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 Page 1


 
 
 
 
             
 
 
SBI Clerk Mains 2018 (Solutions) 
 
 
 
 
Directions (1-5):  
 
 
1. (d)  2. (a)  3. (c) 
 
4. (b)  5. (e) 
 
Directions (6-10): Logic: In the given Input-Output the 
numbers are arranged from both the ends simultaneously. In 
first step- Lowest number is arranged from the left end and 
highest number is arranged from the right end. In second 
step- 2
nd
 lowest number is arranged from the left end and 2
nd
 
highest number is arranged from the right end and so on... 
Also while arranging the numbers, the numbers which are 
arranged from left end are replaced by the addition of the 
digits of that number whereas the numbers which are 
arranged from the right end are replaced by the difference of 
the digits of that number.   
Input: 75  12  10  94  84  32  63  42  54  22 
Step I:  01  75  12  84  32  63  42  54  22   05 
Step II: 03  01  75  32  63  42  54  22   05  04   
Step III: 04  03  01  32  63  42  54  05  04  02 
Step IV: 05  04  03  01  42  54  05  04  02  03 
Step V:  06   05  04  03  01   05  04  02  03  01 
 
6. (b)  7.   (d)  8. (e) 
 
9. (a)  10. (c) 
 
Directions (11-15): 
 
 
11. (b)  12. (a)  13. (d) 
 
14. (c)  15. (e) 
 
Direction (16-20): Logic- The different number codes for all 
the consonant as per the given condition are, 
 
Step 1: The consonants of the word ‘ NORMAL’ are to be coded 
as the number allotted to the consonant: 
 
Step 2: The numbers immediately preceded and followed by 
the vowels are to be coded as per the given conditions;  
So, the code for consonant for word ‘ NORMAL’ is coded as 
‘4O73(M)A2’, numbers 4 and 7 is immediately followed and 
preceded respectively by ‘ O’ so, ‘4’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. Similarly, ‘3’ and ‘2’ is immediately followed 
and preceded respectively by ‘ A’ so, ‘3’ is coded as ‘#1’ and ‘2’ 
is coded as ‘@#’.       
 
Step 3: Now the vowels are to be coded as per the given 
conditions, as ‘O’ comes after ‘M’ in the alphabetical series so 
‘O’ is coded as ‘$$’ and ‘A’ comes before ‘M’ in the alphabetical 
series so, ‘A’ is to be coded as ‘ **’.  
 
So, the final code for the word ‘ NORMAL’ is ‘#1$$@##1**@#’.   
 
16. (b); Therefore, the code for the word ‘NORMAL’ is 
‘#1$$@##1**@#’.   
 
17. (d); Step 1: The consonants of the word ‘ EMBARKS’ are to 
be coded as the number allotted to the consonant: 
   
Step 2: The numbers immediately preceded and 
followed by the vowels are to be coded as per the 
given conditions;  
So, the code for consonant for word ‘ EMBARKS’ is 
coded as ‘ E3(M)1A711’, number ‘3’ is followed by 
vowel so the code for ‘3’ is ‘@#’ and numbers ‘1’ and 
‘7’ is immediately preceded and followed 
respectively by ‘A’ so, ‘1’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. But the numeric code of ‘K’ and ‘S’ is ‘1’ 
is neither followed by nor preceded by any vowel. 
Hence, there code will remain the same. 
REASONING ABILITY 
Page 2


 
 
 
 
             
 
 
SBI Clerk Mains 2018 (Solutions) 
 
 
 
 
Directions (1-5):  
 
 
1. (d)  2. (a)  3. (c) 
 
4. (b)  5. (e) 
 
Directions (6-10): Logic: In the given Input-Output the 
numbers are arranged from both the ends simultaneously. In 
first step- Lowest number is arranged from the left end and 
highest number is arranged from the right end. In second 
step- 2
nd
 lowest number is arranged from the left end and 2
nd
 
highest number is arranged from the right end and so on... 
Also while arranging the numbers, the numbers which are 
arranged from left end are replaced by the addition of the 
digits of that number whereas the numbers which are 
arranged from the right end are replaced by the difference of 
the digits of that number.   
Input: 75  12  10  94  84  32  63  42  54  22 
Step I:  01  75  12  84  32  63  42  54  22   05 
Step II: 03  01  75  32  63  42  54  22   05  04   
Step III: 04  03  01  32  63  42  54  05  04  02 
Step IV: 05  04  03  01  42  54  05  04  02  03 
Step V:  06   05  04  03  01   05  04  02  03  01 
 
