Entrepreneurship Quiz 8 Oct

www.afterschoool.tk AFTERSCHO☺OL's MATERIAL FOR PGPSE PARTICIPANTS

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MANAGEMENT APTITUDE TEST

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


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MANAGEMENT APTITUDE TEST

Dr. T.K. Jain.

AFTERSCHO☺OLCentre for social entrepreneurship

Bikaner M: 9414430763

[email protected]

www.afterschool.tk, www.afterschoool.tk


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What is cube of 12?

• Cube = 12*12*12 = 1728


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What is cube root of 1728?

• Let us undertake factorisation of 1728.

• Let us first divide it by 2 and then by 3 so long as we can.

• 1728, 864, 432 ,216 ,108,54, 27, 9,3, 1

• As we can see that we have 2^6 *3^3 in this factorisation. Thus we can say that cubic root is equal to : 2^2*3 or 12 answer.


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Create a number less than 2500, which contains maximum possible

prime numbers (all different).

• Let list down prime numbers so that we may build one such number…

• 2*3*5*7*11 = 2310

• This number contains 5 prime numbers.

• (prime number is that which cannot be divided by any other number).


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Find HCF and LCM of 78 and 82?

• Let us undertake factorisation of 78 and 82

• 78 = 2*3*13

• 82 = 2*41

• Thus HCF (highest common factor) = 2

• Thus LCM (least common multiple) is

• 2*3*13*41=3198 answer.


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What is cubic root of 658503?

• The number is in 6 digits, therefore its cubic root should be less than 100.

• The starting digit is 6, so its cubic root should be between 85 and 90. thus we will try for each digit between these digits to find cubic root of this number.

• 87*87*87 = 658503. Thus cubic root is 87 answer.


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Convert 12 into binary number system ?

• In binary system, we have only two digits – 1,and 0.

• 12 divided by 2 gives6 divided by 2 gives one as remainder & 2divided by 2 gives 1

• Thus answer is : (last first) 1100

• We can verify it : 1*2^3+1*2^2 +0*2^1+0*2^0

• =8+4+0+0 =12 answer.

• Division method :•

12 0

6 0

3 1

1


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Convert 100 into octal number system..

• Octal number system has its base as 8 (not ten).

• =144• Let us verify it : • 1*8^2 + 4*8^1 + 4*8^0• =64+32+4• =100 answer.

100 4

12 4

1


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Convert 200 into hexa-decimal system

• Hexa decimal system has 16 as its base.

• Thus we will divide by 16 and get the answer .

• =C8, where c=12

Let us verify it :

=c*16^1 + 8*16^0

=12*16 + 8*1

=200 answer

200 8

12


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Convert .0000005 into scientific notation..

• Here we can observe that there are 7 digits after decimal, therefore, we shall use 10^(-7) to multiply the number (which is 5 in this case).

• = 5* 10^(-7 )


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Find cubic root of 13824?

• Let us undertake factorisation of this number. Since this number is divisible by 2 and 3 both (divisible by 2 ,because the last digit is 4, which is even number and divisible by 3 because total of all the digits is 18, which is also divisible by 3). We start with 2. start: 13824 > 6912> 3456>

1728> 864 >432> 216>108 >54> 27 now divide by 3>9> 3

1


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Solution …

• As we can see that we have divided by 2 by 9 times and by 3 by 3 times.

• = 2^9 * 3^3

• Now let us divide the powers by 3.

• We get:

• = 2^3 *3^1

• =24 answer.


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Find the minimum number that may be multiplied to 12345 to make

it a perfect cube? • This number is divisible by both 3 and 5 (its total

is 15, which is divisible by 3 and it ends in 5, so divisible by 5).

• Let start by dividing it by 3. we get 4115. now we divide it by 5, we get 823, which is a prime number. Thus in order to make it a perfect cube, we have to multiply it by the following : = 3^2 * 5^2* 823^2

• We have 3,5, and 823 as just one factors, but to be a cube, we need them three times, therefore we shall multiply them again (2 times) to make it cube. answer.


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Find HCF of 740 and 1184?

• Let us first divide 1184 by 740, we get 444 as remainder. Now divide 740 by 444 and we get 296 as remainder. Now divide 444 by 296 and we get : 148 as remainder. Now we have to divide 296 by 148, we get zero as remainder. The number which got us Zero as remainder is 148, therefore HCF is 148. Answer.


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There are 4 bells, which toll at an interval of 2,3,5,7 minutes

respectively, they start together, how many times in a day of 12 hours, will they toll together?

• We have to find LCM of 2,3,5,7. this is 210. thus after every 210 minutes they will toll together.

