Introduction to maths tricky questions
Math is a study of quantity. Here is an article that provides some tricky questions in math that will help you in cracking various placement tests as well as other class tests.Lets see some of the tricky questions on math and their solutions. I like to share this Solve Math Problems Free with you all through my article.
Solutions to maths tricky questions
1. The average age of a class is 15.8 years. The average age of the boys in the class is 16.4 years while that of the girls is 15.4 years. What is the ratio of boys to girls in the class?
Solution: Let the ratio be k:1
Then, k * 16.4 + 1 * 15.4 = (k+1) * (15.8)
16.4 k + 15.4 = 15.8k + 15.8
16.4k - 15.8k = 15.8 - 15.4
0.6k = 0.4
k = 0.4 / 0.6
or k = `(2)/(3)`
So, required ratio = `(2)/(3)` : 1 = 2 : 3 Answer
2. A sum of Rs. 1000 is lent to be returned in 11 monthly instalments of Rs. 100 each, interest being simple. What is the rate of interest?
Solution: Rs. 100 + SI on Rs. 1000 for 11 months
= Rs. 1000 + SI on Rs. 100 for (1 + 2 + 3 + 4 + ..... + 10) months
= Rs. 1000 SI on Rs. 100 for `(110)/(2)` months
= Rs 1000 + SI on Rs. 100 for 55 months
SI on Rs. 100 for 55 months = Rs. 100
So, Rate =`(100 * 100 * 12)/(100 * 55)` %
= 21.82 % Answer
3. Ramlal is four times as old as his son. Four years later the sum of their ages will be 43 years. What is the present age of the son?
Solution: Let present age of son = x years
Present age of Ramlal = 4x years
Four years later:
Son's age = x + 4
Ramlal's age = 4x + 4
According to the given condition:
x + 4 + 4x + 4 = 43
5x + 8 = 43
5x= 43 - 8
5x= 35
x= `(35)/(5)` = 7
So, present age of son = 7 years Answer
4. A man can row at 5 km/hr in still water and the velocity of the current is 1 km/hr. It takes him 1 hr to row to a place and back. How far is the place?
Solution: Distance = 1 x `(5^2 - 1^2)/(2 * 5)` 52 - 12 / 2 x 5
= `(24)/(10)` = 2.4 km Answer
Having problem with online math word problem solver keep reading my upcoming posts, i will try to help you.
Maths Tricky Questions (Continued)
Some more maths tricky questions with the solutions :-
5. Bucket P has thrice the capacity has bucket Q. It takes 60 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum?
Solution: Let capacity of P be x litre
Then, capacity of Q = x/3 litre
capacity of drum = 60x litres
Required number of turns = `(60x)/(x + x/3)`
= `(3 * 60x)/(3x+x)`
= `(180x)/(4x)` = 45 turns Answer
6. Rajan invests 65% of his intended investment in machinery and 20% on raw material in his business. If he has a balance of $6000 what was the intended investment?
Solution: Let intended investment be x
Investment spent on machinery = 65%
Investment spent on raw materials = 20%
So, total percentage of investment spent = 65% + 20% = 85%
Percentage of investment left = (100% = 85%) = 15%
According to the statement:
15% of x = 6000
`(15x)/(100)` = 6000
15x = 6000 x 100
x = `(600000)/(5)` = 40000
So, intended investment = $40000 Answer
7. By selling an article at $310 a merchant loses 7% on his outlay. Find the percentage of profit or loss when he sells the article at $350 ?
Solution: When article is sold at $310, percentage lost = 7%
So, remaining = 93%
Now, 93 : x = 310 : 350
93 x 350 = 310 * x
x = 105 %
i.e. Profit = (105 - 5) % = 5% Answer
8. Two trains each 300 m in length are running on the parallel linesin opposite directions with the speed of 70 km/hr and 50 km/hr respectively. In how much time will they cross each other completely?
Solution: t = `(300)/((70+50) * 5/18)`
= `(300)/((120) * 5/18)`
= 9 sec Answer
9. One year ago the ratio between father's and son's age was 4:1. The ratio of their ages after 4 years will be 3:1. Find the ratio of their ages after 15 years.
Solution: Let present age of father = x
present age of son = y
According to the statement:
x-1 : y -1 = 4:1
So, 4(y - 1) = x - 1
4y - 4 = x - 1
4y - x = - 1 + 4
-x + 4y = 3 -----> (1)
Also, x + 4 : y + 4 = 3:1
So, 3 (y + 4) = x + 4
3y + 12 = x + 4
3y - x = 4 - 12
-x + 3y = -8 -------> (2)
Subtracting (2) from (1), we get,
4y - x - (-x + 3y) = 3 - (-8)
4y - x + x - 3y = 11
y = 11
Substituting the value of y in equation (1)
-x + 4y = 3
-x + 44 = 3
-x = 3 - 44
- x = -41 or x = 41
So, present age of father = x = 41 years
present age of son = y = 11 years
Note: If you require any more help in Maths Tricky Questions, our group of expert math tutors will help you in doing do.
