Mensuration part 2

Problem 1:Find the area, perimeter and diagonal of the square whose side is 5m?

Solution:

Area of a square = a^{2}=5^{2}=25

Perimeter of a square = 4a =4*5= 20

Diagonal of a square= \sqrt{2}a=\sqrt{2}\times 5=5\sqrt{2}

Problem 2:The altitude of an equilateral triangle is \sqrt{3} cm. What is its perimeter?

Solution:

Altitude means height

h=\frac{\sqrt{3}}{2}a

\sqrt{3}=\frac{\sqrt{3}}{2}a

a=2

Perimeter = 3a=3\times 2=6

Problem 3: The length and breadth of a room are 6 meters and 4.5 meters respectively. Find the floor area of the room?

Solution:

Length = 6

Breadth = 4.5

Area = 6*4.5 = 27

Mensuration part 1

There are two types of problems in mensuration

2 dimensional : it contains length and breadth

area

perimeter

3 dimensional : it contains length, breadth and height

volume

total surface area

curved surface area

Important formulas:

Square:

Area of square = a^{2}

Perimeter of a square is= 4a

Diagno1 = \sqrt{2}a

When diagno1 is given area =\frac{d^{2}}{a}

Rectangle:

Area of rectangle = length * breadth

Perimeter = 2(Length + Breadth)

d=\sqrt{l^{2}+b^{2}}

Triangle:

\angle A+\angle B+\angle C=180

S=\frac{a+b+c}{2}

Area of a triangle = \frac{1}{2}\times base\times height

Scalen triangle

Area of triangle when sides are given = \sqrt{s(s-a)(s-b)(s-c)}

Equilateral triangle:

Area of equilateral triangle = \frac{\sqrt{3}}{4}a^{2}

Height of a triangle = \frac{\sqrt{3}}{2}a

p=3a

Isosceles triangle

A=\frac{b}{4}\sqrt{4a^{2}-b^{2}}

h=\frac{1}{2}\sqrt{4a^{2}-b^{2}}

p=2a+b

RIGHT ANGLE TRIANGLE

Perimeter= Perpendicular + base + hypothesis

Area = \frac{1}{2}\times Perpendicular\times base

Circle :

Area of circle = \Pi r^{2}

Circumference of a circle = 2\Pi r

\Pi =\frac{22}{7}=3.14

Trapezium:

In trapezium opposite sides are equal

Area of trapezium = \frac{1}{2}(a+b)\times h

Parallelogram:

Area of parallelogram = \frac{1}{2}(h_{1}+h_{2})\times D

Perimeter = a+b+c+d

Rhombus:

In rhombus all sides are equal and diagonal are not equal

Area of rhombus = \frac{1}{2}\times d_{1}\times d_{2}

SECTOR:

Length of arc = 2\Pi r\frac{\Theta }{360}

Perimeter = 2r(\Pi\times \frac{\Theta }{360}+1)

Area = \Pi r^{2}\frac{\Theta }{360}

Allegations and mixture part 6

Problem 1: A jar contains a mixture of two liquids A and B in the ratio 3 : 1. When 15 liters of the mixture is taken out and 9 liters of liquid B is poured into the jar, the ratio becomes 3 :4. How many liters of liquid A was contained in the jar?

Solution:

A  :  B

3  :  1

3  :  4

1-4=3

3——->9

1——–>?\frac{9}{3}=3

4——->?4*3=12

12+15=27

27\times \frac{3}{4}=20.25

Problem 2:A vessel is filled with liquid, which is 3 parts water and 5 parts milk. How much of the liquid should be drawn off and replaced by water to make it half water and half milk?

Solution:

w   :   m

3    :    5————-> 3    :   5

1    :    1*5———–>5   :   5

\frac{2}{8+2}=\frac{2}{10}=\frac{1}{5}

Allegations and mixture

\frac{1}{5} is answer

Problem 3: In three classes X, Y and Z an algebra test is conducted. The average score of class X is 83. The average score of class Y is 76 and the average score of class Z is 85. The average score of class X & Y is 79 and the average score of class Y & Z is 81. What is the average score of the classes X, Y and Z.?

Solution:

x : y : z= 3 : 4 : 5
Average score of a class
\frac{3\times 83+4\times 76+85\times 5}{3+4+5}=\frac{978}{12}=81.5

 

Allegations and mixture part 5

Problem 1: There are two mixtures A and B, both contains milk and water .In mixture A ratio of milk and water is 5:4 whereas in mixture B ratio of milk and water is 5:7. What part of mixture B should be taken out and replaced with mixture A so that in mixture B quantity of milk and water becomes equal ?

Solution:

Answer is \frac{2}{3}

Problem 2: Milk and water are in c an A as 4 : 1 and in can B as 3 : 2. For Can C, if one takes equal quantities from A and B, find the ratio of milk to water in can C ?

