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2) Answer: A
Electricity produced in village B in 1st week at
75% efficiency= (75/100)*maximum power
produced =75/100*(total number of windmills*%
of windmills operating)*(power produced by 1
windmill)
=> Electricity produced in village B in 1st week
at 75% efficiency=75/100*15*60/100*1,00,000
=6,75,000 units
Similarly, electricity produced in village B in 2nd
week at 90 %
efficiency=90/100*15*40/100*100000 =5,40,000
units
In a similar way we can determine electricity
produced by windmills in village C in the 3rd and
4th weeks
At 80% efficiency, electricity produced by
windmills in village C in 3rd
week=80/100*25*60/100*1,50,000 =18,00,000
units
At full efficiency, electricity produced by
windmills in Village C in 4th
week=25*80/100*1,50,000=30,00,000 units
Thus, total electricity produced in village
B=6,75,000+5,40,000=12,15,000 units
Total electricity produced in village
C=18,00,000+30,00,000=48,00,000 units
Thus, required percentage=12.15 lakh/48
lakh*100=25.3125 %
sиαρσnє🌱
3) Answer: C
At full efficiency, units consumed in village A in
all the four weeks= Electricity produced in all the
four weeks= (40/100*20 +50/100*20
+30/100*20 +80/100*20)*1,20,000
=40*1,20,000 =48,00,000 units
At full efficiency, units consumed in village B in
all the four weeks= Electricity produced in all the
four weeks= (60/100*15 +40/100*15
+60/100*15 +60/100*15)*1,00,000
=33*1,00,000 =33,00,000 units
At full efficiency, units consumed in village C in
all the four weeks= Electricity produced in all the
four weeks = (40/100*25 +60/100*25
+60/100*25 +80/100*25)*1,50,000
=60*1,50,000 =90,00,000 units
At full efficiency, units consumed in village D in
all the four weeks= Electricity produced in all the
four weeks= (25/100*16 +50/100*16
+50/100*16 +75/100*16)*2,00,000
=32*2,00,000 =64,00,000 units
Units consumed by A and B=48 lakh+33 lakh
=81 lakh units
Units consumed by C and D=90 lakh+64 lakh
=154 lakh units
Required ratio=81/154=81:154
4) Answer: C
When windmills operate at full efficiency, they
produce sufficient power to light up houses for
24 hours.
Thus, power required for 24 hours electricity in
all houses= 75/100*16*200000 =24 lakh units
When houses are lit for 12 hours the electricity
produced must be half i.e., 12 lakh units
3 windmills stopped working
Power produced by 5 windmills at 45%
efficiency=5*45/100*200000=4.5 lakh units
Remaining units functional= (75% of 16)-3-
5=12-8=4
Thus, power to be produced by remaining four
windmills=12-4.5=7.5 lakh
units
Maximum power possible from remaining 4
windmills =4*200000=8 lakh units
=> Required efficiency=7.5/8*100=93.75 %
Also, total windmills operational is 9 and power
produced by them is 12 lakh units
=> Average power produced by each
windmill=12/9=1.33 lakh units.
5) Answer: D
Electricity produced in A by windmills at 90%
efficiency in 2nd
week=50/100*20*90/100*120000=10.8 lakh
units
At full efficiency, electricity produced=Power
produced at 90%
efficiency/(90%)=10.8/(90/100)=10.8/0.9=12
lakh units
In 12 lakh units, houses will remain lit for 24
hours
=> Light up time in 10.8 lakh
units=24/12*10.8=21.6 hrs
Electricity produced in D in 1st week at full
efficiency=25/100*16*200000=8 lakh units
Therefore, required percentage= (10.8-
8)/8*100=35 %
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2 918
1) Answer: A
At maximum efficiency, electricity produced by
windmills in village A= (Maximum power of one
windmill)*(windmills operational)
=1,20,000*(40/100*20) =1,20,000*8 =9,60,000
units.
