ANALYTICAL SKILLS - TIME AND WORK

TIME AND WORK

Introduction;

              This chapter deals with problems related with time and work done by person.   

Important points and formula;

1. If ‘A ‘can finish a piece of work in ‘n’ days then A s 1 day’s  work is =1\n

2. If A is X-times as good a workman as B. then A will take 1\x of the time that B –takes to do a certain work.

3.If M1 person can do W1 times in D1 days for T1 hours and M2 person works in D2 days for T2 hours then=M1D1T1W2=M2D2T2W1

4. If A can finish a work in X-days and B can finish the some work in Y- days then live taken by both to finish the work are xy\X+Y days

5. If A and B together can do a piece of work in X- days and A alone can do it in Y- days then B alone can do it  XY\Y-Z  days.

6. If A,B and C can do a work in X,Y and Z days respectively then all of them working together can finish the work in (XYZ)/(XY+YZ+ZX)

7. If two taps A and B take a and b hours respectively to fill a tank then the two taps together fill 1/a+1/b part of the tank in an hour and the total tank fill in ab/a+b hours.

Solved examples

1.’A’ can do a piece of work in 12 days and ‘B’ alone can it in 15 days how much time both will take to finish the work together?

Sol.

        Work done by’A’ in 1 day =1/12

         Work done by’b’ in 1 day =1/15

There fore both together can finish the mork in 20/3days or 6 2/3days.

Short cut: xy/x+y=12*15/12+15=12*15/27=20/3=6 2/3days.

2. A tank can be filled separately by the two pipes in 12 and 16 minutes respectively if both pipes are opend together, when will the tank be filled?

Short cut: the time taken to fill the tank by both pipe

ab/a+b=12*6/12+16=12*16/28=48/7=6 6/7 minutes

3.16 men can do a piece work in ten days. How many men are needed to complete the work in 40 days?

M1 D1=M2 D2 => M2=M1 D1/D2=16*10/40=4men

4. A and B can do a piece of work in 18 days. B and C in 24 days. A and C in 36 days. In what time can they do it in all work together?

Sol: [(A+B) + (B+C) + (A+C)]‘ s 1day’s work=1/18+1/24+1/36=18

  2(A+B+C)’s 1 day work=1/8

    (A+B+C)’s 1 day work=1/16

There fore all can finish the work in 16 days.

Solve the Problems

1.A takes twice as much time as B or thrice as much time to finish a piece of work ,working together they can finish the work in 2 days, can do the work alone in

a) 4 days    b) 6 days     c) 12 days    d) 15 days

2. X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of work. How long did the work left?

a) 6 days   b) 10 days   c) 15 days   d) 20 days

3. A,B and C can do a piece of work in 20 , 30 and 60 days respectively. In how many days A do the work if the A is assisted by B and C on every third day?

a) 12 days  b) 15 days   c) 16 days  d) 18 days

Revamping of Analytical skills Curriculum

1.    Time and Distance

Introduction:

                      This chapter mainly deals with problem related among Distance, Time and Speed of the objects like body, birds, trains, persons, motor vehicles and submarines like any moving bodies.

Basic Formulas:

1.      Speed=Distance/Time          ;   Speed(s), Distance(d), Time(t)

                      2. s∞1/t   => s1t1=s2t2   =>   s1/s2=t2/t1

                      3. If speeds are constant then   d1/t1=d2/t2    =>   d1/d2=t1/t2

                      4. If reaching times are same

                           d/s=constant     d1/s1=d2/s2  =>   s1/s2=d1/d2

                      5. when a motor car covers a distance at s1 km/hr and an equal distance at         s2 km/hr then the average speed for the total journey is S=(2s1s2)/s1+s2

                    

 Conversion Factor:

       1. from km/hr to m/sec:

         When a motor car running with speed x km/hr its speed in m/sec is (x*5/18)m/sec

          Ex: 54km/hr    54*5/18=15m/sec

      2. from m/sec to km/hr:

            When a car is running with speed y m/sec then its speed in km/hr is y*18/5 km/hr

          Ex: car speed 50m/sec then 50*18/5=180 km/hr.

  Conversion of Time:

        1Hr=60min=3600sec, 1day=24hr=24*60*60sec=86400sec

          Ex: 1Hr.40min => 1hr*40*1/60hr=5/3hr.

                  2min.30sec => 2*30/60=1min

  Note: when the speeds are in the same direction the resultant speed is S=s1-s2

        

          When the speeds are in the opposite direction the resultant speed is S=s1+s2

PROBLEMS:

1.      While covering a distance of 240km, a car driver noticed that after driving for 1hr and 40min, the distance covered by him was 5km of the remaining distance. What was the speed of the car?

