Detail worksheet on simplification is here. Check simplification example problems and solutions here. Refer step by step procedure to find the accurate solution on fundamental expressions and simplifications. Simplify all the algebraic expressions and rules to solve problems.
Solve many mock questions based on the simplification of algebraic expressions and integers from Simplification of Algebraic Expressions Worksheet to get perfection and crack the exam easily. Before solving the below problem, check our previous articles to know the formulae, rules, and methods. Follow the below sections to know various model questions and solutions.
Problem 1:
After striking a floor, a certain ball rebounds 4/5th of the height from which it has fallen. What is the total distance that it travels before coming to rest if it is gently dropped from a height of 120 meters?
Solution:
As given in the question,
The height at which the ball rebounds = 4/5
The height at which the ball drops = 120 meters
Let the distance be n
The total distance it travels to come to rest position = 120 + 120 + n/1 – n
= 120 + 120 + (4/5 )/1-(4/5)
= 120 + 96/ 1-4/5
= 1080 kms
Therefore, the total distance is 1080 kms.
Problem 2:
8 years hence the sum of A’s and B’s age is 70 years. 4 years ago, the ratio between the sum of ages of A and B together and C’s age was 2:1. What is the present age of C?
Solution:
As given in the question,
The sum of A’s and B’s age is 70 years.
A + B = 70
As given, the ratio between the sum of ages of A and B together and C ‘s age was 2:1
The equation will be A + B – 8/C-4 = 2/1
23 = C-4
C = 17
Therefore, the present age of 17 years
Problem 3:
The income of a company doubles after one every year. If the initial income was Rs. 4 lakhs. What would be the income after 5 years?
Solution:
As given in the question,
Incomes double every year
The initial income = Rs 4 lakhs
After the first year, the income will become = 4 * 2 = 8 lakhs
After the 2 nd year, the income will become = 8 * 2 = 16 lakhs
After the 3 rd year, the income will become = 16 * 2 = 32 lakhs
After the 4th year, the income will become = 32 * 2 = 64 lakhs
After the 5th year, the income will become = 64 * 2 = 128 lakhs
Hence, the income after 5 years = 128 lakhs
Problem 4:
Man earns Rs 20 on the first day and spends Rs 15 on the next day. He again earns Rs 20 on the third day and spends Rs 15 on the next day. If he continues to save like this, how soon will he have Rs 60 in his hand?
Solution:
As given in the question,
The amount of money man earns = Rs 20
The amount of money man spent = Rs 15
Therefore, for every 2 days, he saves Rs 5.
For the first 2 days, the amount he saves = Rs 5
For the first 4 days, the amount he saves = Rs 10
For 8 days, he saves = Rs 20
For 16 days, he saves = Rs 40
Therefore, he saves Rs 60 in 18 days
Thus, he takes 18 days to save Rs 60.
Hence, the final solution is 18 days.
Problem 5:
If Raju have 24 hamburgers and he eats half of them in one hour, then his dog eats half of what was left in one hour, how many hamburgers do he have left?
Solution:
As given in the solution,
No of hamburgers Raju have = 24
Amount of burger he eats in one hour = 1/2
Amount of burger his dog eats = 1/2
Therefore, the no of burgers left = 24 * 1/2 * 1/2
= 24 * 1/4 = 6
Hence, Raju has 6 hamburgers left with him.
Thus, the final solution is 6 hamburgers.
Practice Questions & Worksheet on Simplification
Problem 6:
Number of working days of a school is 5 days per week and I spend 5 hours in class each day.. Boring!!! How much time do I spend in class each week?
Solution:
As given in the question,
No of hours I spent in each day = 5
No of working days = 5
Therefore, to find the total time spend in class each week, we apply the simplification rule of multiplication.
Hence, the total time spend = 5 * 5 = 25
I attend 25 hours of class in each week.
Thus, the final solution is 25 hours.
Problem 7:
Find the three consecutive even integers such that 6 times the sum of first and the third is 24 and greater than 11 times the second.
Solution:
As given in the question,
Let the three consecutive even integers = x, x+2,x+4
Also given, 6 times the sum of first and the third is 24 and greater than 11 times the second
Therefore, 6(x+x+4)=11(x+2)+24
6(2x+4)=11x+22+24
2x+24 = 11x + 46
On further simplification, x= 22
Then, x+2 = 24
and x+4 = 26
Therefore, the three consecutive even integers are 22,24 and 26
Hence, The final solution is 22,24 and 26
Problem 8:
Leonard found four consecutive integers such that 3 times the sum of the first and fourth was 114 less than the product of -5 and the sum of the first two. What were the integers?
Solution:
Let the four consecutive integers be x, x+1,x+2,x+3
As given in the question,
3 times the sum of the first and fourth was 114 less than the product of -5 and the sum of the first two
=3(x+x+3) = -5(x+x+1)-144
=3(2x+3) = -5(2x+1)-144
=6x+9 = -10x-5-144
=6x+9 = -10x-119
=16x+9 = -119
=16x/16 = -128/16
= x=-8
As considered, the consecutive integers are x, x+1,x+2,x+3
Therefore, the integers are -8,-7,-6,-5
Problem 9:
The sum of 2 numbers is 36 and their product is 248. What will be the sum of their reciprocals?
