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2 918
#Day_20_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
Ratio between the ages of A and B is 4:5. C is Y
years older than B. Initially they have different
amounts. Ratio of initial amount is the same as
their age ratio. B has initially Rs M. They start a
business together with some amounts. B invest
maximum in the business. After Y years their
profit is Rs P. Each one invests the rest of their
amount in different schemes for Y years. A invest
Rs.14000 in R% rate of simple interest. B invest
at 2R% compound interest and earn Rs. I as an
interest. C invest Rs. F at [R/2] % compound
interest and get Rs.2050 as an interest. After 2
years the sum of age of A and C is 60 years.
Difference of amount of A and C is 8000. Share
of B from in the profit is Rs.3000. Interest earned
by A is Rs.2800. Difference between invested
amount of B and C in interest schemes is
Rs.3000 which is equal to the profit of share of B
in business. After 4Y years the age of C is 40
years.
sиαρσnє🌱
1) Find the sum of age of A after 2Y years, B
after Y/2 years and C after 3Y/2 years.
a) 80
b) 85
c) 94
d) 82
e) 36
2) Find the difference between the total amount
[profit in business + interest] earned by A and C.
a) 310
b) 300
c) 320
d) 350
e) 250
3) B buy a cycle with the interest amount and a
watch with the profit amount. B sold the cycle at
R% profit and sold the watch at 2R% loss. Find
the total selling price.
a) 7840
b) 9660
c) 9620
d) 8460
e) None of these
4) Find the approximate value of R% of M + 9.09% of I + 25% of P =?
a) 5450
b) 5200
c) 5630
d) 5820
e) None of these
5) Find the compound interest if C invests his initial amount at an interest rate 2R% for Y year.
a) 14080
b) 12540
c) 16520
d) 14320
e) None of these
2 918
Detailed Solution:
6.
Quantity I,
Let CP be 100. When sold at 10% loss the SP
is=90.
To make a profit, 20% SP should be increased
by 30.
So, when 45 increase CP of the article is=150
SP when profit is 25%=150*125/100=187.5
Quantity II,
CP of the article=165*100/110=150
So, the marked price is=150*(120/100) *(100/75)
=240
Quantity I < Quantity II (C)
sиαρσnє🌱
7.
Quantity I,
Sum of the 7th and 8th number is=9*31-3*39-
3*23-31=62
Quantity II,
Sum of the 11th and 12th number is=19*37-9*35-
7*42-42/3=80
Quantity I <Quantity II (C)
8.
Side of the cube = (2197)(1/3) = 13
Radius of the cone = 5
Slanting height of cone = 13
4/3 * 22/7 * r3 = 38808
Radius of the sphere = 21
Height of the cylinder = 2 * 13 + 2 = 28
CSA of cone = 22/7 * 5 * 13 = 204.28
SA of sphere = 4 * 22/7 * 21 * 21 = 5544
From quantity I,
Required % = 204.28/5544 * 100 = 3.68%
From quantity II,
Required % = 5/28 * 100 = 17.85%
Quantity I < Quantity II (C)
sиαρσnє🌱
9.
Quantity I,
Their investment ratio is=
14000*6: 12000*8 :16000*6:
16000*8=21:24:24:32
A’s share of profit is=21*24240/101=5040
Quantity II,
Interest earn by A is=16000(1+18/100)^4-
16000=15020.44
Interest earn by B is=20000*21*4/100=16800
Amount invest by C is=[16800+15020]/2=15910
Interest earns by C=15910*5*6/100=4773.06
Quantity I > Quantity II (A)
10.