  Sol.  Let x be the speed of the car in km/hr

         The distance traveled in 1hr and 40min = xkm/hr*1hr.40min*1/60hr

                                                                         = 5/3x km

In the same time we found the distance covered is=5/7(240-5x/3) km

       Distances covered are same, therefore 

               5x/3=5/7(240-5x/3)

                 7x=3(240-5x/3) = (3*240-(5x*3)/3)

                  7x=720-5x

                  7x+5x=720

                   12x=720

                   X=720/12=60km/hr

                   X=60km/hr.

2. Two men starting from the same place walk at the rate of 10km/hr and 8km/hr respectively. Find the separation between them at the end of 7/2 hr   if a) They walk in the opposite direction   b) they walk in the same direction

Sol. A) When they walk in opp. Direction the speeds are added S=s1+s2=10+8=18km/hr

      The separation b/w them is 18km in one hr.

     In 7/2hr they are apart, 18*7/2=63km

 B) When they walk in same direction speeds are subtracted S=s1-s2=10-8=2km/hr

          They are separated in 2km,

       In 7/2 hr they are apart=2*7/2=7km.

Solve the Problems:

1. A train traveling 25km/hr leaves Delhi at 9am and another train traveling 35km/hr starts at 2pm in the same direction. How many km from Delhi will they be together?

a) 430km           b) 437.5km       c) 500km      d) 117km

2. A is faster than B. A and B walk 24km the sum of their speed is 7km/hr and the sum of times taken by them is 14hr then A’s speed is

a) 3km/hr      b) 4km/hr    c) 5km/hr      d) 7km/hr

3. In covering a distance of 30km, shaker takes 2hr more than suman. If shaker doubles his speed then he would take 1hr less than suman.Then shaker’s speed is.

 a) 5kmph    b) 6kmph      c) 6.25kmph       d) 7.5kmph

4. The jogging in a sports complex is 726m in circumference. Sachin and his wife  starts from the same point and walk in opposite direction at 4.5km/hr and 3.75km/hr.they will meet first time in

  a) 4.9min      b) 5.28min    c) 5.5min     d) 6min

5. A can complete a journey in 10hrs. He travels first half of the journey at the rate of 21km/hr and second half at the rate of 24km/hr. The total journey in km

  a) 220km        b) 224km      c) 230km       d) 234km

6. A person travels from P to Q at a speed of 40kmph and returns by increasing his speed by 50%. What is his average speed for both the trips?

  a) 36kmph      b) 45kmph       c) 48kmph      d) 50kmph

ANALYTICAL SKILLS- PARTNERSHIP

Unit - I PARTNERSHIP

      Partnership is a business where two (or more) partners invest a certain amount (known as capital) for a specific period to carry out the business. At the end of the period (usually one year), the business may generate profit or loss. This profit or loss is divided among the partners in the ratio of their investments.

            If A and B are in partnership, at the end of one year (which is convention), the profit or loss would set distributed as:

Profit (or loss) = (Capital invested) × (Period of investment)

NOTE:

1. A partner retires during one year period: - in such case, only the period for which his capital was utilized, is taken into confederation.

2. A partner joins during the one year period: - in such case, only the period for which his capital was utilized, is taken into confederation.

EXAMPLE:

1. A and B started a business in partnership by investing Rs.8000 and Rs.7000 respectively. If at the end of a year, a profit of Rs.22, 500 was earned, what is the share of A?

Solution:

Ratio of amount inverted by A and B = 8000: 7000

                                                                   = 8:7

Profit at the end of the year                 = Rs.22500

Share of A =8/15x22500=8x1500 = Rs.12000/-

2. In partnership business, A has inverted Rs.4200 while B has inverted a certain amount. If out of the overall profit of Rs.600, A’s share is Rs.320, what is the amount inverted by B (in Rs)?

Solution:

               A’s investment              = Rs.4200

               Overall profit                 = Rs.600

A’s share in profit                      = Rs.320

:. B’s share in profit                   = Rs.600- Rs.320=Rs.280

The ratio of profit of A and B   = 320: 280= 8:7

:. The amounts inverted by A and B will be in the ratio Of 8:7

Rs. 4200: B’s investment =8:7

ð  B’s investment = 7x4200/8= Rs.3675

3. chetan and suman started a business in partnership by investing Rs.15000 and Rs.18000 respectively. If at the end of the year, chetan’s share in the profit was Rs.1200, what was the amount of total profit?

