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2 918
Detailed Solution:
Number of pens the shopkeeper bought on
Monday is 20 and the cost price of each pen on
Monday is Rs.12.
Selling price of each pen sold on Monday
=12*125/100=15
Selling price of each pen sold on Tuesday =
15*5/3=25
Cost price of each pen bought on Tuesday =
25*100/125=20
Number of pens sold on Monday is =20-5=15
Number of pens sold on Tuesday is =15*4/5=12
Number of pens boughton Tuesday is =12+3=15
Selling price of pens sold on Wednesday is
=15+5=20
Cost price of each pen bought on Thursday
=20*100/133.33=15
Cost price of pens sold on Wednesday =20-
2=18= Number of pens bought on Wednesday.
Number of pens bought on Thursday
=18*5/3=30
Number of pens sold on Thursday = 30-2*5 = 20
Total number of pens sold on Friday
=5+3+10=18
Total cost price of the pens, which sold on Friday
=5*12+3*20+10*15
=270
Total selling price of all unsold pen is
=18*15=270
sиαρσnє🌱
1) Answer: C
Total selling price of all unsold pen is
=18*15=270
Therefore, he gets neither profit nor loss.
2) Answer: B
Profit amount of pens sold only On Monday is =
(15*15)-(15*12) =45
Profit amount of pens sold only on Tuesday is =
(12*25)-(12*20)=60
Required difference =60-45=Rs.15
3) Answer: B
On Saturday,
Cost price of each pen is =20*100/125=16.
Total cost price of all the pens is =16*20=320
4) Answer: A
Total cost price of Wednesday and Thursday
together = (18*18)+(30*15)= 774
Total selling price of Wednesday and Thursday
together =25*[18+30] = 1200
So, overall profit percentage = [(1200-774)/774]
*100 = 55.038%= 55% (approx.)
5) Answer: A
On Monday,
Total CP is = 15*12=180
Total SP is = 10*18+5*14= 250
So, profit percentage = [(250-180)/180] *100 =
38.88%
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2 918
A shopkeeper bought some pens from Monday
to Thursday at some price and sold them. All the
unsold pens of a particular day are sold on
Friday. On Friday the shopkeeper did not buy
any new pens. Cost price and selling price of
pens of different days may vary. Number of
unsold pens on Thursday is double the number
of pens unsold on Monday. Cost price of each
pen, which was bought on Wednesday is 2 less
than the selling price of each pen sold on
Wednesday. Ratio of the number of pens sold on
Monday and Tuesday is 5:4. 3 pens are unsold
on Tuesday. Ratio of the selling price of the pen
sold on Monday and on Tuesday is 3:5. Selling
price of each pen on Wednesday is Rs.5 more
than the selling price of each pen on Monday.
Shopkeeper sold each pen on Friday at Rs.15.
Ratio of the number of pens bought on
Wednesday and Thursday is 3:5. On Monday,
the shopkeeper sold the pen at 25% profit.
Number of pens bought on Thursday is double
the number of pens bought on Tuesday.
Shopkeeper sold all the pens, which he bought
on Wednesday. 5 pens are unsold On Monday.
Selling price of each pen on Wednesday and
Thursday is the same. Shopkeeper sold each
pen at 25% profit on Tuesday. Number of pens
the shopkeeper bought on Monday is 20 and the
cost price of each pen on Monday is Rs.12.
Shopkeeper makes a profit of 33.33% on
Thursday by selling each pen. Number of pens
brought on Wednesday is equal to the numerical
value of the cost price of each pen of the same
day.
sиαρσnє🌱
1) Find the profit/loss percentage of the pens,
which are sold only on Friday?
a) 2% profit
b) 5% loss
c) No profit/loss
d) 4% profit
e) 8% loss
2) Find the difference in the profit amount of
pens, which is sold only on Monday and that on
Tuesday?
a) Rs.20
b) Rs.15
c) Rs.25
d) Rs.30
e) Rs.12
3) If shopkeeper on Saturday bought same
number of pens, which he bought on Monday
and sold all the pens at Rs.20 and make a profit
of 25%. Then find the total cost price of all the
pens bought on Saturday?
a) Rs.325
b) Rs.320
c) Rs.330
d) Rs.340
e) None of these
4) If all the pens,which were bought on
Wednesday and Thursday together, sold at Rs.
