LIMITATION ACT

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BASIC MATHEMATICS : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 1 www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS BASIC MATHEMATICS AFTERSCHO?OL – DEVELOPING CHANGE MAKERS CENTRE FOR SOCIAL ENTREPRENEURSHIP PGPSE PROGRAMME – World’ Most Comprehensive programme in social entrepreneurship & spiritual entrepreneurship OPEN FOR ALL FREE FOR ALL

BASIC MATHEMATICS : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 2 www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS BASIC MATHEMATICS Dr. T.K. Jain. AFTERSCHO?OL Centre for social entrepreneurship Bikaner M: 9414430763 tkjainbkn@yahoo.co.in www.afterschool.tk, www.afterschoool.tk

11% of a number exceeds 7% of the same by 18, then What is the number? : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 3 11% of a number exceeds 7% of the same by 18, then What is the number? As we can see the difference is (11-7) = 4 4% is equal to 18 Thus 100% should be equal to : 18/4 * 100 = 450 answer.

If x is 90% of y, what percent of x is y? : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 4 If x is 90% of y, what percent of x is y? Let us assume Y to be 100. thus X is 90 Y is 100/90*100 = 111% of X. answer.

If the numerator of a fraction is increased by 20% and its denominator is decreased by 10%, the fraction becomes : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 5 If the numerator of a fraction is increased by 20% and its denominator is decreased by 10%, the fraction becomes Let us assume the numerator has become : 120 Similarly, the denominator has become 90. Thus the new fraction is : 4/3 of the previous fraction. Answer.

The average weight of a group of boys and girls is 38 kg. average weight of the boys is 42 kg, and that of thegirls is 33kg. If the number of boys is 25, find the number of girls? : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 6 The average weight of a group of boys and girls is 38 kg. average weight of the boys is 42 kg, and that of thegirls is 33kg. If the number of boys is 25, find the number of girls? 38 is the mid point, so take difference from both the numbers. We get (42-38) = 4 and (38-33) = 5. thus the ratio of boys and girls is 5: 4. Thus the number of girls is 4/5*25= 20 answer.

The average age of 10 boys is l5years.Of this 3 boys having an average age of 22 years left. Whatis the average age of the remaining boys? : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 7 The average age of 10 boys is l5years.Of this 3 boys having an average age of 22 years left. Whatis the average age of the remaining boys? Total age = 10 *15 = 150 3 left out : 22*3 = 66 so remaining 150-66 =84 The average of remaining =84/7= 12 years answer.

The cost of carpeting a room 15 metres long with a carpet 75 cm wide at 30 paise per metre is Rs. 36. The breadth of the room is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 8 The cost of carpeting a room 15 metres long with a carpet 75 cm wide at 30 paise per metre is Rs. 36. The breadth of the room is Area of room = length * bredth Total length of carpet : 36/.3 = 120 meters Total area of carpet used :=.75 * 120= 90 sqmeters. Thus width of the room = 90/15= 6 meters. Answer. (length = 15, width = 6 meters). Answer.

The length of a rectangle is doubled while its breadth is halved. Whatis the percentage change in area ? : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 9 The length of a rectangle is doubled while its breadth is halved. Whatis the percentage change in area ? Let us assume that length is 20 and width is 10. Length is doubled = 40, but wideth is halved = 5. new area is 200. Thus area remain unchanged.

The length of a rectangle is increased by 60%. By what percent would the width have to be decreased to maintain the same area ? : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 10 The length of a rectangle is increased by 60%. By what percent would the width have to be decreased to maintain the same area ? Let us assume the length be 100, width be 50. We have new length 160. Thus new width is 5000/160 = 31.25. Thus there is reduction of 37.5% in the width. Answer.

A rectangular plot of land with an area 600 sq.m is fenced, length of fencing being 100 m. Length of the plot is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 11 A rectangular plot of land with an area 600 sq.m is fenced, length of fencing being 100 m. Length of the plot is Area = length X bredth L + B = 50 (half of total fence, which is 100 meters) and L * B = 600 The length and width are 30 and 20 respectively. Answer.

A 5 m wide lawn is cultivated all along the outside of a rectangular plot measuring 90 m X 40 m. The total area of the lawn is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 12 A 5 m wide lawn is cultivated all along the outside of a rectangular plot measuring 90 m X 40 m. The total area of the lawn is We can add lawn on either side of length and width. We have new area 100*50 = 5000. The area of rectangle is 90*40 = 3600. Thus area of lawn is 5000-3600=1400 sq.m. answer.