• Thus in a day: (12*60) / (210) = 3.4 or they will toll together 3 times a day (excluding the start).


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A Major puts 65000 soldiers in rows and columns such that there are as

many soldiers in rows as in columns. Some soldiers are left

out. He asks them to go for a walk. How many soldiers went for a

walk? • Let us find the square root of 65000 to find

the number of persons left out.


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Solution…

Divide by 2

6 50 00

4

254

Add 2 to 2= 45 *5

2 50

2 25

Add 5 to 45=504 *4

25 00

20 16

4 84


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solution

• As we can see that the nearby square root is 254 and there are 484 extra soldiers. Thus 484 soldiers went for a walk.


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A Major puts 19500 soldiers in rows and columns such that there are as many

soldiers in rows as in columns. He finds that he needs some more soldiers to complete this squire. He asks some

students to fill up the vacancies. What is the minimum number of students,

which will make this square?


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Solution – find square root of 19500

Divide by 1 1 95 00

1

139

Add 1 back to 1

=23 *3

0 95

69

Add back 3 to 23 = 269*9

26 00

24 21


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Soluton…

• The nearest square root is 139. Therefore the Major must have been making rows and columns of 140 soldiers in each side. The square of 140 *140 = 19600

• Thus the Major needed 100 students to make it a perfect square and had there been 100 more soldiers, he could have achieved his task without asking for the students. Answer.


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What is the number which will divide 40,70,160 to give the same

remainder in each case?

• Let us find the difference of these numbers. The differences are :

• (70-40) = 30

• (160 – 70) = 90

• The HCF of 30 and 90 is 30 thus we can find the number as 30. if we divide all these numbers by 30, we get remainder 10 in each case. Remember the rule.


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Puzzle test

• Goutam has recently acquired four companies viz (BOB), (MOB), (ZOB) and (DOB). It is found that that the sales of DOB are half that Of BOB whereas the profits of DOB are double that of BOB The expenses of ZOB are Rs 3 crores less than that of DOB whereas the profit of MOB is Rs 1. Crore less than that of ZOB The expenses of BOB are three times that of DOB. It is also known that the sales of ZOB are Rs.15 crores or one—fourth that of MOB. ; sales of DOB are Rs.1O crores more than that of ZOB and the expenses of BOB are 90% of its own sales. Which company has earned maximum profit.


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Solution ….

• Sale of BOB = 50 (DOB *2) Crores Exp. 45 (90% OF SALES) Crores, thus profit = 5 Crores.

• Sales of MOB = 60 (4*ZOB) Crores profit = 2 cr. (zOB-1) Exp. Cr .58

• Sales of DOB = 25 (ZOB +10) crores; profit = 10 (B0B*2) cr. Exp = 15 Cr.(1/3*BOB)

• Sales of Zob = 15 Crores Exp. Of ZOB = 12 Crores. (because it is 3 crores less than those of DOB) PROFIT = 3(MOB+1)


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Solution

• Total sales : 150 crores,

• The company with highest profit is DOB with 10 crores of profit.


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Puzzle test • Three companies A, B & C are the largest producers of a variety of

automobiles. Model “Perfect” made by A is priced at Rs.1 lakh, the same model made by B is priced at Rs.2.5 lakhs and by C at Rs.1.96 lakhs. Company C has quoted Rs.2.67 lakhs as its price for model “Roger” which has been priced at Rs.3.98 lakhs and Rs.2,22 lakhs by A and B respectively. A latest model “Futura” being sold by B for Rs.8.92 lakhs is being sold by A and C for Rs.6.08 lakhs and Rs.5.0 lakhs respectively. The total sales of company A are Rs.67.98 Crore whereas for B and C the figures are Rs.88.82 Crore and Rs.65.56 Crore respectively. If next year, company A decides to hike its prices by 5% for all models, it is expected to meet stiff resistance from B and C in the form of a 3% reduction in prices for all their respective models. Rate of growth of auto industry is known to be 3.25% pa (average). profit contribution for model Roger are 15%, l9% and 8% for A, B and C respectively on the selling price For Futura,A, B and C realise l0°/o, 6% and 7.5%. profit on the selling price respectively. Finally, profits are 7%, 8.5% and 12% on the selling price of Perfect for A, ‘B and C respectively. What is total sales by all the companies together.

• What will be the difference in price of perfect and futura (by c) next yr.


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Solution

• Total sales (67.98 + 88.82 + 65.56)

• Price of perfect by A = 1 *1.05 = 105000

• Price of futura by C = 5 *.97= 485000

• Thus the difference is 380000


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Three leaders (a,b,c,) are in the race, a is twice likely to win as b, and b is twice likely to win as c.

what is the probability of c’s victory?

