Aptitude Discussion

Q. |
21 pencils and 29 pens cost Rs 79. But if the number of pencils and pens were interchanged, the cost would have reduced by Rs 8. Find the cost of each pen. |

✖ A. |
Re 1 |

✔ B. |
Re 2 |

✖ C. |
Re 3 |

✖ D. |
Re 4 |

**Solution:**

Option(**B**) is correct

Let the cost of each pencil be Rs $x$ and that of each pen be $y$.

\(21x+29y=79\)-------- (i)

\(29x+21y=71\)-------- (iI)

On solving both equations we get:

$y=2$

**Edit:** For detailed steps needed to solve the equations, check comment by **Deepak.**

**Ashu**

*()
*

Multiply equation (1) by 29 and equation (2) by 21, we get,

$21*29x+29*29y=79*29$ -------- (i)

$29*21x+21*21y=71*21$ -------- (iI)

Now, (i)-(ii),

$y(29^2-21^2)=79*29-71*21$

$\Rightarrow y =\dfrac{79*29-71*21}{29^2-21^2}$

You can now directly solve this and this will give you $y=2$.

But to make make calculations easier let's break it up and solve further.

$\Rightarrow y =\dfrac{[(71+8)*(21+8)]-71*21}{(29-21)(29+21)}$

$\Rightarrow y =\dfrac{[71*21+71*8+8*21+8*8]-71*21}{(8)(50)}$

$\Rightarrow y =\dfrac{71*8+8*21+8*8}{(8)(50)}$

$\Rightarrow y =\dfrac{8*(71+21+8)}{(8)(50)}$

$\Rightarrow y =\dfrac{71+21+8}{50}$

$\Rightarrow y =\dfrac{100}{50}$

$\Rightarrow y =2$

I don't know how to solve these equations.

Can anyone help me?