Speed of train A is=72*5/18=20m/sec
Length of train A =20*16-160=160
Speed of train C is=90*5/18=25 m/sec
We can say, 180+320/Speed of train B =20
Or, speed of train B is=500/20=25 m/sec
Quantity I:
Required time = 160+320/(20+25) + 340/(25 - 20)=78.66 sec
Quantity II:
Distance of train C from platform is l
So, (l+160+180)/20=23
Or, l=460-160-180=120
So, total distance cover by train C to cross
platform is =120+180+140=440
So, time taken by C is=440/25=17.6
Quantity I>Quantity II (A)
2 918
Detailed Solution:
F = 2A
D = 3E
(B + F)/2 = C
D/F = 6/5
C = D – 8
D = 6F/5
D = 6 ∗2A/5 = 12A/5
C = 12A/5 – 8
(B + 2A)/2 = 12A/5 – 8
5B + 10A = 24A – 80
5B = 14A – 80
B = (14A – 80)/5
12A/5 = 3E
4A/5 = E
A + (14A – 80)/5 + (12A/5 – 8) + 12A/5 + 4A/5 +
2A = 90
5A + 14A – 80 + 12A – 40 + 12A + 4A + 10A =
90 * 5
57A = 570
A = 10
B = (14 * 10 – 80)/5 = 12
C = (12 * 10)/5 – 8 = 16
D = 16 + 8 = 24
E = 4 * 10/5 = 8
F = 2 * 10 = 20
sиαρσnє🌱
1)
Black balls in C = 50/100 * 16 = 8
Black balls in D = 50/100 * 24 = 12
Black balls in F = 30/100 * 20 = 6
Required total = 8 + 12 + 6 = 20 (E)
2)
Required percentage = 8/(10 + 16) * 100 = 30.76% (A)
3)
Black balls in B = 3/4 * 12 = 9
White balls in B = 1/4 * 12 = 3
Black balls in D = 1/4 * 24 = 6
White balls in D = 3/4 * 24 = 18
Required percentage = (9 + 6)/(3 + 18) * 100 = 71.42% (B)
sиαρσnє🌱
4)
Required ratio = (16 + 24 + 8):(10 + 12 + 20)
= 48:42
= 8:7 (B)
5)
Number of balls in G = 125/100 * 24 = 30
Required difference = 1/5 * 30 = 6 (C)
2 918
#Day_20_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
6)
Quantity I: A person sold an article at a loss of
10%. Had he sold it for Rs.45 more, he would
have gained 20%. What should be the selling
price to make a profit of 25%?
Quantity II: B allows a 25% discount on the
marked price of an article and still makes a profit
of 20%. If the Price of the article is Rs.165 when
it is sold at 10% profit then find the marked
price?
A. Quantity I > Quantity II
B. Quantity I ≥ Quantity II
C. Quantity I < Quantity II
D. Quantity II ≥ Quantity I
E. Quantity I = Quantity II or relation can't be
established
7)
Quantity I: The average of nine numbers is 31.
The average of the first three numbers is 39 and
that of the next three numbers is 23. If the 9th
number is equal to the average of nine numbers,
then find the sum of the 7th and 8th numbers?
Quantity II: The average of 19 numbers is 37.
The average of the first 9 numbers is 35 and that
of the last 7 numbers is 42. If the 10th number is
1/3rd of the average of the last 7 numbers then
find the sum of the 11th and 12th numbers?
A. Quantity I > Quantity II
B. Quantity I ≥ Quantity II
C. Quantity I < Quantity II
D. Quantity II ≥ Quantity I
E. Quantity I = Quantity II or relation can't be
established
sиαρσnє🌱
8) Slant height of cone equal to the side of a
cube whose volume is 2197 cube units. Radius
of the cone is 5 units. Volume of the cylinder is
equal to the volume of the sphere i.e., 38808.
Height of the cylinder is 2 more than the double
of the slant height of the cone.
Quantity I: Curved surface area of cone is what
percent of surface area of sphere?
Quantity II: Radius of cone is what percent of
height of cylinder?
A. Quantity: I > Quantity: II
B. Quantity: I ≥ Quantity: II
C. Quantity: I < Quantity: II
D. Quantity: II ≥ Quantity: I
E. Quantity I = Quantity II or relation can't be
established
9)
Quantity I: A, B, and C start a business with
Rs.14000,12000, and 16000 respectively. After 4
months A left the business and D joined the
business with Rs.16000. After 6 months C and
after 8 months B left the business. After 10
months A again joined the business with the
same money. After 1 year they total earn
Rs.24240. Find A’s share of profit?
Quantity II: A invests Rs.16000 at 18%
compound interest and B invests Rs.20000 at
21% Simple interest for 4 years. C invest the
average amount of interest earned by A and B
together at 5% simple interest for 6 years. Find
the interest received by C.
