Confusing Maths Problems!!?

1.) All you have to do is sit down and think. If you count by 2's and you have 1 left over, that means it HAS to be an odd number because even numbers go into equally. So we narrowed it down to 31, 33, 35, 37, and 39. Now it said that if you count by 3's there are two leftover meaning that the numbers that will be left are 31, 35, and 37 because 3 goes into 33 and 39 equally. Now It said that when they counted by 3's, there were TWO left over. Well if you count by 3's to 31 you will have 1 leftover. If you count by 3's to 37, you will also have 1 leftover. That only leaves us with 35 because when you count by 3's, you will have 2 left over. 35 <--- This is your answer!

1. If you count the sheep in pairs 1 will be left over. The number of sheep is an odd number. If divided by 3 there will be 2 left over, which means this odd number is 3n+2, where 3n is also an odd number. The first odd multiple of 3 in that range is 33. 33+2=35. 35 satisfies the two conditions, where no other number within range does.

1. Ans = 35.

2. Stamps cost 25 c. each, and $2.50 was spent. So 10 stamps were bought for postcards. So 10 postcards were sent. She wrote to 8 friends, so 8 x 35 c. is $2.80.

3. Mike is 7 years older than Annie. So let us put that as M is 7+A.

Annie is 5 years older than Peta. So that is A-5=P, or M-7-5=M-12=P

The sum of their ages is 86. So M + A + P = 86 = M + (M - 7) + (M - 12) = 3M - 19

86 + 19 = 105 = 3M, so M = 35. Thus Mike is 35, Annie is 28, and Peta (isn't 'peta' map in Malay?) is 23. 35 and 28 and 23 give 86, so nothing is wrong.

4. By cycling 4km at 8km/h, he has spent 30 minutes. He has 5 minutes left to cycle 1km. So his speed must now be 12km/h.

No ÷ 2 Remainder is 1

No ÷ 3 Remainder is 2

here remainder is not same. if remainder is same then

least number in such case = (LCM of divisors + Remainder)

again for your question, though remainder is not same. difference in divisor and remainder is sam.

No ÷ 2 Remainder is 1 difference in divisor and remainder 2 – 1 = 1

No ÷ 3 Remainder is 2 difference in divisor and remainder 3 – 2 = 1

then least number = LCM of divisors – difference).

Least number = 2 × 3 – 1 = 6 – 1 = 5

to find number nearer to 30

6 × 5 – 1 = 29 is < 30

next number = 6 × 6 – 1 = 35

(multiples of LCM are to be found)

annie = peta + 5

pea = annie – 5

annie = mike – 7

mike = annie + 7

a + p + m = 86

a + (a – 5) + (a + 7) = 86

3a + 2 = 86

a = 84/3 = 28

p = 28 – 5 = 23

m = a + 7 = 35

time for 4 km = 4/8 = 1/2 hr = 30 min

distance remained = 1 km time left = 5 min = 5/60 = 1/12 hr

speed = 1 ÷ 1/12 = 1 × 12/1 = 12 km/hr

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answer1.

as there are sheep in the field in between 30 and 40.Therefore,no of sheep could be,

31 to 39.

when 2 sheep are counted at a time then 1 is left

therefore, 31 33 35 37 39

when three sheep are counted then 2 are left

therefore, 32 35 38

32 and 38 are fully divisible by 2(hence can't be the answer)

35 is common in both findings

hence,35 sheep are present in the field.

answer 2

let, x be the no of postcards and y be the no of letters sent by susan.

from question,

x+y=8................(1)

$2.50=250 cents

25x+35y=250..............(2)

solving 1 and 2..........we get

x=3 anmd y=5

therefore, she sent 3 postcards and 5 letters to her friends.

answer3

let, annie,peta and mike be denoted by a,p and m

therefore,from question

a=p+5...........(1)

a=m-7............(2)

a+p+m=86

a+a-5+a+7=86....................from 1 and 2

a=28.

putting value of a in 1 and 2 we get

p=23 and m=35

anwer4

speed=distance/time

8=4/t

t=0.5 hours

t=30 minutes

now, he has to travel 1km in 5 minutes

therefore, speed=1/5=0.2km/min

0.2km/min=0.2*60km/h

12km/h.

therefore, he must go at 12km/h to cover the next 1km in order to complete the journey in 30 min.