Math is a study of quantity. Here is an article that provides some tricky questions in math that will help you in cracking various placement tests as well as other class tests.Lets see some of the tricky questions on math and their solutions. I like to share this Solve Math Problems Free with you all through my article.
Solutions to maths tricky questions
1. The average age of a class is 15.8 years. The average age of the boys in the class is 16.4 years while that of the girls is 15.4 years. What is the ratio of boys to girls in the class?
Solution: Let the ratio be k:1
Then, k * 16.4 + 1 * 15.4 = (k+1) * (15.8)
16.4 k + 15.4 = 15.8k + 15.8
16.4k - 15.8k = 15.8 - 15.4
0.6k = 0.4
k = 0.4 / 0.6
or k = `(2)/(3)`
So, required ratio = `(2)/(3)` : 1 = 2 : 3 Answer
2. A sum of Rs. 1000 is lent to be returned in 11 monthly instalments of Rs. 100 each, interest being simple. What is the rate of interest?
Solution: Rs. 100 + SI on Rs. 1000 for 11 months
= Rs. 1000 + SI on Rs. 100 for (1 + 2 + 3 + 4 + ..... + 10) months
= Rs. 1000 SI on Rs. 100 for `(110)/(2)` months
= Rs 1000 + SI on Rs. 100 for 55 months
SI on Rs. 100 for 55 months = Rs. 100
So, Rate =`(100 * 100 * 12)/(100 * 55)` %
= 21.82 % Answer
3. Ramlal is four times as old as his son. Four years later the sum of their ages will be 43 years. What is the present age of the son?
Solution: Let present age of son = x years
Present age of Ramlal = 4x years
Four years later:
Son's age = x + 4
Ramlal's age = 4x + 4
According to the given condition:
x + 4 + 4x + 4 = 43
5x + 8 = 43
5x= 43 - 8
5x= 35
x= `(35)/(5)` = 7
So, present age of son = 7 years Answer
4. A man can row at 5 km/hr in still water and the velocity of the current is 1 km/hr. It takes him 1 hr to row to a place and back. How far is the place?
Solution: Distance = 1 x `(5^2 - 1^2)/(2 * 5)` 52 - 12 / 2 x 5
= `(24)/(10)` = 2.4 km Answer
Having problem with online math word problem solver keep reading my upcoming posts, i will try to help you.
Maths Tricky Questions (Continued)
Some more maths tricky questions with the solutions :-
5. Bucket P has thrice the capacity has bucket Q. It takes 60 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum?
Solution: Let capacity of P be x litre
Then, capacity of Q = x/3 litre
capacity of drum = 60x litres
Required number of turns = `(60x)/(x + x/3)`
= `(3 * 60x)/(3x+x)`
= `(180x)/(4x)` = 45 turns Answer
6. Rajan invests 65% of his intended investment in machinery and 20% on raw material in his business. If he has a balance of $6000 what was the intended investment?
Solution: Let intended investment be x
Investment spent on machinery = 65%
Investment spent on raw materials = 20%
So, total percentage of investment spent = 65% + 20% = 85%
Percentage of investment left = (100% = 85%) = 15%
According to the statement:
15% of x = 6000
`(15x)/(100)` = 6000
15x = 6000 x 100
x = `(600000)/(5)` = 40000
So, intended investment = $40000 Answer
7. By selling an article at $310 a merchant loses 7% on his outlay. Find the percentage of profit or loss when he sells the article at $350 ?
Solution: When article is sold at $310, percentage lost = 7%
So, remaining = 93%
Now, 93 : x = 310 : 350
93 x 350 = 310 * x
x = 105 %
i.e. Profit = (105 - 5) % = 5% Answer
8. Two trains each 300 m in length are running on the parallel linesin opposite directions with the speed of 70 km/hr and 50 km/hr respectively. In how much time will they cross each other completely?
Solution: t = `(300)/((70+50) * 5/18)`
= `(300)/((120) * 5/18)`
= 9 sec Answer
9. One year ago the ratio between father's and son's age was 4:1. The ratio of their ages after 4 years will be 3:1. Find the ratio of their ages after 15 years.
Solution: Let present age of father = x
present age of son = y
According to the statement:
x-1 : y -1 = 4:1
So, 4(y - 1) = x - 1
4y - 4 = x - 1
4y - x = - 1 + 4
-x + 4y = 3 -----> (1)
Also, x + 4 : y + 4 = 3:1
So, 3 (y + 4) = x + 4
3y + 12 = x + 4
3y - x = 4 - 12
-x + 3y = -8 -------> (2)
Subtracting (2) from (1), we get,
4y - x - (-x + 3y) = 3 - (-8)
4y - x + x - 3y = 11
y = 11
Substituting the value of y in equation (1)
-x + 4y = 3
-x + 44 = 3
-x = 3 - 44
- x = -41 or x = 41
So, present age of father = x = 41 years
present age of son = y = 11 years
Note: If you require any more help in Maths Tricky Questions, our group of expert math tutors will help you in doing do.
