Question

A farmer bought a cow and a calf together for Rs.14,000. After some time, he sold the
cow at a profit of 20% and the calf at a profit of 13%. On the whole he got a profit of
17%. Find the cost price of the cow and the calf.please give clear image and good handwritted letter which can understanded​

Answer 1

$\sf\small\underline\red{Let:-}$

$\tt{\implies Cost\: price\:_{(cow)}=x}$

$\tt{\implies Cost\:price\:_{(calf)}=14000-x}$

$\sf\small\underline\red{Given:-}$

$\sf{\implies Cost\: price\:_{(cow+Calf)}=Rs.14000}$

$\sf{\implies Profit\:on\:cow=20\%}$

$\sf{\implies Profit\:on\:calf=13\%}$

$\sf{\implies whole\: profit\:on\:sale=17\%}$

$\sf\small\underline\red{To\:Find:-}$

$\tt{\implies Cost\: price\:_{(cow)}=?}$

$\tt{\implies Cost\:price\:_{(calf)}=?}$

$\sf\small\underline\red{Solution:-}$

To calculate cost price of cow and calf ,at first we have to find out the selling price of cow and calf separately. Then calculate the cost price of cow and calf. Simply by using formula to calculate sp.

$\sf\small\underline\red{Formula\:used:-}$

$\tt{\implies SP=\dfrac{100+G\%}{100}\times\:CP}$

$\tt{\implies SP\:_{(cow)}=\dfrac{100+20}{100}\times\:x}$

$\tt{\implies SP\:_{(cow)}=\dfrac{120}{100}\times\:x}$

$\tt{\implies SP\:_{(cow)}=\dfrac{120x}{100}}$

$\tt{\implies SP\:_{(calf)}=\dfrac{100+13}{100}\times\:(14000-x)}$

$\tt{\implies SP\:_{(calf)}=\dfrac{113}{100}\times\:(14000-x)}$

$\tt{\implies SP\:_{(calf)}=\dfrac{1582000-113x}{100}}$

• Now calculate total CP and SP :-]

$\tt{\implies SP\:_{(Cow)}+SP\:_{(calf)}}$

$\tt{\implies \dfrac{120x}{100}+\dfrac{1582000-113x}{100}}$

$\tt{\implies \dfrac{120x+1582000-113x}{100}}$

$\tt{\implies \dfrac{7x+1582000}{100}}$

• Here total cp of cow and calf:-]

$\tt{\implies CP\:_{(cow)}+CP\:_{(calf)}=14000}$

• Calculate cp here for cow and calf:-]

$\tt{\implies Total\:_{(SP)}=\dfrac{100+G\%}{100}\times\: Total\:_{(CP)}}$

$\tt{\implies \dfrac{7x+1582000}{100}=\dfrac{100+17}{100}\times\:14000}$

$\tt{\implies \dfrac{7x+1582000}{100}=\dfrac{117}{100}\times\:14000}$

$\tt{\implies 7x+1582000=117\times\:14000}$

$\tt{\implies 7x+1582000=1638000}$

$\tt{\implies 7x=1638000-1582000}$

$\tt{\implies 7x=56000}$

$\tt{\implies x=8000}$

$\sf\large{Hence'}$

$\tt{\implies Cost\: price\:_{(cow)}=x}$

$\bf{\implies Cost\: price\:_{(cow)}=Rs.8000}$

$\tt{\implies Cost\: price\:_{(calf)}=14000-x}$

$\tt{\implies Cost\: price\:_{(calf)}=14000-8000}$

$\bf{\implies Cost\: price\:_{(calf)}=Rs.6000}$

Answer 2

${ \large{ \red{ \underline{ \pmb{ \frak{Givey...}}}}}}$

➼ A farmer bought a cow and a calf together for Rs.14,000.

➼ he sold the cow at a profit of 20% and the calf at a profit of 13%

➼ On the whole he got a profit of 17%.

