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2. According to a plan, a drilling team ............How many meters of drilling was the plan for each day?
Soln:
Let the initial plan to drill be: $x$ m/day.
Let the original no. of days required to drill = t days.
=>$xt$=270
ATQ:
$3x+(x+8)(t-1-3)=280$
$3x+xt-4x+8t-32=280$
$3x+270-4x+8t-32=280$
$-x+8t=42$
$-x+8*\frac{270}{x}=42$
$\frac{-x^2+8*270}{x}=42$
$-x^2+8*270=42x$
$x^2+42x-8*270=0$
$x=-72,30$
Ans. 30 m/day
13. A tank holds 100 gallons of water. Its inlet is 7 inches..........proportional to the diameter)
Soln.
Capacity of tank (Work to be done) = 100 gallons
Rate at which the inlet pipe fills the tank = 5 gallons/min
=> Rate at which the outlet pipe empties the tank =5*2=10 gallons/min [Since the rate of filling or emptying is directly proportional to the diameter.]
The time taken by the outlet to empty the tank= $\frac{100}{10}$=10 min.
Ans. 10 min.
14. A tank of capacity 25 liters has an inlet and an outlet tap............can be the outlet flow rate in liters/min.
Soln.
Let the inlet fill the tank at I liters/min.
Let the outlet empty the tank at O litres/min.
The capacity of the tank is 25 liters.
I and O take 5 minutes to fill the tank.
Therefore, rate at which I and O combined fill the tank of capacity 25 liters
= $\frac{25}{5}=5$ liters/min.
In other words,
$$I-O=5 liters/min.$$
Let the new rate at which the outlet empties the tank = O' liters/min, which is equal to 2O
Now, since the tank does not get filled, therefore the rate at which I and O' combinedly fill the the tanke less than 0, this will ensure the tank never gets filled.
In other words,
$$I-O'<0$$
$$I-2O<0$$
$$(5+O)-2O<0 ; [Since I-O=5]$$
$$5-O<0$$
$$O>5$$
Therefore from the given options, the outlet flow could be 6 liters/min.
15. X takes 4 days to complete one-third ...........Y to complete the remaining work done.
Soln.
X takes $\frac{1}{3}rd$ of a day to complete the work.
=> X completes the work in 12 days.
Y takes $\frac{1}{6}th$ of a day to complete the work.
=> Y completes the work in 18 days.
Z takes $\frac{1}{6}th$ of a day to complete the work.
=> Z completes the work in 10 days.
Let the total work be LCM(12,18,10)=180 units.
Rate of doing work for:
A=180/12=15 units/day.
B=180/18=10 units/day.
C=180/10=18 units/ day.
A,B,C's one day work =15+10+18=43units.
ATQ:
$3*(15+10+18)+10*x=180$
$10x=180-129-51$
$x=5.1$
Ans. 5.1 days
17. Three diggers dug a ditch 324 m deep.....................................does the first digger dig per shift?
Soln.
Let the 3 diggers be A,B,C.
Let they dig a m/day, b m/day, c m/day respectively.
ATQ:
$$(a+b+c)*6=324----(1)$$
Let duration of each shift be t
$$(c-b)*t =(b-a)*t$$
$$c-b =b-a-----(2)$$
also,
$$10c=14a$$
$$or, 5c=7a.------(3)$$
Solving (1),(2),(3) we get a=15
Here they have considered only one shift takes place per day
Ans. 15m
18. A,B and C working together completed a job in 10 days.......................fastest worker to complete the entire work?
Let the rate of work done by A, B, and C be a units/day, b units/day, c units/day respectively.
Let the total unit of work done be 1 units.
For the first three days, C worked along with A and B.
Therefore,
$$3(a+b+c)=\frac{37}{100}\tag 1$$
Remaining work=$1-\frac{37}{100}=\frac{67}{100}$
The remaining work was completed by A and B in 7 days.
Therefore,
$$7(a+b)=\frac{63}{100}$$
$$\implies (a+b)=\frac{9}{100} \tag 2$$
Also, the work done by A in five days is equal to the work done by B in four days.