A. Quantity: I > Quantity: II
B. Quantity: I ≥ Quantity: II
C. Quantity: I < Quantity: II
D. Quantity: II ≥ Quantity: I
E. Quantity I = Quantity II or relation can't be
established
sиαρσnє🌱
10) Speed of train A is 72 km/hr and the train
crosses a 160 m long bridge at 16 sec. Length of
train B is double of train A. Train B crosses train
C when train C is standing in 20 sec. Speed of
train C is 90 km/hr and length of train C is 180 m.
Quantity I: Find the total time taken by train A to
cross train B at first then train C when train A and
train B moving opposite direction and train A and
train C in the same direction?
Quantity II: After crossing a 140m long platform
train A cross train C in 23 sec. Train C is
standing some distance away from the platform.
At the moment train A crosses train C, train C
starts moving towards the platform. Find the time
taken by train C to cross platform?
A. Quantity: I > Quantity: II
B. Quantity: I ≥ Quantity: II
C. Quantity: I < Quantity: II
D. Quantity: II ≥ Quantity: I
E. Quantity I = Quantity II or relation can't be
established
2 918
#Day_20_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
Total number of balls in six boxes A, B, C, D, E
and F is 90. The number of balls in F is twice of
the number of balls in A and the number of balls
in D is thrice the number of balls in E. The
average number of the balls in B and F is equal
the number of balls in C. The ratio of the
number of balls in D to F is 6:5 and the number
of balls in C is 8 less than that of D.
sиαρσnє🌱
1) The boxes C, D and F contain white and black
balls. If the number of White balls in C, D and F
is 50%, 50% and 70% respectively, then what is
the total number of Black balls in C, D and F
together?
a) 18
b) 22
c) 28
d) 34
e) None of these
2) The total number of balls in E is approximately
what percent of the total number of balls in C and
A together?
a) 30.76%
b) 31.86%
c) 32.66%
d) 29.86%
e) 25.96
3) The ratio of the number of White to Black balls
in B and D is 1:3 and 3:1 respectively. If the
number of Black balls in B and D together is
approximately what percent of the number of
White balls in B and D together?
a) 75.23%
b) 71.42%
c) 81.78%
d) 79.51%
e) 88.32%
4) What is the ratio of the number of balls in C, D
and E together to the number of balls in A, B and
F together?
a) 7:6
b) 8:7
c) 9:8
d) 6:5
e) None of these
5) If the number of balls in G is 25% more than
the number of balls in D and the ratio of the
number of red to black balls in G is 3:2, then find
the difference between the number of red and
black balls in G?
a) 4
b) 7
c) 6
d) 8
e) 5
2 918
Repost from Gk +imp sub pdfs [ɴɪʜᴀʀ]❤️
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#𝚁𝚎𝚐𝚊𝚛𝚍𝚜
𝙽𝚒𝚑𝚊𝚛 & 𝚃𝚎𝚊𝚖
2 918
Detailed Solution:
1.
Efficiencies of A, B, C, and D in work W1 are
=12, 20, 10, 8 units.
Efficiencies of A, B, C, and D in work W2 are
=12, 30, 30, 24 units.
In two days, A and B together did [12+20]
*2=64-unit work of W1.
In two days, C and D together did [30+24] *2=
108-unit work in W2.
Left work in W1 is=120 - 64=56 which is
completed by A and C together.
So, A and C complete the rest of the work=
56/22=2.54 days
Left work in W2 is=240-108=132 which is
completed by B and D together.
B and D complete the rest of the work in
=132/54=2.44 days.
Required difference= 0.10 days. (C)
sиαρσnє🌱
2.
Efficiency of A, B, C, and D in W3 is 10,15,25,6.
A and B together did 25*2=50 units of work in
two days.
Next 2 days C did 50 units of work. So, left work
is=150-100=50
So, 50 units of work C and D together did=50/
(25+6) =1.61 days.
So total work completes in=2+2+1.61=5.61 days (D)
sиαρσnє🌱
3.
Total unit work is 180 units.
Efficiency of A, B, and C in W4 is=10, 20, 15.
They complete the whole work in =180/45=4
days.
In 4 days, A did =40 units of work.
So, A received=9000*40/180=2000
So, interest earned by A is =
(2000*6*6)/(12*100)=60 (B)
4.