${ \large{ \red{ \underline{ \pmb{ \frak{To \: find... }}}}}}$

➼ the cost price of the cow and the calf for which the farmer bought ( Intial pay )

${ \large{ \red{ \underline{ \pmb{ \frak{ Full \: solution...}}}}}}$

➱ Here,

• We have given the total C.P of the calf and cow, and the total amount of profit gained,

Now,

• ↠ Let the C.P of the cow be x
• ↠ The C.P of the calf be y

According to the first condition,

• ★ x + y = Rs.14,000 ----- ( equation 1 )

As per condition 2,

↝ he sold the cow at a profit of 20%

↠ So,

• S.P of cow will be : -

$\longrightarrow \tt \: s.p = x + \frac{20}{100} \times x$

↝ And then, he sold the calf for the profit 13%

↠So,

• S.P of calf will be : -

$\longrightarrow \tt \: s.p = y + \frac{13}{100} \times y$

➱ As said ,

• ↠On the whole he got a profit of 17%

So,

• S.P of the cow and the calf will be,

$\longrightarrow \tt \: s.p =( x + \frac{20}{100} \times x) + (y + \frac{13}{100} \times y) = (14000 + \frac{17}{100} \times 14000) \\ \\ \\ \longrightarrow \tt \: s.p = \frac{100x + 20x}{100} + \frac{100y + 13y}{100} = \frac{1400000 + 2380000}{100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \tt \: s.p = \frac{120x}{100} + \frac{113x}{100} = 16380 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \tt \: s.p = 1.2x + 1.3y = 16380 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• 1.2x + 1.13y = 16380 ----- ( equation ii )

➱Now,

• Let's Multiply equation 1 by 1.2 and subtract it from equation 2

↠Equation 1

$\longrightarrow \tt \:x + y = 14000$

Let's multiply it now,

$\longrightarrow \tt \:( x + y = 1400) \times 1.2 \\ \\ \\ \longrightarrow \tt 1.2x + 1.y = 16800 \: \: \:$

Now let's subtract it from equation 2

$\tt 1.2x + 1.13y = 16380 \\ \\ \tt- 1.2x - 1.2y = 16800$

• After subtracting we get,

$\longrightarrow \tt - 0.07y = - 420 \\ \\ \\ \longrightarrow \tt y = \frac{ - 420}{ - 0.07} \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \tt y = { \boxed{ \tt{6000}}} \: \: \: \: \: \: \: \: \: \: \: \:$

• Hence, the cost price of the equals 6000

➼ Now, let's substitute the value of y in equation 1

$\longrightarrow \tt x + y = 14000 \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \tt x + 6000 = 14000 \\ \\ \\ \longrightarrow \tt x = 14000 - 6000 \\ \\ \\ \longrightarrow \tt x = { \boxed{ \tt{8000}}} \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Hence, the cost price of the cow equals Rs. 8,000

${ \large{ \red{ \underline{ \pmb{ \frak{Therefore... }}}}}}$

➼ The cost price of the cow are calf are $\pmb{6,000}$and $\pmb{ 8,000}$ respectively

${ \large{ \red{ \underline{ \pmb{ \frak{Additional \: Info... }}}}}}$

Formulae : -

$\begin{gathered}\begin{gathered}\small\boxed{\begin{array}{cc}\large\sf\dag \: {\underline{More \: Formulae}} \\ \\ \bigstar \: \sf{Gain = S.P – C.P} \\ \\ \bigstar \:\sf{Loss = C.P – S.P} \\ \\ \bigstar \: \sf{Gain \: \% = \Bigg( \dfrac{Gain}{C.P} \times 100 \Bigg)\%} \\ \\ \bigstar \: \sf{loss \: \% = \Bigg( \dfrac{loss}{C.P} \times 100 \Bigg)\%} \\ \\ \bigstar \: \sf{S.P = \dfrac{100+Gain\%}{100} \times C.P} \\ \\ \bigstar \: \sf{C.P =\dfrac{100}{100+Gain\%} \times S.P} \\ \\ \bigstar \: \sf{C.P =\dfrac{100}{100-loss\%} \times S.P}\end{array}}\end{gathered}\end{gathered}$