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# Accounting And Bookkeeping For Business And Management 13 October

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### Accounting And Bookkeeping For Business And Management 13 October

1. 1. MANAGEMENT APTITUDE TEST FOR CAT,GMAT,CET,UPMAT,RMAT,XAT 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. 2. MANAGEMENT APTITUDE TEST FOR CAT,GMAT,CET,UPMAT,RMAT,XAT 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. 3. The integers 1, 2 ,….. 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a+b-1 is written. What will be the number left on the board at the end? <ul><li>(1)820 </li></ul><ul><li>(2) 821 </li></ul><ul><li>(3) 781 </li></ul><ul><li>(4) 819 </li></ul><ul><li>(5) 780 </li></ul>
4. 4. SOLUTION… <ul><li>We can observe a pattern, the numbers are : (from second number onwards) 2,4,7,11,16….. We can convert this into a formula : the number * next number / 2 – previous number, we get the answer. For example, for the 5 th number, we get : (5*6)/2 – 4 = 11 </li></ul><ul><li>Applying this logic, our answer is 781. it is important to understand the question and prepare a formula to solve the question and verify the solution for any number randomly. </li></ul><ul><li>For the first two : (1+2) -1= 2, for second (2+3)-1 =4, so on, which is also the same as from our formula. Ans. </li></ul>
5. 5. What are the last two digits of 7^(2008)? <ul><li>(1) 21 </li></ul><ul><li>(2) 61 </li></ul><ul><li>(3) 01 </li></ul><ul><li>(4) 41 </li></ul><ul><li>(5) 81 </li></ul>
6. 6. Solution… <ul><li>7 ^ (4*502) </li></ul><ul><li>(7^4)^502 </li></ul><ul><li>7^4 = 2401 </li></ul><ul><li>We can also write it as : (2400 +1) ^502 </li></ul><ul><li>Expanding this we get : </li></ul><ul><li>two part, the first being divisible by 2400 and the second part as 1^502 </li></ul><ul><li>Power of 1^502 gives 1. Any number divisible by 2400 must be divisible by 100, therefore the last digit will be 00+1 = 01 answer. </li></ul>
7. 7. In a triangle ABC, tbe lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC? <ul><li>(1) 17.05 </li></ul><ul><li>(2) 27.85 </li></ul><ul><li>(3) 22.45 </li></ul><ul><li>(4) 32.25 </li></ul><ul><li>(5) 26.25 </li></ul>D B C A
8. 8. Solution… <ul><li>Area of triangle : </li></ul><ul><li>½ * base * height </li></ul><ul><li>Area of triangle : product of three sides / 4*radius of circumscribing circle </li></ul><ul><li>Thus we can say: </li></ul><ul><li>½*base (BC)*3 = product of three sides / 4R </li></ul><ul><li>=base *R=(17.5*9*base*2)/4*3 </li></ul><ul><li>=R = (17.5*9*2)/12 </li></ul><ul><li>=26.25 answer. </li></ul>
9. 9. How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed? <ul><li>(1) 499 </li></ul><ul><li>(2) 500 </li></ul><ul><li>(3) 375 </li></ul><ul><li>(4) 376 </li></ul><ul><li>(5) 501 </li></ul>
10. 10. SOLUTION…. <ul><li>FOR THE FIRST DIGIT WE HAVE 3 OPTIONS (1,2,3,) </li></ul><ul><li>FOR EACH OF NEXT DIGITS, WE HAVE 5 OPTIONS (0,1,2,3,4) </li></ul><ul><li>THUS TOTAL OPTIONS ARE : </li></ul><ul><li>3*5*5*5 = 375 </li></ul><ul><li>+ WE ALSO HAVE 4000 AS ONE OPTION, THUS OUR TOTAL NUMBER OF OPTIONS ARE 376. ANSWER. </li></ul>
11. 11. SOLUTION…. <ul><li>We can make only 4 digit numbers, therefore the starting number will be 1,2 or 3. (because 4 may give a number greater than 4000). For second digit, we have 0,1,2,3,or 4 (any number is permitted). </li></ul>
12. 12. Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer, then how many such triangles exist? <ul><li>(1) 5 </li></ul><ul><li>(2) 21 </li></ul><ul><li>(3) 10 </li></ul><ul><li>(4) 15 </li></ul><ul><li>(5) 14 </li></ul>
13. 13. SOLUTION… <ul><li>If we assume X to be the longest side, then it can be maximum : 15+8 = (less than 23). </li></ul><ul><li>Thus it can be 16,17,18,19,20,21,22. however, 17 gives right angle (15^+8^2 = 17^2), therefore it cannot be 16 or 17. </li></ul><ul><li>Thus we have 5 options. </li></ul><ul><li>Taking 15 as the longest side, the minimum value of X will be = (15-8 =7 (more than 7), thus we have 8,9,10,11,12,13,14 as options. Again, taking right angle specifications : 15^2 -8^2 =161, square root of which is 12.66. </li></ul><ul><li>Thus available options are : 8,9,10,11and 12. thus we have 5 options here. </li></ul><ul><li>Thus we have total 10 options. </li></ul>
14. 14. Ram plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Ram must leave A and still catch the train is closest to <ul><li>(1) 6:15am </li></ul><ul><li>(2) 6:30am </li></ul><ul><li>(3) 6:45 am </li></ul><ul><li>(4) 7:00am </li></ul><ul><li>(5) 7:15am </li></ul>
15. 15. Solution….. <ul><li>First picturise the situation and prepare a graph. </li></ul><ul><li>Sin 30 = ½ </li></ul><ul><li>Perp. / Hyp.= ½ </li></ul><ul><li>Per / 500 = ½ </li></ul><ul><li>Per = 250 </li></ul><ul><li>AC= 500^2-250^2 </li></ul><ul><li>AC = 433 </li></ul>90 30 60 Now =A c =b Hypoteneuse = 500 Per=250
16. 16. Solution…. <ul><li>The train will start at 8 AM, it will reach C by : 250/50 = 5 hours = 1 PM </li></ul><ul><li>Thus Ram must reach by 12.45 </li></ul><ul><li>Ram has to take : 433/70 =6.2 hours to cover his distance. Thus Ram must start: </li></ul><ul><li>12.45 – 6.12 = 6.33 </li></ul><ul><li>Ram must start by 6.33. therefore the answer is (2). Ans. </li></ul>
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