In 8 days, everybody works 2 times in W2.
Efficiency of A, B, C and D in W2 is 12,30,30,24.
Total efficiency of A, B, C and D together in W2
is=12+30+30+24=96
Total work complete in 8 days is=96*2=192
So, percentage of work left= (240-192)/240 * 100=20% (A)
sиαρσnє🌱
5.
Efficiency of A, B, C, and D in W3 is 10,15,25,6.
W3 complete in=150/56=2.67 days
Efficiencies of A, B, C, and D in work W1 are
=12,20,10,8 units.
W1 complete =120/50=2.4 days.
Required difference=0.27 (A)
2 918
Detailed Solution:
1.
Efficiencies of A, B, C, and D in work W1 are
=12, 20, 10, 8 units.
Efficiencies of A, B, C, and D in work W2 are
=12, 30, 30, 24 units.
In two days, A and B together did [12+20]
*2=64-unit work of W1.
In two days, C and D together did [30+24] *2=
108-unit work in W2.
Left work in W1 is=120 - 64=56 which is
completed by A and C together.
So, A and C complete the rest of the work=
56/22=2.54 days
Left work in W2 is=240-108=132 which is
completed by B and D together.
B and D complete the rest of the work in
=132/54=2.44 days.
Required difference= 0.10 days. (C)
sиαρσnє🌱
2.
Efficiency of A, B, C, and D in W3 is 10,15,25,6.
A and B together did 25*2=50 units of work in
two days.
Next 2 days C did 50 units of work. So, left work
is=150-100=50
So, 50 units of work C and D together did=50/
(25+6) =1.61 days.
So total work completes in=2+2+1.61=5.61 days (D)
sиαρσnє🌱
3.
Total unit work is 180 units.
Efficiency of A, B, and C in W4 is=10, 20, 15.
They complete the whole work in =180/45=4
days.
In 4 days, A did =40 units of work.
So, A received=9000*40/180=2000
So, interest earned by A is =
(2000*6*6)/(12*100)=60 (B)
4.
In 8 days, everybody works 2 times in W2.
Efficiency of A, B, C and D in W2 is 12,30,30,24.
Total efficiency of A, B, C and D together in W2
is=12+30+30+24=96
Total work complete in 8 days is=96*2=192
So, percentage of work left= (240-192)/240 * 100=20% (A)
sиαρσnє🌱
5.
Efficiency of A, B, C, and D in W3 is 10,15,25,6.
W3 complete in=150/56=2.67 days
Efficiencies of A, B, C, and D in work W1 are
=12,20,10,8 units.
W1 complete =120/50=2.4 days.
Required difference=0.27 (A)
2 918
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2 918
#Day_18_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
There are four works [W1, W2, W3 & W4]. Work
W1, W2, W3, and W4 are of 120, 240, 150, &
180 units. A alone completes the W1 in 10 days.
Ratio of efficiency of B to C in W1 is 2:1. C
completes W1 in as many days as he takes to
complete W4. Ratio of time taken by C to D to
complete W1 is 4:5. B and C take the same
number of days to complete W2. Ratio of
efficiency C to D in W2 is 5:4. A takes 100% and
33.33% more days to complete W2 than he takes
to complete W1 and W3 respectively. Ratio of
the number of days taken by D to complete W2
to W3 is 2:5. D takes 50% more days to
complete W1 than A to complete the same Work.
Ratio of time taken by A to complete W1 to W4 is
5:9. D takes 50% fewer days to complete W2
than the number of days taken by A to complete
the same work. D takes 66.66% more days to
complete W4 than the time taken by C to
complete the same work. A takes double the
number of days to complete W4 than B takes to
complete the same work. Ratio of efficiency A to
B in W3 is 2:3. Number of days taken by B to
complete W1 is the same number of days taken
by C to complete W3.
Note:- Each one have different capacity for
different work(i.e..W1,W2,W3,W4)
sиαρσnє🌱
1) A and B started working together in W1 and C
and D started working together in W2. After
working for two days B was replaced by C in W1
and C was replaced by B in W2. Find the
difference in the number of days to complete the
rest of the work in W1 and W2.
