IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –

Quantitative Aptitude

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

Directions (1-5): Read the following information carefully and answer the questions.
The given table chart shows the number of poetry books printed by a publisher in three different years
namely 2011, 2012 and 2013.
Year Number of poetry books

printed

2011 702

Till 2012 1640

Till 2013 1910

The given table chart shows the percentage of poetry books printed.
Year % of poetry books printed

out of the total number of

books printed

2011 M%

2012 N%

2013 K%

Note:-
1) Total number of books printed = Number of poetry books printed + Number of story books printed
2) K = N – M
3) N + K = 95
4) N + M = 115
1) Which of the following represents the I) 2K – 2
difference between the number of story books II) M + 3
printed in 2011 and 2013? III) N/2 + 8

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

a) Only I follows together and the total number of story books

b) Only II and III follows printed in all the three years together?
c) Only I and II follows a) 3M + 60
d) Only I and IIIfollows b) N + 3M
e) None of these c) 8M – 6N
2) The number of poetry books printed in 2014 is d) K2 – 6N – 45
equal to 10M + K + 5 and the number of story e) None of these
books printed in 2014 is 60% of the total number
of books printed in 2014. The number of story 5) Out of the total number books printed in 2012,
books printed in 2011 is how much more than 55% of the books are English language books
the number of story books printed in 2014? and the rest of the books are other languages
a) N + 68 books. The number of English poetry books
b) 3M printed in 2012 is 428. Find the difference
c) 6K – 18 between the number of other language poetry
d) M + K and story books printed in 2012?
e) None of these a) 3M + 4N + 2
b) 6N + 3
3) The number of poetry books printed in 2015 is c) 17K + 8
12 less than the 7 times the difference between d) 9M - 12
the of story books printed in 2011 and 2013. The e) None of these
number of poetry books printed in 2015 is what
percentage more than the number of poetry 6) Sum of two numbers A and B is 61 and when
books printed in 2013? B is divided by A the remainder will be 7 and the
a) 6M – 8K quotient will be 2. C n – 2 is equal to the largest
b) K – 5 negative integer in the number systemand C is a
c) M – N positive integer which is greater than 2. Which of
d) 1.2M – 10 the following value comes between C n – n + A
e) None of these and B – A?
a) 29
4) Find the difference between the total number b) 17
of poetry books printed in all the three years c) 27
d) 23

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

e) None of these Directions (9-10): Read the following information

carefully and answer the questions.
7) A right angled cylindrical can of radius R cm Three friends Shivam, Tarun and Chetan start
and height (R + 7) cm is melted to form N their journey from City A to City B at 8 AM, 9 AM
number of hemispheres and the volume of each and 10 AM respectively and the speed ratioof
hemisphere can is 11R 3/42. The contents that Shivam, Tarun and Chetan is 1:1.25:0.5
can be stored in the cylindrical can will be same respectively.
as the contents that can be stored in all the 9) If Shivam meets Tarun after t hours from the
hemispheres. Which of the following options can start of the journey and after t hours Tarun starts
be the values of N? (Both N and R are intege rs) returning back to City A and meet Chetan, then
1) 16 2) 17 3) 8 4) 13 at what time Tarun and Cheta n meet each
a) Only (1) follows other?
b) Both (3) and (4) follows a) 3.00 PM
c) Only (3) follows b) 3.20 PM
d) Both (1) and (4) follows c) 2.30 PM
e) None of these d) 2.45 PM
e) None of these
8) The cost price of an article is Rs.X. If the price
of the article is marked by Y% and discount of 10) If the speed of Shivam is 4 km/hr, then find
25% is offered, then the profit obtained is Rs.(Y the distance covered by Tarun in 4 hours?
+ 20). If the article is marked by (y + 5)% and a) 16 km
the same discount is offered, then the profit b) 28 km
obtained is Rs.(Y + 65). Which of the following c) 24 km
option represents the relationship between X d) 20 km
and Y? e) None of these
a) 0.75X = Y/2
b) 0.25X = 1.2Y - 132 11) Anmol, Preethi and Sana together started a
c) 0.5X = 15Y business with an investment of Rs.(X – 1200),
d) X = 20Y Rs.X and Rs.(X + 1800) respectively and at the
e) None of these end of one year they obtained a total profit of
Rs.P. Preethi invested her p rofit in a scheme
which offers an interest rate of 18% per annum

