Pocket Money

Systems of linear equations.

Let x be the number of $20 notes and y be the number of $10 notes. Total number of notes: x + y = 50

x number of $20 notes and y number of $10 notes sum up to $800. ($20)x + ($10)y = $800 or simply 20x + 10y = 800.

Solving for x and y, we got x=30 and y=20

So, there are thirty $20 notes and twenty $10 notes.

x = no of 20 $ notes

y = no of 10 $ notes

total notes = x + y = 50

total money = 20 x + 10 y = 800

solve: 20 x + 10 y = 800 and x + y = 50 (or 10x +10y = 500)

x = 30 and y = 20

x + y = 50 10x +20y = 800 solve for x & y to get the values x=20 & y = 30

let x=no. of $10 let y=no. of $20 x+y=50 10x+20y=800 x=20 y=30

solution:

let x & y be no. of $10 and $20 notes respectively

1. x + y = 50 notes ; x = 50 - y

2. $10x + $20y = $800

3. 10( 50 - y ) + 20y = 800 ; one variable equation

4. 500 - 10y + 20y = 800 ; distributive property

5. 10y = 300 ; y = 30 ;simplification

6. x = 50 - y = 50 - 30 = 20; substitution

7. therefore there are 20-$10 and 30-$20 ..........check

x+y=50 and 20x+10y=800 solving both equations we get x=30 & y=20

Let:

$10\$$ notes $= x$ notes

$20\$$ notes $= (50-x)$ notes

Total amount $=10x+20(50-x)=800$

$10x+1000-20x=800$

$-10x=800-1000$

$-10x=-200$

$x=20$

Thus, the answer is:

$10\$$ notes $= x=\boxed{20}$ notes

$20\$$ notes $= (50-x)=50-20=\boxed{30}$ notes

Let the no. of $20 notes be 'x' Therefore, the no. of $10 notes= (50-x)

According to question, 20 x + 10 (50-x)=800 => 20 x + 500 - 10 x = 800 => 20 x - 10 x = 800 - 500 => 10*x = 300 => x = 300/10 => x = 30

So, we get:- No. of $20 notes = 30 No. of $10 notes = 50-30 = 20

Hence, he has thirty $20 notes and twenty $10 notes.

use hit and trial as options are given

option 1 is correct Bcz 20 *10=200 , 20 * 30=600 , 600+200=800

20 notes of 10$ and 30 notes of $20

20 $10=$200 30 $20=$600 $200+$600=$800

Check options, 1st is almost Wrong ( 30$ note ) . ...... keep add all options , One of them makes 800$ total . That's right answer ! :)

Multiply number of notes and the values.Then add them.

you may solve it by checking from the choices or solve it using simultaneous equations. x for ten dollar notes, y for 20 dollar note. we have: x + y = 50 and 10x + 20y = 800....x = 20 (ten dollar notes), y = 30 (twenty dollar notes)

Let number of $20 notes be x Therefore, number of $10 notes= 50-x (since total number of notes= 50) Value of these notes= $800 Therefore, 20(x) + 10(50 - x) = 800 20x + 500 - 10x = 800 x(20-10) =800 -500 10x = 300 x = 30 therefore, No. of $20 notes = x = 30 No. of $10 notes = 50-x = 20

x = no of 20 $ notes

y = no of 10 $ notes

total notes = x + y = 50

total money = 20x + 10y = 800

solve: 20x + 10y = 800 and x + y = 50 (or 10x +10y = 500)

x = 30 and y = 20

let x be $10 notes let y be $20 notes therefore x + y =50 (1) 10x + 20y =800 (2) in 1: x=50 - y substitute 50 - y for x in 2 10(50-y) + 20y =800 500 - 10y +20y =800 10y=800- 500 10y=300 y=30. substituting 30 for y in 1 x+30= 50 x=50-30 x=20 Hence 20 $10 notes and 30 $20 notes

its very simple [30$X20=600$] and [20$X10=200$] [600$+200$=800$]

just multiply the two given notes with their corresponding number beside them and after doing that add their sum if u can get $800...then if u got $800 then its solve

Systems of linear equations.

Let x be the number of $20 notes and y be the number of $10 notes. Total number of notes: x + y = 50

x number of $20 notes and y number of $10 notes sum up to $800. ($20)x + ($10)y = $800 or simply 20x + 10y = 800.

Solving for x and y, we got x=30 and y=20

So, there are thirty $20 notes and twenty $10 notes. you can check the option also..:)

20 multiply with 30 we get 600 dollars and 20 multiply with 10 we get 200 600+200=800

20x+10*(50-x)=800 ... 20x+500-10x=800 10x=300 x=30

systems of linear equations let x be the number of $20 notes and y be the number of $10 notes hence total no.of notes: x + y = 50 as we know that x represents the no.of $20 notes and y represents no.of $10 notes and their sum is equal to $800 therefore 20x + 10y = 800 now by solving, we got x=30 & y=20 so there are 30 $20 notes and 20 $10 notes

30 20=20 10 600+200=800

Let the no. of 20$ notes be x and the no. of 10$ notes be y . It is said that he has saved 800$ therefore we can form the equation 20x + 10y =800. The total no. of notes is 50 therefore the equation x + y = 50 is formed . Solving the two equations we get x=30 and y=20.

Solve these two linear equations: 20x+10y=800 and x+y=50, where x=the number of $20 notes and y=the number of $10 notes

Let the number of $10 notes be x. Therefore no. of 20$ notes =(50-X) ATP $10xX+$20(50-X)=$800 10X+1000-20X=800 10X=200 X=20 Therefore, $10 notes=20 $20 notes=30

Let x be the no. of $10 bills Let y be the no. of $20 bills

Total Money equal to: 10x +20y = 800 Total no. of bills: x + y = 50 x = 50 -y

10(50 - y) + 20y =800 500 - 10y + 20y = 800 10y + 500 = 800 10y = 300 y = 30 (no. of $20 bills) 50 - 30 = 20 ( no. of $10 bills)

It was easy to solve :) just make two variables and make a equation...and you will get the answer!

20 $10=$200 & 30 $20=$600 => $200+$600=$200

x+y=50 20x+10y=800+ -10x-10y=-500/10x=300/x=30number of $20 notes/y=50-30=20number of $10 notes

10 * 20 = 200 20 * 30 = 600 so 200 + 600 = 800

20x+10y=800 and x+y=50 solve these equations we find x=30 & y= 20

20x+10y=800 x+y=50 By solving above equation for x & y, we get y=20 & x=30

20x + 10y = 800 x + y = 50

Solving these two equations we get , there are 30 notes of 20$ and 20 notes of 10$

Let x notes of 20 $ and y notes of 10 $ therefore 20X x + 10Xy =800 ......(1) and x + y =50 ........(2) multiply (2) by 10 to eliminate y by subtracting (3) from (10 10 X x + 10 Xy = 500 , thus 10 x = 300 ,so x ==30 and y =20
there are 20 notes of 10 $ and 30 notes of 20 $ Ans

K.K.GARG,India