• Let us start with c with 1. b has now 2 as possibility value and a has 4 as possibility

• Thus C’s probability is 1/(1+2+4) = 1/7 answer.


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There are 4 hotels in a city. If three men check in hotels in a day, what is the probability that are staying in

different hotels each? • .375


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Log 64 512

• We can write it as

• Log 2^6 2^9

• Thus we can say that the value of log is 1.5 answer.


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In a class 30%students offered english 20% offered Hindi and 10% offered both. If a student is selected

randomly, what is the probability that he has both english and hindi

and for either eng or hindi? • 10% answer.

• (30+20-10) = 40% for either english or hindi.


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A speaks truth in 75% cases, and B speaks truth in 80% cases. In what % cases they likely to contradict

each other ?• When A is true and B false : .75*.2 = .15

• When A is false and B is true = .25*.8= .2

• Thus (.15+.2) in 35% cases they will contradict. Answer.


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There are three numbers in A.P. whose sun, is 24 and product is

480. Find the numbers.• Let us assume the numbers to be (a-d), a, (a+d)• Let us add them, we get 3a = 24, a = 8 ( middle

number is 8)• Products : (8-d) *8* (8+d) = 480• (64-d^2) = 60 or d^2 = 4 or D = 2• Thus the series is 6,8,10 answer.


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The first term and the last term of an A.P. are

7 and 51 respectively, the sum of the terms of the A.P. is 348, find

the common difference.

• Sum = (n/2) (first term + last term)• 348 = (n/2) (58)• n/2 = 6 or n = 12• The difference between 51 and 7 = 44. let us

divide it by 11 to get common difference of 4. Ans.


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The 5th term and the 21st term of a series in an A.P. are 10 and 42

respectively. Find the 31st term.

• A + 4d = 10

• A + 20 d = 42

• 16d = 32 or d = 2, thus a = 2 (a+8 = 10)

• Thus 31st term is : a +30d = 2+60 = 62 answer.


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Find the arithmetic mean of an A.P. with 33 terms if its first term is 1

and common difference is 2.

• We have to find (33+1) /2 = 17th term.

• Thus the term is : 1+(17*2) = 35.

• Thus A.M. is 35 answer.


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Find the number of terms in an arithmetic progression with the first

term being 3 and the last . term being 67, given that the common

difference is 4.

• A + (n-1)d = 67

• 3+ (n-1) 4 = 67

• (n-1) = 64 / 4

• N = 16 +1 = 17 answer.


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148 is split into four parts which are in arithmetic progression such that the product of the second and

third parts is 8 more than the product of the first and last part.

What are the four parts?

• Let us assume the parts are : (a-3d),(a-d),(a+d) and (a+3d) • (a-3d)(a+3d) +8 = (a-d) (a+d)• A^2 – 9d^2 +8 = a^2 –d^2 • 8d^2 = 8 or d = 1. • (a-3d)+(a-d)+(a+d) +(a+3d) =148 or 4a = 148 or a = 37 • The numbers are : 34,36,38,40 answer.


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A cube is coloured red onthree adjacent faces. It is then cut into four

cuboids of equal size.Each of these cuboids is coloured green on

all the uncoloured faces and is then cut into four cuboids of equal size.

How many cuboids have three red faces each, & 2 & 3 green faces?

• Only one has red colour on 3 faces, and 2 have 3 green faces,


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Solution – those with circle with 2 faces and those with sq. with 3

faces.


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Solution…

• Similarly the cuboids with 2 faces are : 9.


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What will come at the question mark?


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Answer…

• K 8


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How many squares are there?


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Solution …

• This shape is not clearly visible, but it has 6 squares. And triangles are 28.


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What is the sum of area of shaded region? All radius are equal.


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Solution ..

• Exterior angle = 360/5 = 72

• Interior angle : 180 – 72 = 108

• The ratio of 108 to 360 = 5 * 108/360 =

• = 1.5 * pi * r^2. answer.


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What is the average production ofmopeds in terms of the average

production of all the two wheelers?


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Solution

• Total production in 4 years:

• Scooter = 100

• Bike = 100

• Moped = 90

• Cycles = 60

• Total of all these is : 350.

• Moped is : 90/350 *100 = 25.7% answer.


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What is the difference in total production of the years which have

1. highest production of bikes 2. highest production of scooters?

• Highest production of scooter is in 1991 – when total production is 80

• Highest production of bikes is in 1992, when total production is 100 bikes.

• Thus the difference is 20 units. Answer.


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In which year the percentage change in the nymber of students over the previous year is the maximum?


