A farmer sold a calf and cow for Rs.760, there by, making a profit 25% on the calf and 10% on the cow. By selling them for Rs. 767.50, he would have realized a profit of 10% on the calf and 25% on the cow. Find the cost of cow.

M4maths Previous Puzzles

13 Aug 2013 Farmer A farmer sold a calf and cow for Rs.760, there by, making a profit 25% on the calf and 10% on the cow. By selling them for Rs. 767.50, he would have realized a profit of 10% on the calf and 25% on the cow. Find the cost of cow.: OPtion
1) 250
2) 270
3) 310
4) 350
5) 360
6) 390
7) 420
8) 450
9) 500
10)None of these; Correct Option: 4 Best Solution (48)

Jigyasa Kumar 11 years AGO

Let C.P. of a calf be Rs 'x' and C.P. of a cow be Rs 'y'.
Now according to the question;
125x/100 + 110y/100 = 760 ..............(1)

110x/100 + 125y/100 = 767.50 ..............(2)

Now by solving (1) and (2), we get,
x = 300 and y = 350
therefore,
the cost of calf is Rs 300.
the cost of cow is Rs 350.

Like? Yes (24) | No (5)

Aarush Singh 11 years AGO

Like? Yes (13) | No (6)

siddhartha 11 years AGO

so accoring to first statement
125% of calf + 110% of cow = 760
110% of calf + 125% of cow = 767.50
so writing eq's

125*calf + 110*cow = 76000
110*calf + 125*cow = 76750

by solving we get calf = 300rs cow = 350rs

Like? Yes (9) | No (2)

Ankit Raj 11 years AGO

Like? Yes (7) | No (2)

NIDHI 11 years AGO

1.25x+1.10y=760;

110x-1.25y=767.50;

so, x=350. cost of cow =350.

Like? Yes (9) | No (4)

Sahil 11 years AGO

answer is 350

let cost of calf is x, cost of cow is y
so according to the question there vl b 2 equations

(x+25% of x) + (y +10% of y)=760
(x+10% of x) + (y +25% of y)=767.50

so
125x+110y=76000
110x+125y=76750

after solving we will get x=300, y=350

Like? Yes (8) | No (3)

MUKESH KUMAR 11 years AGO

1.25x+1.10y=760;

110x-1.25y=767.50;

so, x=350.

Like? Yes (6) | No (3)

Vinay Sipani 11 years AGO

calf -> x , cow -> y

(x + x/4) + (y + y/10) = 760
=> 25x + 22y = 15200

(x + x/10) + (y + y/4) = 767.5
=> 22x + 25y = 15350

=> x = 300,y = 350

Like? Yes (3) | No

ravi singh 11 years AGO

let cost of calf and cow is X and Y

then,
125*X/100 + 110*Y/100 = 760

and

110*X/100 + 125*Y/100 = 767.50

by solving these two equation we get,

X=300 and Y=350

hence answer 350

Like? Yes (3) | No

mandeep 11 years AGO

let x be the price of calf and y be the price of cow.
when he sells calf for 25% profit,it means he sells it for (x+x/4).
and by selling the cow at profit of 10%,he sells it for (y+y/10)
he sells both of them at the price of 760.
so adding above eqtns.
(x+x/4)+(y+y/10)=760
similarly in second case
we get the eqtn.
(x+x/10)+(y+y/4)=767.50
on solving these two equtions,'x'=350 is obtained.

Like? Yes (2) | No

ASHA.S 11 years AGO

let the cost price fo calf be x and cost price be y.
so 25percent profit means selling price 1.25x
so the total selling price 1.25x+1.10y 760.0 and 1.10x+1.25y 767.50
sovving we get y equal to 450

Like? Yes (2) | No

Chandni Maurya 11 years AGO

ans=350 Rs
Let cost of calf and cow are x and y resp.
according to ques,
(x+x*25/100)+(y+y*10/100)=760
=> 5x/4+11y/10=760
=> 25x+22y=15200 ---(1)
and
(x+x*10/100)+(y+y*25/100)=767.50
=> 11x/10+5y/4=767.50
=> 22x+25y=15350 ---(2)
solving equ. (1) & (2).
x=300 (cost of calf)
y=350 (cost of cow)

Like? Yes (2) | No

sridevi 11 years AGO

C.P of calf =x
C.P of cow = y

125x/100 + 110y/100 = 760
110x/100 + 125y/100 = 767.50

solving this we will get

x = 300 and y = 350
The cost of cow is Rs 350.

Like? Yes (2) | No

Anuja Subramanian 11 years AGO

let cp of cow be a and calf be b.
framing eqns,
Eqn 1, 10a+25b=760..
Eqn 2, 25a+10b=767.50..
solving eqns 1 and 2
a=350,b=300..

