Time and Work: Quiz

Directions: Study the following information and answer the questions given below:

1. 2 men or 5 women or 7 boys can finish a work in 469 days, then the number of days taken by  7 men, 5 women and 2 boys to finish the work is-

1. 134
2. 106
3. 100
4. 98

2.  6 children and 2 men complete a certain piece of work in 6 days. Each child takes twice the time taken by a man to finish the work. In how many days will 5 men finish the same work-?

1. 6
2. 8
3. 9
4. 15

3.(x-2) men can do a piece of work in x days and (x+7) men can do 75%of the same work in (x-10) days. Then in how many days can (x+10) men finish the work?

1. 27
2. 12
3. 25
4. 18

4.  A man , a woman and a girl worked for a contractor for the same period. A man is twice efficient as a woman and a woman is thrice efficient as girl Rs. 10000 were given to all of them. What is the sum of money received by a woman and a girl together?

1. 5500/-
2. 4500/-
3. 4000/-
4. 6000/-

5.A piece of work can be done by 10 men and 6 women in 18 days. Men work 9 hr per day while women work 7.5 hr per day. Per hr efficiency of a woman is 2/3 of a man’s. In how many days 10 men and 9 women complete the work?

1. 16
2. 20
3. 30
4. 25

6.Tap A can fill a tank in 20 hr, B in 25 hr but tap C can empty a full tank in 30 hr. Starting with A, followed by B and C each tap opens alternatively for one hr period till the tank gets filled up completely. In how many hr the tank will be filled up completely?

1. 51 11/15
2. 52 2/3
3. 24 4/11
4. None of these

7. If one pipe A can fill a tank in 20 min, then 5 pipes, each of 20%efficiency of A, can fill the tank in-

1. 80 min
2. 100 min
3. 20 min
4. 25 min

8. A, B and C three weavers have to supply an order of 100 Sweater. A can weave a Sweater in 2 hr, B in 3 hr and C in 4 hr respectively. It is known that even being a joint contract each one weaver help to the rest weavers. In how many hr they will complete the order irrespective of day or night?

1. 93 hr
2. 100 hr
3. 92 4/13 hr
4. 94 hr

9. A tank is connected with 4 pipes A, B, C and D of which two are filling the tank and other two are emptying it. The time taken by A, B, C and D to finish their jobs are 10hr, 15hr, 20 hr and 30 hr respectively. All four pipes are opened. When the tank was empty, it took 12 hr to fill it completely. Which two are the outlet pipes?

1. A and B
2. C and D
3. A and C
4. B and D

10.Himesh and fatima are two different persons, but when they worked together they complete it in 10 days. Had Himesh worked at half of his efficiency and Fatima at five times of his efficiency it would have taken them to finish the job in 50% of the scheduled time. In how many days Fatima can complete the job working alone?

1. 15 days
2. 24 days
3. 15 days
4. 30 days

SOLUTION

1. 2 men = 7 boys => 1man= 7/2 days

5 women= 7boys=> 1 women=7/5boys

7men+5women+2boys=7×7/2+5×7/5+2=67/2boys

Now,     B1*D1= B2*D2

 7*469=67/2*D2

D2= 98days

2. 6C+2M=6days

36C+12M=1days

Again

1M=2C

36+12*2=1day

60 children can do the work in 1day

Now, 5men= 10Children

10 children can do the work in 6days.

3. ¾*(x-2)x=(x+7)(x-10)

X­­­­­­­2 – 6x-280=0

X=20 and x= -14

So, the acceptable value is x=20

Total work= (x-2)*x=18*20=360unit

Now       360=30*K         (30=20+10)

K=12days

4. Efficiency of a man :  woman   : girl  = 6:3:1

Share of a woman and girl =   (3+1)/(6+3+1)*10000

=4/10*10000=Rs.4000

5. Work of a man for 1 hour= 3/2 women’s work for 1 hour

Again, work of a man for 1 da   =(3/2*9/7.5)women’s  work for 1 hour

Work of a man for 1 day= 9/5 women’s work for 1 day

1man=9/5women

10Men+6Women= 10* 9/5 + 6=24women

10Men+9 Women= 10*9/5 + 9= 27 women

Now,

D1*W1= D2*W2

                                                    18*24= D2*27

D2= 16 days

6. Efficiency of A=5%

Efficiency of B=4%

Efficiency of C= -3.33%

It means in every 3 consecutive hours taps A,B and C can fil    5.66% =(5+4-3.33)

Therefore in 51 hours= (3*17) taps A,B and C can fill 96.33%=(5.66*17)

The remaining part i.e, 3.66%=(100-96.33) can be filled up by A in 11/15 hours =(3.66/5),sine it is now A’s turn

Hence the total time required= 51+11/15=51 11/15

7. 20 minute

8. LCM of 2,3 and 4= 12

In 12 hours A will make 6 Sweter

In 12 hours B will make 4 Sweter

In 12 hours C will make 3 Sweter

i.e, in 12 hours they will weave 13 sweter

so, in 84 hours they will weave 91 sweter

Now, in 9 hours A will make 4 sweter

In 9 hours B will make 3 sweter

In 9 hours C will make 2 sweter

So. They will complete 100 sweter in 93 hours.

Note: Since, they cannot share each others work so B will take completely 9 hours to make 3 sweter,even when A and C stay idle for the last 1 hour till B completes his own work.