A. 0.69 days
B. 0.35 days
C. 0.10 days
D. 0.28 days
E. 0.98 days
2) A and B started working in W3. After
completing 1/3rd of the work they left the work
and C joined the work. C worked for two days
then D joined him and finished the rest of the
work. Find the number of days taken to complete
the whole work.
A. 2 days
B. 2.61 days
C. 1.61 days
D. 5.61 days
E. 4 days
3) Rs.9000 allocated for the work W4. A, B and
C together complete the W4. Amount received
by A invest in a scheme for 6 months at 6%
Simple Interest Per annum. Find the amount of
interest received by him?
A. Rs.50
B. Rs.60
C. Rs.20
D. Rs.27
E. None of these
4) All of them work in W2 in an alternative format
starting with A, then B, then C and last D. After 8
days how much percent of work is left in W2?
A. 20%
B. 25%
C. 15%
D. 24%
E. None of these
5) Find the difference of time taken by A, B, C,
and D to complete the W1 and W3.
A. 0.27
B. 0.26
C. 0.35
D. 0.96
E. None of these
2 918
#Day_18_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
There are four works [W1, W2, W3 & W4]. Work
W1, W2, W3, and W4 are of 120, 240, 150, &
180 units. A alone completes the W1 in 10 days.
Ratio of efficiency of B to C in W1 is 2:1. C
completes W1 in as many days as he takes to
complete W4. Ratio of time taken by C to D to
complete W1 is 4:5. B and C take the same
number of days to complete W2. Ratio of
efficiency C to D in W2 is 5:4. A takes 100% and
33.33% more days to complete W2 than he takes
to complete W1 and W3 respectively. Ratio of
the number of days taken by D to complete W2
to W3 is 2:5. D takes 50% more days to
complete W1 than A to complete the same Work.
Ratio of time taken by A to complete W1 to W4 is
5:9. D takes 50% fewer days to complete W2
than the number of days taken by A to complete
the same work. D takes 66.66% more days to
complete W4 than the time taken by C to
complete the same work. A takes double the
number of days to complete W4 than B takes to
complete the same work. Ratio of efficiency A to
B in W3 is 2:3. Number of days taken by B to
complete W1 is the same number of days taken
by C to complete W3.
Note:- Each one have different capacity for
different work(i.e..W1,W2,W3,W4)
sиαρσnє🌱
6) A and B started working together in W1 and C
and D started working together in W2. After
working for two days B was replaced by C in W1
and C was replaced by B in W2. Find the
difference in the number of days to complete the
rest of the work in W1 and W2.
A. 0.69 days
B. 0.35 days
C. 0.10 days
D. 0.28 days
E. 0.98 days
7) A and B started working in W3. After
completing 1/3rd of the work they left the work
and C joined the work. C worked for two days
then D joined him and finished the rest of the
work. Find the number of days taken to complete
the whole work.
A. 2 days
B. 2.61 days
C. 1.61 days
D. 5.61 days
E. 4 days
8) Rs.9000 allocated for the work W4. A, B and
C together complete the W4. Amount received
by A invest in a scheme for 6 months at 6%
Simple Interest Per annum. Find the amount of
interest received by him?
A. Rs.50
B. Rs.60
C. Rs.20
D. Rs.27
E. None of these
9) All of them work in W2 in an alternative format
starting with A, then B, then C and last D. After 8
days how much percent of work is left in W2?
A. 20%
B. 25%
C. 15%
D. 24%
E. None of these
10) Find the difference of time taken by A, B, C,
and D to complete the W1 and W3.
A. 0.27
B. 0.26
C. 0.35
D. 0.96
E. None of these
2 918
Detailed Solution: (6 to 10)
6. Answer: E
Let speed of train A and train B are ‘3x’ m/s and
‘2x’ m/s respectively.
Also let length of train A and train B are ‘5y’ m
and (5y + 50) m respectively.
And length of the platform = 80% of ‘5y’ = ‘4y’ m
Since, time taken by train A to cross the platform
is 9 seconds.
So, (5y + 4y)/3x = 9
y = 3x --------------(1)
Since, train A will cross train B in 7 seconds
while running in the opposite direction.
