*Simplifications**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] +...............+[1/9-1/10]

= [ ½ – 1/10]

= 4/15 = 2/5

6.The value of 999 of 995/999* 999 is:

Sol: [1000- 4/1000]*999 = 999000-4

= 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

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