Accounting And Bookkeeping For Business And Management 13 OctoberPresentation 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?
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)?
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?
D B C A
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
=R = (17.5*9*2)/12
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?
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.
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?
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
(3) 6:45 am
First picturise the situation and prepare a graph.
Sin 30 = ½
Perp. / Hyp.= ½
Per / 500 = ½
Per = 250
AC = 433
90 30 60 Now =A c =b Hypoteneuse = 500 Per=250
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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