Solution:

\frac{x-\frac{3}{5}}{\frac{4}{5}-x}=\frac{1}{1}

x-\frac{3}{5}=\frac{4}{5}-x

2x=\frac{7}{5}

x=\frac{7}{10}

here 7 is milk and 3 parts are water

7  :  3 is answer

Problem 3: A jar contains a mixture of two liquids A and B in the ratio 4:1. When 10 liters of mixture is taken out and 10 liters of liquid B is poured into the jar, the ratio becomes 2:3. How many liters of liquid A was contained in the jar?

Solution:

A  :  B = 4X  :  X——->5X

In 10 liters 8 parts is milk and 2 parts is water

\frac{4x-3}{x-2+1}=\frac{2}{3}

12x-24=2x+16

10x=40

x=4

4*4=16

Al ligation method

1  :  1

1——->10

1——-> Is remaining mixture

1———->10+10=20

20\times \frac{4}{5}=16

20\times \frac{1}{5}=4

Shortcut

A  :  B

4  :  1—————>4  :  1

2  :  3 *2———–>4  :   6

6-1=5

5——>10

1——->?2

10+10=20

20\times \frac{4}{5}=16

20\times \frac{1}{5}=4

Allegations and mixture part 4

Problem1: Some amount out of Rs.7000 was lent at 6% per annum and the remaining was lent at 4% per annum. If the total simple interest from both the fractions in 5 years was Rs.1600, the sum lent at 6% per annum was?

Solution:

S.I=\frac{PTR}{100}

1600=\frac{7000\times 5\times R}{100}

R=\frac{32}{7}

2  :  5

\frac{2}{7}\times 7000=2000

Problem 2: The blowing average of a cricketer was 12.4. he improves his bowling average by 0.2 points when he takes 5 wickets for 26 runs in his last match. The number of wicket taken by him before the last match was ?

Solution:

Blowing Average =\frac{Runs}{Wicket}

Let number of wickets = x

Runs=12.4X

12.2=\frac{12.4x+26}{x+5}

Solve the equation

Shortcut:

\frac{26}{5}=5.2

7  :  0.2—–>70  :  2—–>35  :  1

1——–>5

35——>? 35*5 = 175

Problem 3:Two mixtures A and B contain milk and water mixed in the ratio 8 : 5 and 5 : 2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing milk and a water in the ratio 9 : 4?

Solution:

\frac{2}{91}:\frac{1\times 7}{13\times 7}

\frac{2}{91}:\frac{7}{91}

2  :  7

Allegation and mixture part 3

Problem 1:The cost of Type 1 material is RS.15 per kg and Type 2 material is RS 20 per kg. If both type 1 and Type 2 are mixed in the ratio of 2:3, then what is the price per kg of mixed variety of material?

Solution:

Problem2 :In an examinations, a students gets 3 marks for every correct answer and loses 1 mark for every wrong answer. If he scores 120 marks in a paper of 100 questions, how many of his answers were correct assuming that he attempts all the questions?

Solution:

Let the number of correct answers = x

number of wrong answers = 100-x

3x + (100-x)-1 = 100

Solve the equation

Shortcut:

\frac{120}{3}=40

\frac{1}{4}\times 60=15

40+15=55

Problem 3:A man possessing Rs. 8400 lent a part of it at 8% simple interest and the remaining at 20/3% simple interest .His total income after 1 ½ years was Rs. 882. Find the sum lent at 8% rates ?
Solution:
S.I=\frac{PTR}{100}
882=\frac{8400\times R\times 3}{2\times 100}=7%
4————->8400
1————–>?\frac{8400}{4}=2100

Allegation and mixtures part 2

Problem 1:A mixture of 70 liter of wine and water contains 10% water. How much water should be added to the mixture to make 37% water in the resulting mixture?

Solution:

Basic method

Wine              Water

90%                  10%    ——>  ( 9   :   1)*7 —–>63  :  7

63%                  37%  ———>  (63 : 37)*1 —–> 63   :  37

63 + 7 =70

30 liter of water is added

shortcut:

Mixture        Water

7                    3

Wine            Water

4                        3

7——->70

1———>?\frac{70}{7} = 10

3——->?3*10 =30

Problem 2:In a zoo, there are rabbits and pigeons. If heads are counted, there are 200 and if legs are counted, there are 580. How many pigeons are there?

Solution:

pigeons = x

Rabbit = y

x+y=200

2x+4y = 580

Solve this equation

Shortcut:

\frac{580}{200}=2.9 legs/heads

pigeon =11

Rabbit = 9

\frac{11}{20}\times 200=110

Problem 3: In what ratio must a grocer mix tea at RS. 60 a kg and 65 a kg so that by selling a mixture at RS. 68.20 a kg, he may gain 10%?

