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Free Answer to Math Riddles
Math Riddles:
Math riddles are basically based on the mathematical basic.Most of the Riddles questions are of logical thinking, tricky and reasoning type. Solving riddles helps in improving math skill level.
Free Answer to Math Riddles:
Solution: By making use of BODMAS operation we simplify the given sum.
Free Answer to Math Riddles:
1. Find the value by making using mathematical code language.
21 ÷ 10*20  4 + 2
Solution: By making use of BODMAS operation we simplify the given sum.
= 21 ÷ 10 * 20  4 + 2
= 21 * 2  4 + 2
= 42 – 4 + 2
= 42  2
= 40
Solution: Let the first number be a and the second number be b.
= $\frac{a^{2}b^{2}}{ab}$ = $\frac{a}{b}$ + $\frac{b}{a}$
$\frac{4}{5}$ + $\frac{5}{4}$
= $\frac{41}{20}$
3. By what percent is 64 less than 80.
= 20%
4. The average weight of 18 students in a class is 15 kg. If one student leaves the class, the average weight decreases by 0.8 kg. Find the weight of the student in kg.
Solution: Let the weight of the student who left the class be x kg.
2. There are two numbers. Five times the first and four times the second are equal. Find the ratio of the sum of the squares of the two numbers and their product.
Solution: Let the first number be a and the second number be b.
By making use of proportions.
5a = 4b
$\frac{a}{b}$ = $\frac{4}{5}$
= $\frac{a^{2}b^{2}}{ab}$ = $\frac{a}{b}$ + $\frac{b}{a}$
$\frac{4}{5}$ + $\frac{5}{4}$
= $\frac{41}{20}$
3. By what percent is 64 less than 80.
Solution: Required percentage = $\frac{8064}{80}$*100
= $\frac{16}{80}$*100
= 20%
4. The average weight of 18 students in a class is 15 kg. If one student leaves the class, the average weight decreases by 0.8 kg. Find the weight of the student in kg.
Solution: Let the weight of the student who left the class be x kg.
Given total weight of 18 students = 270 kg
Total weight of 17 students = (17) (150.8)
= 241.4 kg
weight of student = (total weight of 18 students)  (Total weight of 17 students)
= 270 – 241.4
= 28.6 kg
Solution: Given B travels 50 m in 10 seconds.
= $\frac{50}{10}$ = m/s
Time taken by B to complete the race = $\frac{1000}{5}$ = 200 s
Time taken for A = 200  10 = 190 s
Speed of A = $\frac{1000}{190}$ = 5 $\frac{5}{19}$ m/s
5. In a 100 m race. A beats B by 50 m or 10 seconds. Find the speed of A.
Solution: Given B travels 50 m in 10 seconds.
Speed = $\frac{distance\ traveled}{time\ taken}$
= $\frac{50}{10}$ = m/s
Time taken by B to complete the race = $\frac{1000}{5}$ = 200 s
Time taken for A = 200  10 = 190 s
Speed of A = $\frac{1000}{190}$ = 5 $\frac{5}{19}$ m/s
Best Results From Yahoo Answers Youtube
From Yahoo Answers
Question:I need the answer and the steps it took to get the answer to this problem:
My digits are three,
But just what can they be?
Take a third of my first
From a half of my third:
And a negative, right there,
Would, of course be absurd
Now add twice my middle
To continue the riddle.
If you figured it right,
Then you can't have got seven;
For, believe me, I know
That you must have eleven.
There's no catch in this,
But don't take it amiss
When I add one thing more,
Just to make it quite clear:
You should know that a five
Is no part of me here.
