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Try this beautiful Problem from Algebra based on Pen & Note Books from PRMO 2019.

A pen costs Rs.$13$ and a notebook costs Rs.$17$. A school spends exactly Rs.$10000$ in the year $2017 - 18$ to buy $x$ pens and $y$ notebooks such that $x$ and $y$ are as close as possible (i.e. $|x-y|$ is minimum). Next year, in $2018-19,$ the school spends a little more than Rs.$10000$ and buys $y$ pens and $x$ notebooks. How much more did the school pay?

- $9$
- $40$
- $34$

Algebra

Equation

multiplication

But try the problem first...

Answer:$40$

Source

Suggested Reading

PRMO-2019, Problem 16

Pre College Mathematics

First hint

Given A pen costs Rs.\(13\) and a note book costs Rs.\(17\)

Nunber of pens \(x\) and numbers of notebooks \(y\)

According to question

\(13 x+17 y=10,000\) ..........................(1)

\(17 x+13 y=10000+P\)......................(2)

Where $\mathrm{P}$ is little more amount spend by school in $2018-19$.we have to find out the value of \(P\).Can you find out?

Can you now finish the problem ..........

Second Hint

Adding (1) & (2)

$x+y=\frac{20,000+P}{30}$

Subtract (1) form (2)

$x-y=\frac{P}{4}$

Let $P=4 a$

$\begin{aligned} \text { So, } x+y=& \frac{20,000+4 a}{30} \ &=666+\frac{10+2 a}{15} \end{aligned}$

Therefore \(a\)=\(10\)

Can you finish the problem........

Final Step

Since $P=4 a$ \(\Rightarrow P=40\)

So school had to pay Rs. $40$ extra in $2018 - 2019$

- https://www.cheenta.com/ordered-pairs-prmo-2019-problem-18/
- https://www.youtube.com/watch?v=gGT15ls_brU

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