Around a rectangular garden of length 10 m and width 5 m, a road I m wide is laid. The cost of metalling the road at Rs. 200 per sq. m is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 13 Around a rectangular garden of length 10 m and width 5 m, a road I m wide is laid. The cost of metalling the road at Rs. 200 per sq. m is Let us add 1 meters on either side of length and width. We get 12*7 = 84. the area of garden is 50. Thus the area of road is 84-50 = 34 sq.meters. The cost is : 34*200 = Rs. 6800 answer.

The area of a rectangular field is 52000 sq m This rectangular area has been drawn on a map to the scale 1 cm to 100 m. The length is shown as 3.25 cm on the map. The breadth of the rectangular field is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 14 The area of a rectangular field is 52000 sq m This rectangular area has been drawn on a map to the scale 1 cm to 100 m. The length is shown as 3.25 cm on the map. The breadth of the rectangular field is The length is 3.25*100= 325 meters. The width is : 52000/325= 160 meters. thus the width is 1.6 CM answer.

What is the square root of .4? : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 15 What is the square root of .4? It is not .4, it is .40 Thus square root of .40 is .632 … answer.

The cost of fencing a square garden is Rs. 400 at the rate of Re. I per metre. The area of the garden in square metres is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 16 The cost of fencing a square garden is Rs. 400 at the rate of Re. I per metre. The area of the garden in square metres is One wall is of 100 meters (because 4 walls are of 400 meters). The area of square field is : 100*100 = 10000 Sq. meters.

If the side of a square be increased by 50%, the percent increase in area is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 17 If the side of a square be increased by 50%, the percent increase in area is Suppose the side of square 100 Area : 100 * 100 = 10000 Now we increase the side by 50% : 150 * 100 = 15000 The increase is by 5000 or 50%. If both the sides are increased, the new are will be : 150 * 150 = 22500 the growth is of 125% increase. Answer.

If the area of an equilateral triangle is 24 .5 sq. m, then its perimeter is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 18 If the area of an equilateral triangle is 24 .5 sq. m, then its perimeter is Area of equilateral triangle = sqrt(3)/4 * side * side = 24.5 Side *side = 24.5 * 4 / (sqrt (3) Side * side = 98 / sqrt (3) = 56 Side = 7.52 Perimter = 7.52 * 3 = 22.5 answer.

The ratio of the area of a square to that of the square drawn on its diagonal is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 19 The ratio of the area of a square to that of the square drawn on its diagonal is Let us assume the side of square is 1. its diagonal is sqrt (2) Area of square is 1, then area of square on diagonal is 2. Thus the ratio is 1: 2 answer. Sqrt(2) 1 (side)

The base of a right angled triangle is 5 metres and hypotenuse is 13 metres. Its area will be : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 20 The base of a right angled triangle is 5 metres and hypotenuse is 13 metres. Its area will be Altitude = sqrt (13^2 – 5^2) = sqrt (169 – 25) =sqrt (144) = 12 Area = ½ * base * height =1/2 * 5 * 12 = 30 answer. 13 5

The area of an equilateraltriangle whose side is 8 cms, is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 21 The area of an equilateraltriangle whose side is 8 cms, is Area of equilateral triangle = sqrt(3)/4 * side * side = =sqrt (3) / 4 * 8 * 8 =27.3 approx. answer.

A room 5.44 m x 3.74 m is to be paved with square tiles. The leastnumber of tiles required to cover the floor is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 22 A room 5.44 m x 3.74 m is to be paved with square tiles. The leastnumber of tiles required to cover the floor is Let us find HCF 544 and 374. for this divide 544 by 374, we get remainder of 170, now divide 374 by 170, we get 34 as remainder, now divide 170 by 34, we get 0 as remainder. Thus one side of tile should be 34 cm. the number of tiles is (544*374)/ (34*34) = 176 answer.

The sides of a triangular board are 13 metres, 14 metres and 15 metres.The cost of painting it at the rate of Rs. 8.75 per sq. m. is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 23 The sides of a triangular board are 13 metres, 14 metres and 15 metres.The cost of painting it at the rate of Rs. 8.75 per sq. m. is (13 + 14 + 15) / 2 = 21 = sqrt((21-13) * (21- 14) * (21-15) *21) =84 Total cost = 84 * 8.75 = 735 rupees. Answer.