Solution:  If Rs.15000 earns a profit of Rs.1200, then Rs.33000 earn a profit of 33000x1200/15000=2640.

4. In a partnership business, A has inverted 2000 for 5months, while B has invested Rs.3500 for a certain period. If out of the total annual profit of Rs.1440, B’s share has been Rs.840, for how many months has he kept his investment in the business?

Solution:

A has invested Rs.2000 for 5 months

B has invested Rs. 3500 for ‘m’ months.

Total profit at the end of the year= Rs.1440, B’s share in it is= RS.840

A’s share in profit = Rs. (1440-840) =Rs.600

The ratio of profit of A and B = 600:840=5:7 The ratio of investment of A and B is the same as that of the profits    2000x5/3500x5=5/7

:. m= 7x2000x5/3500x5=4

Hence, B kept the investment for 4 months only.

Practice:

1. A started business with an investment Rs.6000. four months later, B joined him with his investment Rs.60000. If at the end of the year Rs. 35000 profit is made, what is the share of A in the same?

a)      Rs.14000  b) Rs.21000  c) Rs.15000 d) Rs.20000 e) none of these

                                         Ans: b) Rs.21000

2. A and b invest Rs 3000 and Rs. 2000 respectively in a business. A was a sleeping partner. At the end of the month, both set Rs.150 each. What was B’s remuneration for his work?

        a) Rs.60   b) Rs.40  c) Rs.50  d) Rs.30 e) None of these

                                              Ans: c) Rs.50

                               DATA INTERPRETATION AND ANALYSIS

 Directions (1-5) : study the following table and answer the questions  that follow:

The table shows the percentage of unemployed male and female youths and the total population of different states in two years.

State	2008			2009
	M	F	T	M	F	T
P	11	16	35	9	11	39
Q	9	7	19	10	8	24
R	8	11	25	10	13	41
S	12	6	30	7	7	34
T	7	6	32	5	6	24
U	8	5	26	8	9	35

M- percentage of unemployed male youth over total population

F- percentage of unemployed female youth over total population

T- Total population of the state in lakhs.

1) what was the total number of unemployed youths in state Q in 2009?

a) 351000   b) 447000 c)432000 d) 521000 e) none

                          Ans: c) 432000

The percentage of unemployed male youths in state Q in 2009 is 10% and the percentage of unemployed female youths in state Q in 2009 is 8%. Thus , the total number of unemployed youths in 2009 in state Q  will be (10+8) = 18%  of the total population.

The total population of state Q in 2009 is 24 lakhs

:. The  number of unemployed  youths  in state Q in 2009 :- 18/100x24 lakhs = 4.32 lakhs = 432000

2. How many female youths were unemployed in state T in 2008?

a) 192000  b) 241000  c) 224000  d) 301000  e) none.

                                 Ans: a) 192000

The percentage of unemployed female youths in state T in 2008 is 6%. The total population of state T in 2008    6/100x32lakhs =1.92lakhs =192000

3. what was the difference between the no. of unemployed male youths in state s in 2008 and 2009 ?

 a) 109000   b) 122000  c) 212000 d) 98000 e) none

                                      Ans : b) 122000

The number of unemployed male youths in state s in 2008 =12% of 30lakhs

=12/100x30lakhs = 3.6lakhs= 360000

The no of unemployed male youths in state S in 2009 = 7% of 34lakhs

                                                                                               = 7/100x34lakhs = 2.38lakhs = 238000

:. The required difference = 360000 -238000 = 122000.

4. What was the respective ratio of the unemployed male youths in state P in 2008 and state R in 2008

a)      11:47  b) 23:49  c) 52:47  d)  2:3  e)  none of these

                              Ans : e) none of these

The no of unemployed male youths in state P in 2008 = 11% of 35lakhs

                                                                                               =11/100x35lakhs = 3.85lakhs =385000.

The no of unemployed male youths in state R in 2008 = 8% of 25lakhs

                                                                                                = 8/100x25lakhs =2lakhs =200000

:. The required ratio = 385000 = 200000 = 77:40

As the answer is not given in any of the assume options, the answer is (2).

5. The no. of unemployed female youths in state (4) u in 2008 is ap0proxiamately what percent of  the no of unemployed female youths in state U in 2009?

a)      20%  b) 35%  c) 55% d) 40% e) none

                                                                Ans: d) 40% 

The no of unemployed female youths in state U in 2008 = 5% of 26 lakhs

                                                 =5/100x26lakhs= 1.30lakhs -130000.