9. Efficiency of A=10%

Efficiency of B= 6.66%

Efficiency of C= 5%

Efficiency of D= 3.33%

Efficiency of A+B+C+D=8.33     (TIME=12HOURS)

Now, go through options and consider A and B as inlet pipes and C and D as outlet pipes, then

(10+6.66)-(5+3.33)=8.33

Which is required hence it is certain that C and D are outlet pipes.

10. Efficiency of Himesh and Fatima (Combined)=10%

Consider option (d)

Efficiency of Fatima=3.33%

Efficiency of Himesh=6.66%

Now, the new efficiency of Fatima=16.66%

And the new efficiency of Himesh = 3.33%

Therefore, newly combined efficiency of Himesh and Fatima=20%

Therefore, required number of days by Himesh and Fatima

Working together=5

Since 5 is half of 10, hence the option (D) is correct.

Permutations and Combinations

Important Concepts and Formulas 1. Multiplication Theorem (Fundamental Principles of Counting)

If an operation can be performed in $m$ different ways and following which a second operation can be performed in $n$ different ways, then the two operations in succession can be performed in $m\times n$ different ways.

2. Addition Theorem (Fundamental Principles of Counting)

If an operation can be performed in $m$ different ways and a second independent operation can be performed in $n$ different ways, either of the two operations can be performed in $(m+n)$ ways.

3. Factorial

Let $n$ be a positive integer. Then $n$ factorial can be defined as
$n!=n(n-1)(n-2)\cdots 1$

Examples
$5!=5\times 4\times 3\times 2\times 1=1203!=3\times 2\times 1=6$
Special Cases
$0!=11!=1$

4. PermutationsPermutations are the different arrangements of a given number of things by taking some or all at a time.

Examples
All permutations (or arrangements) that can be formed with the letters a, b, c by taking three at a time are (abc, acb, bac, bca, cab, cba)
All permutations (or arrangements) that can be formed with the letters a, b, c by taking two at a time are (ab, ac, ba, bc, ca, cb)

5. CombinationsEach of the different groups or selections formed by taking some or all of a number of objects is called a combination.

Examples
Suppose we want to select two out of three girls P, Q, R. Then, possible combinations are PQ, QR and RP. (Note that PQ and QP represent the same selection.)
Suppose we want to select three out of three girls P, Q, R. Then, only possible combination is PQR

6. Difference between Permutations and Combinations and How to identify them

Sometimes, it will be clearly stated in the problem itself whether permutation or combination is to be used. However if it is not mentioned in the problem, we have to find out whether the question is related to permutation or combination.
Consider a situation where we need to find out the total number of possible samples of two objects which can be taken from three objects P, Q, R. To understand if the question is related to permutation or combination, we need to find out if the order is important or not.
If order is important, PQ will be different from QP, PR will be different from RP and QR will be different from RQ
If order is not important, PQ will be same as QP, PR will be same as RP and QR will be same as RQ
Hence,
If the order is important, problem will be related to permutations.
If the order is not important, problem will be related to combinations.
For permutations, the problems can be like "What is the number of permutations the can be made", "What is the number of arrangements that can be made", "What are the different number of ways in which something can be arranged", etc.
For combinations, the problems can be like "What is the number of combinations the can be made", "What is the number of selections the can be made", "What are the different number of ways in which something can be selected", etc.

pq and qp are two different permutations, but they represent the same combination.

Mostly problems related to word formation, number formation etc will be related to permutations. Similarly most problems related to selection of persons, formation of geometrical figures, distribution of items (there are exceptions for this) etc will be related to combinations.

7. Repetition

The term repetition is very important in permutations and combinations. Consider the same situation described above where we need to find out the total number of possible samples of two objects which can be taken from three objects P, Q, R.
If repetition is allowed, the same object can be taken more than once to make a sample. i.e., PP, QQ, RR can also be considered as possible samples.
If repetition is not allowed, then PP, QQ, RR cannot be considered as possible samples.
Normally repetition is not allowed unless mentioned specifically.

8. Number of permutations of n distinct things taking r at a time

Number of permutations of n distinct things taking r at a time can be given by

ⁿP_r = $\frac{n!}{(n-r)!}$ $=n(n-1)(n-2)...(n-r+1)$ where $0\le r\le n$

Special Cases
ⁿP₀ = 1
ⁿP_r = 0 for $r>n$
ⁿP_r is also denoted by P(n,r). ⁿP_r has importance outside combinatorics as well where it is known as the falling factorial and denoted by (n)_r or n^r

Examples
⁸P₂ = 8 × 7 = 56
⁵P₄= 5 × 4 × 3 × 2 = 120

9. Number of permutations of n distinct things taking all at a time
Number of permutations of n distinct things taking them all at a time
= ⁿP_n = n!

10. Number of Combinations of n distinct things taking r at a time
Number of combinations of n distinct things taking r at a time ( ⁿC_r) can be given by 
ⁿC_r = $\frac{n!}{(r!)(n-r)!}$ $=\frac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$ where $0\le r\le n$
Special Cases
ⁿC₀ = 1
ⁿC_r = 0 for $r>n$
ⁿC_r is also denoted by C(n,r). ⁿC_r occurs in many other mathematical contexts as well where it is known as binomial coefficient and denoted by $(\frac{n}{r})$

Examples
⁸C₂ = $\frac{8\times 7}{2\times 1}$ = 28
⁵C₄= $\frac{5\times 4\times 3\times 2}{4\times 3\times 2\times 1}$ = 5