1.

n . . . n (mod 2) . . . n (mod 3) . . . n (mod 6)

0 . . . . . . 0 . . . . . . . . . 0 . . . . . . . . . . 0

1 . . . . . . 1 . . . . . . . . . 1 . . . . . . . . . . 1

2 . . . . . . 0 . . . . . . . . . 2 . . . . . . . . . . 2

3 . . . . . . 1 . . . . . . . . . 0 . . . . . . . . . . 3

4 . . . . . . 0 . . . . . . . . . 1 . . . . . . . . . . 4

5 . . . . . . 1 . . . . . . . . . 2 . . . . . . . . . . 5

30 < x < 40

x = 1 (mod 2), x = 2 (mod 3) -----> x = 5 (mod 6) (from table above)

x = 35

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2.

x = # of postcards (@ $0.25 each)

y = # of letters (@ $0.35 each)

x + y = 8

0.25x + 0.35y = 2.50

Solving, we get

x = 3 postcards

y = 5 letters

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3.

Annie's age = x

Peta's age = x - 5 (Annie is 5 years older, so Peta is 5 years younger)

Mike's age = x + 7 (Annie is 7 years younger, so Mike is 7 years older)

x + (x-5) + (x+7) = 86

3x + 2 = 86

3x = 84

x = 28

Annie is 28 yrs old

Peta is 23 yrs old

Mike is 35 yrs old

--------------------

4.

4 km / 8km/hr = 1/2 hr = 30 min

Mike has 5 minutes left to cover last km

1 km / 5 min = 1 km / (5/60) hr = 12 km/hr

i think ive explained...

1)35

take the multiples of 3 bet 30 and 40

they are 30,33,36,39

by the sum 2 left over if counted by threes... thus the ans must be one of 32,35,38,41

even numbers not possible since when counted by 2's 1 is left... so the ans must be 35

2)let postcards be x and letters be y

now, x+y=8 and also 25x+35y=250

equating both, 10y=50

and y=5

therefore, x=3

letters--- 5 and postcards----3

3)28

take annie's age as x...

for an equation..

peta's age--- x-5 mike's age---x+7

x-5+x+x+7=86

3x+2=86

3x=84

x=28

annie--- 28

peta------23

mike-----35

4)12km/h

he cycled 4km in 8km / h which would take 30 mins

therefore only 5 mins and 1km remaining

1 km in 5 mins means speed is 12km/hr

If 210 are relatives and family members of both the bride and groom, I would subtract that 210 from 330, which gives 120. That's the number of guests left to invite who are not related to both. I would then divide the 120 by 2, and I would say that each should invite 60. Another possibility is that since 210 people might be considered invitees of both the bride and the groom, you might add 210 to the 60, giving each side 270 guests (including those who are receiving invitations from both the bride and the groom.) The problem isn't worded very well, and I think it is made to trick you. I think my second paragraph is the answer to the trick. If there is no trick intended, it still isn't well-written, but then 60 would be the answer. Go prepared with both answers.

1.

35 sheep

3.

take annie age = a

then age of peta = a-5

age of mike = a+7

given sum of their ages = 86

a+a-5+a+7 = 86

3a = 84

a = 28

annie age is = 28 year

age of peta = 23

age of mike = 35.

4.

he wanted to complete 5km in 35 minutes.

when he run with the speed 8 km/h for 4km he use 30 minutes

then he must complete remaining 1km in 5minutes i.e (5/60)hour

velocity = (1km) / (5 minute)

= (1km) * 60/ 5

= 12 km / h.

there are between 30 and 40 sheep on a field . if i count them by twos there is 1 left over. and if i count them by threes there is 2 left over. how many sheep were there?

When counted by twos there is 1 left over i.e number of sheep is odd number.When counted threes there is 2 left over i.e number of sheep is multiple of 3 increasd by 2 . This number lies between 30 and 40.

Only number satisfying this condition is 35

35=2×17+1

35=3×11+2