Total cost of 45 books and 65 copies is 1290 and that of 65 books and 45 copies is 1500. Cost of 1 b

poojasaisri   9 years AGO

45b+65c=1290 eqn 1
65b+45c=1500 eqn 2
eqn 1+eqn 2 =>
110b+110c=2790 =>b+c=25.35 eqn3
eqn 1-eqn 2 =>
20b-20c=210 =>b-c=10.5 eqn4
eqn3+eqn4
2b=35.85 =>b=17.93

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Shiva Teja   9 years AGO

cost of 45 books and 65 copies is 1290
(45*B)+(65*C)=1290
cost of 65 books and 45 copies is 1500
(65*C)+(45*B)=1500

Buy solving the two equations we get,
Price of one book=17.931

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ravi singh   9 years AGO

Answer is 17.93

Let, represent Book=B
Copy=C
Then,
45*B + 65*C = 1290______eq(1)

And
65*B + 45*C = 1500_____eq(2)
From eq(1) & eq(2)

B=17.93

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chepuru sai priyanka   9 years AGO

let book=X and copies=Y then,
45X+65Y=1290
65X+45Y=1500
if u solve X=17.93
y=7.43
so 1 book will be 17.93

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MRITUNJAY   9 years AGO

45B+65C=1290
65B+45C=1500
after solving this equation b=17.93 and c=7.43
so answer will be 17.93
option 4.

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Surit Datta   9 years AGO

Solving as linear equation we get
PRICE OF BOOK IS 17.93 Rupee.

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Pranati Kundu   9 years AGO

let cost of 1 book be x and cost of 1 copy be y
45x+65y=1290
65x+45y=1500
solving this we get x=17.93

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Ayush Gupta   9 years AGO

45x+65y=1290
65x+45y=1500
Solve these equations and get
x=17.93(price of 1 book)
y=7.43 (price of 1 copy)

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RAKESH   9 years AGO

1/2 * [(1290+1500)/ (45+65) + (1500-1290)/(65-45)] = 17.93

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Vijendra singh   9 years AGO

let x as a book and y as a copy
45x +65y=1290
65x +45y=1500
then x=17.93 ans

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THUMMALA VINEESHA   9 years AGO

45x+65y=1290
65x+45y=1500
by solving these two equations we will get x=17.931 that is the value of 1 book

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umesh   9 years AGO

45b+65c=1290......(1)*(9/13)
65b+45c=1500......(2)

solving the eqns we get 17.93

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Karthick Raja   9 years AGO

45B+65C=1290 -----> EQN1
65B+45C=1500 ----->EQN2

(EQN1 x45) - (EQN2 x65)

2200B=39450
B=39450/2200
B=17.93

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Venkatesh kadali   9 years AGO

45b+65c=1290
65b+45c=1500
solve this two equation now we get one book is=17.93

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Prayani Singh   9 years AGO

45b+65c=1290
65b+45c=1500
solving above 2 equations:
20b-20c=210
b-c=10.5
replace c=B-10.5 in equation 2:
110b-472.5=1500
b=17.926

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sompon temtheerakij   9 years AGO

let books = A , copies = B
45A+65B = 1290
65A+45B = 1500
by solve these two equations
A = 17.93

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ANANTHI.P   9 years AGO

45B+65C=1290
65B+45C=1500
ON SOLVING C=7.43
SUBSTITUTE IN ANY OF THE 2 EQUATIONS;
B=17.93

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sravani   9 years AGO

45b+65c=1290
65b+45c=1500

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Meenakshi   9 years AGO

45B+65C=1290-->(1)
65B+45C=1500--(2)
C=1290-45B/65
substitute C value in equation (2)
65B+45(1290-45B/65)=1500
845B+11610-405B=19500
440B=19500-11610
440B=7890
B=7890/440
B=17.9318
cost of one book=17.93

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Radhika   9 years AGO

Hit and Trial Method
We'll choose from options to calculate the cost of copies first i.e. when we take the cost of book as 17.93 than we get the cost of copies as 7.43
1290-45*17.93=65*cost of copies
from this we get the cost of copies than put it in the 2nd equation if it comes 1500 than our ans is correct.