These 9,60,000 units of electricity will be
consumed by 360 houses in 24 hours
Let the power produced by the windmills be X
units such that 300 houses will be lit for 16
hours a day in 1st week.
(960000/(360*24)) = (X/ (300*16))
X= 960000*300*16/ (360*24) =5,33,333.33
units
Thus, efficiency of the
windmills=5,33,333.33/9,60,000*100%=55.555
%
Also, power deficit=(9,60,000-5,33,333.33)
units=4,26,666.67 units =4,26,667
units(approx)
2 918
The given information is about windmills installed
in four villages A, B, C and D and electricity
produced by these windmills.
Number of windmills in A, B, C and D are 20, 15,
25 and 16 respectively. At maximum
efficiency(maximum power generation possible),
electricity produced by every windmill in one day
in A, B, C and D are 1.2 lakh units, 1 lakh units,
1.5 lakh units and 2 lakh units respectively. The
number of houses in villages A, B, C and D are
360, 480, 625, and 512 respectively. Not all
windmills are operational at a time in any of the
four villages. Number of windmills operations by
A in the 1st week, 2nd week, 3rd week and 4th
week are 40%, 50%, 30% and 80% respectively.
Number of windmills operations by B in the 1st
week, 2nd week, 3rd week and 4th week are 60%,
40%, 60% and 60% respectively. Number of
windmills operations by C in the 1st week, 2nd
week, 3rd week and 4th week are 40%, 60%, 60%
and 80% respectively. Number of windmills
operations by D in the 1st week, 2nd week, 3rd
week and 4th week are 25%, 50%, 50% and 75%
respectively.
Also, the windmills are not operational at full
efficiency all the time. At full efficiency, they
produce sufficient electricity to light up all houses
for 24 hours. They work at different efficiencies
and generate only a part of the maximum power
they can produce. The efficiency of a windmill
can be defined as:
Efficiency= (Power Output Generated/Maximum
Power Generation Possible)*100
e.g., If a windmill is capable of generating 2000
units/hr at maximum efficiency and runs 50%
efficiency, it will generate only 50% of 2000 units
in one hour.
sиαρσnє🌱
1) The windmills in village A in 1st week are
operating at efficiency such that 300 houses get
electricity for only 16 hours. What is the efficiency
at which it is working and how much more
electricity should be produced for 24 hours
electricity in all houses? (approximately)
A. 55.55 %, 4,26,667
B. 66.67 %, 3,19,968
C. 43.33 %, 5,54,592
D. 53.33%, 5,55,456
E. None of these
2) Total electricity produced in village B in 1st
and 2nd week at 75% and 90% efficiency
respectively is what percentage of electricity
produced in C in 3rd and 4th weeks at 80% of
maximum efficiency respectively?
A. 25.3125 %
B. 17.67%
C. 31.25%
D. 36%
E. None of these
3) What is the ratio of units consumed in village
A and B in four weeks at full efficiency to the ratio
of units consumed in village C and D in the same
period at full efficiency?
A. 27:52
B. 52:42
C. 81:154
D. 127:81
E. None of these
4) Due to a storm in village D in week 4th, 3
windmills stopped working while 5 windmills
began working at 45% efficiency. How much
efficiency should therest of the windmills start
working so that houses remain light for at least
12 hours and what is the average electricity
produced by each functional windmill?
A. 87.5 %, 1.5 lakh units
B. 90.33% ,1.35 lakh units
C. 93.75%, 1.33 lakh units
D. 94.50%, 1.40 lakh units
E. None of these
5) Fill in the following blanks with the data given
in options:
The electricity produced by windmills in A in 2nd
week at 90% efficiency lights up all the houses
for ………… hrs a day and is ……………% more
than electricity produced in D in 1st week at full
efficiency.
