Simplifications problems
Posted by
Ravi Kumar at Monday, November 24, 2008
Share this post:

Simplifications problems
Simplifications problems
Introduction:
'BODMAS' rule: This rule depicts the correct sequence in which
the operations are to be executed, so as to find out the value of
a given expression.
Here B stands for Bracket, O for Of, D for Division, M for
Multiplication, A for Addition and S for Subtraction.
First of all the brackets must be removed, strictly in the
order () , {} , [].
After removing the brackets, we want use the following operations:
1.Of 2. Division 3. Multiplication 4. Addition 5. Subtraction
Modulus of a real number:
Modulus of a real number is a defined as
a = a, if a>0 or a, if a < 0;
Problems:
1.(5004 /139) – 6= ?
Sol: Expression = 5004/ 139 – 6 = 36 – 6 = 30;
2.What mathematical operations should come at the place of ? in the
equation : (2 ? 6 – 12 / 4 + 2 = 11) ?
Sol: 2 ? 6 = 11 + 12 / 4 – 2
= 11 + 3 – 2
= 12
2 * 6 = 12
3.( 8 / 88) * 8888088 = ?
Sol : (1/11) * 8888088 = 808008
4.How many 1/8's are there in 371/2 ?
Sol: (371/2) /(1/8)= (75/2) /(1/8) = 300
5.Find the values of 1/2*3 +1/3*4 +1/4*5+ .................+1/9*10 ?
Sol: 1/2*3 +1/3*4+1/4*5+ ..................+1/9*10
= [½ 1/3] +[ 1/3 – ¼] + [¼ 1/5] +...............+[1/91/10]
= [ ½ – 1/10]
= 4/15 = 2/5
6.The value of 999 of 995/999* 999 is:
Sol: [1000 4/1000]*999 = 9990004
= 998996
7.Along a yard 225m long, 26 trees are planted at equal distance, one
tree being at each end of the yard. what is the distance between two
consecutive trees ?
Sol: 26 trees have 25 gaps between them.
Hence , required distance = 225/ 25 m= 9m
8.In a garden , there are 10 rows and 12 columns of mango trees. the
distance between the two trees is 2 m and a distance of one meter is
left from all sides of the boundary of the length of the garden is :
Sol: Each row contains 12 plants.
leaving 2 corner plants, 10 plants in between have 10 * 2 meters and
1 meter on each side is left.
length = (20 + 2) m = 22m
9.Eight people are planning to share equally the cost of a rental car,
if one person with draws from the arrangement and the others share
equally the entire cost of the car, then the share of each of the
remaining persons increased by?
Sol: Original share of one person = 1/8
new share of one person = 1/7
increase = 1/7 – 1/8 = 1/56
required fractions = (1/56)/(1/8) = 1/7
10.A piece of cloth cost Rs 35. if the length of the piece would
have been 4m longer and each meter cost Re 1 less , the cost
would have remained unchanged. how long is the piece?
Sol: Left the length of the piece be x m.
then, cost of 1m of piece = Rs [35 / x]
35/ x – 35 /x+4 = 1
x + 4 – x = x(x+ 4)/35
x2 + 4x – 140 = 0
x= 10
11.A man divides Rs 8600 among 5sons, 4 daughters and 2 nephews.
If each daughter receives four times as much as each nephew, and
each son receives five as much as each nephew. how much does each
daughter receive ?
Sol:
Let the share of each nephew be Rs x.
then, share of each daughter Rs 4x.
share of each son = 5x Rs
so, 5 *5x+ 4 * 4x + 2x =8600
2x + 16x + 25x= 8600
43x = 8600
x = 200
share of each daughter = 4 * 200 = Rs 800
12.A man spends 2/5 of his salary on house rent, 3/10 of his salary
on food, and 1/8 of his salary on conveyance. if he has Rs 1400 left
with him, find his expenditure on food and conveyance?
Sol: Part of the salary left = 1[2/5 +3/10+1/9]
= 1 33/40
=7/40
Let the monthly salary be rs x
then, 7/40 of x = 1400
x= [1400*40]/7
x= 8000
Expenditure on food = 3/10*8000 =Rs 2400
Expenditure on conveyance= 1/8*8000 =Rs 1000