Question

5 pencils and 7 pens together cost rs 50, whereas 7 pencils and 5 pens together cost rs 46. Find the cost of one pencil and that of one pen

Answer 1

Answer:

let the no of pencils be x and no of pens be y

so according to question in first case

5x+7y=50

and in second case

7x+5y=46

∴7(5x+7y)=50*7

5(7x+5y)=46*5

35x+49y=350  (1)

35x+25y=230  (2)

(-)     (-)      (-)

24y=120

y=120/24

y=5

put y=5 in equation 2

35x+25*5=230

35x+125=230

35x=230-125

35x=105

x=105/35

x=3

value of one pen is Rs 5 and value of one pencil is Rs 3

Step-by-step explanation:

Answer 2

Answer:

Answer:-

$\red{\bigstar}$ Cost of 1 pencil

$\large\leadsto\boxed{\sf\purple{Rs. \: 3}}$

$\red{\bigstar}$ Cost of 1 pen

$\large\leadsto\boxed{\sf\purple{Rs. \: 5}}$

• Given:-

Cost of 5 pencils and 7 pens = Rs. 50

Cost of 7 pencils and 5 pens = Rs. 46

• To Find:-

Cost of 1 pencil and 1 pen = ?

• Solution:-

Let the price of pencil be 'x' and the price of pen be 'y'.

According to the question:-

→ $\sf{5x + 7y = 50} \: \: \: \longrightarrow\bf\red{[eqn.i]}$

and

→ $\sf{7x + 5y = 46} \: \: \: \longrightarrow\bf\red{[eqn.ii]}$

Multiplying eqn[i] by 5:-

→ $\sf{(5x + 7y = 50) \times 5}$

→ $\sf{25x + 35y = 250}\: \: \: \longrightarrow\bf\red{[eqn.iii]}$

Multiplying eqn [ii] by 7:-

→ $\sf{(7x + 5y = 46) \times 7}$

→ $\sf{49x + 35y = 322}\: \: \: \longrightarrow\bf\red{[eqn.iv]}$

Subtracting equation [iii] from [iv]:-

→ $\sf{(49x + 35y) - (25x + 35y) = 322 - 250}$

→ $\sf{49x + 35y - 25x - 35y = 72}$

→ $\sf{24x = 72}$

→ $\sf{x = \dfrac{72}{24}}$

→ $\boxed{\bf\green{x = 3}}$

Substituting value of x in [eqn.i]:-

→ $\sf{5x + 7y = 50}$

→ $\sf{5 × 3 + 7y = 50}$

→ $\sf{15 + 7y = 50}$

→ $\sf{7y = 50 - 15}$

→ $\sf{7y = 35}$

→ $\sf{y = \dfrac{35}{7}}$

→ $\boxed{\bf\green{y = 5}}$

Therefore,

The cost of 1 pencil is Rs. 3

The cost of 1 pen is Rs. 5