A. 18 hrs, 33 1/3%
B. 21.6 hrs,33 1/3 %
C. 18 hrs, 35 %
D. 21.6 hrs, 35 %
E. None of these
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From Group 1 and 2
4m * Y = Xm * 42/4
4y = 10.5X ---(1)
From Group 1 and 3
4m * Y = 7m * (X+4)
4y = 7X+28 –--(2)
Substitute the value 4y in equation--(1)
7X+28 = 10.5X
X = 8
In Group 2, apply the value X
8m * 42/4 = 84m –--(3)
Combine with group 1
4m * Y = 84m
Y = 21
Apply X and Y values in all the groups
Group 1
4m * 21 days
3w * 36 days
7c * 27 days
4m * 21 = 3w * 36 = 7c * 27
28m = 36w = 63c
Total units= L.C.M (28,36,63) = 252 units
Efficiency ratio of Men =252/28 = 9
Efficiency ratio of Women = 252/36 = 7
Efficiency ratio of Children = 252/63 = 4
Efficiency ratio of Men, Women, Children = 9 : 7
: 4
Total work = 4*21*9 = 756 units
Group 2
8m * 42/4 days
12w * 9 days
14c * 13.5 days
84m = 108w = 189c
Total units= L.C.M (84,108,189) = 756 units
Efficiency ratio of Men =756/84 = 9
Efficiency ratio of Women = 756/108 = 7
Efficiency ratio of Children = 756/189 = 4
Efficiency ratio of Men, Women, Children = 9 : 7
: 4
Total work = 4*21*9 = 756 units
sиαρσnє🌱
1) Answer: B
From the common solution we obtained Y = 21
and X =8
So, 21*5 – 8*7 = 49
49*49 = 2401
2) Answer: E
From the common solution we obtained,
Total work = 756 units
Efficiency ratio = 9 : 7 : 4
2 men * 9 = 18 units/day
6 women * 7 = 42 units/day
11 children * (4*3/2) = 66 units/day
60x + 66x = 756
X = 6 days
3) Answer: A
From the common solution we obtained Y = 21
and X =8
P = X+4 = 12 days
Q = X+7 = 15 days
R = Y+21 = 25 days
(P+Q)-- 1 day work = 1/12 + 1/15 = 3/20They
worked for 3 days = 3*3/20 = 9/20
Remaining work done by R = 1 – 9/20 = 11/20
Share of R = 11/20 of 4500
Share of R = Rs.2475
4) Answer: B
From the common solution we obtained Y = 21
and X =8
(X+4) * (3Y) = 12*63
12*63*1.75/18 = 73.5 days
5) Answer: A
From the common solution we obtained Y = 21
and X = 8
Let ‘d’ be number of days to complete the
remaining work after 22 men increased.
7X * 2.5 + (7X+22) * d = (Y+9) * 6X
56*2.5 + 78 * x = 30*48
d = 16 2/3 days
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2 918
Group 1: 4 men can complete the work in Y days.
3 women can complete the work in Y+15 days. 7
children can complete a work in Y+6 days.
Group 2: X men can complete the work in 42/4
days. 1.5X women can complete the work in 9
days. 3.5X children can complete a work in 13.5
days.
Group 3: 7 men complete the work in X+4 days.
36/5 women can complete the work in Y-6 days.
Note: In all groups, people work on the same
project. And their efficiency is the same in all the
groups.
sиαρσnє🌱
1) Find the square value of Y*5 – X*7.
A. 1296
B. 2401
C. 4096
D. 6561
E. None of these
2) If 2 men and 6 women started working
together but after X days they left and the
remaining work done by 11 children with 50%
more efficiency in the same number of X days.
Find the value of X.
A. 8
B. 7
C. 12
D. 9
E. None of these
3) P, Q and R can complete a work in (X+4),
(X+7) and (Y+4) respectively. They received
Rs.4500 for the completion of the work. P and Q
started the work, after 3 days both of them left
the work and the remaining work was finished by
R. Find the share of R.
A. Rs.2475
B. Rs.2765
C. Rs.2835
D. Rs.2585
E. None of these
4) X+4 men can do a piece of work in 3Y days.