Therefore,
$$5a=4b\tag 3$$
Substituting $(3)$ in $(2)$
$$\frac{4.b}{5}+b=\frac{9}{100}$$
$$\implies \frac{9b}{5}=\frac{9}{100}$$
$$\implies b=\frac{1}{20} \tag {IV}$$
Substituting $(IV)$ in $(3)$ we get:
$$a=\frac{4}{5}*\frac{1}{20}=\frac{1}{25} \tag{V}$$
Substituting $(IV)$ and $(V)$ in $(1)$, we get:
$$3(\frac{1}{25}+\frac{1}{20}+c)=\frac{37}{100}$$
$$3(\frac{9}{100} +c)=\frac{37}{100}$$
$$3(\frac{9+100c}{100})=\frac{37}{100}$$
$$27+300c=37$$
$$300c=10$$
$$c=\frac{1}{30}$$
Therefore:
$a =\frac{1}{25}$ units per day
$b=\frac{1}{20}$ units per day
$c =\frac{1}{30}$ units per day
Therefore, the fastest worker is B and he completes the entire work in 20 days.
Ans. 20 days.
19. A and B completed a work together in 5 days. Had A worked............A alone to do the work?
Soln:
Let the rate of doing work by A and B be a units/day and b units/day respectively.
ATQ:
$$(a+b)*5=1$$
Multiplying the above equation by 2 yields:
$$(5a+5b=1)*2 \tag 1$$
Also from the question:
$$(2a+\frac{b}{2})*4=1$$
$$\implies (4a+b)*2=1$$
Multiplying the above equation by 5 yields:
$$\implies (8a+2b=1)*5 \tag 2$$
From (1) and (2):
$$10a+10b=2$$
$$40a+10b=5$$
Therefore
$$30a=3$$
$$\implies a=\frac{1}{10}$$
Therefore A can complete the work in 10 days.
Ans. 10 days.
20. Two typists of varying skills can do a job ............................typing job working alone?
Soln:
Let the rate of working of two typists A and B (say) be a units/min and b units/min respectively.
Let the total amount of work be 1 unit.
A and B can do the job in 6 minutes.
$$\implies (a+b)*6=1\tag a$$
Multiplying the above equation by 5 yields
$$(6a+6b=1)*5 \tag 1$$
ATQ:
$$4a+6b=\frac{4}{5}$$
$$10a+15b=2$$
Multiplying the above equation by 2 yields
$$(10a+15b=2)*2 \tag 2$$
From $(1)$ and $(2)$
$$30a+30b=5$$
$$20a+30b=4$$
This implies:
$$10a=1$$
$$\implies a=\frac{1}{10} \tag b$$
From $a$ and $b$
$$6b=1-6\frac{1}{10}$$
$$6b=\frac{2}{5}$$
$$b=\frac{1}{15}$$
Therefore
$a=\frac{1}{10}$
$b=\frac{1}{15}$
Thus the slower typist is B and he completes the work in 15 minutes.
Ans. 15 minutes.
21. Three cooks have to make 80 idlis...................160 idlis for a marriage party the next day?
Soln.
Let the rate at which the three cooks A,B and C prepare idlis be a burger/min, b burger/min, c burger/min respectively.
"They are known to make 20 pieces of idlies every minute working together"
$$\implies a+b+c=20$$
$\implies (b+c)=(20-a) \tag 1$
Let the amount of time in which the first cook, A worked alone be t.
Considering the statement:"The first cook began working alone and made 20 pieces having worked for some time more than 3 minutes"
$\implies at=20$
According to the statement in the question:
$$at+(b+c)(8-t)=80$$
$$\implies 60+(b+c)(8-t)=80$$
Since $b+c=20-a$, the above equation becomes:
$$60+(20-a)(8-t)=80$$
$$(20-a)(8-t)=60$$
$$160-20t-8a+at=60$$
$$-20t-8a+at=-100$$
$$-20t-8a+20=-100$$
$$-20t-8a=-120$$
$$20t+8a=120$$
$$5t+2a=30$$
$$5\frac{20}{1}+2a=30$$
$$\frac{100+2a^2}{a}=30$$
$$2a^2-30a+100=0$$
$$a^2-15a+50=0$$
$$\implies a=5,10$$
CASE 1: a=5 idlis/min
A prepares 5 idlis in 1 min.