Solution:

110%——>68.20

100%——->?\frac{6820\times 100}{110}=62

Alligations and mixtures part 1

Problem 1: In what ratio two varieties of milk costing  Rs 16 and RS 18 per liter respectively mixed, so the mixture cost RS.16.60 per liter ?

Solution:

Basic metod:

16——->x——–>16x

18——->y———>18y

16x+18y = 16.60(x+y)

16x+16y = 16.60x+16.60y

0.60x = 1.40y

\frac{x}{y}=\frac{14}{6}=7:3

shortcut:

Problem 2: The average weight of 30 students of a class is 45 kg . The average weight of girls is 37 kg. and that of the boy is 49 kg. Find the number boys in the class ?

Solution:

Basic method:

Boys =x

Girls = (30-x)

x*49 + (30-x)*37 = 30*45

Solving this equation is time taking process.

Shortcut:

2+1=3

3——->30

1———>?\frac{30}{3}=10

2———->?2*10=20 BOYS

Problem 3:A shopkeeper sold 40kg of goods partly at 7% profit and rest at 5% loss, suffering 2% loss on the whole transactions. What is the quantity sold at profit?

Solution:

CP = 100

1+3=4

4——–>40

1———>?\frac{40}{4}=10 KG sold for profit

3———->?3*10 = 30

Ages part 4

Problem 1:Ram got married 10 years ago. His present age is 1\frac{1}{3} times of the age at the time of marriage. Ram’s sister was 4 year younger to him at the time of his marriage. Find out the present age of Ram’s sister?

Solution:

1\frac{1}{3}——->\frac{4}{3}

R            M

4             3

1——–>10

4———>?4*10=40

3———>?3*10=30

40-4=36 is  the present age of Ram’s sister

Problem 2:If the two digits of the age of Arun are reversed then the new age so obtained is the age of his wife. 1/11 part of the sum of their ages is equal to the difference between their ages. If arun is older than his wife then find the difference between their ages?

Solution:

Assume age of Arun =10x+y

Age of his wife = 10y+x

\frac{1}{11}(11x+11y)=9x-9y

10y=8x

\frac{x}{y}=\frac{10}{8}=\frac{5}{4}

A=54     w = 45

Difference between the ages is 9

Problem 3:Arun,s brother is 3 years ealder to him. his father was 28 years of age when his sister was born while his mother was 26 years of age whaen he was born, The ages of arun,s father when his brother was born?

Solution:

Father age =28——->28+4=32

Mother age = 26——>26+4=30

Sister age=0——–>0+4 = 4

brother age = 0

The age of arun’s father when the brother born is 32

Problem 4: If 6 years are subtracted from the present age of Randheer and the remainder is divided by 18 then the present age of his grandson Anup is obtained. If the anup is 2 years younger to mahesh whose age is 5 years, then whats the age of Randheer?

Solution:

Age of mahesh is= 5

Age of Anup is = 5-3 =2

\frac{R-6}{18}=3

R= 54+6 =60

Ages part 3

Problem 1:16 years ago, my uncle was 8 times more older than me. After 8 years from today, my uncle will be thrice as old as I will be at that time. Eight years ago, What was the ratio of my age and my uncle’s age?

Solution:

U^{-16}:M^{-16}=9:1——->difference is 8 (9-1=8)

U^{+8}:M^{+8}=3:1———->difference is (3-1=2)

U^{-16}:M^{-16}=9:1 * 1———–>9:1

U^{+8}:M^{+8}=3:1  *  4————>12:4

12-9 = 3          4-1 = 3

8-(-16)=24

3———>24

1———->?\frac{24}{3}=8

Age after 8 years

12———>?12*8=96

4———->?4*8=32

Age 8 years ago 8+8=16

96-16=80

32-16=16

M      :       U

16     :        80

1        :         5

Problem 2:If Sita’s mother was 4 times as old as sita 10 years ago. After 10 years mother will be twice as old as sita. How old is Sita’ s mother at present?

Solution:

M^{-10}:S^{-10}=4:1——–>difference is 3 (4-1 = 3)

M^{+10}:S^{+10}=2:1——–>difference is 1 (2-1=1)

M^{-10}:S^{-10}=4:1 * 1——–>4:1

M^{+10}:S^{+10}=2:1 *  3——–>6:3

6-4=2  ,   3-1=2

10-(-10) = 20

2———->20

1———–>\frac{20}{2}=10

Age after 10 years

6———–>?6*10 = 60

3————>?3*10 = 30

Sita’s mother present age = 60-10 = 50

Problem 3: Jayesh is as more younger to Amit as Jayash is older to prashant. If the sum of the ages of Amit and Prashant is 48 years. What is the age of Amit in years?

Solution:

A-J=X

J-P=X

A-J = J-P

A+P = 2J

48 = 2J

J = 24

We can derive only jayaesh age based on the given data

we can not determined Amith and prasanth age