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

for 5 years and she obtained an interest of

Rs.3600. The total profit obtained by all the 13) If P, Q and R are the wrong term in Series I,
three is Rs.4800 more than twice the profit II and III respectively, then which of the following
obtained by P reeti. Which of the following is correct?
statements are true? a) R > Q > P
i) Anmol obtained 18.75% of their total profit. b) P > Q < R
ii) Sana obtained 50% of their total profit. c) Q > P < R
iii) Value of X is multiple of 18. d) P > R > Q
a) Both (i) and (ii) follows e) P > R < Q
b) Both (ii) and (iii) follows 14) X and Y are the correct terms to replace the
c) Both (i) and (iii) follows wrong terms in Series I and Series II
d) All the three follows respectively and Z is equal to the square of the
e) None of these largest root of the equation a 2 – 15a = -9 2 + 5 2 .
Which of the following options are correct?
Directions (12-14): Read the following i) Z = X + Y
information carefully and answer the questions. ii) 10Y + 3 = X + Z
The following question consists of three series, iii) X = YZ
each following a different pattern and in each a) All the three follows
series one of the term is wrong. b) Only (ii) follows
Series I : 24, 31.5, 46.5, 69, 98, 136.5, 181.5 c) Both (i) and (iii) follows
Series II : 5926, 886, 166, 46, 22, 18, 14 d) Only (i) follows
Series III : 11, 18, 44, 107, 231, 445, 788 e) None of these
12) If the starting term in Series IV is 23 and it
follows the same logic as that of Series III and L 15) The following question contains two
and M are the 4 th and 7 th term in Series IV equations as I and II. You have to solve both
respectively, then find the difference between L equations and determine the relationship
and M? between them and give the answer as follows
a) 721 I) √(100x 4 + 125x 4 ) + 7x + 4 1/2 = -4x
b) 674 II) 3 √(64y 3 )* 2y + 19y + 7 2 = -3y + √1600
c) 625 If the smallest root of equation II is multiplied by
d) 681 2, then which of the following options follows?
e) 700 i) Resultant > -4

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

ii) Resultant + 21 (1/2) = Multiple of 5 e) All except C

iii) Resultant i s less than the smallest root of
equation I 18) Following questions contains three
a) Both (i) and (iii) follows statements as i, ii, and iii. You have to determine
b) Only (iii) follows which statement/s is/are necessary to answer
c) None of the above the question and give answer as,
d) Only (ii) follows Mixture A and Mixture B contains x% of milk and
e) All the three follows b% of water respectively and Mixture A and B
contains a% of water and y% of milk
Direction (16-17): The following question respectively. Ratio of the quantity of mixture A to
contains four equations as A, B, C and D. You B is 3:2. If both the mixtures are mixed together,
have to solve all equations and determine the then the final quantity of milk becomes 23% of
relationship between them and give the answer the final quantity of mixture. Find the final
as follows quantity of milk? (x + y = 45)
A) (a * a) – 3a - √(4a 2 )= - 6 i) x – y = 10
B) b 2 - √(81b 2 ) = - 4 * 5 ii) Initial quantity of mixture A is 60 liters and the
C) c 2 * √(625c 6 ) ÷ 5c 3 + 4 * 7 = 39c initial quantity of milk in mi xture A is 15 liters.
D) d 2 – 3 * 5d = 7 * - 8 iii) 15 liters of mixture A is mixed with mixture B ,
16) Find the LCM of the largest roots of all the then the total quantity of milk in mixture B
four equations. becomes 40 liters.
a) 860 a) The data in Both statement (i) and (ii) are
b) 720 necessary to answer the question.
c) 760 b) The data in all the three statements together
d) 840 are necessary to answer the question.
e) None of these c) The data statement (ii) alone is sufficient to
answer the question.
17) For which of the foll owing equations the d) The data in statement (i), (ii) and (iii) are not
difference between their roots is 1? sufficient to answer the question.
a) All the four equation e) The data statement (iii) alone is sufficient to
b) Both B and C answer the question.
c) Both A, C and D
d) Only D