Therefore A prepares 20 idlis in $\frac{20}{5}$=4 min which is >3
Therefore the condition that "The first cook began working alone and made 20 pieces having worked for some time more than 3 minutes" is satisfied. Therefore a=5 idlies/min is possible.
CASE 2: a=10 idlis/min
A prepares 10 idlis in 1 min.
Therefore A prepares 20 idlis in $\frac{20}{10}$=2 min which is <3
Therefore the condition that "The first cook began working alone and made 20 pieces having worked for some time more than 3 minutes" is not satisfied. Therefore a=10 idlies/min is NOT possible.
Therefore $a=5$ idlies/min.
If A makes 5 idlies in 1 minute.
Then A will make 160 idlies in 160/5=32 min. [Unitary method].
Ans. 32 min
23. Each of A,B and C need a certain unique time to do a certain work..........How much woes C do per hour?
Soln.
Let the total work be 1 unit.
Let the rate at whic A,B and c work be a units/hr, bunits/hr, c units/hr respectively.
Now,
C completes c units of work in 1 hour.
$\implies$ C completes 1 unit of work in $\frac{1}{c}$ hour.
Therefore, C takes $\frac{1}{c}$ hours to complete the entire work.
According to question:
"C needs 1 hour less than A to complete the work"
This implies:
$$a(\frac{1}{c}+1)=1$$
$$\frac{1}{c}=\frac{1}{a}-1=\frac{1-a}{a}$$
$$\implies c=\frac{a}{1-a} \tag 1$$
According to question:
"Working together, they require 30 minutes to complete 50% of the job"
This implies:
$$(a+b+c)*\frac{1}{2}=\frac{1}{2}$$
$$\implies (a+b+c)=1 \tag 2$$
According to question: "The work also gets completed if A and B start working together and A leaves after 1 hour and B works for furthur 3 hours."
This implies:
$$(a+b)*1+b*3=1$$
$$\implies a+4b=1 \tag 3$$
Substituting $(1)$ in $(2)$, we get:
$$a+b+\frac{a}{1-a}=1$$
$$\frac{a-a^2+b-ab+a}{1-a}=1$$
$$2a=a^2+b-ab=1-a$$
$$3a-a^2+b-ab-1=0$$
Substituting value of a from $(3)$ in the above equation, we get:
$$3(1-4b)-(1-4b)^2+b-(1-4b)b-1=0$$
$$3-12b-(1+16b^2-8b)+b-b+4b^2-1=0$$
$$4b^2-12b+2-1-16b^2+8b=0$$
$$-12b^2-4b+1=0$$
$$12b^2+4b-1=0$$
$$b=\frac{-6}{12},\frac{2}{12}=\frac{1}{6}$$
Now,
$$a=1-4b=1-\frac{4}{6}=\frac{1}{3}$$
$$c=1-\frac{1}{3}-\frac{1}{6}=\frac{3}{6}=\frac{1}{2} work/hour$$
Therefore, C does 50% of the work/hour.
Ans. 50%
25.Mini and Vinay are quiz masters......make in x minutes?
Soln:
Let Mini and Vinay take M and V minutes to frame a question respectively.
Mini takes M mins to frame 1 question.
Therefore Mini takes $x$ mins to frame $\frac{x}{M}$ questions.
Vinay takes V mins to frame 1 question.
Therefore Vinay takes $x$ mins to frame $\frac{x}{V}$ questions.
ATQ:
$$\frac{x}{M}=\frac{x}{V}+y \tag 1$$
Multiplying the above equation by 2 yields:
$$\frac{2x}{M}=\frac{2x}{V}+2y \tag{1a}$$
Now, in the second scenario:
Mini takes (M-2) mins to frame 1 question.
Therefore Mini takes $x$ mins to frame $\frac{x}{(M-2)}$ questions.
Vinay takes V-2 mins to frame 1 question.
Therefore Vinay takes $x$ mins to frame $\frac{x}{(V-2)} $ questions.
