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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.

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

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 = $\frac{80-64}{80}$*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) (15-0.8)
= 241.4 kg
weight of student = (total weight of 18 students) - (Total weight of 17 students)
= 270 – 241.4
= 28.6 kg

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

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

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 3-pound and 5-pound 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.