25, then find the overall profit percentage in
these two days? (approx.)
a) 55%
b) 65%
c) 45%
d) 40%
e) None of these
5) If the shopkeeper sold 5 pens on Monday at
Rs. 14 and sold the remaining pens at Rs.18,
find the new profit percentage?
a) 38.88%
b) 42.35%
c) 52.36%
d) 78.35%
e) None of these
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2 918
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2 918
Detailed Solution:
q^2 - 67q - 360 = 0
= q^2 – 72q + 5q - 360 = 0
= (q - 72) * (q + 5) = 0
Where q = 72
By using the formula,
Final quantity = initial quantity (1 – amount
replaced/total quantity)n times
Final quantity = 72 * 72 – 5124 = 60 litres
60 = a * (a - 72/a) * (a – 96/a)
60a = a^2 – 168a + 6912
a^2 – 228a + 6912 = 0
(a – 192) * (a - 36) = 0
a = 192, where 36 is not possible
Initial quantity of alcohol = 192 litres
Alcohol and water in mixture X = 10x and 216
litres respectively.
Alcohol and water in mixture Y = 144 and 23x
litres respectively.
10x + 216 = 4
23x + 144 = 5
50x + 1080 = 92x + 576
x = 12
Alcohol and water in mixture X = 120 and 216
litres respectively.
Alcohol and water in mixture Y = 144 and 276
litres respectively.
Raghu took 50% of mixture X and whole of
mixture Y and mixed them to form another
mixture Z,
60 + 144: 108 + 276 = 204:384
Alcohol and water added by Ravi to mixture Z,
204 + 3 * 72 = 420 litres
384 + 3 * 72 – 120 = 480 litres
Sumathi has 420 + 480 = 900 litres of alcohol
Final quantity of alcohol ‘m’= 900 * (900 -
180)/900 * 3/5 = 432 litres
sиαρσnє🌱
1) Answer: A
From the common explanation,
The value of a = 192
The value of q = 72
Respective ratio = 192:72 = 8:3
2) Answer: B
From the common explanation,
Quantity of alcohol in mixture Y = 144 litres
Quantity of water in Mixture Y = 276 litres
Respective percentage = (276 - 144) * 100/144 =
91.67%
3) Answer: C
From the common explanation,
Ratio of alcohol and water in Mixture X = 5:9 (U)
Ratio of alcohol and water in Mixture Z = 7:8 (V)
By using allegation,
5/14 7/15
L
7 5
12L = 5/2 + 7/3
L = 29/72
So, ratio of alcohol and water in Mixture L =
29:43
Final mixture is equal to alcohol value of mixture
Y,
Quantity of alcohol in mixture Y = 144 litres
Quantity of alcohol in Mixture L = 144 * 29/72 =
58 litres
Quantity of water in Mixture L = 144 * 43/72 = 86
litres
Respective profit = 144 * 25 – 58 * 50 = Rs.700
4) Answer: D
From the common explanation,
The value of m= 432
The value of a = 192
The value of q = 72
(m – a) * (3q - 120) = (432 - 192) * (3 * 72 – 120)
= 23040
5) Answer: E
From the common explanation,
Quantity of water of Sumathi has in her final
mixture = (900 - 432) = 468 litres
Quantity of water in Mixture Y = 276 litres
Required difference = 192 litres
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2 918
Ranjan had ‘a’ litres of alcohol. He took q litres of
alcohol and replaced with same amount of water.