Like? Yes (2) | No

shamily.k 11 years AGO

Let the cost price of calf and cow be X and Y
Now frame the eqn,
25X+10Y=760
10X+25Y=767.50
Solve the above eqn we get
X=300 Y=350

Like? Yes (2) | No

richa upadhyay 11 years AGO

n(1+25/100)+m(1+10%100)=760
n(1+10%100)+m(1+25%100)=767.50
calculating we get, n=300
m=350

Like? Yes (2) | No

Ashutosh Anshu 11 years AGO

Like? Yes (2) | No

FRANKLIN VIJAY 11 years AGO

125x/100 + 110y/100 = 760
110x/100 + 125y/100 = 767.50
By solving we get
x = 300 and y = 350
Cow cost is 350

Like? Yes (1) | No

Kalidindi Mounika 11 years AGO

Let C.P. of a calf be Rs 'x' and C.P. of a cow be Rs 'y'.
Now according to the question;
125x/100 + 110y/100 = 760 ..............(1)
110x/100 + 125y/100 = 767.50 ..............(2) Now by solving (1) and (2), we get,
x = 300 and y = 350

Like? Yes (1) | No

Bingumalla Abinay 11 years AGO

x be calf and y be cow
1.25x+1.1y=760
1.1x+1.25y=767.5

solving both equations we get y(cow) value =350

Like? Yes (1) | No

kiruthika 11 years AGO

let us suppose that the cost of cow is x and the cost of calf is y.
cost price of cow = selling price+profit
= x+0.25x
= 1.25x
cost price of calf= selling price +profit
= y+0.10y
= 1.10y
so the equations becomes
1.25x+1.10y=760........(1)
1.10x+1.25y=767.50........(2)
solving these we get
x=350,y=300
so x is the cost of cow.then the ans is 350

Like? Yes (1) | No

ritesh kumar pandey 11 years AGO

let cost of calf is x, that of cow is y.
then 1.25x + 1.1y = 760
1.1x +1.25y = 767.5
on solving we get y = 350..
that is the answer
cost of cow is 350

Like? Yes (1) | No

Jasmine Khan 11 years AGO

let x be the the actual cost of the calf and y be the cost of the cow
125% of x+ 110% of y=760
125x=110y=76000.........(1)
again by the second condition
110% of x+ 125% of y=767.5
110x+125y=76750........(2)
now solve 1 and 2 we get
y=350
cost of the cow

Like? Yes (1) | No

Rats 11 years AGO

let cost of calf=x, cost of cow=y
1.25x+1.1y=760.....(1)
1.1x+1.25y=767.50..(2)
on solving (1)&(2)
y(cow)=350

Like? Yes (1) | No

Devendra Marghade 11 years AGO

If cost of calf=x and cow=y
1.25x+1.1y=760 ---(i) &
1.1x+1.25y=767.5 ---(ii)
Solving x=300, y=350

Like? Yes (1) | No

Ch Kumaraswamy 11 years AGO

first time sp=760=(125/100)*(cp of calf)+(110/100)*(cp of cow)
second time sp=765.50=(110/100)*(cp of calf)+(125/100)*(cp of cow)
by taking calf=x1,cow=x2..and solving
x1=x2-50,by sub in first equation we get x2=350..
ie cow cost=350.

Like? Yes (1) | No

SANJEET KUMAR JHA 11 years AGO

Considering the costs of calf and cow 'x '& 'y' respectively and calculating the selling prices by taking 25% & 10% profits,we get one equation in x & y.Again we have second equation from the second condition.Solving both the equation,we get the final answer.

Like? Yes (1) | No

Mayuresh Srivastava 11 years AGO

Let cost price of calf and cow are f and w.
Their first time selling price are f1,w1 and second time f2,w2.
Now f1=f+(f/4) and w1=w+(w/10)
f2=f+(f/10) and w2=w+(w/4)
and also f1+w1=760----1st and f2+w2=767.50----2nd
Put values of f1,f2,w1 and w2 in equation 1st and 2nd and solve.It will give w=350.

Like? Yes (1) | No

Uthra PV 11 years AGO

By solving,
1.25 calf + 1.1 cow = 760
and
1.1 calf +1.25 cow = 767.50

Like? Yes (1) | No

malarvezhi.m 11 years AGO

Let C.P. of aalf be Rs 'x' and C.P. of a cow be Rs 'y'.
Now according to the question;
125x/100 + 110y/100 = 760 ..............(1)

110x/100 + 125y/100 = 767.50 ..............(2)

Now by solving (1) and (2), we get,
x = 300 and y = 350
therefore,
the cost of calf is Rs 300.
the cost of cow is Rs 350.

Like? Yes (1) | No

Balakumaran N.A 11 years AGO

Solution:
Let C.P. of a calf be Rs 'x' and C.P. of a cow be Rs 'y'.
Now according to the question;
125x/100 + 110y/100 = 760 ..............(1)

110x/100 + 125y/100 = 767.50 ..............(2)

Now by solving (1) and (2), we get,
x = 300 and y = 350
therefore,
the cost of calf is Rs 300.
the cost of cow is Rs 350.

Like? Yes (1) | No

Banumathy 11 years AGO

form equations using the statements :
let x -> c.p of calf ; y -> c.p of cow
from 1st statement :
(25/100*x)+x + (10/100*y)+y = 760 ----> eqn.1
from 2nd statement :
(10/100*x)+x + (25/100*y)+y =767.50 ----> eqn.2
simplifying and solving eqn 1 nd 2 we get,
x= 300 and y= 350
hence ans is 4)350.

