y value 0 to get one's value.

So multiplying (i) with 3 and (ii) with 2 we will get x part 0 by equating both equations [ (i) - (ii) ] then only we get 5y=85 further values.

Also, we can multiply (i) with 2 and (ii) with 3.
Then by equating both, we will get 5x=65 then x=65/5..... x=13.

The purpose of multiplying the equations is to make either x or y equal, then by equating we will get either x or y =0.
You can take any no. For multiply, but make values of both x or both y equal.

Hope you will understand.

Here,

Let the fares for city A to B = x.

Let the fares for city A to C = y.

In 1st condition.

2x+3y = 77....(i).

In 2nd condition.

3x+2y = 73....(ii).

Multiply equation (i) by 3 & equation (ii) by 2.

6x+9y = 231....(iii).

6x+4y = 146....(iv).

Subtract (iii) from (iv).

5y = 85.

Y = 17.

Put the value of why in equation (i).

2x = 77-51.

x = 13.

So,

Fares for the city A to B = x = 13.

Fares for the city A to C = y = 17.

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.

2x + 3y = 77.
3x + 2y = 73.

Solution :
2x + 3y = 77 (i)multiplying by 3.
6x + 9y = 231.

3x + 2y = 73 (ii)multiplying by 2.
6x + 4y = 146.

Now subtract its,
6x + 9y = 231.
6x - 4y = -146.

0x + 5y = 85.
5y = 85.
y = 85/5.
Ans y = 17.

Now put y value on the equation (i).

6x + 9(17) = 231.
6x + 153 = 231.
6x + 153 - 231 = 0.
6x - 78 = 0.
x = 78/6.
x = 13.

Solution set is { 13 , 17 }.

2X + 3Y = 77 ------> 1
3X + 2Y = 73 ------> 2

1.multiply with 3.
2.multiply with 2.

6X + 9Y = 231 ------> 3
6X + 6Y = 146 ------> 4

Subtract 3 & 4.

6X + 9Y = 231
6X + 6Y = 146
-----------------------
0 + 5y = 85
y = 85/5
y = 17.

Put y value in 1 or 2.

2X + 3(17) = 77,
2X + 51 = 77,
2X = 77 - 51 = 26.
X = 26/2 = 13.
X = 13 and Y = 17.

@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].

Let Rs. x be the cost of city B from city A and Rs. x + 4 be the cost of city C from city A.

Then, 2x + 3x + 12= 77= 5x + 12.

Which means that, 77 - 12= 65= 5x.

Thus, x= 65 ÷ 5= 13.

Because A to B is Rs. x, that means that it costs, Rs. 13 and,
Because A to C is Rs. x + 4, that menas that is costs Rs. 13 + 4, which is equal to 17.

2x + 3y = 77 multiply this line with 3 so it becomes
6x + 9y = 231-----------(i)

3x + 2y = 73 multiply this line with 2 so it becomes
6x + 4y = 146 -------------(ii)

now solve both the equestion (i) and (ii)
6x + 9y = 231
6x + 4y = 146
----------------
ox + 5y = 85

y = 85/5 = 17

in the same way do for X ..

How 3 and 2 is there?

Why do we multiply with 3&2 why not 2&3? everyone know 13&17 is the correct answer. But how & where do we get that 3 & 2 why not 2&3 anyone who can explain this for me? Yes, we know if we multiply by 3 & 2 we get an answer but how do we got that?