He again took (q + 24) litres and replaced with
same amount of water such that the resultant
mixture contains (q^2 – 5124) litres alcohol and
rest is water. Raman has two mixtures X and Y
of alcohol and water, only. Mixture X contains 3q
litres water and Mixture Y contains 2q litres of
alcohol, while the quantity of water in mixture Y
is (a - 62)% more than the quantity of alcohol in
mixture X. Total quantity of mixture Y is 25%
more than the total quantity of mixture X. Raghu
took 50% of mixture X and whole of mixture Y
and mixed them to form another mixture Z. He
sold mixture Z to Ravi who then added 3q litres
of milk and (3q - 120) litres of water to mixture Z
such that the ratio of alcohol to water in the
mixture becomes 7:8. Sumathi has alcohol equal
to the new quantity of mixture Z and she took
180 litres of alcohol and replaced with water also
took 40% of the mixture and replaced with water
such that the final mixture contains ‘m’ litres of
alcohol where ‘q’ is calculated from the equation
q^2 – 67q - 360 = 0 where ‘q’ is a positive integer.
sиαρσnє🌱
1) What is the ratio of value of ‘a’ to that of ‘q’?
a) 8:3
b) 7:5
c) 5:4
d) 9:5
e) none of these
2) Find water in Mixture Y is how much percent
more/less than that of alcohol in the same
Mixture?
a) 83.67%
b) 91.67%
c) 89.33%
d) 97.67%
e) none of these
3) If mixture U and Mixture V are mixed in the
ratio 7:5 to form final mixture L. Cost price of
alcohol is Rs.50 per litre while the final mixture of
L is sold at Rs.25 per litre, then what is the profit
amount earned where the final quantity of mixture
is equal to alcohol quantity of mixture Y and also
the respective ratio of alcohol and water
of mixture U and mixture V is equal to the ratio of
alcohol and water in Mixture X and final Mixture
of Z?
a) Rs.800
b) Rs.900
c) Rs.700
d) Rs.1000
e) none of these
4) Find the value of (m-a) * (3q-120)?
a) 25450
b) 26480
c) 24640
d) 23040
e) none of these
5) Find the difference between the quantity of
water of Sumathi has in her final mixture and the
quantity of water in Mixture Y?
a) 188 litres
b) 176 litres
c) 190 litres
d) 212 litres
e) none of these
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2 918
Detailed Solution:
Let number of boys and number of girls in class
P are 3p and 5p respectively.
Total students in class Q = 90
Total students in class P = 90 + 30
3p + 5p = 120
p = 15
Number of boys in class P = 3p = 45
Number of girls in class P = 5p = 75
Average number of boys and girls in class P = A
= (45 + 75)/2= 60
So,A = 60
Total boys in classes P and Q together = 120
Number of boys in class Q = 120 – 45 = 75
Number of girls in class Q = 90 – 75 = 15
Percentage of number of girls out of number of
boys in class Q = B% = (15/75) * 100= 20%
So,B = 20
Number of boys in class R = 80% of 75 = 60
Average number of girls in classes P, Q, and R =
60
Number of girls in class R = 3 * 60 – 75 – 15 =
90
Average number of boys in classes P, Q, R, and
S = 55
Number of boys in class S = 55 * 4 – 45 – 75 –
60 = 40
Number of girls in class S = 150% of 40 = 60
Difference between number of boys and number
of girls in class S = C = 60 – 40= 20
So,C = 20
Average number of students (boys and girls) in
class S = D = (40 + 60)/2= 50
So,D = 50
Total students in class T = 100
Total students in all the given classestogether =
100+100+150+90+120 = 560
Let total boys and total girls in school is 15x and
13x respectively.