Solution:
As per the given question,
A+B=36
A * B = 248
To find the sum of their recriprocals, we have to simplify the equation 1/A+1/B
=B+A/BA = 36/248 = 9/62
Therefore the sum of their reciprocals = 9/62
Problem 10:
In a poultry farm having hens and pigs, Rohan can see 84 heads and 282 legs. How many hens are there?
Solution:
As per the given question,
Every hen has 1 head and 2 legs
Every pig has 1 head and 4 legs
Therefore, hens + pigs = 84 is the first equation
2(Hens) + 4 (Pigs) = 282 is the second equation
On simplifying both the equations, we get
4Hens + 4Pigs – 2Hens – 4Pigs =336-282
2Hens = 54
Hens = 27
Therefore, 27 hens are present in the poultry farm.
The final solution is 27 hens.
Problem 11:
The summation of 5 consecutive numbers is found out to be 335. If we add the largest and smallest number what will we get?
Solution:
Let the numbers be x,x+1,x+2,x+3,x+4,x+5 = 335
5x + 10 = 335
5x = 325
x = 65
Therefore, the numbers are 65,66,67,68,69
To find the sum of largest and smallest number, we have to add 65 and 69
Therefore, the solution is 134
Hence, the final solution of adding the largest and smallest number is 134.
Problem 12:
Since Raj was not paying attention in class, instead of multiplying M by 3/4, he divided by 3/4. This led to the difference of 14 between the two answers. What is the value of M?
Solution:
As given in the question,
The number that has to be multiplied = 3/4
The number Raj divided = 3/4
The difference of two answers = 14
To find the value of M, we simplify the equation as
M*3/4 – M/3/4 = 14
3M/4 – 4M/4 = 14
9M-16M/12 = 14
-7M/12 = 14
M=-24
Hence, the final solution is -24
Problem 13:
Raman has 2 urns. Both these urns have the same pebbles. If 20 pebbles from urn B are shifted to urn A, then the number of pebbles in both urns get interchanged. But if 10 pebbles from urn A are put into urn B, then the number of pebbles in urn B is twice the number of pebbles in urn A. How many pebbles do A and B respectively have?
Solution:
As given in the question,
No of urns Raman has = 2
Also given,
20 pebbles from urn B are shifted to urn A, then the number of pebbles in both urns get interchanged and 10 pebbles from urn A are put into urn B, which are twice the number of pebbles in urn A
On simplifying the equation, we get
B-20 = A
2(A-10) = (B+10)
2A-20 = B+10
2A-B = 30
B-A+2A-B = 20+30
A = 50
Therefore, B = A + 20 = 50 + 20 = 70
No of pebbles A and B have are 50 and 70
Therefore, the final solution is 50 and 70 respectively.
Problem 14:
On exchanging the digits in unit’s and ten’s place, the difference between original and new numbers become 27. The digits in the unit place are 2 times the digit in the hundred’s place. The digit in the ten’s place is 3 times the digit in the hundred’s place. What is 75% of the original number?
Solution:
As given in the question,
U = 2H
H = 3H
(3*100+4*10+7*1)
(H*100+T*10+U*1)
100H +30H + 2H = 132H
100H + 10T + U
100H + 20H + 3H = 123H
132H -123H = 27
9H = 23
H = 3
Also given,
U = 6
T = 9
HTU = 396
Therefore, 75% of the original number = 396
Thus, the final solution is 396
Problem 15:
In an area 2% of families have 5 children each. But 8% have no children at all. Amongst the rest of the families 18% have 4 children and 27% have only one child. How many families live in the area, if 297 families have either 2 or 3 children each?
Solution:
As given in the question,
Percentage of families having 5 children = 2%
Percentage of families having no children = 8%
Total percentage of families = 2% + 8% = 10%
Consider there are 100% families
Therefore, no of families with above data = 100=10 = 90 families
Percentage of families having 4 children = 18%
Percentage of families having 1 child = 27%
Total no of families in those 90 families = 45% of 90
To find the families with 2 or 3 children
100%-45% = 55% of 90
55/100* 90 = 99/2 = 44.5 families
For 100% of families, 44.5 families have children
To find the families who have either 2 or 3 children
= 297 * 100/49.5 = 600
Therefore, 600 families have either 2 or 3 children.
Problem 16:
A group wanted to renovate their club. Each member contributed an amount equal to twice the number of members in the club. But the government contributed same amount as the number of members. If each member had contributed the same amount as the number of numbers and the government had given the twice amount of the members, then they would have Rs.210 less. How many members are there?
Solution:
M*2M+M
(2M2 + M) – (M2 + 2M)=210
M2 – M = 210
(M – 15) (M + 14) = 0
M * M = 2M
M2 + 2M
M-15 = 0
M = 15
Therefore, there were 15 members in the group who wanted to renovate their club.
The final solution is 15 members.