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- 1. Advanced Mathematics for GMAT CAT MAT 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 www.afterschoool.tk AFTERSCHO☺OL's MATERIAL FOR PGPSE PARTICIPANTS
- 2. Advanced Mathematics for GMAT CAT MAT Dr. T.K. Jain. AFTERSCHO ☺ OL Centre for social entrepreneurship Bikaner M: 9414430763 [email_address] www.afterschool.tk , www.afterschoool.tk www.afterschoool.tk AFTERSCHO☺OL's MATERIAL FOR PGPSE PARTICIPANTS
- 3. There are 2 tangents of 21 inches from point P to the circle with centre C inclined at an angle of 60 degree. Find the circumference of the circle? <ul><li>W will always be 90 degree. WP = 21 </li></ul><ul><li>WP=sqrt(3)/2 * CP </li></ul><ul><li>21=sqrt(3)/2 *CP </li></ul><ul><li>CP =42/ (sqrt(3) =24.24 </li></ul>P c W 60 degree
- 4. Solution continued… <ul><li>When the angle is 60 degree, CW should be half of CP. Thus radius should be ½*24 </li></ul><ul><li>Radius should be 12.12 and thus cirmuference should be : 2* 22/7*12.12 </li></ul><ul><li>= 76.18 ans. </li></ul>
- 5. In the next graph, you find RS =12, radius of the circle is 70. What is the distance of WX, when W is intersection of RS and diameter AX. <ul><li>Let us assume the centre to be C. C to R distance is = radius = 70. let us assume mid point of RS to beW. Now WR= 6. </li></ul>r s X A c W
- 6. solution <ul><li>WC^2= 70^2 – 6^2 </li></ul><ul><li>WC^2 = 4900-36 = 4864 </li></ul><ul><li>WC = sqrt (4864)= 69.7 </li></ul><ul><li>We already know that CX is 70 (radius). Thus length of WX = 139.7 </li></ul>
- 7. What is the probability of throwing a number greater than 3 with an ordinary die whose faces are numbered from 1 to 6 ? <ul><li>There are only 3 digits greater than 3 : 4,5,6 out of 6 digits. Thus probaility is 3/6 or ½ (probability = the number of favourable outcomes / total number of outcomes). </li></ul>
- 8. In a class of 12 students, 5 are boys and the remaining are girls. Find the probability that a student selected is a girl ? <ul><li>There are 7 girls – so the probability of selection of a girl is 7/12 answer. </li></ul>
- 9. A bag contains 7 white and 2 black balls. Find the probability of drawing a white ball. <ul><li>7/9 answer. </li></ul>
- 10. In a given race, the odds in favour of four horses, A, B, C, D are 1: 3, 1:4, 1: 5 and 1: 6 respectively. Assuming that a dead heat is impossible, find the chance that one of them will win the race. <ul><li>Probability that none of them wins : </li></ul><ul><li>¾ * 4/5 * 5/6 * 6/7= .428 </li></ul><ul><li>The probability that one of them will win the race = 1- .428 = .572 answer. </li></ul>
- 11. X^3 + 5x^2 +10k leaves remainder - 2x, when divided by (X^2 +2), what is k? <ul><li>Let us divide this number by (x^2 +2) </li></ul><ul><li>X^3 + 5x^2 +10k </li></ul><ul><li>- X^3 - 2x </li></ul><ul><li>_________________ 5x^2 – 2x +10k -5X^2 – 10 </li></ul><ul><li>----------------------------- </li></ul><ul><li>- 2x -10 +10k </li></ul><ul><li>Thus the remainder is : </li></ul><ul><li>- 2x – 10 + 10k= -2x </li></ul><ul><li>10k = 10 </li></ul><ul><li>K = 1 answer. </li></ul>
- 12. If (x-2) is a factor of (X^2 + 3qx – 2q), what is the value of q? <ul><li>As X – 2 is a factor, so take X = 2 </li></ul><ul><li>(X^2 + 3qx – 2q) = 0, and taking X=2, </li></ul><ul><li>Thus we will have the following equation: </li></ul><ul><li>4 + 6q – 2q = 0 </li></ul><ul><li>4q = -4 </li></ul><ul><li>q = -1 answer. </li></ul>
- 13. X^100 + 2x^99 + k is divisible by (x+1), what is the value of k? <ul><li>Let us take X = -1. </li></ul><ul><li>1 + 2 + k = 0 </li></ul><ul><li>K = -3 answer. </li></ul>
