Discussion :: Arithmetic Reasoning - Section 1 (Q.No.5)
5B = 125 | Ashutosh said: (Feb 18, 2012) | |
| Explain me the last 2 lines. | |
| Archana said: (Apr 6, 2012) | |
| B=C+2=>C=B-2 2B=D+E=>D=2B-E (B+10)+B+(B-2)+2B-E+E=133 B+10+B+B-2+2B=133 5B+8=133 5B=133-8 B=125/5 B=25 |
|
| Archana Santhosh said: (Mar 4, 2013) | |
| A+B+C+D+E=133 (B+10)+B+(B-2)+2B=133 5B+8(i.e 10-2)=133 5B=133-8 5B=125 B=125/5 B=25. |
|
| Marcos said: (Aug 15, 2013) | |
| How do you get from, B-3 = E. B+3 = D. to 2B = D+E. Do you just add them together? |
|
| Bidya said: (Jan 28, 2014) | |
| What was the day on 100 November 1581 please give me solve? | |
| Nikolas said: (Apr 21, 2014) | |
| We have, B - 3 = E . B + 3 = D. B=C+2=>C=B-2. Then, A+B = D + E+10. Which we can be done into, A+B = B-3 + B+3 + 10. A+B= 2B + 10. Then lets add all. A+B+C+D+E=133. 2B+10+B-2+B+3+B-3=133. 5B=133-8. B=125/5. B=25 -> we get the answer. |
|
| Neha said: (Jan 15, 2015) | |
| A+B+C+D+E = 133. B+3 = D(1). B-3 = E(2). B = C+2, C = B-2(3). A+B = D+E+10 = (B+3)+(B-3)+10(4). From 1, 2, 3 & 4. We get... B+3+B-3+10+B-2+B+3+B-3 = 133. 5B+8 = 133. 5B = 125. B = 25. |
|
| Raheem said: (Apr 14, 2016) | |
| n = 96. => 4n-4 = 96. => 4(n-1) = 96. => n-1 = 96/4. => n-1 = 24. => n = 24+1. => n = 25. |
|
| Lavanya said: (Aug 24, 2016) | |
| Please explain this clearly. | |
| Shudhi said: (Sep 30, 2016) | |
| How A + B = D + E + 10? It must be D + E = A + B + 10. |
|
| Pavan Pj said: (Dec 27, 2016) | |
| Nice answer @Neha. | |
| Manimegalai said: (Jan 5, 2017) | |
| Not getting this, Please explain step by step clearly. | |
| Saradha said: (Mar 28, 2017) | |
| I'm not getting this. Please explain me with simple steps. | |
| Siva said: (May 22, 2017) | |
| Total no of cards=133. i.e A+B+C+D+E=133 ---> (1) Then B-3=E ---> (2) B+3=D ---> (3) B=2+C ie C=B-2 ---> (4) A+B=10+D+E ---> (5) Now, we know the values of E and D ie eq(2) and eq(3) replace those values in eqn(4) A+B = 10+B+3+B-3, A+B = 10+2B, A = 10+2B-B, A = B+10. Now we know all values except B. Replace the all Known values in eq(1). then B+10+B+B-2+B+3+B-3=133 5B+8 =133, 5B = 133-8, 5B = 125, B = 125/5, B = 25. |
|
| Soma Datta said: (Jan 19, 2018) | |
| Total no of cards=133. i.e A+B+C+D+E=133 ---> (1) Then B-3=E ---> (2) B+3=D ---> (3) B=2+C ie C=B-2 ---> (4) A+B=10+D+E ---> (5) Now, we know the values of E and D ie eq(2) and eq(3) replace those values in eqn(4) A+B = 10+B+3+B-3, A+B = 10+2B, A = 10+2B-B, A = B+10. Now we know all values except B. Replace the all Known values in eq(1). then B+10+B+B-2+B+3+B-3=133, 5B+8 =133, 5B = 133-8, 5B = 125, B = 125/5, B = 25. |
|
| Gajjela said: (Jul 19, 2018) | |
| @All. 1. A says to B "If you give me 3 cards you will have as many as E has means, B has more cards than E. If you add 3 cards to E, then he equals to B. So, B=E+3. 2. A says to B "If I give you 3 cards you will have as many as D has means, D has more cards than B. So, D=B+3. 3. From 1 and 2, you will get ( D+E)=2B. 4. A and B together have 10 cards more than what D and E have (A+B) have more cards than (D+E). So,( A+B)=(D+E)+10=2B+10 (from point 3). 5. B has two cards more than what C has. So, B=C+2 => C=B-2. 6. Total no.of cards be 133. then, (A+B)+(C)+(D+E)=133, (2B+10)+(B-2)+(2B)=133, (5B)=125, B=25. So, the Answer is option [C]. |
|
| Sonia said: (Aug 6, 2018) | |
| Thanks @Gajjela. | |
| Ashutosh Priyadarshan said: (Nov 4, 2020) | |
| B = x. A + B = 2x + 10, A = x + 10. E = x - 3. D = x + 3. C = x - 2. So, x + 10 + x + x - 2 + x + 3 + x - 3 = 133. => 5x = 125. => x = 25. |
|
Post your comments here:
Name *:
Email : (optional)
» Your comments will be displayed only after manual approval.