### Discussion :: Arithmetic Reasoning - Section 1 (Q.No.11)

Dasharath said: (Jan 7, 2012) | |

I didn't get this point explain be clearly. |

Sujit Chowra said: (Jul 30, 2013) | |

Given A= B+3, and C = A+3. C = B+3+3, C= B+6 here B = D So, 6 years older C than D. |

Sanjay said: (Jun 7, 2016) | |

I am not understanding this clearly. |

What said: (Jun 27, 2016) | |

Not getting this. Can anyone help me to get it? |

Gtaja said: (Jul 10, 2016) | |

I couldn't get this. Help me to understand this. |

Chris said: (Dec 2, 2016) | |

Easiest way for me to solve this was by doing the following: A - 3 = B and A + 3 = C, B = D. Let A = 14 14 - 3 = B, so B = 11, therefore D = 11 as B = D 14 + 3 = C, so C = 17. 17(C) - 11(D) = 6, therefore, C is 6 years older than D. |

Srikanth said: (Jan 22, 2017) | |

I didn't get it. Someone help me by the explanation. |

Jamel Telesford said: (Jun 19, 2017) | |

Is there a way for you to do this with only one variable? |

Jayanthi said: (Mar 20, 2018) | |

A=B+3........(1) A=B-3..........(2) B=D...............(3) B+3=B-3, B=-3-3, B=6, 6=D, both D and C are 6. |

Avani said: (Dec 15, 2019) | |

Assume a random number for A. Let it be A = 10. Then B will be 7 (A is 3 years older than B is given in question). Then C will be 13 (A is younger than C by 3 Years is also given). Since B = D. Find different between B and C. ie, 13-7 = 6. Hence we get the diff of C and D = 6. |

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