Answers:I got it! ok here we go: Let's say the number is abc First, consider the first verse: "Take a third of my first (1/3 a) From a half of my third: (1/2c) And a negative, right there, Would, of course be absurd" This means that : 1/2c  1/3 a is Not negative So, 1/2c  1/3a > 0 => 1/2c > 1/3a Alright, there are 6 combinations of a and c that fit this criteria . Assuming a is divisible by 3, a must be 3, 6, or 9. Assuming c is divisible by 2, a must be 2, 4, 6, or 8. Consider the condition 1/2c > 1/3a, The possibilities of (a, c) are as follows: 1. (3,4) => (1<2) 2. (3,6) => (1<3) 3. (6,6) => (2<3) 4. (3,8) => (1<4) 5. (6,8) => (2<4) 6. (9,8) => (3<4) Ok, so now consider this part of the riddle: "Now add twice my middle To continue the riddle." This means that we now have: 1/2c  1/3a + 2b Then: "If you figured it right, Then you can't have got seven; For, believe me, I know That you must have eleven." So this means that we can set our equation equal to 11: 1/2c  1/3a + 2b = 11 So, plug each of these sets of values into the equations: 1. (3,4) => 2  1 + 2b = 11 2. (3,6) => 3  1 + 2b = 11 3. (6,6) => 3  2 + 2b = 11 4. (3,8) => 4  1 + 2b = 11 5. (6,8) => 4  2 + 2b = 11 6. (9,8) => 4  3 + 2b = 11 Simplify: 1. (3,4) => 1 + 2b = 11 2. (3,6) => 2 + 2b = 11 3. (6,6) => 1 + 2b = 11 4. (3,8) => 3 + 2b = 11 5. (6,8) => 2 + 2b = 11 6. (9,8) => 1 + 2b = 11 Continue to simplify: 1. (3,4) => 2b = 10 2. (3,6) => 2b = 9 3. (6,6) => 2b = 10 4. (3,8) => 2b = 9 5. (6,8) => 2b = 8 6. (9,8) => 2b = 10 Continue to simplify: 1. (3,4) => b = 5 2. (3,6) => 2b = 9 3. (6,6) => b = 5 4. (3,8) => 2b = 9 5. (6,8) => b = 4 6. (9,8) => b = 5 Now, we know that: "You should know that a five Is no part of me here." So no digits can be 5, which rules out pairs 1, 3, and 6! We are left with: 2. (3,6) => b = 4.5 4. (3,8) => b = 4.5 5. (6,8) => b = 4 Since b can obviously not be 4.5, since it must be a single digit, then # 5 is our choice! 5. (6,8) => b = 4 So, a = 6, c = 8, and b = 4 abc = 648
Answers:I got it! ok here we go: Let's say the number is abc First, consider the first verse: "Take a third of my first (1/3 a) From a half of my third: (1/2c) And a negative, right there, Would, of course be absurd" This means that : 1/2c  1/3 a is Not negative So, 1/2c  1/3a > 0 => 1/2c > 1/3a Alright, there are 6 combinations of a and c that fit this criteria . Assuming a is divisible by 3, a must be 3, 6, or 9. Assuming c is divisible by 2, a must be 2, 4, 6, or 8. Consider the condition 1/2c > 1/3a, The possibilities of (a, c) are as follows: 1. (3,4) => (1<2) 2. (3,6) => (1<3) 3. (6,6) => (2<3) 4. (3,8) => (1<4) 5. (6,8) => (2<4) 6. (9,8) => (3<4) Ok, so now consider this part of the riddle: "Now add twice my middle To continue the riddle." This means that we now have: 1/2c  1/3a + 2b Then: "If you figured it right, Then you can't have got seven; For, believe me, I know That you must have eleven." So this means that we can set our equation equal to 11: 1/2c  1/3a + 2b = 11 So, plug each of these sets of values into the equations: 1. (3,4) => 2  1 + 2b = 11 2. (3,6) => 3  1 + 2b = 11 3. (6,6) => 3  2 + 2b = 11 4. (3,8) => 4  1 + 2b = 11 5. (6,8) => 4  2 + 2b = 11 6. (9,8) => 4  3 + 2b = 11 Simplify: 1. (3,4) => 1 + 2b = 11 2. (3,6) => 2 + 2b = 11 3. (6,6) => 1 + 2b = 11 4. (3,8) => 3 + 2b = 11 5. (6,8) => 2 + 2b = 11 6. (9,8) => 1 + 2b = 11 Continue to simplify: 1. (3,4) => 2b = 10 2. (3,6) => 2b = 9 3. (6,6) => 2b = 10 4. (3,8) => 2b = 9 5. (6,8) => 2b = 8 6. (9,8) => 2b = 10 Continue to simplify: 1. (3,4) => b = 5 2. (3,6) => 2b = 9 3. (6,6) => b = 5 4. (3,8) => 2b = 9 5. (6,8) => b = 4 6. (9,8) => b = 5 Now, we know that: "You should know that a five Is no part of me here." So no digits can be 5, which rules out pairs 1, 3, and 6! We are left with: 2. (3,6) => b = 4.5 4. (3,8) => b = 4.5 5. (6,8) => b = 4 Since b can obviously not be 4.5, since it must be a single digit, then # 5 is our choice! 5. (6,8) => b = 4 So, a = 6, c = 8, and b = 4 abc = 648
Question:Please show your work. And if you would need to tell me whether you use multiplication or adding or subtracting or dividing.