According to the question:
15x+13x = 560
x = 20
Total boys in the school = 15x = 300
Total girls in the school = 13x = 260
Total students in class T = 100
Number of boys in class T = 300 – (45 + 75 + 60
+ 40) = 80
Number of girls in class T = 260 – (75 + 15 + 90
+ 60)= 20
Part of number of girls out of number of boys =
1\E = 20/80= 1/4
So, E = 4
sиαρσnє🌱
1) Answer: C
According to question,
Equation: x^2 – 3Ex + B = 0
x^2 – 12x + 20 = 0
x^2 – 2x – 10x + 20 = 0
x (x – 2) – 10 (x – 2) = 0
(x – 2) (x – 10) = 0
x = 2 and 10
Roots of the equation = (2, 10)
2) Answer: E
According to question,
A – E = 60 – 4 = 56 = 2^3 * 7^1
C + E = 20 + 4 = 24 = 2^3 * 3^1
LCM of 56 and 24 = 2^3 * 3^1 * 7^1
n = 168
HCF of 56 and 24 = 2^3
m = 8
Now,
n ÷ m= 168 ÷ 8
= 21
3) Answer: D
Difference between total number of students in
classes Q and R = 150 – 90 = 60
Difference between the values of B and E = 20 –
4 = 16
Required percentage = (60/16) * 100
= 375%
4) Answer: C
Sum of values of A, B, C, D, and E together = 60
+ 20 + 20 + 50 + 4 = 154
154 = 2^1 * 7^1 * 11^1
Total number of factors = (1 + 1) * (1 + 1) * (1 +
1) = 2^3 = 8
5) Answer: B
Sum of values of A, B, C, D, and E = 60 + 20 +
20 + 50 + 4 = 154
Total number of boys in the school = 300
Total number of girls in the school = 260
Difference = 300 – 260 = 40
x = 154 ÷ 40
x = 3.85
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2 918
Information given below is about the total number
of students (Boys and Girls) in five different
classes P, Q, R, S, and T of a school.
Ratio of boys to girls in class P is 3: 5, average of
number of boys and girls in class P is __(A)__.
Total students in class Q is 90 which is 30 less
than the total students in class P. Total girls in
class Q is __(B)__ percent of total boys in that
class while total boys in classes P and Q
together is 120. Total boys in class R are 80% of
total boys in class Q and average number of girls
in classes P, Q, and R are 60. Difference
between number of boys and number of girls in
class S is __(C)__ while average number of boys
in classes P, Q, R, and S is 55. Total girls in
class S are 50% more than the total boys in that
class and average number of students (boys and
girls) in class S is __(D)__. Fraction of the part of
number of girls out of number of boys in class T
is 1 ÷ __(E)__and ratio of number of boys to
number of girls in all the schools together is 15:
13 and total students in class T is 100.
sиαρσnє🌱
1) Find the roots of the equation x^2 – 3Ex + B =
0?
a) (4, 6)
b) (4, 5)
c) (2, 10)
d) (1, 20)
e) (6, 6)
2) If HCF and LCM of (A – E) and (C + E) is ‘m’
and ‘n’ respectively, then find the value of n ÷ m?
a) 15
b) 20
c) 24
d) 18
e) 21
3) Difference between total number of students
in classes Q and R is what percentage of
difference between the values of B and E?
a) 275%
b) 325%
c) 225%
d) 375%
e) 350%
4) Find the total number of factors of sum of
values of A, B, C, D, and E together?
a) 12
b) 4
c) 8
d) 16
e) 9
5) Find the value of ‘x’, if x = Sum of values of
A, B, C, D, and E ÷ Difference between total
number of boys and total number of girls in the
school.