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

Directions (19-21): Read the following c) 70

information carefully and answer the questions. d) 80
X men can complete a work in Y days and Y e) None of these
women can complete the same work in X days Directions (22-23): Read the following
and 24 men and 12 women together can information carefully and answer the questions.
complete the same work in 53 (1/3) days. A hollow cylinder (C1) with a radius of 14 m and
19) If 15 men and 12 women work together, then is filled with water and a solid cylinder (C2) with
in how many days the work will be completed? a radius of 7 m is put inside C1 and then amount
a) 71 (1/9) days of 9240 m 3 water is spilled from C1.
b) 70 (5/8) days 22) Find the height of the hollow cylinder, if the
c) 60 (1/11) days height of both the cylinders is same?
d) 78 (9/11) days a) 60 m
e) None of these b) 48 m
c) 50 m
20) B boys can complete a work in D days. B d) 72 m
boys started the work and worked for 0.45D e) None of these
days and after 0.45D days all the boys le ft the
work. 20 men and 24 women together complete 23) If the radius of the C2 is increased by 3.5
the remaining work in 0.6D days. Find the value mand the height of both the cylinders is same,
of D? then find the amount of excess water spilled
a) 40 from C1?
b) 45 a) 11200 m3
c) 50 b) 13000 m3
d) 54 c) 12400 m3
e) None of these d) 11550 m3
e) None of these
21) 10 men, 14 women and (G + 24) girls
together can complete a work in 16 days and G 24) Following question contains three
girls alone can complete the whole work in 26 statements as i, ii, and iii. You have to determine
days. Find the value of G? which statement/s is/are necessary to answer
a) 88 the question and give answer as,
b) 65

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

Find the difference between the marked price of Find the ratio of the speed of the boat in still
article A and selling price of article B? water to the speed of the current?
i) The cost price of article A is Rs.224 less than The upstream speed of the boat is 24 km/hr and
the selling price of article B an d the discount % the boat can cover P km in upstream in Q hours
offered on article A is 30% and the profit and it covers 144 km in downstream in the same
obtained on article B is 40%. time. Time taken by the boat to cover 216 km in
ii) The selling price of article A is Rs.140 less upstream is R hours more than the time taken
than that of article B and the price of article A is by the boat to cover the same distance in
marked by Rs.264. downstream.
iii) Discount given in article B is Rs.56 more than i) It takes 18 hours for the boat to cover R 2 km in
the profit obtained on article A. downstream.
a) The data in statement (i) alone is sufficient to ii) P – Q 3 + R 3 > 102
answer the question. a) The data in statement I alone is sufficient to
b) All the three statements are necessary to answer the question, while the data in statement
answer the question. II alone is not sufficient to answer the question
c) The data in Both statement (ii) and (iii) are b) The data in statement II alone is sufficient to
necessary to answer the question. answer the question, while the data in statement
d) The data in statement (ii) alone is sufficient to I alone is not sufficient to answer the question
answer the question. c) The data either in statement I alone or in
e) The data in Both statement (i) and (ii) are statement II alone is sufficient to answer the
necessary to answer the question. question
d) The data given in both statements I and II
25) Following question contains two statements together are not sufficient to answer the
as i, and ii. You have to determine which question
statement/s is/are necessary to answer the e) The data given in both statements I and II
question and give answer as, together are necessary to answer the question.

Directions (26-30): Read the following information carefully and answer the questions.
The given pie chart shows the percentage distribution of the total runs scored by three batsman namely
A, B and C

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

The given pie chart shows the percentage distribution of the number of balls faced by three batsman A,
B and C.

Note:-
1) The strike rate of A and B is 25% and 33 1/3% respectively.
2) The number of balls faced by B is 180 and C faced half the number of balls faced by A.
3) If A had faced the same number of balls as that of B but scored the same runs as that of his original
score, then his strike rate would be 2 times the strike rate of B.
4) Strike rate = (Total number of runs scored)/(Number of balls faced) * 100

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

5) The central angle of the runs scored by C is 198 o

26) Find the difference between the central iii) The strike rate of A is more than that of the
angle of the runs scored by A and the central strike rate of C.
angle of the number of balls faced by B? a) All the three follows
a) 200 b) Only (i) follows
b) 160 c) Both (ii) and (iii) follows
c) 360 d) Both (i) and (ii) follows
d) 240 e) None of these
e) None of these
29) If the strike rate of D is 40% more than the
27) If C scored runs by only scoring 4’s and 6’s strike rate of A and ratio of the number of balls
and hit 25 4’s in the match, then find the number faced by B to D is 9:10, then find the runs
of dot balls faced by C? scored by D?
a) 195 a) 65
b) 205 b) 70
c) 184 c) 75
d) 200 d) 84
e) None of these e) None of these