Then, find how many days X+10 men can finish
175℅ of the work.
A. 72.5 days
B. 73.5 days
C. 74.5 days
D. 75 days
E. None of these
5) (Y+9) men can complete a work in 6X days.
7X men started the same work after 5/2 days,
Find the number of days required to complete the
remaining work after 22 men increased.
A. 16 2/3 days
B. 14 2/7 days
C. 12.5 days
D. 8.33 days
E. None of these
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2) Answer: D
Batting average of I =
[{(620+780)/2}/(12*125/100)] = 46.66
Bowling average of I = 240/15=16
Required sum = 46.66+16=62.66
3) Answer: B
New average of E = [430+160]/15=39.33
Required difference = 43-39.33= 3.67
4) Answer: A
Required sum = [43+55.71]/2 + [51.66+45.33]/2=
97.85
5) Answer: A
New average of D = [300+50]/[12+3+3] =
350/18= 19.44
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2 918
Detailed Solution:
E scored 620 – 190=430 runs.
G scored430+350=780 runs.
H scored780 – 100=680 runs
E played 12*(100-16.666)/100=10 matches
H played =12*5/4=15 matches
G played 15 – 1=14 matches
Let, A took x wicket and B took 3x wicket.
C took 3x*2/3=2x wicket.
D took 2x wickets.
3x-2x=6
Or, x=6
So, A took 6 wickets, B took 18, C took 12 and D
took 12 wickets respectively.
Let runs conceded by B and D is 400a and 300a
respectively.
Runs conceded by C are 300a*100/120=250a
Runs conceded by A are 250a-100
So, 300a=2*[250a-100]
Or, a=200/200=1
Runs conceded by B and D are 400 and 300
respectively
Runs conceded by C are 250.
Runs conceded by A are 250 – 100=150.
sиαρσnє🌱
1) Answer: C
Required difference = [25+22.22] – [25+20.83]
=1.39
2 918
There are four bowlers A, B, C and D and four
batters E, F, G, H.
Bowling average of a bowler = [No of runs run
conceded/No of wickets taken] and batting
average of a batter is = [No of runs scored by
batter/No of matches played].
Ratio of the number of wickets taken by A and B
is 1:3. F scored 190 runs more than E. Runs
conceded by C are 100 more than runs
conceded by A. G scored 350 runs more than E.
C and D took equal number of wickets. Number
of wickets taken by C is 33.33% less than the
number of wickets taken by B. D conceded twice
runs of the number of runs conceded by A. E
played 16.66% less matches than played by F.
Ratio of number of matches played by F and H is
4:5. G scored 100 runs more than H where H
played 1 match more than G. F scored 620 runs
in 12 matches. Ratio of runs conceded by B and
D is 4:3. B took 6 wickets more than D. Number
of runs conceded by D is 20% more than number
of runs conceded by C.
sиαρσnє🌱
1) Find the difference between the total bowling
average of A and C together and total bowling
average of B and D together?
a) 1.78
b) 1.25
c) 1.39
d) 1.45
e) 1.20
2) Another player I scored runs of average the
runs scored by F and G together and played
25% more matches than F played. He also took
15 wickets and conceded 240 runs. Find the sum
of bowling and batting average of I?
a) 54.32
b) 60.32
c) 64.33
d) 62.66
e) 55.33
3) In the next 5 matches, E scored 160 runs.
Find the difference between the new and old
batting average of E?
a) 2.35
b) 3.67
c) 4.35
d) 3.33
e) None of these
4) Find the sum of average of the batting
average of E and G together and batting average
of F and H together?
a) 97.85
b) 92.36
c) 91.33
d) 90.37
e) None of these
5) In the next two matches, D conceded only 50
runs and took 3 wickets in each match. Find the
new average of D?
a) 19.44
b) 12.33
c) 19.37
d) 18.55
e) None of these
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