a) 4.25
b) 3.85
c) 3.75
d) 4.05
e) 3.95
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2 918
Detailed Solution:
Total number of vacancies in both the
Companies together = 14,400
Number of vacancies in company X = 14,400 ×
5/12 = 6000
Number of vacancies in company Y = 14,400 ×
7/12 = 8400
Vacancies for male in company X = 6000 ×
70/100 = 4200
Vacancies for female in company X = 6000 –
4200 = 1800
Vacancies for male in department P in company
X = 4200 × 3/5 = 2520
Vacancies for male in department Q in company
X = 1/8 × (4200 – 2520) = 210
Vacancies for male in department R in company
X = 7/8 × (4200 – 2420) = 1470
Vacancies for female in department Q in
company X = 1800 × 24/100 = 432
Vacancies for female in department P in
company X = 5/8 × (1800 – 432) = 855
Vacancies for female in department R in
company X = 1800 – 432 – 855 = 513
Vacancies for male in company Y = 8400 ×
80/100 = 6720
Vacancies for female in company Y = 8400 –
6720 = 1680
Vacancies for male in department P in company
Y = 6720 × 65/100 = 4368
Vacancies for male in department R in company
Y = 1470 × 12/10 = 1764
Vacancies for male in department Q in company
Y = 6720 – 4368 – 1764 = 588
Vacancies for female in department P in
company Y = 4368 × 25/100 = 1092
Vacancies for female in department Q in
company Y = 1/4 × (1680 – 1092) = 147
Vacancies for female in department R in
company Y = 3/4 × (1680 – 1092) = 441
sиαρσnє🌱
1. Answer: B
For question 1:
The total number of female vacancies in the
department P in both the Company X and Y
together = 855 + 1092 = 1947
2. Answer: D
Vacancies for male in department R in company
X = 1470
Vacancies for male in company X = 4200
So, the required percentage = 1470/4200 × 100
= 35%
3. Answer: C
Total of male and female vacancies in
department Q = 588 + 147 = 735
Total of male and female vacancies in
department R = 1764 + 441 = 2205
Hence, the required ratio between them =
735:2205 = 1:3
4. Answer: B
Average number of female vacancies in
department R in both companies together = (513
+ 441)/2 = 477
Average number of male vacancies in
department Q in both companies together =
(210 + 588)/2 = 399
Thus, the required difference between them =
477 – 399 = 78
5. Answer: B
Vacancies for female in department Q in
company Y = 147
Total vacancies for female in company Y = 1680
So, the required percentage = 147/1680 × 100 =
8.75%
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2 918
The data is given about the total number of
vacancies in the department P, Q and R of
Companies X and Y. The total number of
vacancies in both the companies together is
14,400. The respective ratio of number of
vacancies in Companies X and Y is 5: 7. Each
vacancy is splitted into three departments P,Q,R
of the respective companies.
In company X, 70% of the total vacancies are for
males. 3/5th of the total number of male
vacancies are in department P. Out of the
remaining male vacancies 1/8th is in department
Q. Out of the total female vacancies, 24% are in
department Q and 5/8th of the remaining female
vacancies is in department P.
In company Y, 80% of the total vacancies are for
males. 65% of the total male vacancies are in
department P. Number of male vacancies in
department R in company Y is 20% more than
the male vacancies in department R in company
X. Number of female vacancies in department P
in company Y are less than the number of male
vacancies in department P in the same company
by 75%. Out of the remaining female vacancies,
1/4th is in department Q.
sиαρσnє🌱
1) What is the total number of female vacancies
in the department P in both the Companies X
and Y together?
a) 1893
b) 1947
c) 2021
d) 2385
e) None of these
2) Find what percentage of the total number of
male vacancies in company X are in the
department R?
a) 46%
b) 21%
c) 19%
d) 35%
e) None of these
3) In Company Y, what is the respective ratio
between the total number of vacancies both male
and female in the department Q together to that
of total number of vacancies both male and
female in the department R together in the same
company?
a) 3:2
b) 4:5
c) 1:3
d) 7:5
e) None of these
4) Find what is the difference between the
average number of male vacancies in the
department Q in both the companies together
and the average number of female vacancies in
the department R in both the companies
together.
a) 103
b) 78
c) 121
d) 85
e) None of these
5) What percentage of the total number of female
vacancies in the Company Y are in the
department Q?
a) 6.35%
b) 8.75%
c) 7.25%
d) 9.55%
e) None of these
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2 918
The Ages of A and B are 4x and 5x.
Age of C is 5x + Y.
According to the question, 4x + 5x + Y + 4 = 60,
or, 9x + Y = 56
So, 5x + Y + 4Y = 40, or, X + Y = 8
So, X = 6 and Y = 2
Ratio of age of A, B and C 24:30:32 = 12:15:16
Ratio of total amount of A, B and C is 12:15:16.
We can say 4 unit = 8000
Or, 1-unit = 2000.