28) If A had played the same number of balls as 30) If the number of 1’s played by C is 40 and
B and B had played the same number of balls as runs scored by C by hitting 2’s is 50, then find
C, then which of the following statements is/are the number of runs scored on hitting 4’s and 6’s
correct? by C?
i) Present strike rate of B is the same as the a) 155
previous strike rate of A. b) 145
ii) If B strikes 160 more balls, then B would c) 175
score a century. d) 165
e) None of these

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

Directions (31-33): Read the following information carefully and answer the questions.
The given table chart shows the cost price and the marked price of two articles A and B.
Articles Cost price in Rs Marked price in Rs

A 600X 16Y

B 700X 20Y

The given table charts shows the relationship between X and Y in two different shops P and Q
Shop Relationship between X

and Y

P Y = 50X

Q Y = 100X

The given table chart shows the discount offered while selling articles A and B in Shop P and Q at
various times
Time Discount percentage

1.00 Pm 15X

2.00 Pm 10X

3.00 Pm 12X

4.00 Pm 11.5X

5.00 Pm 20X

Note:-
Article A and B is sold in both the shops. X is an integer

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

31) Shop P sold article A at 2.00 Pm and 34) Who among the following A or C is elder?
obtained a profit of Rs.80. Find the Selling price i) Age of A after 6 years is equal to the sum of
of article B in Shop Q at 4.00 Pm? the 2/3rd of the present age of C and 1/5 th of the
a) Rs.3840 present age of B.
b) Rs.4200 ii) B’s son is 18 years younger than B and A is
c) Rs.3080 12 years elder than B’s son.
d) Rs.4000 iii) Double the present age of A is equal to the
e) None of these age of C after 12 years.
a) Both statement (i) and (ii) together are
32) If the profit obtained in article A by Shop Q necessary to answer the question.
at 3.00 Pm is Rs.616X, then find the difference b) Only statement (i) is necessary to answer the
between the cost price of article B in Shop P and question.
the marked price of article B in Shop P? c) Both statement (ii) and (iii) together are
a) Rs.800 necessary to answer the question
b) Rs.450 d) All the three statements are necessary to
c) Rs.500 answer the question.
d) Rs.600 e) None of these
e) None of these
35) Total interest obtained by A and B is Rs.(x +
33) The profit obtained by Shop Q at 4.00Pm by 5832). A invested Rs.x in SI at 40% per annum
selling article A is what percentage of the profit for 3 years and B invested the same amount in
obtained by Shop Q at 5.00 Pm by Selling article CI at the same rate of interest for the same time
B? period. If A invested Rs.2x in SI at 28% rate of
a) (1000 – 184X)/(1300 – 400X)% interest per annum for 6 years, then find the
b) (1300 – 400X)/(1000 – 184X)% interest obtained by A?
c) (1200 – 100X)/(1200 – 128X)% a) Rs.x2
d) (1280X – 3000X)(400X – 800)% b) Rs.(2x + 300)
e) None of these c) Rs.6x
d) Rs.(1.5x + 220
e) Rs.(3.4x – 120)

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

Ans wer With Explanation

Directions (1-5): Year Total Number of Number of

Number of poetry books printed in 2011 = 702 number of poetry story books
Number of poetry books printed in 2012 = 1640
books books printed
– 702 = 938
printed printed
Number of poetry books printed in 2013 = 1910
– 1640 = 270 2011 1560 702 858
K=N–M
N + K = 95 2012 1340 938 402