The no of employed female youths in state U in 2009 = 9% of 35lakhs

                                       =9/100x35lakhs = 3.15lakhs = 315000.

130000 = x/100x315000

X= 130000 x 100 / 315000 = 41.27% = 40%

||.  Directions (Q.no.6-10) : Study the following pie-charts and answer the questions that follow.

The pie chart shows the percentage of various products produced.

The following Pie chart shows the income obtain selling the product.

Total income obtained by selling all the products = Rs. 50 million.

6. If the expenditure incurred in production of product N per tonne was Rs. 5000, what was the percent profit earned ?

a)      15%  b) 25% c) 55% d) 45% e) none of these

7. what is the approximate per selling price of all the six products together ?

a)      Rs 14,320 b) Rs. 15770  c)Rs.16670  d) data in adequate  e) none of these.

8. Which product has got the highest selling price per tonne ?

a)      product L  b) product M c) product N d) product O e) none of these

9. What is the selling price of product Q per tonne?

a)      Rs 21800 b) Rs.23450  c) RS.19760  d) Rs. 24740 e) none

10. If the quantity of product N produced increased by 20% in the next year and selling price per tonne remains the same, then what would be the income generated by selling product N in the next year ?

Explanation (6-10)

Product	Quantity of Product Produced in Tonnes	Income obtained By Selling the Product	Income obtained per Tonnes
L	510	1,50,00,000	Rs. 29,411.76
M	420	60,00,000	Rs. 14,285.71
N	480	30,00,000	Rs. 6,250
O	360	50,00,000	Rs. 13,888.89
P	720	70,00,000	Rs. 9,722.22
Q	510	1,40,00,000	Rs. 27,450.98

6) (b) 25%

The expenditure incurred in production of product N is Rs. 5000 per tonne

The income obtained by selling products N per Tonne = 3000000/480 = 6250

As can be seen, the profit obtained per tanne by selling product N:-

                                       =6250 – 5000 = 1,250.

:. The required percentage profit :-

1250/5000x100 = 25%

7) © Rs.16670

Total quantity produced=3000 tonnes

Total income obtained = Rs.50 million

Selling price per tanne = 50000000/3000 = Rs 16666.67

                                                                              ~ Rs. 16670

8. a) product L

 As can be seen from the table above, the selling price of product Q is Rs. 27450 per tanne. As the answer is not given in any of the options.

11. d) Rs. 3600000

 The quantity of product N produced = 480 tanned The quantity produced in the next year : 120/100x480 = 576 tannes

Per tanne selling price of product N remains the same i.e. Rs. 6250.

:. Amount obtained by selling N in the next year =6250 x 576 = 3600000.

                                                               UNIT  |||

                                     AVERAGE  AND  PROBLEMS ON AGES

 Average is a very simple but effective way of representing an entire group by a single value.

Average of a group is defined as

                        Average = sum of all items in the group / no of items in the group

If  the value of each item is increased by the same value ‘p’ then the  average of the group or items will also increase by p.

If the value of each item is decreased by the same  value ‘p’ then the average of the group or items will also multiply by ‘p’.

If the value of each item is divided by the same value p (p#0), then the average of the group or items will also be divided by p

The average of group0 of items will always lie between the smallest value in the group and largest value in the group i.e., the average will be grater than the smallest value and less than the largest value in the group.

Allegation rule :

This states that “ when different quantities of the same or different ingredients, of different costs (one cheap and the other dear) are mixed together to produce a mixture of a mean cost , the ratio of their quantities are inversely proportional to the difference in their cost from the mean cost “

 Quantity of cheap/ Quantity of dear = price of dear –mean price/ mean price – price of mean

Average value : These are four different means.

1)      arithmetic mean 2) harmonic mean 3) geometric mean  4) weighted mean

Examples :

1. find the average of all prime numbers between 30 and 50.

Solution : the prime numbers between 30 and 50 are : 31,37,41, 43 and 47

:. Required average   = 31+37+41+43+47/5 =199/5 =39.8

2. The average age of a class of 39 students is 15 years if the age of the teacher be included, then the average increases by 3 months. Find the age of the teacher.

Solution: Total age of 39 persons = (39x15) years = 585 years

Average age of 40 persons = 15 years 3months

                                                =61/4 years

Total age of 40persons = 61/4x40 years

                                           = 610 years

:. Age of the teacher     = (610-585) years = 25 years.

3. Distance between two stations A and B is 778 km. Han covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km per hour. Find his average speed during the whole journey.