Like? Yes (1) | No

Tapesh mishra 11 years AGO

Let cost of calf is x and cow is y.
1.25X+1.10Y=760
1.10X+1.25Y=767.5
BY Solving it
Y is 350.

Like? Yes (1) | No

pravallika nathi 11 years AGO

1.25x+1.1y=760
1.1x+1.25y=767.50
solve y for cost of cow

Like? Yes (1) | No

Sanampreet Singh 11 years AGO

Let x and y are the cost of calf and cow repectively.
From first statement it is clear that (x+x*0.25)+(y+y*0.10)=760 i.e. 25x+22y=15200
ANd from second statement it is clear that
(x+x*0.10)+(y+y*0.25)=767.50 i.e. 22x+25y=15350
After solving both equations we will get y=350.

Like? Yes (1) | No

Anil Kumar Bharti 11 years AGO

Exp:
Let C.P. of a calf be Rs 'x' &
C.P. of a cow be Rs 'y'.
Now according to the question;
125x/100 + 110y/100 = 760 ..............(1)

110x/100 + 125y/100 = 767.50 ..............(2)

Now by solving (1) and (2), we get,
x = 300 and y = 350
Hence
the cost of calf is Rs 300.
the cost of cow is Rs 350. Ans

Like? Yes (1) | No

Grishma Shah 11 years AGO

let the cost price of calf be 'x' and that of cow be 'y'.
Eq1 formed: 125x/100 + 110y/100=760
i.e. 125x + 110y = 76000
Eq2 formed: 110x + 125y =76750

solving the 2 equation its will give
x= 300 and y= 350

Like? Yes (1) | No

Balakrishnan M 11 years AGO

CP of Calf = x
CP of Cow= y
x+(25/100)x+y+(10/100)y=760
==> 50x+44y=30400
x+(10/100)x+y+(25/100)y=767.5
44x+50y=30700
solve this eqn,
we get Cost of cow = 350
Cost of calf =300

Like? Yes (1) | No

ShAg 11 years AGO

let original cost of calf = x, and that of cow = y; so our equations are:
25x + 22y = 15200
22x + 25y = 15350

Like? Yes (1) | No

ch n m chaitanya 11 years AGO

760=125calf+110cow
767.5=110calf+125cow

by solving cow=3.5
100*cow=350;

Like? Yes (1) | No

JAGYANDUTTA CHOUDHURY 11 years AGO

Ans:4) 350
1.25calf + 1.1cow = 760--------- i)
1.10calf + 1.25cow= 767.50------ii)

By solving eqn i) and ii) cow = Rs.350

Like? Yes (1) | No

Pargat Pal Singh 11 years AGO

If x is cost of cow and y is cost of calf, then
1.1x+1.25y=760
1.25x+1.1y=767.50

solving these eqns, we get
x=350 Rs

Like? Yes (1) | No

Udhaya kumar 11 years AGO

let x as calf and y as cow
FROM GIVEN DATA
X+(25X/100)+Y+(10Y/100)=760
X+(10X/100)+Y+(25Y/100)=767.5
BY SOLVING THESE
WE GET X=300,Y=350.................
SO ANSWER IS 350.......

Like? Yes (1) | No

Anilkumar 11 years AGO

let cost of calf=x
cost of cow =y
(125/100)*x+(110/100)*y=760--(1)
(110/100)*x+(125/100)*y=767.50--(2)
solving 1 and 2 we get
x=300
y=350

Like? Yes (1) | No

Prem Kumar 11 years AGO

Like? Yes (1) | No

KARTHIGA DEVI.J 11 years AGO

Let cost of calf be x and cow be y.
x+25x/100 + y + 10y/100 =760
==> 125x+110y=76000...........(1)
x+10x/100 + y+ 25y/100 =767.50
==>110x+125y=76750.............(2)
solving 1 and 2,
x=300 and y=350
Hence cost of cow is Rs.350.

Like? Yes (1) | No

KUNAL CHATTOPADHYAY 11 years AGO

solve the equations
(5/4)x+(11/10)y=760
(11/10)x+(5/6)y=767.50
find and solve y
y=350

Like? Yes (1) | No

yogesh 11 years AGO

calf cost x
cow cost y
x(125/100) + y(110/100)= 760.... (1)
1.25x + 1.1y=760

now
x(110/100) + y(125/100)= 767.50
1.1x + 1.25y= 767.50 ........ (2)

(1.25*1.25 - 1.1*1.1)Y= 767.50 * 1.25 - 760 * 1.1
.3525Y = 123.375
y= 123.375/.3525
Y= 350
cost of cow = 350

Like? Yes (2) | No (1)

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Time: 00:01:33

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A farmer sold a calf and cow for Rs.760, there by, making a profit 25% on the calf and 10% on the co

M4maths Previous Puzzles

13 Aug 2013 Farmer