- 14. (x-1) is a factor of (X^3 – k), what is the value of k? <ul><li>Taking x = 1, we have following solution: </li></ul><ul><li>1 – K = 0 </li></ul><ul><li>K = 1 answer. </li></ul>
- 15. (X+2) and (X-1) are factors of (x^3 +10x^2 +mx +n), the values of m and n are ? <ul><li>Let us take X = -2 and 1. we have two solutions with these two values: </li></ul><ul><li>-8 +40 -2m + n =0 or -2m+n = -32 </li></ul><ul><li>1 + 10 + m +n = 0 or 2m +2n = -22 </li></ul><ul><li>3n = -54 or n = -18, and m = 7 answer. </li></ul>
- 16. What will be the remainder when (X^4 – 3x^3 +2x^2 -5x +7) is divided by (x-2)? <ul><li>Let us take X = 2. </li></ul><ul><li>16 – 24 + 8 – 10 + 7 </li></ul><ul><li>= -3 answer. </li></ul>
- 17. What will be the remainder if (5x^3 + 5x^2 – 6x +9) is divided by (x+3)? <ul><li>Let us take X = -3 </li></ul><ul><li>-135 +45 +18 + 9 </li></ul><ul><li>= -63 answer. </li></ul>
- 18. What will be the remainder if (X^11 + 1) is divided by (x +1)? <ul><li>Let us take X = -1 </li></ul><ul><li>-1 + 1 = 0 </li></ul><ul><li>Thus the remainder is zero. </li></ul><ul><li>Answer. </li></ul>
- 19. In order to make (X^4 - 11X^2Y^2 + Y^4) a perfect square, what should we add in it? <ul><li>(X^2 – Y^2)^2 = X^4 - 2X^2Y^2 + Y^4, </li></ul><ul><li>Thus in order to make it perfect square, we have to add 9x^2Y^2 to this number, so that it becomes a perfect square. Answer </li></ul>
- 20. What is the sum of (X^2 +1) and reciprocal of (x^2 -1)? <ul><li>We have to obtain sum of : </li></ul><ul><li>(X^2 +1) + ( 1/ (X^2 -1) </li></ul><ul><li>We get ( X^4 – 1 + 1) / ( X^2 – 1) </li></ul><ul><li>Solving further, we get </li></ul><ul><li>(X^4 / (X^2 – 1)) </li></ul><ul><li>Answer. </li></ul>
- 21. WHAT IS THE NUMBER OF NON-CONGRUENT RECTANGLES THAT WE CAN FIND ON A CHESS BOARD? <ul><li>There are 8 rectangles on 8 rows – which are non- congruent. We can change the sizes of these rectangles. By taking minimum any 2 rectangles, we can make ½(64-8) = 28 more rectangles. Thus we have total 8 + 28 = 36 non-congruent rectangles. </li></ul>
- 22. How many words can you make taking all the letters of ZENITH. What will be the rank of ZENITH in the list of these words? <ul><li>6! = 720 words can be made. The letters from Z will start from 600 onwards. 4! Or 24 letters start with ZE. Thus there will be 6+6+4 letters more before ZENITH. Thus its rank will be 616 from the beginning. </li></ul>
- 23. Example… <ul><li>ZENITH </li></ul><ul><li>There are 600 words starting with either e,N,I, T,H. </li></ul><ul><li>Thus there are 120 words starting with Z </li></ul><ul><li>Out of them first 24 are having ZE. In ZE, we can have words like Zen.. Or Zet.. Or Zei.. Thus counting the words like Zet (2 words) and the words like ZENTHI, ZENTIH. Thus removing them, we can 616 as the number. </li></ul>
- 24. What is the number of words fored from : AAAAA, BBBB,CCC,DD,F? <ul><li>There are 15 letters, so we have 15! Ways of making these letters. </li></ul><ul><li>There are also many similar letters. </li></ul><ul><li>Thus we have 15!/(5! *4! * 3! *2! * 1!) </li></ul>
- 25. What is the number of words fored from : AAAAA, BBBB,CCC,DD,F with all Cs taking different places? <ul><li>There are 12 letters exclusing Cs. Thus Cs can take 12 +1 = 13 places between or before them. Thus we have 13c3 options. Thus we have 13c3 * (12! / (5!*4!*2!)) options. </li></ul>
- 26. The probability of a student A solving a problem is 0.6 and that of another student B solving the problem is 0.3. What is the probability that at least one will solve it right? <ul><li>A will not solve = (1-.6)=.4, B will not solve = .7 </li></ul><ul><li>Both of them will not solve = .4 *.7 = .28 </li></ul><ul><li>One of them / both may solve it = 1-.28=.72 </li></ul>