N + N – M = 95
2013 1080 270 810
2N – M = 95 ---- (1)
N + M = 115 ---- (2) 1. Answer: C
From (1) and (2), Difference between the number of story books
3N = 210 printed in 2011 and 2013 = 858 – 810 = 48
N = 70% From (I),
M = 115 – 70 = 45% 2K – 2 = 2 * 25 – 2 = 50 – 2 = 48
K = 70 – 45 = 25% From (II),
Number of story books printed in 2011 = 702 * M + 3 = 45 + 3 = 48
55/45 = 858 From (III),
Number of story books printed in 2012 = 938 * N/2 + 8 = 70/2 + 8 = 35 + 8 = 43
30/70 = 402
Number of story books printed in 2013 = 270 * 2. Answer: A
75/25 = 810 Number of poetry books printed in 2014 = 10M +
Total number of books printed in 2011 = 702 + K + 5 = 10 * 45 + 25 + 5 = 450 + 30 = 480
858 = 1560 Number of story books printed in 2014 = 480 *
Total number of books printed in 2012 = 938 + 60/40 = 720
402 = 1340 Required difference = 858 – 720 = 138
Total number of books printed in 2013 = 270 + N + 68 = 70 + 68 = 138
810 = 1080

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

3. Answer: B B – 2A = 7 ---- (1)

Number of poetry books printed in 2015 = (858 – A + B = 61 ---- (2)
810) * 7 – 12 = 48 * 7 – 12 = 336 – 12 = 324 From (1) and (2),
Required percentage = (324 – 270)/270 * 100 = 3A = 54
54/270 * 100 = 20% A = 18
K – 5 = 25 – 5 = 20 B = 61 – 18 = 43
The largest negative integer is – 1
4. Answer: D C is a positive integer which is greater than 2
Required difference = (858 + 402 + 810) – 1910 Cn – 2 = - 1
= 2070 – 1910 = 160 Cn = 1
K2 – 6N – 45 = 252 – 6 * 70 – 45 = 625 – 420 – n = 0 (anything power 0 will get 1)
45 = 160 Cn – n + A = 1 – 0 + 18 = 19
B – A = 43 – 18 = 25
5. Answer: A
Total number of English books printed in 2012 = 7. Answer: D
1340 * 55/100 = 737 22/7 * R 2 * (R + 7) = N * 11R 3/42
Total number of other language books printed in 2 * (R + 7) = N * R/6
2012 = 1340 – 737 = 603 12R + 84 = NR
Number of English language poetry books 12 + 84/R = N
printed in 2012 = 428 From (1),
Number of English language story books printed 12 + 84/R = 16
in 2012 = 737 – 428 = 309 84/R = 4
Number of other language Poetry books printed R = 21 cm
in 2012 = 938 - 428 = 510 Option (1) is possible
Number of other language Story books printed in From (2),
2012 = 603 – 510 = 93 12 + 84/R = 17
Required difference = 510 – 93 = 417 84/R = 5
3M + 4N + 2 = 3 * 45 + 4 * 70 + 2 = 135 + 280 + Option (2) is not possible
2 = 417 From (3),
12 + 84/R = 8
6. Answer: D Option (3) is not possible
B = 2A + 7 From (4),

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

12 + 84/R = 13 9. Answer: A
84/R = 1 Shivam and Tarun meet each other after 9 AM.
R = 84 cm In one hour Shivam covers 40 km.
Option (4) is possible Time taken by Shivam and Tarun to meet each
other = 40/(50 – 40) = 40/10 = 4 hours
8. Answer: C Chetan starts at 10 AM and he covers 60 km in
X * (100 + Y)/100 * 75/100 = X + Y + 20 3 hours (10 AM to 1 AM) and the distance
X * (100 + Y)/100 * 3/4 = X + Y + 20 covered by Tarun till 1 PM is (50 * 4) = 200 km
(300X + 3XY)/400 – 20 = X + Y ---- (1) Chetan and Tarun meet at a distance of 140 km
X * (100 + Y + 5)/100 * 75/100 = X + Y + 65 (200 – 60)
X * (105 + Y)/100 * 3/4 = X + Y + 65 Time taken by Chetan and Tarun to meet each
(315X + 3XY)/400 – 65 = X + Y ---- (2) other = 140/(50 + 20) = 140/70 = 2 hours
From (1) and (2), They meet at 3.00 PM
(315X + 3XY)/400 – 65 = (300X + 3XY)/400 – 20
15X/400 = 45 10. Answer: D
X = 1200 Speed of Tarun = 4 * 5/4 = 5 km/hr
From (1), Distance covered by Tarun = 5 * 4 = 20 km
(300 * 1200 + 3 * 1200 * Y)/400 – 20 = 1200 + Y
(360000 + 3600Y)/400 = 1220 + Y 11. Answer: A
360000 + 3600Y = 488000 + 400Y Profit obtained by Preethi = 3600 * 100/18 * 1/5
3200Y = 128000 = Rs.4000
Y = 40 Total profit = 4800 + 2 * 4000 = 4800 + 8000 =
0.5X = 15Y => 0.5 * 1200 = 15 * 40 => 600 = Rs.12800
600 P = Rs.12800
Option (C) follows X/(X – 1200 + X + X + 1800) = 4000/12800
X/(3X + 600) = 5/16
Directions (9-10): 16X = 15X + 3000
Ratio of the speed of Shivam, Tarun and Chetan X = 3000
= 1:1.25:0.5 = 4:5:2 Investment ratio of Anmol, Preethi and Sana =
Let the speed of Shivam be 40 km/hr and the (3000 – 1200) : 3000 : (3000 + 1800) =
speed of Tarun be 50 km/hr and the speed of 1800:3000:4800 = 3:5:8
Chetan be 20 km/hr. Profit % of Anmol = 3/16 * 100 = 18.75%