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Solution …

• Let us compare the year 1992 and 1994

• In 1992 we have 50/400 = 12.5%

• In 1994, we have 75/500 = 15%

• Thus we witness the highest change in 1994. answer.


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If 60% of the students pass every year, the number of students

passing in 1995 differs from those not passing in 1991 by

• 60% of 1995 gives us 390

• 40% of 1991 gives us 160

• The difference is 230.

• Answer.


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For how many years is the number of students more than the average number of students during the five

years?

• Sum of the five years : 2575 divided by 5 gives average of 515, thus in only 2 years we have more production than average during these 5 years.


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The average annual percentage growth rate of the number of

students for the given period is

• Difference in last and first year :

• 650 – 400 = 250 / 4 = 62.5

• Thus average growth is 37.5 per year.

• 62.5 /400 *100 =15.6%

• Another method is to look at growth rates of each year individually and then find the average of the same:


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which year the percentage change in the number of students over the

previous year is the minimum?‘

• From 1992 to 1993

• Growth from 450 to 500

• Thus 50/450 *100 = 11.11%


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ABOUT AFTERSCHO☺OL

Afterschoool conducts three year integrated PGPSE (after class 12th along with IAS / CA / CS) and 18 month PGPSE (Post Graduate Programme in Social Entrepreneurship) along with preparation for CS / CFP / CFA /CMA / FRM. This course is also available online also. It also conducts workshops on social entrepreneurship in schools and colleges all over India – start social entrepreneurship club in your institution today with the help from afterschoool and help us in developing society.


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Why such a programme?

• To promote people to take up entrepreneurship and help develop the society

• To enable people to take up franchising and other such options to start a business / social development project

• To enable people to take up social development as their mission

• To enable people to promote spirituality and positive thinking in the world


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Who are our supporters?

• Afterschoolians, our past beneficiaries, entrepreneurs and social entrepreneurs are supporting us.

• You can also support us – not necessarily by money – but by being promotor of our concept and our ideas.


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About AFTERSCHO☺OL PGPSE – the best programme for developing great

entrepreneurs• Most flexible, adaptive but rigorous programme• Available in distance learning mode• Case study focused- latest cases • Industry oriented practical curriculum• Designed to make you entrepreneurs – not just

an employee• Option to take up part time job – so earn while

you learn • The only absolutely free course on internet


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Workshops from AFTERSCHO☺OL

• IIF, Delhi• CIPS, Jaipur• ICSI Hyderabad Branch• Gyan Vihar, Jaipur• Apex Institute of Management, Jaipur• Aravali Institute of Management, Jodhpur• Xavier Institute of Management, Bhubaneshwar • Pacific Institute, Udaipur• Engineering College, Hyderabad


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Flexible Specialisations:

• Spiritualising business and society• Rural development and transformation• HRD and Education, Social Development• NGO and voluntary work• Investment analysis,microfinance and inclusion • Retail sector, BPO, KPO• Accounting & Information system (with CA / CS /CMA)• Hospital management and Health care• Hospitality sector and culture and heritage• Other sectors of high growth, high technology and social

relevance


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Salient features:• The only programme of its kind (in the whole world)• No publicity and low profile course• For those who want to achieve success in life – not just a

degree• Flexible – you may stay for a month and continue the rest of

the education by distance mode. / you may attend weekend classes

• Scholarships for those from poor economic background• Latest and constantly changing curriculum – keeping pace

with the time• Placement for those who are interested• Admissions open throughout the year • Latest and most advanced technologies, books and study

material


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Components • Pedagogy curriculum and approach based on IIM Ahmedabad and ISB

Hyderabad (the founder is alumnus from IIMA & ISB Hyderabad)• Meditation, spiritualisation, and self development • EsGotitial softwares for business• Business plan, Research projects• Participation in conferences / seminars• Workshops on leadership, team building etc. • Written submissions of research projects/articles / papers• Interview of entrepreneurs, writing biographies of entrepreneurs• Editing of journals / newsletters• Consultancy / research projects • Assignments, communication skill workshops• Participation in conferences and seminars• Group discussions, mock interviews, self development diaryng • Mind Power Training & writing workshop (by Dr. T.K.Jain)


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Pedagogy

• Case analysis,• Articles from Harvard Business Review • Quiz, seminars, workshops, games, • Visits to entrepreneurs and industrial visits• PreGotitations, Latest audio-visuals• Group discussions and group projects• Periodic self assessment• Mentoring and counselling• Study exchange programme (with institutions out of

India)• Rural development / Social welfare projects


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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.


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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.


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Basic values at AFTERSCHO☺OL

• 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


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Entrepreneurship Quiz 8 Oct