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lovely verma   9 years AGO

45x+65y=1290...............(1)
65x+45y=1500.................(2)
now we multiply by 9 in !st Equation And 13 in second equation
+405x+585y=11610.........(3)
+845x+585y=19500.........(4)
(-) (-) (-)
-------------------------------------------
After Subtraction we get:-
-440x=-7890
x=789/44
x=17.925
x=17.93 Ans

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pragya sahu   9 years AGO

let b=book
c=copy
45b+65c=1290.........1st
65b+45c=1500.........2nd
so,
c=7.4318182
when put in 1st eq
we get the value of b=17.93

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Ankita Anand   9 years AGO

let teh cost of a book is Rs x and of a copy is Rs y.
now 45x+65y=1290--(i)
and 65x+45y=1500--(ii)
now multiplying i by 45 and multiplying ii by 65 and subtracting i from ii we get;
2200x=39450
so x=17.93

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Divyabala R   9 years AGO

Simple Equation type question consider Books as X,Copies as Y which gives 45X + 65Y = 1290,
65X + 45Y = 1500

on solving this u l get X = 17.93

which is the answer. :)

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FOLSHIYA SATHYARAJ   9 years AGO

Divide the numbers b

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SANTOSH   9 years AGO

45b+65c=1290
65b+45c=1500
-------------------------
b=17.93

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SAKTHEESVARI S   9 years AGO

we get 2 eqn,45x+65y=1290;
65x+45y=1500;
solving these two, we get value of x (i.e) cost of 1 book=17.93 Rs

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Dharmveer Prajapat   9 years AGO

45x + 65 y =1290
65 x + 45 y = 1500
solve this you will get x=17.93

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Preethi   9 years AGO

books 17.93*45=806.85 copies 17.93 *65=483.15

books 17.93*65=1165.45 copies 17.93 *45=334.55.

each book= 17.93
copies = 7.43

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Penchal kumar   9 years AGO

45*x+65*y=1290
65*x+45*y=1500

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vivek   9 years AGO

45b +65c =1290..........(i)
65b +45c =1500..........(ii)
eq (i)+(ii),
b+ c= 25.36
eq (ii)- (i),
b- c=10.5
so, b = 17.93

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karthikeya varma   9 years AGO

45B+65C=1290-------(1)
65B+45C=1500-------(2)
-------------------------------------
SOLVE THE GIVEN EQUATIONS YO WILL GET
B=17.93
C=7.43
SO HERE AFTER THE COST OF SINGLE BOOK IS B=17.93

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g.soujanya   9 years AGO

By solving the two equations we can get the answer as 17.93

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Kavitha Lakshmi A N V   9 years AGO

boks are more than copies then price is more.

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T LAVANYA   9 years AGO

45b+65c=1290
65b+45c=1500
c=(1290-45b)/65
then
65b+45(1290-45b)/65=1500
b=17.93

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Rohit Gurjar   9 years AGO

45b+65c=1290
65b=45c=1500

solve this Equation Book
b=17.93

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DHANIKONDA RATNA BHAVANI   9 years AGO

Let X be Books
Y be Copies
(45X+65Y=1290 ) * 65
(65X+45Y=1500 ) * 45
2925x+4225y-2925x-2025y=83850-67500
2200y=16350
y(copies)=7.431818
45x=1290-65(7.431818)
45x=1290-806.93183

x(books)=17.93181844

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deep   9 years AGO

45x+65y=1290.....(1)
65x+45y=1500...(2)
solve these equation
n we get x=17.9

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padam   9 years AGO

make the equation and solve for x and y book price is 17.93

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Udhaya kumar   9 years AGO

let cost of book and copy. as x and y
then by given conditions
45x+65y = 1290
65x+45y = 1500
by solving x = 17.93...

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Payal Garg   9 years AGO

45b+65c =1290
65b +45c=1500

Solving above two , we get c=7.4 and b= 17.9

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sai   9 years AGO

45B + 65C=1290
65B+45C=1500
subtract we get 20B-20C=210
b-c = 10.5
now solve any two with above
45b+65(b-10.5)= 1290
1 book cost = 17.93

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prashanth   9 years AGO

117x+81(7.43)=2700
x=2700-81(7.43)/117=17.93

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Rahul Lone   9 years AGO

45books + 65 copies=1290
65 books+ 45 copies=1500
by solving
cost of 1 books =17.93

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Aashish gohel   9 years AGO

make equation such that
x= cost of books
y= cost of copies
45x +65y= 1290
65x+45y=1500
solve this both
you will get x=17.93
that is cost of book

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Rajesh Kumaresan   9 years AGO

45X+65Y=1290
65X=45y=1500
solving these 2 equations we get X=17.93

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T.Aravinth kumar   9 years AGO

45x+65y=1290
65x+45y=1500
solve
x=17.93

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Ram   9 years AGO

45x+65y=1290; 65x+45y=1500
By simplying we get x=17.93

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P R KANAKA MAHA LAXMI   9 years AGO

let us consider x is the cost of one book and y is the cost of one copy
45x+65y=1290
65x+45y=13500
by solving we get x=17.93

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Piyush Kumar   9 years AGO

45b + 65c = 1290 ---- (1)
.'. 9b + 13c = 258 ---- (2)
65b + 45c = 1500 ---- (3)
Adding 1 and 3 we get,
11b + 11c = 279 ---- (4)

Now, (4) - (2)
2b - 2c = 21 --- (5)

From (4) we get:
2b + 2c = 558/11 ----(6)

from (5) and (6) we get
b = 789/44 = 17.93 ----(Ans)

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