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

So Statement (i) follows

Profit % of Sana = 8/16 * 100 = 50% 13. Answer: B
So Statement (ii) follows P = 98, Q = 18andR = 445
X is not a multiple of 18 P>Q<R
So Statement (iii) does not follow
14. Answer: B
Directions (12-14): X = 99 , Y = 16
Series I: a2 – 15a = -81 + 25
24 + 7.5 = 31.5 a2 – 15a + 56 = 0
31.5 + 15 = 46.5 a2 – 8a – 7a + 56 = 0
46.5 + 22.5 = 69 a(a – 8) – 7(a – 8) = 0
69 + 30 = 99 a = 8, 7
99 + 37.5 = 136.5 Z = 82 = 64
136.5 + 45 = 181.5 From (i),
Series II: Z=X+Y
64 = 99 + 16
Option (i) does not follow.

Series III: From (ii),

11 + (23 – 1) = 18 10Y + 3 = X + Z

18 + (33 – 1) = 44 10 * 16 + 3 = 99 + 64

44 + (43 – 1) = 107 163 = 163

107 + (53 – 1) = 231 Option (ii) follows.

231 + (63 – 1) = 446 From (iii),

446 + (73 – 1) = 788 X = YZ

12. Answer: D 99 = 16 * 64

23 + (23 – 1) = 30 99 = 1024

30 + (33 – 1) = 56 Option (iii) does not follow.

56 + (43 – 1) = 119
119 + (53 – 1) = 243 15. Answer: B

243 + (63 – 1) = 458 √(100x4 + 125x4) + 7x + 41/2 = -4x

458 + (73 – 1) = 800 √225x4 + 7x + 2 = - 4x

Required difference = 800 – 119 = 681 15x2 + 11x + 2 = 0

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

15x2 + 6x + 5x + 2 = 0 5c(c – 7) – 4(c – 7) = 0

15x(x + 2/5) + 5(x + 2/5) = 0 c = 4/5, 7
x = -1/3, -2/5 D) d2 – 3 * 5d = 7 * - 8
3√(64y3)* 2y + 19y + 72 = -3y + √1600 d2 – 15d + 56 = 0
4y * 2y + 19y + 49 = - 3y + 40 d2 – 8d – 7d + 56 = 0
8y2 + 22y + 9 = 0 d(d – 8) – 7(d – 8) = 0
8y2 + 18y + 4y + 9 = 0 d = 7, 8
8y(y + 9/4) + 4(y + 9/4) = 0 16. Answer: D
y = -1/2, -9/4 Required LCM = (3, 5, 7, 8) = 840
Resultant = -9/4 * 2 = -4.5
i) -4.5 < -4 So i) does not follow 17. Answer: E
ii) -4.5+ 21 (1/2) = (-9 + 43)/2 = 34/2 = not a Difference between the roots of equation A = 3 –
multiple of 5 2=1
iii) Resultant is less than the smallest root of Difference between the roots of equation B = 5 –
equation I (iii) follows 4=1
Difference between the roots of equation C = 7 –
Direction (16-17): 4/5 not equal to 1
A) (a * a) – 3a - √(4a2)= - 6 Difference between the roots of equation D = 8 –
a2 – 3a – 2a = - 6 7=1
a2 – 5a + 6 = 0
a2 – 3a – 2a + 6 = 0 18. Answer: C
a(a – 3) – 2(a – 3) = 0 From statement (i),
a = 2, 3 2x = 55
B) b2 - √(81b2) = - 4 * 5 x = 55/2
b2 – 9b + 20 = 0 y = 45 – 55/2 = 35/2
b2 – 5b – 4b + 20 = 0 So statement (i) alone is not sufficient to answer
b(b – 5) – 4(b – 5) = 0 the question
b = 4, 5 From statement (ii),
C) c2 * √(625c6) ÷ 5c3 + 4 * 7 = 39c % of milk in mixture A = 15/60 * 100 = 25%
c2 * 25c3 ÷ 5c3 + 28 = 39C x = 25
5c2 – 39C + 28 = 0 y = 45 – 25 = 20
5c2 – 35c – 4c + 28 = 0 Initial quantity of mixture B = 60 * 2/3 = 40 liters