So, (5y + 5y + 50)/ (3x + 2x) = 7
(2y + 10)/x = 7 --------------(2)
From equations (1) and (2):
6x + 10 = 7x
x = 10, y = 30
Speed of train B = 2 * 10 = 20 m/s
Length of train B = 5 * 30 + 50 = 200 m
Length of the platform = 80% of 150 = 120 m
So, the value, which will be filled in the blank =
(200 + 120)/20 = 16 seconds (E)
sиαρσnє🌱
7.
Since, ratio of efficiency of pipe A to B is 2: 1
respectively.
So, ratio between their times will be = 1: 2
Let time taken by pipe A alone and pipe B alone
to fill the tank is ‘x’ minutes and ‘2x’ minutes
respectively.
Also let time taken by pipe C alone to empty the
tank is ‘y’ minutes.
So, (1/x) – (1/y) = (1/22.5) -------------------(1)
And, (1/2x) – (1/y) = (1/90) -------------------(2)
From equations (1) and (2):
(1/x) – (1/2x) = (1/22.5) – (1/90)
1/2x = 1/30
x = 15
Time taken by pipe A alone to fill the tank = 15
minutes
Time taken by pipe A alone, when works with
125% efficiency:
15 * (100/125) = 12 minutes
Time taken by pipe B alone to fill the tank = 2 *
15 = 30 minutes
Time taken by pipe B alone, when works with
83(1/3) % efficiency:
30 * (300/250) = 36 minutes
Part of tank filled by pipes A and B together with
their new efficiencies in 1 minutes = (1/A) + (1/B)
= (1/12) + (1/36) = 1/9
So, time taken by pipes A and B together to fill
the tank with their new efficiencies = 9 minutes (D)
8. Answer: B
Let A’s present age B’s present age are ‘5x’
years and ‘6x’ years respectively.
So, (5x + 2)/ (6x + 4) = 4/5
25x + 10 = 24x + 16
x = 6
A’s present age = 5 * 6 = 30 years
And A will get 35 years old after 5 years.
B’s present age = 6 * 6 = 36 years
And B will get 40 years old after 4 years.
Let A’s investment and B’s investment are ‘p’
rupees and (p + 2000) rupees.
So,
[p * 5]/ [(p + 2000) * 4] = 9/8
40p = 36p + 72000
p = 18000
So, B’s investment = 18000 + 2000 = Rs.20000 (B)
sиαρσnє🌱
9. Answer: E
Logics in the series are:
Series I:
15 + 52 = 40
40 + 62 = 76
76 + 72 = 125
125 + 82 = 189 (Not 190)
189 + 92 = 270
A = 190
Series II:
18 * 2 – 1 = 35
35 * 2 – 1 = 69
69 * 2 – 1 = 137 (Not 139)
137 * 2 – 1 = 273
273 * 2 – 1 = 545
B = 139
Hence,
A > B (E)
10.
Logics in the series are:
Series I:
100 + 23 = 123
123 + 29 = 152
152 + 31 = 183
183 + 37 = 220
220 + 41 = 261
Series II:
20 + 23 = 28
28 + 33 = 55
55 + 43 = 119
119 + 53 = 244
244 + 63 = 460
I: Missing term of series I is 220 which is not
divisible by 13.
II: Missing term of series II is 55 which is
divisible by 11.
III: Missing term of series I ÷ Missing term of
series II = 220 ÷ 55 = 4
Hence, only II is TRUE. (C)
2 918
Detailed Solution: ( 1 to 5)
Spherical Ball:
Volume of the spherical iron ball =
4/3 𝜋𝑟^3 = 4/3 *
22/7 * 42 * 42 * 42 = 310464 cm^3.
Cylindrical Shape:
LSA of cylindrical shape = 2𝜋𝑟ℎ
=> 2 * 22/7 * 5 * h = 330
=> h = 2310/220
=> h = 10.5
Volume of the cylindrical shape = 𝜋(𝑟^2)ℎ = 22/7 *
5 * 5 * 10.5 = 825 cm^3.
Conical Shape:
Base perimeter of the cone = 37(5/7) cm^2.
=>2𝜋𝑟 = 264/7
=> 2 * 22/7 * r = 264/7
=> r = 6 cm
Given, height of the conical shape is 133(1/3) %
more than its radius.
h = 700/3% * r
=> h = 7/3 * r
=> h = 7/3 * 6 = 14 cm
Volume of the conical shape =
1/3 𝜋(𝑟^2)ℎ= 1/3 * 22/7 * 6 * 6 * 14 = 528 cm^3.