- 27. Veeru and Jai are two friends. If Veeru tells the truth 40% of the time and Jai 30% of the time, the probability that they will contradict each other when they are narrating the same incident is? <ul><li>V true, J false = .4*.7 = .28 </li></ul><ul><li>V false, J true = .6 * .3 = .18 </li></ul><ul><li>They will contradict = .28 +.18 = .46 answer </li></ul>
- 28. If one diagonal of a rhombus is equal to a side, what is the ratio between two adjacent angles of the rhombus? <ul><li>This will be the situation when we combine two equilateral triangle to form a rhombus. Thus two angles of the rhomus will be 60 each and remaining two angles will be 120 each. Thus ratio will be 1:2 ANSWER. </li></ul>
- 29. PUZZLE <ul><li>A,B,C,D, are 4 villages on a straight road in that order, given that B is equidistant from A and C while C is equidistant from B and D Two girls Reena and Sheena decide to watk frorn A to D, covers the distance walking at 3 km/hr and Sheena covers the first lap (upto B) at either l.5 or 4.5 km/ hr and then walked at the other speed till she meets Reena. Thereafter they walk together to D at Reena ‘s speed.What is Sheena’s speed from A to B? </li></ul>
- 30. Solution… <ul><li>The challenge is to ensure that the speed should be such that average of which is 3 KM per hour. Let us assume that the distance from A to B and B to C and C to D is 9 KM each. if the speed is 4.5, it will take her 2 hours to reach B and then X / 1.5 and Y / 3 (total = 27/3 = 9 hours). </li></ul><ul><li>Other possibility is : 9/1.5 + 18 (at 4.5 and 3 per hour) </li></ul>
- 31. About AFTERSCHO☺OL <ul><li>PGPSE - World’s most comprehensive programme on social entrepreneurship – after class 12 th </li></ul><ul><li>Flexible – fast changing to meet the requirements </li></ul><ul><li>Admission open throughout the year </li></ul><ul><li>Complete support from beginning to the end – from idea generation to making the project viable. </li></ul>
- 32. Branches of AFTERSCHO☺OL <ul><li>PGPSE programme is open all over the world as free online programme. </li></ul><ul><li>Those who complete PSPSE have the freedom to start branches of AFTERSCHO☺OL </li></ul><ul><li>A few branches have already started - one such branch is at KOTA (Rajasthan). </li></ul>
- 33. Workshop on social entrepreneurship <ul><li>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. </li></ul><ul><li>These workshops are great moments of learning, sharing, and commitments. </li></ul>
- 34. FREE ONLINE PROGRAMME <ul><li>AFTERSCHO☺OL is absolutely free programme available online – any person can join it. The programme has four components : </li></ul><ul><li>1. case studies – writing and analysing – using latest tools of management </li></ul><ul><li>2. articles / reports writing & presentation of them in conferences / seminars </li></ul><ul><li>3. Study material / books / ebooks / audio / audio visual material to support the study </li></ul><ul><li>4. business plan preparation and presentations of those plans in conferences / seminars </li></ul>
- 35. 100% placement / entrepreneurship <ul><li>AFTERSCHO☺OL has the record of 100% placement / entrepreneurship till date </li></ul><ul><li>Be assured of a bright career – if you join AFTERSCHO☺OL </li></ul>
- 36. Pursue professional courses along with PGPSE <ul><li>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. </li></ul><ul><li>Come and join AFTERSCHO☺OL </li></ul>

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