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In a certain beanbag game, players toss beanbags at a board with a hol

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Bunuel
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In a certain beanbag game, players toss beanbags at a board with a hol [#permalink] New post   18 Mar 2020, 08:59
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In a certain beanbag game, players toss beanbags at a board with a hole in it. If the beanbag falls into the hole, the player scores 5 points. A beanbag that lands on the board but misses the hole is worth 2 points. If a toss misses the board entirely, 1 point is deducted from the player’s score. Julianna had 20 tosses and missed the board with 3 throws. Her total score was 52. How many times did she toss the beanbag into the hole?

A. 5
B. 6
C. 7
D. 8
E. 9
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C
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Re: In a certain beanbag game, players toss beanbags at a board with a hol [#permalink] New post   18 Mar 2020, 09:27
OA is B
HE SCORED 52
And tossed 20 times in total.
He missed 3 times
So he scored from 17 shots.
Let X be the no. Of shots in hole.
Y be the no. Of shots on board.
So, equation will be
5x+2y=52
Put the options one by one.
5(5)+2(12)=49 (5+12=17shots)
5(6)+2(11)=52 (correct)
5(7)+2(10)=55 (incorrect)
5(8)+2(9)=58 (incorrect)
5(9)+2(8)=61 (incorrect)

OA is B i.e. 6 times

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Re: In a certain beanbag game, players toss beanbags at a board with a hol [#permalink] New post   18 Mar 2020, 09:35
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The required equation is \(5*x + 2*y - 3*1 = 52\)
\(5*x + 2*y = 55\)
x has to be an odd integer. (5,7 or 9 could be the answer)

checking for 7, \(5*7+2*y = 55\); y =10
so total number of tosses = 7 + 10 + 3 = 20; matches with given information.
Ans: C
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Re: In a certain beanbag game, players toss beanbags at a board with a hol [#permalink] New post   18 Mar 2020, 09:59
Bunuel wrote:
In a certain beanbag game, players toss beanbags at a board with a hole in it. If the beanbag falls into the hole, the player scores 5 points. A beanbag that lands on the board but misses the hole is worth 2 points. If a toss misses the board entirely, 1 point is deducted from the player’s score. Julianna had 20 tosses and missed the board with 3 throws. Her total score was 52. How many times did she toss the beanbag into the hole?

A. 5
B. 6
C. 7
D. 8
E. 9


from given info
we know
5x+2y-3=52
x+y=17
5x+2y=55
solve for x
we get x= 7
IMO C
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Re: In a certain beanbag game, players toss beanbags at a board with a hol [#permalink] New post   20 Mar 2020, 06:52
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Bunuel wrote:
In a certain beanbag game, players toss beanbags at a board with a hole in it. If the beanbag falls into the hole, the player scores 5 points. A beanbag that lands on the board but misses the hole is worth 2 points. If a toss misses the board entirely, 1 point is deducted from the player’s score. Julianna had 20 tosses and missed the board with 3 throws. Her total score was 52. How many times did she toss the beanbag into the hole?

A. 5
B. 6
C. 7
D. 8
E. 9


Let x = the number of tosses into the hole and y = the number of tosses onto the board. Since 3 tosses missed the board, without those tosses the score would have been 55.

Thus, we can create the equations:

x + y = 17

y = 17 - x

and

5x + 2y = 55

Substituting, we have:

5x + 2(17 - x) = 55

5x + 34 - 2x = 55

3x = 21

x = 7

Answer: C
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Re: In a certain beanbag game, players toss beanbags at a board with a hol [#permalink] New post   20 Mar 2020, 08:04
yashikaaggarwal wrote:
OA is B
HE SCORED 52
And tossed 20 times in total.
He missed 3 times
So he scored from 17 shots.
Let X be the no. Of shots in hole.
Y be the no. Of shots on board.
So, equation will be
5x+2y=52
Put the options one by one.
5(5)+2(12)=49 (5+12=17shots)
5(6)+2(11)=52 (correct)
5(7)+2(10)=55 (incorrect)
5(8)+2(9)=58 (incorrect)
5(9)+2(8)=61 (incorrect)

OA is B i.e. 6 times

Posted from my mobile device


Why don't we subtract 3 from total score of 52, because it is mentioned that 1 point is deducted whenever the beanbags land outside the board. Shouldn't it be the score of 49 after 3 shots?
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Re: In a certain beanbag game, players toss beanbags at a board with a hol [#permalink] New post   01 May 2023, 02:55
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Re: In a certain beanbag game, players toss beanbags at a board with a hol [#permalink]
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