Cuboid Shape:
Height of the cuboid is 3 more than its length
and breadth is 2 more than the 25% of the
height and the total surface area of the cuboid is
356.5 cm^2.
ATQ, ℎ = 3+ 𝑙 and
𝑏 = 25/100 ∗ ℎ + 2 =>𝑏 = 1/4 ∗ (3 + 𝑙) + 2
=>𝑏 = (3 + 𝑙 + 8)/4
=>𝑏 = (11 + 𝑙)/4
Total surface area of the cuboid = 356.5 cm^2.
=>2(𝑙𝑏 + 𝑏ℎ + 𝑙ℎ) = 356.5
=>2(𝑙 ∗ (11 + 𝑙)/4 + (11 + 𝑙)/4 ∗ (3 + 𝑙) + 𝑙 ∗ (3 + 𝑙)) = 356.5
=>2/4 (11𝑙 + 𝑙^2 + 33 + 11𝑙 + 3𝑙 + 𝑙^2 + 4(3𝑙 + 𝑙^2)) = 356.5
=>37𝑙 + 6𝑙^2 + 33 = 713
=>6𝑙^2 + 37𝑙 − 680 = 0
=>6𝑙^2 − 48𝑙 + 85𝑙 − 680 = 0
=>6𝑙(𝑙 − 8) + 85(𝑙 − 8) = 0
=>(6𝑙 + 85)(𝑙 − 8) = 0
=>𝑙 = −85/6 (𝑜𝑟) 8
We know that the value of the length is nonnegative,
then the value of the length of the
cuboid shape is 8 cm.
Also, ℎ = 3 + 8 = 11 𝑐𝑚
𝑏 =(11 + 8)/4
= 4.75 𝑐𝑚
Therefore, Volume of the cuboid shape =
𝑙 ∗ 𝑏 ∗ ℎ = 8 * 4.75 * 11 = 418 cm^3.
Given, ratio between the number of cylindrical,
conical and cuboid shapes is 32:29:24.
Let us take, the number of cylindrical, conical
and cuboid shapes made by the iron ball is 32x,
29x and 24x.
ATQ,
Volume of the spherical iron ball = 32x * Volume
of cylindrical shape + 29x * Volume of the
conical shape + 24x * Volume of the Cuboid
Shape
=> 310464 = 32x * 825 + 29x * 528 + 24x * 418
=> 310464 = x * (26400 + 15312 + 10032)
=> x = 310464/51744
=> x = 6
Therefore, the number of cylindrical, conical and
cuboid shapes made by the iron ball is 32(6),
29(6) and 24(6) = 192, 174 and 144.
sиαρσnє🌱
1.
Therefore, the number of conical shapes made
by using the spherical iron ball is 174. (D)
2. Answer: A
We already find the radius and height of the
conical shape is 6 cm and 14 cm.
Now, we have to find the slant height of the
conical shape to find the lateral surface area.
𝑙 = √(ℎ^2 + 𝑟^2)
=>𝑙 = √(14^2 + 6^2) = √232 = 2√58 𝑐𝑚
LSA of conical shape = 𝜋𝑟𝑙 = 22/7 ∗ 6 ∗ 2√58 ≅ 287.22 𝑐𝑚^2 (A)
3.
TSA of conical shape = 𝜋𝑟(𝑙 + 𝑟) = 22/7 ∗
6(2√58 + 6) = 400.366
≅ 400.37 𝑐𝑚^2
Total number of conical shapes = 174
Required cost = 174 * 400.37 * 5 ≅
3,48,322 (C)
4.
Volume of Cylindrical shape = 825 cm^3
Total number of cylindrical shapes made by
Spherical iron ball is 192
Required volume = 192 * 825 =158400 cm^3. (D)
5.
CSA of the cuboid shape = 2ℎ (𝑙 + 𝑏) = 2 ∗ 11 ∗
(8 + 4.75) = 280.5 𝑐𝑚^2
Total number of cuboid shapes made by the
spherical iron ball is 144.
Required CSA = 144 * 280.5 = 40392 cm^2. (D)