Solution :  Required average speed

                = 2xy/x+y km/hr ( harmonic mean)

                = 2x84x56 / 84+56  km/hr

               = (2x84x56) /140    km/hr  = 67.2 km/hr.

4. A batsman makes a score of 87 runs in the 17^th innings and these increases his average by 3. find the average after 17^th innings.

   Solution : Let the average after 17^th innings =x

     Then. Average after 16^th innings = (x-3)

:. 16 (x-3) + 87 = 17x

 => x= 87-48=39

5. The average age of the husband , his wife and their sons ten years ago was 25 years and that of the husband and wife six years ago was 37 years. What is the present ages of son ( in years ) ?

Solutions : Total of present ages of husband, wife and son = (25x3) +10+ 10 +10 = 105 years.

Average age of husband and  his wife six years ago was 37 years

Total of present ages of husband and wife = (37x2) +6 +6 = 86 years.

:. Son’s present age = 105 – 86 = 19 years.

5. The average age of A and B is 30 years. If c were to replace A, the average age would be 40 years and if he were to replace B, the average age would be 35years. What is A’s age?

Solutions :

Average age of A and B = 30 years

Total age of A and B = 30x2 = 60 years

Average age of B and C = 40 years

Total age of B and C = 40x2= 80 years

Total age of  A and c = 35x2 = 70 years

After adding (1) and (3) we get

2 A + B + C = 130

But we know that (B+C) = 80

:. 2A + 80 = 130

 A = 130 – 80/2 = 25 years.

Hence, a’ s age is 25 years.

Practice:

1.       Ram secured 50,50 and 60 marks  in 3 out of 4 papers. If he secured 50 members in aggregate, how many marks did he secure in the fourth paper?

a)      40  b) 45 c) 50 d) 55 e) none of these

                                     Ans a) 40

2.       A father is now 7 times older than his son but two years hence he will be only five times older. What is the father’s present age (in years)?

                                       a) 28   b) 30   c) 31  d) data inadequate  e) none of these

                                       Ans a) 28.

Dr. D.Giri

Reader in Statistics

S.V.A. Govt college (M)

SRIKALAHASTI -517644,  Mobile : 9703102294

ANALYTICAL SKILLS - MENSURATION

MENSURATION

Introduction : - This chapter deals with Measurments of areas, volumes of triangles,

Rectangles, circles, parallelogram, cubes etc..

Important formulas :

• Area of the rectangle =1/2 *base*height
• Area of equilateral triangle = √3/ 4 (side)²
• Semi perimeter of the triangle of sides a, b ,c    is   s= (a +b +c) / 2
• Area of the triangle = √(s(s-a)(s-b)(s-c))   
• Area of the parallelogram = base * height
• Area of the rectangle = length*breadth
• Area of the square = (a)² = ½ (diagonal)²
• Diagonal = √2 a
• Perimeter = 4a

Trapezium :  area = ½(sum of the parallel sides) * distance between them

Rhombus :  area = ½ (product of diagonals)

Circle:  area = ╥r², ╥ = 22/7 = 3.14

               Circumference = 2╥r

                Length of an area = 2╥r Ө / 360

                 Area of a sector = ╥r²Ө / 360

Volume and area of solid structure:

Cuboids:  volume of the cuboid = l b h

                 Longest diagonal of the cuboid = √ l²+b²+h²

Total surface area = 2(lb+bh+lh)

Area of the room (length * breadth)

Area of 4 walls of the room = 2(l+b)*height

Cube:

Ø  Volume of the cube = a³

Ø  Longest diagonal = √3a

Ø  Total surface area = 6a²

Ø  Total length of all edges = 12a

Cylinder:

v  Volume = ╥r²h

v  Base area = ╥r²

v  Curved surface area = 2╥rh

v  Total surface area = 2╥rh+2╥r²

Pyramid:  volume = 1/3 * area of the base * height

Cone:  if the radius of the base of the cone is r and height is h and slant height (l)
 

Then l²≠ h²+r²

·         Volume of the cone = ⅓ ╥ r²h

·         Area of the curved surface = ╥rl

·         Total surface area = ╥rl+╥r² = ╥r(l+r)

Sphere :

• Volume = 4/3 ╥r³              r-is the radius
• Curved surface area = 4╥r²

Hemi sphere:

·         Volume = 2/3 ╥r³

·         Curved surface area = 2 ╥r²

·         Total surface area = 3╥r²

·         Volume of spherical shell = 4/3 ╥(R³-r³)

·         Volume of the metal in a hallow pipe = ╥(R²-r²)h

·         Total surface area of an open pipe = 2╥[(Rh+rh+(R²-r²)]