Therefore ATQ:
$$\frac{x}{M-2}=\frac{x}{V-2}+2y \tag 2$$
We need to find $\frac{x}{M}$
Equating 2y from $(1)a$ and $2$ yields:
$$\frac{x}{M-2}-\frac{x}{V-2}=\frac{2x}{M}-\frac{2x}{V}$$
$$\frac{1}{M-2}-\frac{1}{V-2}=\frac{2}{M}-\frac{2}{V}$$
$$\frac{V-2-M+2}{(M-2)(V-2)}=\frac{2V-2M}{MV}$$
$$\frac{V-M}{MV-2M-2V+4}=\frac{2V-2M}{MV}$$
$$(V-M)MV=2(V-M)(MV-2M-2V+4)$$
$$MV=2MV-4M-4V+8$$
$$MV-4M-4V+8=0$$
$$MV-4V-4M+8=0$$
$$V(M-4)-4M+8=0$$
$$V=\frac{4M-8}{M-4}$$
Substituting value of V in (1):
$$x(\frac{1}{M}-\frac{1}{V})=y$$
$$x(\frac{1}{M}-\frac{M-4}{4M-8})=y$$
$$x(\frac{(4M-8)-M(M-4)}{M(4M-8)})=y$$
$$\frac{4M-8-M^2+4M}{4M^2-8M}=\frac{y}{x}=k$$
$$8M-M^2-8=4kM^2-8kM$$
$$M^2(4k+1)+M(-8M-8)+8=0$$
$$M^2(4k+1)-8(k+1)M+8=0$$
$$M=\frac{8(k+1) \pm \sqrt {64(k+1)^2-4(4k+1).8}}{2(4k+1)}$$
$$M=\frac{8(k+1) \pm 4\sqrt {4(k+1)^2-2(4k+1)}}{2(4k+1)}$$
$$M=\frac{4(k+1) \pm 2\sqrt {4(k+1)^2-2(4k+1)}}{(4k+1)}$$
________________________________________________
ROUGH WORK:
Now,
$$4(k+1)^2-2(4k+1)$$
$$=4(k^2+1+2k)-2(4k+1)$$
$$=4k^2+4+8k-8k-2=4k^2+2$$
_______________________________________________
$(1)$ becomes:
$$M=\frac{4(k+1) \pm 2\sqrt{4k^2+2}}{4k+1}$$
$$M=\frac{4(\frac{y}{x}+1) \pm 2\sqrt{4(\frac{y}{x})^2+2}}{4\frac{y}{x}+1}$$
$$M=\frac{4\frac{y+x}{x} \pm 2\sqrt{\frac{4y^2+2x^2}{x^2}}}{\frac{4y+x}{x}}$$
$$M=\frac{4(x+y) \pm 2 \sqrt{2x^2+4y^2}}{x+4y}$$
We basically have to find $\frac{x}{M}$
$$\frac{x}{M}=x*\frac{x+4y}{4(x+y) \pm 2 \sqrt{2x^2+4y^2}}$$
$$\frac{x}{M}=x*\frac{x+4y}{4(x+y) \pm 2 \sqrt{2x^2+4y^2}}*\frac{4(x+y) \mp 2 \sqrt{2x^2+4y^2}}{4(x+y) \mp 2 \sqrt{2x^2+4y^2}}$$
$$\frac{x}{M}=x*\frac{(x+4y)*(4(x+y) \mp 2\sqrt{2x^2+4y^2})}{(4(x+y))^2-(2\sqrt{2x^2+4y^2})^2} \tag 4$$
____________________________________________
ROUGH WORK:
$$(4(x+y))^2-(2\sqrt{2x^2+4y^2})^2=8x(x+4y)$$
____________________________________________
$(4)$ becomes:
$$\frac{x}{M}=\frac{x.(x+4y)*(4(x+y) \mp 2\sqrt{2x^2+4y^2})}{8x(x+4y)}$$
$$\frac{x}{M}=\frac{2(x+y) \mp \sqrt{2x^2+4y^2}}{4}$$
This matches with option (a)[Taking the '-' sign]. Hence $\frac{1}{4}[2(x+y)-\sqrt{2x^2+4y^2}{}]$ is the answer.
26. A tank of 3600 cu m capacity is being filled with water..................pump discharging the tank.
Soln.
According to question, if delivery of pump filling the tank is 100 units/minuite, then delivery of the pump discharging the tank is 120 units/minuite.
In other words:
$$filling-rate:discharging-rate$$
$$100:120$$
$$5:6$$
So the ratio of filling rate to discharging rate is 5:6.
Let the filling pipe fill the tank at the rate of 5x $m^3$/minute.
