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Ko'proq ko'rsatish![Gk +imp sub pdfs [ɴɪʜᴀʀ]❤️](https://cdn.tlmtr.io/_next/image?url=https%3A%2F%2Fimg.tlmtr.io%2Fc%2F1CoE2V%2F6077874367817166978%3Fty%3Dx&w=32&q=75){kind=small}
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2. For tank Q, when Pipe A and B opened simultaneously and on every third hour pipe C opened alone, so tank is filled in 21/2 hours. Find the rate of emptying by pipe C in cm3/hour, if
radius of tank is 24 cm?
a) 1296 cm^3/h
b) 960 cm^3/h
c) 864 cm^3/h
d) 1080 cm^3/h
e) None of these
sиαρσnє🌱
3. For tank R, when pipe A and pipe C (working with its minimum possible efficiency) opened together, it is not possible to fill the tank. Pipe A and pipe B together can fill 480 cm3 in three
hours. Find the length of tank, which is 33.33% more than its breadth.
a) 36 cm
b) 24 cm
c) 48 cm
d) 32 cm
e) None of these
4. For tank S, Pipe A and Pipe C together can empty 192 cm3 of tank and time taken by pipe B to fill the tank alone is 50 hours, then find the time taken by pipe A alone to fill 60% volume of tank S?
a) 48hours
b) 60 hours
c) 72 hours
d) 64 hours
e) None of these
5. There is hemispherical tank whose volume is 5/72 times as that of volume of S. Find the radius of hemispherical tank?
a) 12 cm
b) 8 cm
c) 10 cm
d) 9 cm
e) None of these
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#Day_10_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
The table given in pic shows the partial information about the radius and height of four tanks – P, Q, R,
and S. It also gives details about Pipe A alone to fill the tank in hr, and % of time taken by Pipe B to fill
the tank is less/more than empty the tank by Pipe C. Use π = 3.
sиαρσnє🌱
1. Pipe B is opened alone to fill tank P. After filling completely, it was closed for 45 minutes and then filled another tank, which was half of capacity of tank P and again closed for 45 minutes, after that it filled a tank, which had same capacity as that of tank P. Total time taken
for tank P to fill all such tanks was106.5 hours. When all three pipes (A, B and C)are opened together, they can fill 540 cm3 in two hours and radius of tank P is K cm. Find the value of (2K + 14)?
a) 48
b) 50
c) 60
d) 42
e) None of these
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Detailed Solution:
6M + 11Z = 75
M = (75 – 11Z)/6
Z should be odd, because 75 is also odd.
If Z = 1, M = 64/6 [Not possible]
If Z = 3, M = 42/6 = 7 [Possible]
If Z = 5, M = 20/6 [Not possible]
So, M = 7, Z = 3
Also,
N^7 = N^3
Possible value of N = 0, 1
If N = 0
Number of vacant seats in R1 of bus B1 = (M +
N – 8) = 7 + 0 – 8 = - 1
Not possible, so value of N = 1
Number of vacant seats in R1 of Bus B1 = (7 + 1
– 8) = 0
So, number of passengers in R1 of Bus B1 = 16
Number of total seats in B2 = [2M + Z – 3N]
= 2 x 7 + 3 – 3 x 1 = 14
Number of passengers in R2 of bus B1 and B2 =
3:2 [3a and 2a]
Let number of passengers in R1 of B2 = b
So, Total passengers in (R1 + R2) of B1 = 16 +
3a
Total passengers in (R1 + R2) of B2 = 2a + b
Now,
16 + 3a = 2 x (2a + b)
16 – a = 2b
Value of b = (16 – a)/2
Value of a must be even,
If a = 2
Value of b = (16 – 2)/2 = 7
Number of passengers in R3 of B2 = 7 x 2
= 14 [not possible, because no seats are full in
any of the three rounds of B2]
If a = 4
Value of b = (16 – 4)/2 = 6
Then number of passengers in R3 of B2 = 2 x 6
= 12 [possible]
If a = 6, then number of passengers in R2 of B1
= 3a = 3 x 6 = 18 [not possible].
So, number of passengers in R3 of B1 = 38 – 16
– 12 = 10
Buses R1 R2 R3 Total
B1 16 12 10 38
B2 6 8 12 26
Total 22 20 22 64
sиαρσnє🌱
1. Revenue generated by bus B1 for return journey
= 60% x [16 x 80 + 12 x 60 + 10 x 120]
= Rs.1920 (C)
2. For bus B2,
Number of female passengers = 6 x 1/2 + 8 x
3/4 + 12 x 2/3 = 17 (D)
3. Required sum = 6 + 10 = 16 (C)
4. Required % change = (16 – 12)/12 x 100
= 33.33% (A)
5. Total number of passengers = 64 (B)
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#Day_9_Target_SBI_Clerk_mains_Gk_imp_sub_pdfs_Nihar
There are two buses B1 and B2, both takes three rounds R1, R2, and R3 in a day.
Bus B1: Bus is 16-seater and number of vacant seats in R1 is (M + N – 8). Number of
passengers in R2 is 50% more than that in bus B2 of same round. Total number of passengers
in all three rounds are 38.
Bus B2: Bus is [2M + Z – 3N] seater and no seats are full in any of the round. Number of
passengers in R1 and R2 together of bus B1 is twice as that of bus B2. Number of passengers in
R1 is half as that of R3.
Note: 6M + 11Z = 75, Where M, and Z are distinct natural numbers.
