# during a festival celebration in a school, the coordinator had bought a number of brownies and put in a bag. if they were to be equally divided among 17 children, there are 13 brownies left. if they were to be equally divided among 20 children, there are 7 brownies left. obviously, this can be satisfied if any multiple of 160 brownies are added to the bag. what is the remainder when the minimum feasible number of brownies in the bag is divided by 8?

get during a festival celebration in a school, the coordinator had bought a number of brownies and put in a bag. if they were to be equally divided among 17 children, there are 13 brownies left. if they were to be equally divided among 20 children, there are 7 brownies left. obviously, this can be satisfied if any multiple of 160 brownies are added to the bag. what is the remainder when the minimum feasible number of brownies in the bag is divided by 8? from screen.

## If a box of sweet is divided among 24 children, they will gets 5 sweets each. How many would each get, if the number of children is reduced by 4?

Click here👆to get an answer to your question ✍️ If a box of sweet is divided among 24 children, they will gets 5 sweets each. How many would each get, if the number of children is reduced by 4?

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## If a box of sweet is divided among 24 children, they will gets 5 sweets each. How many would each get, if the number of children is reduced by 4?

No. of students =24Medium Open in App Solution Verified by Toppr

no. of sweets each gets =5

therefore, total no. of sweets 24×5=120

the no. of children when 4 is gone =24−4=20

Hence,each child will get

20 120 =6sweets.

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## After distributing the sweets equally among 25 children, 8 sweets remained. If the number of children had been 28, 22 sweets would have been left after equally distributing. What was the total number of sweets?

Answer (1 of 13): Let each child among 25 children get x sweets (where x is a whole number and not a fraction). Then, total number of sweets will become 25x + 8. Now, assume each child among 28 children get y sweets (where y is a whole number and not a fraction). Then, total number of sweets w...

After distributing the sweets equally among 25 children, 8 sweets remained. If the number of children had been 28, 22 sweets would have been left after equally distributing. What was the total number of sweets?

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Sort Sangi Pateriya

Btech in Mechanical Engineering, College of Technology, Pantnagar7y

Originally Answered: After distributing the sweets equally among 25 children, 8 sweets remain.had the number of children been 28,22 sweets would have been left after equally distributing.ehat was the total number of sweets?

358

Let total no. of sweets= 25x + 8

Then (25x+8)-22 will be divisible by 28

=25x-14 is divisible by 28

= 28x-(3x+14) is divisible by 28

= 3x+14 is divisible by 28

x=14

So total no. of sweets = 25(14)+8= 358

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Tyler Skywalker

Researcher, 13 yrs.Author has 340 answers and 365.1K answer views6y

So x x mod25=8 mod25=8 , and x x mod28=22 mod28=22

First get x by itself in

xmod25=8 xmod25=8

The equation would be

x=y(25)+8 x=y(25)+8

Next get x by itself in

x x mod28=22 mod28=22

The equation would be

x=z(28)+22 x=z(28)+22 It is given that N⊆y N⊆y , and N N ⊆z ⊆z

Set the equations so they are equal,

25y+8=28z+22 25y+8=28z+22 Then, 25y=28z+14 25y=28z+14 25y=(2z+1)∗14 25y=(2z+1)∗14 This is true iff y=14 y=14 , and 2z+1=25 2z+1=25 Now solve z. 2z=24 2z=24 z=12 z=12 Substituting. x=28(12)+22 x=28(12)+22 x=336+22 x=336+22 x=358 x=358

So, the total number of candy is 358

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Ketan Rana

1.6 k edits on Wikipedia at Wikipedia (2017–present)Author has 335 answers and 1.7M answer viewsAug 1

Let each child among 25 children get x sweets (where x is a whole number and not a fraction). Then,

total number of sweets will become 25x + 8.

Now, assume each child among 28 children get y sweets (where y is a whole number and not a fraction). Then,

total number of sweets will become 28y + 22.

As the total number of sweets is same in both the cases, therefore:

25x + 8 = 28y + 22

Simplifying, we get,

25x - 28y = 14

=> x = (28y + 14)/25

=> x = (25y + 3y + 14)/25

=> x = 25y/25 + (3y + 14)/25

now, for x to be a whole number, (3y + 14) must be perfectly divisible by 25 as 25y is already perfectly divisible

Rajendra Rajput

I'm a maths lover. I like to solve hard puzzle or problem of maths.Author has 699 answers and 2M answer views6y

N/25 = 8

N could be 33,58,83,...358,...

N/28 = 22

N could be 50,78,...358

The very first no. Conmon in both terms is 358

So answer is 358. Related questions

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Parveen Kaur

Maths is fun... :)Author has 397 answers and 2.4M answer views6y

Hope it helps.... :)

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Sivapuram Raghavendra

Studied at Keshava Reddy Talent School7y

Originally Answered: After distributing the sweets equally among 25 children, 8 sweets remain.had the number of children been 28,22 sweets would have been left after equally distributing.ehat was the total number of sweets?

## On a school's annual Day, sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus, the remaining children got 6 extra sweets. How many sweets was each child originally supposed to get ?

On a school's annual Day, sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus, the remaining children got 6 extra sweets. How many sweets was each child originally supposed to get ?

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On a school's annual Day, sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus, the remaining children got 6 extra sweets. How many sweets was each child originally supposed to get ?

Question

A 24 B 18 C 15 D

Cannot be determined

E None of these Open in App Solution

The correct option is **D** 15

Let the total number of sweets be x.

According to the question,

x 112 − 32 − x 112 = 6 ⇒ x 80 − x 112 = 6 ⇒ 7 x − 5 x 560 = 6 ⇒ 2 x 560 = 6 ⇒ x = 560 × 6 2 = 1680 ∴ Required answer = 1680 112 = 15 Suggest Corrections 5

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