Amount of A, B, C is Rs.24000, Rs.30000 and
Rs.32000 respectively.
So, M = 30000
A invest Rs.14000 in R% interest for 2 years.
So, 2800 = 14000 * 2 * R/100
Or, R = 10%
Let C invest Rs. F at R/2 = 5% rate of interest.
So, F * (1 + 5/100) ^2 – F = 2050
Or, 41F/400 = 2050
Or, F = 20000
So, C invest in business = 32000 – 20000 =
12000
B’s investment in business is maximum, so B
invest amount 12000 + 3000 = 15000
So, investment ratio = 10:15:12
So, [15 * P/37] = 3000
Or, P = 7400
B invests Rs.15000 in 2R% = 20% compound
interest in 2 years.
So, B’s interest amount is
= 15000 x (1 + 20/100) ^2 – 15000 = 6600 = I
sиαρσnє🌱
1. Answer: C
Required sum of age is = 24 + 2 * 2 + 30 + 2/2 +
32 + 3 * 2/2 = 94
2. Answer: D
A’s share of profit is = 7400 * 10/37 = 2000
C’s share of profit = 7400 – 2000 – 3000 = 2400
So, required difference = 2000 + 2800 – 2400 –
2050 = 350
3. Answer: B
B’s share of profit = 3000
B’s interest amount = 6600
Total selling price = 6600 * 110/100 + 3000 *
80/100 = 9660
4. Answer: A
R% of M + 9.09% of I + 25% of P
= 30000 * 10/100 + 9.09 * 6600/100 + 7400 *
25/100
= 5450
5. Answer: A
Required interest = 32000 x (1 + 20/100) ^2 –
32000 = 14080
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2 918
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
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2 918
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🌝 IBPS EXAM CALENDAR 2023-24
🌝IBPS RRB RECRUITMENT XII 2023
👉OFFICE ASSISTANT AND OFFICER SCALE-I PRELIMS EXAM = 5, 6, 12, 13, 19 AUG 2023
👉OFFICE ASSISTANTS MAIN EXAM =16 SEP 2023
👉OFFICER SCALE-I MAINS EXAM =10 SEP 2023
👉OFFICER SCALE-II, III SINGLE EXAM = 10 SEP 2023
🌝IBPS CLERK 2023 RECRUITMENT XIII
👉PRELIMS EXAM =26, 27 AUG, 2 SEP 2023
👉MAINS EXAM =7 OCT 2023
🌝IBPS PO/MT 2023 RECRUITMENT XIII
👉PRELIMS EXAM =23,30SEP,1 OCT 2023
👉MAINS EXAM =5 NOV 2023
🌝IBPS SO 2023 RECRUITMENT XIII
👉PRELIMS EXAM = 30,31 DEC 2023
👉MAINS EXAM =28 JAN 2024
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2 918
Repost from QUANTessential👑
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),
We get,
15a – 12a = 60
So, value of a = 20
So, 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
sиαρσnє🌱
1) Answer: C
According to question,
Required value = (2L + 50) = 2 x 10 + 50 = 70
2) Answer: D
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
3) Answer: A
Speed of train T = 3 x 80 = 240 km/h
Required time = 480/240 = 2 hours
4) Answer: B
Required ratio = [180/80 + 60/60]: [240/80] =
13:12
5) Answer: E
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
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2 918
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),
We get,
15a – 12a = 60
So, value of a = 20
So, 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
sиαρσnє🌱
1) Answer: C
According to question,
Required value = (2L + 50) = 2 x 10 + 50 = 70
2) Answer: D
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
3) Answer: A
Speed of train T = 3 x 80 = 240 km/h
Required time = 480/240 = 2 hours
4) Answer: B
Required ratio = [180/80 + 60/60]: [240/80] =
13:12
5) Answer: E
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
💥Join : @Quant_Genius
2 918
Repost from QUANTessential👑
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.
sиαρσnє🌱
1) Find the value of (2L + 50).
a) 60
b) 75
c) 70
d) 80
e) None of these
2) 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
3) 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
4) 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
5) 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
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