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

Quantity of milk in mixture B = 40 * 20/100 = 8 BD = 1920

liters 26.4D + 1920 * 0.45 = 1920
Total quantity of milk in the final mixture = 15 + 8 26.4D + 864 = 1920
= 23 liters 26.4D = 1056
So statement (ii) alone is sufficient to answer the D = 40
question
From statement (iii), 21. Answer: D
Quantity of mixture A = 3a (G + 24)/(G * 26) + 10/1920 + 14/1920 = 1/16
Quantity of mixture B = 2a (G + 24)/(G * 26) + 24/1920 = 1/16
Quantity of milk in mixture B = 2a * y/100 (G + 24)/(G * 26) + 1/80 = 1/16
Quantity of water in mixture B = 2a * b/100 (G + 24)/(G * 26) = 1/16 – 1/80
2a * y/100 + 15 * x/100 = 40 (G + 24)/(G * 26) = (5 – 1)/80
So statement (iii) alone is not sufficient to (G + 24)/(G * 26) = 4/80
answer the question (G + 24)/(G * 26) = 1/20
10G + 240 = 13G
Directions (19-21): 3G = 240
Efficiency of men and women are equal (XY = G = 80
YX)
Time taken by one man to complete the work = 22. Answer: A
Time taken by one woman to complete the work If a certain quantity of water spilled out means
= a days the volume of the water spilled will be equal to
24/a + 12/a = 3/160 the volume of the cylinder that is immersed in
36/a = 3/160 another cylinder.
a = 1920 days Let the height of the cylinder be h m.
19. Answer: A 22/7 * 7 * 7 * h = 9240
Time taken by 15 men and 12 women together h = 60 m
to complete the work = 15/1920 + 12/1920 =
27/1920 = 9/640 = 71 (1/9) days 23. Answer: D
Volume of C2 = 22/7 * 10.5 * 10.5 * 60 = 20790
20. Answer: A m3
(20m + 24w) * 0.6D + B * 0.45D = 1920 Excessive amount of water spilled from C1 =
44 * 0.6D + B * D * 0.45 = 1920 20790 – 9240 = 11550 m3

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

Let the speed of the boat in still water be B

24. Answer: E km/hr and the speed of the current be C km/hr.
From statement (i), B – C = 24 km/hr
Cost price of article A = Rs.224 + selling price of P/24 = Q ---- (1)
article B 144/(B + C) = Q ---- (2)
The discount % offered on article A is 30% and 216/24 – 216/(B + C) = R
the profit obtained on article B is 40%. 216/(B + C) = 9 – R ---- (3)
So statement (i) alone is not sufficient to answer From statement (i),
the question R2/(B + C) = 18
From statement (ii), So, Statement I alone is not sufficient to answer
Selling price of article A + 140 = Selling price of the question.
article B From statement (ii),
The price of article A is marked by Rs.264. P – Q3 + R 3> 102
So statement (ii) alone is not sufficient to answer So, Statement II alone is not sufficient to answer
the question the question.
From statement (iii),
Discount given in article B is Rs.56 more than Directions (26-30):
the profit obtained on article A. Let the number of runs scored by B be m.
So statement (iii) alone is not sufficient to 33 1/3 = m/180 * 100
answer the question 100/3 = m/180 * 100
From (i) and (ii), m = 60
Let the selling price of article be Rs.x Let runs scored by A be n.
Cost price of article A = x – 224 33 (1/3) * 2 = n/180 * 100
Marked price of article A = x – 224 + 264 = x + 200/3 = n/180 * 100
40 n = 120
Selling price of article A = (x + 40) * 70/100 Number of balls faced by A = 120 * 100/25 =
Difference between the marked price of article A 480
and selling price of article B = x + 40 – x = Rs.40 Number of balls faced by C = 480/2 = 240
So both (i) and (ii) are necessary to answer the Runs scored by C = (120 + 60) * 198/(360 –
question. 198) = 180 * 198/162 = 220
Batsman Total runs Number of balls
25. Answer: D