Area of four walls of a room is 77 square metres. the length and breadth of the room are 7.5 metres and 3.5 metres respectively. The height of the room is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 24 Area of four walls of a room is 77 square metres. the length and breadth of the room are 7.5 metres and 3.5 metres respectively. The height of the room is 2(7.5*height) + 2(3.5*height) = 77 15 H + 7 H = 77 22 H = 77 or H = 77/22 = 3.5 answer.

The cost of papering the four walls of a room is Rs. 480. Each one of the length, breadth and height of another room is double that of this room. The cost of papring the walls of this new room is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 25 The cost of papering the four walls of a room is Rs. 480. Each one of the length, breadth and height of another room is double that of this room. The cost of papring the walls of this new room is The cost will be 4 times, thus cost will be 1920 .

The length of a rope by which a cow must be tethered in order that she may be able to graze an area of 9856 sq.m. is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 26 The length of a rope by which a cow must be tethered in order that she may be able to graze an area of 9856 sq.m. is Area of circle : pi * radius * radius Radius * radius = 9856 / (22/7) Radius = 56 meters. Answer.

The length of minute hand on a wall clock is 7 cms . The area swept by the minute hand in 30 minutes is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 27 The length of minute hand on a wall clock is 7 cms . The area swept by the minute hand in 30 minutes is Area of half circle = ½ * pi * radius * radius = ½ * 22/7 * 7 * 7 = 77 sq. cm.

An equilateral triangle, a circle, a regular hexagon and a square have the same perimeter . Which one will enclose the minimum area? : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 28 An equilateral triangle, a circle, a regular hexagon and a square have the same perimeter . Which one will enclose the minimum area? Let us assume the perimeter to be 300 Triangle : sqrt(3) / 4 * 10000 area = 4330 Area of square = 75 * 75 = 5625 Area of circle = 2pi *r = 300, r = 150/pi Area = (150 * 150 ) / (22/7) = 7159 Area of hexagon = 6 *((sqrt(3) / 4 ) * 50 * 50 ) =6495 Thus triangle has minimum area. Answer.

Four circular cardboard pieces, each of radius 7 meters are placed in such a way that each piece touches two other pieces. The area of the space enclosed by the four pieces is : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 29 Four circular cardboard pieces, each of radius 7 meters are placed in such a way that each piece touches two other pieces. The area of the space enclosed by the four pieces is There is a square and a circle. The area of square is 14*14 = 196. The area of circle is : pi*7*7 = 154 Thus the area enclosed is 196-154= 42 sq.meters. Answer

Example: : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 30 Example: Area to calculate Radius of a circle = 7 Side of a square formed from the midpoints of the circles = 14 ( 7 + 7)

About AFTERSCHO?OL : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 31 About AFTERSCHO?OL PGPSE - World’s most comprehensive programme on social entrepreneurship – after class 12th Flexible – fast changing to meet the requirements Admission open throughout the year Complete support from beginning to the end – from idea generation to making the project viable.

Branches of AFTERSCHO?OL : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 32 Branches of AFTERSCHO?OL PGPSE programme is open all over the world as free online programme. Those who complete PSPSE have the freedom to start branches of AFTERSCHO?OL A few branches have already started - one such branch is at KOTA (Rajasthan).

Workshop on social entrepreneurship : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 33 Workshop on social entrepreneurship We conduct workshop on social entrepreneurship – all over India and out of India also - in school, college, club, association or any such place - just send us a call and we will come to conduct the workshop on social entrepreeurship. These workshops are great moments of learning, sharing, and commitments.

FREE ONLINE PROGRAMME : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 34 FREE ONLINE PROGRAMME AFTERSCHO?OL is absolutely free programme available online – any person can join it. The programme has four components : 1. case studies – writing and analysing – using latest tools of management 2. articles / reports writing & presentation of them in conferences / seminars 3. Study material / books / ebooks / audio / audio visual material to support the study 4. business plan preparation and presentations of those plans in conferences / seminars

100% placement / entrepreneurship : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 35 100% placement / entrepreneurship AFTERSCHO?OL has the record of 100% placement / entrepreneurship till date Be assured of a bright career – if you join AFTERSCHO?OL

Pursue professional courses along with PGPSE : 

2/6/2009www.afterschoool.tk AFTERSCHO?OL's MATERIAL FOR PGPSE PARTICIPANTS 36 Pursue professional courses along with PGPSE AFTERSCHO?OL permits you to pursue distance education based professional / vocational courses and gives you support for that also. Many students are doing CA / CS/ ICWA / CMA / FRM / CFP / CFA and other courses along with PGPSE. Come and join AFTERSCHO?OL