Discussion :: Arithmetic Reasoning - Section 1 (Q.No.2)
| Sartaj said: (Jul 21, 2011) | |
| I cant get this, please explain. | |
| Ladyno said: (Jul 25, 2011) | |
| I am truly lost, please help. Where does (i) and (ii) fit in? | |
| M. Iyyammal said: (Jul 29, 2011) | |
| I cant get this, please explain (1) and (ii). | |
| Vijaya said: (Aug 2, 2011) | |
| 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 .. |
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| Antony said: (Nov 11, 2011) | |
| New method 3 tickets A to B 2 tickets A to c that's 77/5 = 15.4 3 tickets A to B 2 tickets A to C that's 73/5 = 14.6 so 15.4+14.6 = 30 Ans: 13 + 17 = 30 --- (B) |
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| Bujji said: (Aug 21, 2012) | |
| It is wrong antony. In options 17 and 13 is also there. So your method is wrong. | |
| Gangadhar said: (Sep 8, 2012) | |
| Vijay method is correct. | |
| Deep said: (Sep 14, 2012) | |
| Vijay why did you first multiply by 3 and then by 2 why not the reverse... How would you manage the negative output when you multiply first line by 2 and second line by 3 i.e 4x + 6y 9x + 6x ------------- Minus = 5x ? Don't know if I made sense with my question. |
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| Satwant said: (Sep 30, 2012) | |
| 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 |
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| Madhu said: (Jan 19, 2013) | |
| @Vijay You are method is so good. easy way to understand..in Vijay method if any x,y coefficient should made equal to the values. Because we find x,y values..thats why he multiple 3 with first equation and 2 with second equation. |
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| Anam Charan Patra said: (Apr 29, 2013) | |
| Please explain why multiply 3 series (i) and 2 multiply series (ii)? | |
| Ramkumar said: (Jun 8, 2013) | |
| Its very simple. It is a optional based solution . There forms : 2x+3y = 77. 3x+2y = 73 (then we find x and y). Simply check with A B C D options. Now option B is: 13, 17 is satisfies the above equations. |
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| Jitendra said: (Aug 14, 2013) | |
| New method by @Antony is really good and a short-cut way to find the solution. | |
| M.Haseeb Sahal Shah said: (Aug 16, 2013) | |
| 2x+3y = 77. 2x = 77 -3y. x = 77 - 3y /2 ... (1). 3x+2y = 73. Put x value, 3(77 - 3y /2)+ 2y = 73. 3(77 - 3y)= 2(73-2y). 231-9y = 146-4y. 231-146 = -4y+9y. 85 = 5y. 85/5 =y. 17 = y. x = 77 - 3y /2 ... (1). x = 77 - 3(17) /2. x = 77 - 51 /2. x = 26/2. x = 13. B is answer 13, 17. |
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| Dazedandconfused said: (May 14, 2014) | |
| For some reason, this seems complicated to me. What is the purpose of these tests? | |
| Sajid Khan said: (Jun 17, 2014) | |
| 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 }. |
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| Aneri Desai said: (Aug 18, 2014) | |
| s2b(2) & s2a(3) = 77. s2b(3) & s2a(2) = 73. s2b=x=? & s2a=y=? 4x+6y = 154. 9x+6y = 219. -5x = -65. x = 13. y = 17. |
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| Reghu R said: (Oct 18, 2014) | |
| 2x+3y = 77 --------- i. 3x+2y = 73 --------- ii. --------- i*2 => 4x+6y = 154. ii*3 => 9x+6y = 219. 5x=65; x=65/5; x=13; 26+3y=77; 3y=77-26; 3y=51; y=17. Answer = Rs.13 and Rs.17. |
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| Pvk said: (Nov 26, 2014) | |
| Antony method super and correct substitute the x, y value in 1, 2 equation. Only 13, 17 satisfying. | |
| Josh said: (Apr 20, 2015) | |
| How do they get 26+3y? | |
| Bkazondarnabas said: (May 14, 2015) | |
| Thank you for reminding you the formula of solving simultaneous question. | |
| Durga said: (Aug 4, 2015) | |
| Its very nice of getting answers by these methods. | |
| Charan said: (Aug 21, 2015) | |
| Why Should we multiply 2x+3y = 77 with 3 and the second one with 2? Can anyone explain? |
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| Amit Jain said: (Oct 2, 2015) | |
| 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. |
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| Rashed said: (Jun 14, 2016) | |
| Hi, I'm not much good in maths but can anybody tell me how we are multiplying these equations with 3 and 2? Where these 3 and 2 comes from? | |
| Rashid said: (Jun 15, 2016) | |
| Can anybody explain to me why we are multiplying I and ii equations with 3 and 2 respectively? Where these 3 and 2 comes from. | |
| Ianz said: (Jun 27, 2016) | |
| Its great, by just comparing and sharing knowledge we can learn lot. | |
| Russ said: (Jul 5, 2016) | |
| I don't understand how you got 2 and 3? Where does the 2 and 3 come in place? Please explain it. | |
| Vivek said: (Jul 7, 2016) | |
| We do for same term of both equation. | |
| Srinath said: (Aug 20, 2016) | |
| 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. |
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| Gourav said: (Oct 7, 2016) | |
| Why not option D? | |
| Naveenkumar said: (Oct 9, 2016) | |
| How to take the equation? | |
| Tushar Kanti Das said: (Nov 10, 2016) | |
| Your solution is awesome. I got it easily through your process, Thanks @Satwant. | |
| Kenneth said: (Feb 17, 2017) | |
| 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. |
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| Kamaho said: (Feb 23, 2017) | |
| 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? |
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| Priyan said: (Feb 27, 2017) | |
| How to get 85? Not understanding it. |
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| Siyad said: (Jun 12, 2017) | |
| @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. |
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| Umar said: (Aug 15, 2017) | |
| 2x+3y=77-->(1) 3x+2y=73-->(2) 73(2x+3y)=77(3x+2y), 146x+219y=231x+154y, 65y=85x, 13y=17x. |
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| Chandini said: (Sep 16, 2017) | |
| 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. |
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| Sunil Gampa said: (Sep 19, 2017) | |
| Please explain clearly how to take x y values in an equation? | |
| Vanshika said: (Jun 1, 2018) | |
| How do we get to know that 1st equation will be multiplied 3 and second by 2? | |
| Vanshika said: (Jun 2, 2018) | |
| I can't understand is why the 1st equation is multiplied by 3 and 2nd by 2. | |
| Gajjela said: (Jul 16, 2018) | |
| @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]. |
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| Prince Thakur said: (Sep 17, 2019) | |
| 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. |
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| Jennie said: (Jun 5, 2020) | |
| Why should we multiply the first equation and the second equation with 3and 2? please explain. | |
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