The document discusses an afterschool program called Afterschoool that offers mathematics and aptitude tests as well as a social entrepreneurship program. It provides examples of reasoning questions and puzzles asked in the tests. It also details plans to open branches in major Indian cities and develop case studies on social entrepreneurs through collaboration with entrepreneurs. The basic values promoted at Afterschoool are sharing knowledge, learning from mistakes, asking questions, and embracing change.

1. MATHEMATICS & APTITUDE TESTS 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

2. MATHEMATICS & APTITUDE TESTS Dr. T.K. Jain. AFTERSCHO☺OL Centre for social entrepreneurship Bikaner M: 9414430763 [email_address] www.afterschool.tk www.afterschoool.tk

3. I CUT A PIECE OF PAPER IN 4 EQUAL PARTS. NOW I CUT EXACTLY ONE OF THESE PARTS INTO FOUR EQUAL PARTS. NOW AGAIN IF I KEEP ON REPEATING THE SAME FOR INFINITE NUMBER OF TIMES, WHAT CAN THE THE NUMBER OF PIECES OF PAPER AT ANY POINT OF TIME? OPTIONS: 2049, 2050, 2048.

4. SOLUTION . . . First time we have 4 parts. Now we have 4+3 = 7 parts. Next time we have 6+4 = 10 parts, and next we have 9+4 = 13 parts. Thus we find a series. The series is 4,7,10,13 . .. . There is a gap of 3 in any two pairs. This is AP. Thus the condition (X-A) must be divisible by D. A = 4. X = take from options, D = 3. Let us try from 2048. (2048-4) is not divisible by 3. Let us try 2050, (2050-4) = 2046 is divisible by 3, therefore this is the answer.

5. What is the remainder if 7^7 + 7^77+7^777…7^77777777 is divided by 8? If we divide 7 by 8, we have -1. let us put is like this : -1^7 + -1^77+-1^777…-1^77777777 We know that -1 with odd power will only give -1 as the answer. Thus total of the series is -8. let us divide this -8 by 8 and we get zero as ramainder (there is no remainder). Answer.

6. There are two sets of lines. First set has Y = (X+N) and second set Y=(-X+N). N = {0,1,2,3,4}. The number of squares that can be formed will be ? we can have Y = X + 0,1,2,3,4, and on the same axis we will have Y = -x + 0,1,2,3,4. Thus these two sets of lines will be perpendicular to each other. These lines will be parallel to each other also (as per rules of geometry). Thus we can find the squares made by these lines.

7. Solution continued. . . . We will have in all 5 lines in both sets. Thus we can find out the number of squares made by these lines. There are 4*4 = 16 squares of same size made by them. Then we can combine the squares and count 4 squares at a time. Thus we can have 9 squares. Next we can take 9 squares and we can make 4 such squares and then we can take all squares together and we have 1 such squares. Total we have 30 squares. Answer.

8. Ranu is twice as old as Manu will be when Sanu is as old as Ranu is now. The youngest one is who? Let us assume Ranu to be 10. Sanu is younger to Ranu. Sanu has yet to become 10. Thus Ranu > Sanu. Sanu is approaching Ranu, therefore Sanu is just younger to Ranu. When Sanu will be of the present age of Ranu, then Ranu will be twice Manu, than means, Manu is very young today. Let us assume that Sanu is now 8. When Sanu will be 10, Ranu will be twice Manu. Thus Ranu will be 12, Manu will be 6 at that time. Thus youngest one is Manu now (4 years). Thus Youngest is Manu and Eldest is Ranu. Ans.

9. We have a square ABCD. There are mid points PQRS. We draw lines from A to R (on DC) and from R to P (on AB) then to C. Similarly we draw line from D to Q (on BC) to S (on AD) to B. the areas coming together are shaded. What is the ratio of shaded to unshaded area? Take any quarter. Find the shaded area. The same ratio applies to the total picture. The answer is ¼ .

10. How many number of solutions are possible on log 2 (X+5) = 6-X log 2 (X+5) = 6-X We can write this as : 2^(6-x) = X+5 Let us put values in X and find the result. When we put the value 3, we get the answer. Thus only one solution is possible.

11. Find the number of terms in expansion of (A+B+c+d)^9? If we have only two terms (A+B)^9, then we get 10 terms in the expansion. Which is (9+1)C1. Here we use the formulae of combination. Combination is NCK, where N is total number, K is numbers selected. If we have (A+B+C)^9, we have (9+2)C2 combinations, similarly, if we have (A+B+C+d)^9, we have 12C3 combinations. Answer.

12. If the difference in the number of arrangements of 3 items out of some dissimilar items and number of selections of 3 items from same items is more than 100, what is the minimum number of total items? Arrangement = permutation Selection = combination. Np3 – NC3 >100. (if N=6, we get 100). Let us try with different numbers. If we take 7, we get more than 100, thus minimum such number is 7. answer.

13. Sinatra is century’s musical ___, the _____ between the carefully crafted pop of its beginning and the looser. Options: discovery, difference B. equipoise, pivot. Answer: try to apply both the words, only second option gives you the answer. (because both the words must be useful in the given blanks).

14. Branches AFTERSCHO☺OL will shortly open its branches in important cities in India including Delhi, Kota, Mumbai, Gurgaon and other important cities. Afterschooolians will be responsible for managing and developing these branches – and for promoting social entrepreneurs.

15. Case Studies We want to write case studies on social entrepreneurs, first generation entrepreneurs, ethical entrepreneurs. Please help us in this process. Help us to be in touch with entrepreneurs, so that we may develop entrepreneurs.

16. Basic values at afterschoool Share to learn more Interact to develop yourself Fear is your worst enemy Make mistakes to learn Study & discuss in a group Criticism is the healthy route to mutual support and help Ask fundamental questions : why, when, how & where? Embrace change – and compete with yourself only

17. www.afterschoool.tk social entrepreneurship for better society