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

scored faced
29. Answer: B
A 120 480
Strike rate of D = 25 * 140/100 = 35%
Number of balls faced by D = 180 * 10/9 = 200
B 60 180
Runs scored by D = 35 * 200/100 = 70
C 220 240
30. Answer: C
26. Answer: C
Number of 2’s hit by C = 50/2 = 25
Required difference = 120/(120 + 60 + 220) *
The number of runs scored on hitting 4’s and 6’s
360 – 180/(480 + 180 + 240) * 360 = 120/400 *
by C = 240 – 40 – 25 = 175
360 – 180/900 * 360 = 108 – 72 = 360

Directions (31-33):
27. Answer: A
Shop Article A Article B
Number of 6’s hit by C = (220 – 25 * 4)/6 = (220
– 100)/6 = 120/6 = 20 Cost Marked Cost Marked
Number of dot balls faced by C = 240 – 25 – 20 price price price price
= 195
P 600X 800X 700X 1000X

28. Answer: D
Q 600X 1600X 700X 2000X
Number of balls played by A = 180
Present strike rate of A = 120/180 * 100 = 33
(1/3) % 31. Answer: C
Number of balls played by B = 240 800X * (100 – 10X)/100 = 600X + 80
Present strike rate of B = 60/240 * 100 = 25% 8X * (100 – 10X) = 600X + 80
Present strike rate of B is the same as the 800X – 80X2 = 600X + 80
previous strike rate of A. 2X2 – 5X + 2 = 0
Statement (i) follows. 2X2 – X – 4X + 2 = 0
Runs to be scored by B if he plays 160 more X(2X – 1) – 2(2X – 1) = 0
balls = 60/240 * 400 = 100 X = 2, 1/2
Statement (ii) follows. X=2
Strike rate of C = 220/240 * 100 = 91.67%
Statement (iii) does not follow.

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

Selling price of article B in Shop Q at 4.00 Pm = So, Statement II alone is not sufficient to answer
2000 * 2 * (100 – 11.5 * 2)/100 = 4000 * 77/100 the question.
= Rs.3080 From (i) and (ii),
B – 18 = A – 12
32. Answer: D B=A+6
1600X * (100 – 12X)/100 = 600X + 616X A + 6 = 2C/3 + (A + 6)/5
16X * (100 – 12X) = 1216X A + 6 = (10C + (3A + 18))/15
100 – 12X = 76 15A+ 90 = 10C + 3A + 18
12X = 24 10C – 12A = 72
X=2 C is elder than A
Required difference = 1000 * 2 – 700 * 2 = 2000 So both statements (i) and (ii) are required to
– 1400 = Rs.600 answer the question.
From statement (iii),
33. Answer: A Double the present age of A is equal to the age
Required percentage = [(1600X * (100 – of C after 12 years
11.5X)/100 – 600X)]/[(2000X * (100 – 20X)/100 – So, Statement III alone is not sufficient to
700X) = (1000 – 184X)/(1300 – 400X)% answer the question.

34. Answer: A 35. Answer: E

From statement (i), (3 * x * 40)/100 + x * (1+40/100)3 – x= x + 5832
A + 6 = 2C/3 + B/5 6x/5 + 218x/125 = x + 5832
So, Statement I alone is not sufficient to answer (150x + 218x)/125 = x + 5832
the question. 368x = 125x + 729000
From statement (ii), 243x =729000
B’s son + 18 = B x = Rs.3000
B’s son + 12 = A Interest obtained by A = 2 * 3000 * 28 * 6/100 =
Rs.10080 = Rs.(3.4x – 120)

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 IBPS PO Mains Memory Based Paper 2022 Held on 26th November 2022 –
Quantitative Aptitude

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