Let the discharging pipe empty the tank at the rate of 6x $m^3$/minute.
let the time taken by the discharging pump to discharge the entire tank be t minuites.
Then, according to question:
$$5x(t+12)=720 \tag 1$$
and
$$6x(t)=3600 \tag2$$
$(1)$ reduces to:
$$xt+12x=720$$
$$\implies 600+12x=720$$
$$=>12x=120$$
$$\implies x=10$$
Therefore the discharging pipe discharges the tank at the rate of $6x$=6*10=60$m^3$/minuite.
Ans. 60$m^3$/minuite.
28.A tank of 425 litres capacity has been filled with water..........How long was the seond pipe open?
Soln:
Let the rate at which the pipes A and B (say) fill the tank be a work/hour and b work/hour respectively.
According to question:
$$a(t+5)+bt=425 \tag 1$$
$$2at=b.(t+5) \tag 2$$
$$(a+b)*17=425$$
$$(a+b)=25 \tag 3$$
$(1)$ becomes:
$$(a+b)*t+5a=425$$
$$25t+5a=425$$
$$5t+a=85 \tag{1a}$$
$(2)$ becomes:
$$2at=(25-a)(t+5)$$
$$2at=25t+125-at-5a$$
$$3at-25t+5a-125=0$$
Replacing a from $1a$ we get:
$$3(85-5t)-25t+5(85-5t)-125=0$$
$$15t^2-205t-300=0$$
Taking the positive value of t, we get:
t=15,$\frac{-4}{3}$(Rejected)
Therefore the second pipe was open for 15 hours.
29. Two men and a woman are entrusted........if they all worked together?
Soln.
Let the rate at which two men and one woman(A,B,C) work be a work/day, b work/day, c work/day.
According to question:
$$a .t= (b + c ).t \tag{1}$$
$$a = b + c \tag{1}$$
$$\frac{1}{b} = \frac{1}{b+c} + 3 \tag{2}$$
$$\frac{1}{a}=\frac{2}{b}-8 \tag{3}$$
From $(1)$ and $(2)$
$$\frac{1}{b}=\frac{1}{a}+3 \tag 4$$
From $(3)$ and $(4)$
$$\frac{1}{b}-3=\frac{2}{b}-8$$
$$\frac{1}{b}=5 \tag5$$
From $4$ and $5$
$$5=\frac{1}{a}+3$$
$$\frac{1}{a}=2$$
From $(1)$
$$c=\frac{1}{2}-\frac{1}{5}=\frac{3}{10}$$
Now,
$$(a+b+c)*t=1$$
$$(\frac{1}{2}+\frac{1}{5}+\frac{3}{10})*t=1$$
$$(\frac{5+2+3}{10}.t)=1$$
$$t=1$$
Therefore, the two men and the woman would take one hour to complete the task.
(The answer given in the book is wrong).
30. The Bubna dam has four inlets.............How much time will it take all the four inlets to fill up the dam?
Soln.
Let the four inlets be A, B, C and D.
Let the rate at which A,B,C,D work be a work/minuite, b work/minuite , c work/minuite and d work/minuite respectively.
A,B,C take 12 minuites to complete the job.
B,C,D take 15 minuites to do the job.
A,D take 20 minuites to do the job.
Let the quantum of work be LCM(12,15,20)=60 units of work.
A,B,C combined work at the rate $\frac{60work}{12}$=5 work/minuite.
In other words:
$a+b+c=5 work/minuite \tag1$
B,C,D combined work at the rate $\frac{60work}{15}$=4 work/minuite.
In other words:
$b+c+d=4 work/minuite \tag2$
A,D combined work at the rate $\frac{60work}{20}$=3 work/minuite.
In other words:
$a+d=53work/minuite \tag3$
Adding $1$ and $2$
$$a+2(b+c)+d=4+5=9 work/minuite$$
$$2(b+c)+(a+d)=9 work/minuite$$
$$2(b+c)=9-3=6 work/minuite$$
$$b+c=3 work/minuite \tag 4$$
From $3$ and $4$:
$$a+b+c+d=3+3=6 work/minuite$$
Therefore A,B,C,D complete 60 units of work working at the rate of 6 work/minuite in $\frac{60work}{6 work/minuite}$
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