Accounting And Bookkeeping For Business And Management  13 October
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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 Accounting And Bookkeeping For Business And Management 13 October Presentation Transcript

  • 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
  • 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
  • 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?
    • (1)820
    • (2) 821
    • (3) 781
    • (4) 819
    • (5) 780
  • SOLUTION…
    • 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
    • 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.
    • 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.
  • What are the last two digits of 7^(2008)?
    • (1) 21
    • (2) 61
    • (3) 01
    • (4) 41
    • (5) 81
  • Solution…
    • 7 ^ (4*502)
    • (7^4)^502
    • 7^4 = 2401
    • We can also write it as : (2400 +1) ^502
    • Expanding this we get :
    • two part, the first being divisible by 2400 and the second part as 1^502
    • 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.
  • 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?
    • (1) 17.05
    • (2) 27.85
    • (3) 22.45
    • (4) 32.25
    • (5) 26.25
    D B C A
  • Solution…
    • Area of triangle :
    • ½ * base * height
    • Area of triangle : product of three sides / 4*radius of circumscribing circle
    • Thus we can say:
    • ½*base (BC)*3 = product of three sides / 4R
    • =base *R=(17.5*9*base*2)/4*3
    • =R = (17.5*9*2)/12
    • =26.25 answer.
  • 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?
    • (1) 499
    • (2) 500
    • (3) 375
    • (4) 376
    • (5) 501
  • SOLUTION….
    • FOR THE FIRST DIGIT WE HAVE 3 OPTIONS (1,2,3,)
    • FOR EACH OF NEXT DIGITS, WE HAVE 5 OPTIONS (0,1,2,3,4)
    • THUS TOTAL OPTIONS ARE :
    • 3*5*5*5 = 375
    • + WE ALSO HAVE 4000 AS ONE OPTION, THUS OUR TOTAL NUMBER OF OPTIONS ARE 376. ANSWER.
  • SOLUTION….
    • 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).
  • Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer, then how many such triangles exist?
    • (1) 5
    • (2) 21
    • (3) 10
    • (4) 15
    • (5) 14
  • SOLUTION…
    • If we assume X to be the longest side, then it can be maximum : 15+8 = (less than 23).
    • 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.
    • Thus we have 5 options.
    • 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.
    • Thus available options are : 8,9,10,11and 12. thus we have 5 options here.
    • Thus we have total 10 options.
  • 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
    • (1) 6:15am
    • (2) 6:30am
    • (3) 6:45 am
    • (4) 7:00am
    • (5) 7:15am
  • Solution…..
    • First picturise the situation and prepare a graph.
    • Sin 30 = ½
    • Perp. / Hyp.= ½
    • Per / 500 = ½
    • Per = 250
    • AC= 500^2-250^2
    • AC = 433
    90 30 60 Now =A c =b Hypoteneuse = 500 Per=250
  • Solution….
    • The train will start at 8 AM, it will reach C by : 250/50 = 5 hours = 1 PM
    • Thus Ram must reach by 12.45
    • Ram has to take : 433/70 =6.2 hours to cover his distance. Thus Ram must start:
    • 12.45 – 6.12 = 6.33
    • Ram must start by 6.33. therefore the answer is (2). Ans.
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