PROBLEM:
Lima beans come in 3pound and 5pound bags which cost $1.15 and $1.63 respectively. How many of each should you buy to have at least 17 pounds of lima beans at the lowest cost?
Thanks SO much.
Answers:Hi 1.63*3=4.89+1.15=$6.04 Your answer is actually 3 5pound bags and 1 3pound bag. The other guy didn't read the question preperly.
Answers:Hi 1.63*3=4.89+1.15=$6.04 Your answer is actually 3 5pound bags and 1 3pound bag. The other guy didn't read the question preperly.
Question:Three old ladies go to a second hand shop to buy a TV.
They pick one out and it costs 30. They pay the shopkeeper 10 each, and arrange for free delivery.
The next day, the shop owner gives the delivery driver 3, and asks him to give it back to the ladies, because he feels a bit sorry for them.
On his journey, the driver decides that he'll keep 1 for himself, because he's a bit of a liar and a thief.
So on his arrival, the driver drops off the TV and gives the old ladies 2 back.
Okay, so two of the old ladies have paid 9 each, and one has paid 10, yes?
The driver kept 1, yes?
So where has the other 1 gone? The above only adds up to 29, but in the beginning there was 30.
Please help. It hurts my bain.
Answers:The correct answer for this is that none of the ladies paid 9, they all paid equal amounts therefore when the driver has 3 to give back it is expected that each will recieve 1 however he keeps 1 so there is 2 to be divided by the 3 ladies, this means that they each get 66.6 recurring pence, meaning they've each paid 9.6666666666666 (recurring) 9.666666 times 3 is 29. (well 9.66 X 3 is 28.98 so two ladies will get an extra penny) plus the 1 the driver took makes 30. Hope this clears it up for you and everyone else.
Answers:The correct answer for this is that none of the ladies paid 9, they all paid equal amounts therefore when the driver has 3 to give back it is expected that each will recieve 1 however he keeps 1 so there is 2 to be divided by the 3 ladies, this means that they each get 66.6 recurring pence, meaning they've each paid 9.6666666666666 (recurring) 9.666666 times 3 is 29. (well 9.66 X 3 is 28.98 so two ladies will get an extra penny) plus the 1 the driver took makes 30. Hope this clears it up for you and everyone else.
Question:Suppose 2 circles intersect at 2 points. The radius of one circle is 13 units, the radius of the other circle measures 8 units. What is the absolute value of the difference in the area measures of the non overlapping regions of the two circles. State the answer in terms of pi...
In answer, write pi instead of symbol
please dont just give me the answer...explain how you got it!
Answers:The previous answer is right, the possible answers are numerous. However we can narrow it down some. If the radius of the two circles are 8 and 13 and they intersect at 2 points, not 1 or 0, then we can say that the distance between the center of the two circles is greater than 138=5 and less than 13+8=21. These two distances are the points where the circles intersect at exactly 1 point. Hope this gets you started. You should be able to answer this in terms on variables based on the distance of >5 and <21.
Answers:The previous answer is right, the possible answers are numerous. However we can narrow it down some. If the radius of the two circles are 8 and 13 and they intersect at 2 points, not 1 or 0, then we can say that the distance between the center of the two circles is greater than 138=5 and less than 13+8=21. These two distances are the points where the circles intersect at exactly 1 point. Hope this gets you started. You should be able to answer this in terms on variables based on the distance of >5 and <21.
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