NM = NZ
sиαρσnє🌱
16) If fare for passenger is Rs. 80, Rs, 60, and Rs. 120 for round R1, R2, and R3 respectively
for B1. In return journey fare reduced by 40%. Find the revenue generated from Bus B1 for
return journey?
a) Rs. 2400
b) Rs. 2560
c) Rs. 1920
d) Rs. 3200
e) None of these
17) For B2, if 50%, 25% and 33.33% of passengers in R1, R2 and R3 are males, then
find number of female passengers?
a) 15
b) 19
c) 16
d) 17
e) None of these
18) Find total passengers in R1 of B2 and R3 of B1 together?
a) 15
b) 17
c) 16
d) 18
e) None of these
19) Number of vacant seats in B2 is how much % more or less than number of passengers in R2
of B1?
a) 33.33%
b) 25%
c) 50%
d) 66.66%
e) None of these
20) Find total number of passengers in both buses together?
a) 60
b) 64
c) 56
d) 62
e) None of these
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Detailed Solution:
Both trains P and Q started at same time and
reached at destination at same time when travel
with their original speed.
So, speed of both trains P and Q is same.
After covering 180 km, speed of P reduced by 25%,
so there is delay of 75 minutes (6.00 pm to 7.15 pm).
So, ratio of speed of train P = 4:3 [4a, 3a]
Let the remaining distance after 180 km = d km
Now,
d/3a – d/4a = 5/4
So, d = 5/4 x 12a = 15a
Value of d = 15a…………. (1)
After covering 240 km from station A, speed is
reduced by 50%, so changed speed = 4a x 1/2
= 2a
Now,
(d – 60)/2a – (d – 60)/4a = 3
(d – 60) x 1/4a = 3
(d – 60) = 12a…………… (2)
From (1) and (2),
15a – 12a = 60
a = 20
Original speed of train P = Train Q = 20 x 4 =
80 km/hr
Value of d = 15 x 20 = 300 km
So, distance between A and B = 180 + 300 = 480
km
Value of Z = 80
(2Y + 30) = 80
Value of Y = 25
Speed of train R = 25 + 15 = 40 km/h
Speed of train S = 80/2 – 10 = 30 km/h
Train R and Train S,
If there is no reduction in speed, after 3 pm they
together cover remaining distance in = 3 hours
If there is reduction in speed, after 3 pm they
together cover the remaining distance in 4 hours
40 minutes = 14/3 hours
So, ratio of time = 3:14/3 = 9:14
Distance is constant, so ratio of speed = 14:9
Initial relative speed = (40 + 30) = 70 km/h
Final relative speed = 70/14 x 9 = 45 km/h
Now,
40 – L + 30 – 1.5L = 45
2.5L = 25
Value of L = 10
1. Time taken by train U to cover distance between
A and B = 40% x 480/96 + 30% x 480/72 + 30% x 480/36
= 8 hours
In 8 hours, train T covered 400 km, so speed of
train T = 400/8
= 50 km/h (E)
2. Required ratio = [180/80 + 60/60]: [240/80]
= 13:12 (B)
3. Required value = (2L + 50) = 2 x 10 + 50
= 70 (C)
4. Speed of train P = 80 km/h
Speed of train T = 64 km/h
Distance between A and B = 480 km
Time taken by the faster train P to cover 480km =
480/80 = 6 hours
Hence in 6 hours, distance travelled by T = 6 x 64
= 384 km
After reached station B, train P changed its
direction and running towards station A
Hence, relative speed = 80+64 = 144 km/hr
Required time = 6 + (480-384)/144 = 6 hours 40
minutes (D)
5. Speed of train T = 3 x 80 = 240 km/h
Required time = 480/240 = 2 hours (A)
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#Day_7_Target_SBI_PO_Clerk_mains_Gk_imp_sub_pdfs_Nihar
Train P started at station A at a speed of Z km/h, after travelling 180 km its speed is reduced by 25% so it reaches station B at 7:15 pm. Train Q started at station A at the same time as train P at a speed of (2Y + 30) km/h, after travelling 240 km its speed is halved and reaches station B at 9 pm. Train R stared at 1 pm from station C running towards station D at a speed of (Y + 15) km/h, while on same time, train S started from station D running towards station C, at speed of (Z/2 – 10) km/h. Train R and train S expected to collide at 6 pm. If at 3 pm, the speed of train R is decreased by L km/h and speed of train S is decreased by 1.5L km/h.now, they are expected to meet at
7:40 pm.
Note:Train P and train Q cover the whole distance at speed of Z km/h and (2Y + 30) km/h respectively and both the trains reach station B at 6 pm.
1. If another train U covered 40% of distance between A and B at speed of 96 km/h, half of the rest distance between A and B is 72 km/h and rest is 36 km/h. The total time taken by train U to reach station B is equal to the time taken by train T to travel 400 km. Find the speed of train T?
a) 48 km/h
b) 45 km/h
c) 54 km/h
d) 60 km/h
e) None of these
2. Find the respective ratio of time taken by train P and train Q to cover half of the distance between A and B?
a) 6:5
b) 13:12
c) 16:15
d) 18:11
e) None of these
3. Find the value of (2L + 50).
a) 60
b) 75
c) 70
d) 80
e) None of these
4. If two trains P running at Z km/h and train T running at (3L + Y + 9) km/h start simultaneously from station A towards station B (the faster of them changed its direction after reached station B), then find the time taken for their first meeting? (Ignore the length of trains).
a) 7 hours 20 minutes
b) 6 hours 20 minutes
c) 7 hours 15 minutes
d) 6 hours 40 minutes
e) None of these
5. If speed of train T is 200% more than that of train P, find the time taken by train T to cover the distance between A and B.
a) 2 hours
b) 3 hours
c) 4.5 hours
d) 2.5 hours
e) None of these