6. (b)  7.   (d)  8. (e) 
 
9. (a)  10. (c) 
 
Directions (11-15): 
 
 
11. (b)  12. (a)  13. (d) 
 
14. (c)  15. (e) 
 
Direction (16-20): Logic- The different number codes for all 
the consonant as per the given condition are, 
 
Step 1: The consonants of the word ‘ NORMAL’ are to be coded 
as the number allotted to the consonant: 
 
Step 2: The numbers immediately preceded and followed by 
the vowels are to be coded as per the given conditions;  
So, the code for consonant for word ‘ NORMAL’ is coded as 
‘4O73(M)A2’, numbers 4 and 7 is immediately followed and 
preceded respectively by ‘ O’ so, ‘4’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. Similarly, ‘3’ and ‘2’ is immediately followed 
and preceded respectively by ‘ A’ so, ‘3’ is coded as ‘#1’ and ‘2’ 
is coded as ‘@#’.       
 
Step 3: Now the vowels are to be coded as per the given 
conditions, as ‘O’ comes after ‘M’ in the alphabetical series so 
‘O’ is coded as ‘$$’ and ‘A’ comes before ‘M’ in the alphabetical 
series so, ‘A’ is to be coded as ‘ **’.  
 
So, the final code for the word ‘ NORMAL’ is ‘#1$$@##1**@#’.   
 
16. (b); Therefore, the code for the word ‘NORMAL’ is 
‘#1$$@##1**@#’.   
 
17. (d); Step 1: The consonants of the word ‘ EMBARKS’ are to 
be coded as the number allotted to the consonant: 
   
Step 2: The numbers immediately preceded and 
followed by the vowels are to be coded as per the 
given conditions;  
So, the code for consonant for word ‘ EMBARKS’ is 
coded as ‘ E3(M)1A711’, number ‘3’ is followed by 
vowel so the code for ‘3’ is ‘@#’ and numbers ‘1’ and 
‘7’ is immediately preceded and followed 
respectively by ‘A’ so, ‘1’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. But the numeric code of ‘K’ and ‘S’ is ‘1’ 
is neither followed by nor preceded by any vowel. 
Hence, there code will remain the same. 
REASONING ABILITY 
 
 
 
 
             
 
      
Step 3: Now the vowels are to be coded as per the 
given conditions, as ‘E’ comes before ‘M’ in the 
alphabetical series so ‘E’ is coded as ‘ **’ and ‘A’ comes 
before ‘M’ in the alphabetical series so, ‘A’ is to be 
coded as ‘ **’.  
   
  So, the final code for the word ‘EMBARKS’ is 
‘**@##1**@#11’.   
 
18. (a); The code for ‘’SMITTLE’ is ‘1#1**@#2#1**’. 
   
 
19. (a); The code for ‘ANNUAL’ is ‘**@##1$$**@#’. 
   
 
20. (a);  
   
Direction (21-25): 
Compartment A Compartment B Compartment C 
R N K 
Q P V 
M Y L 
S X Z 
 
21.  (d)  22. (a)  23. (b) 
 
24.  (b)  25. (d) 
 
26.  (b); Both course of action should be followed. For I, As the 
given statement states that ministry has forecasted 
the situation of flood, so the people of city A should 
move to city B as a preventive action. Also as an 
impact of flood here will be a lack of necessities so the 
people of city A should preserve these things. 
 
Direction (27-30): 
 
27.  (d)  28. (d)  29. (c) 
 
30.  (c)   
 
Direction (31-34):  
Subjects Time slot 
Math 6:30am-9:30am 
English 9:30am-11:00am 
Hindi 11:00am-12:30pm 
Chemistry 12:30pm-2:30pm 
Biology 2:30pm-5:30pm 
Physics 5:30pm-8:30pm 
 
31.  (b)  32. (b)  33. (a) 
 
34.  (b) 
 
Direction (35-37): 
 
35.  (c)   
  
 
36. (d);   
   
 
37. (c);   
   
 
Direction (38-41):  
Months/Dates 8th 17th 25th 
January (31) L P N 
March (31) Q S K 
April (30) O M R 
Q>N>L>S>R>K>M>O>P 
 
38.  (b)  39. (d)  40. (e) 
 
41.  (d) 
 
Directions (42-44): 
 
42. (c); A man leave GIP at ‘®Å’ me 6.10, he takes 20 min to 
reach WOW but he reaches 15 min late me he reach 
Wow at 6.45 me ‘®£’. 
 
43.  (d); Airplane departure time is ‘£µ’ me 9.25, A person 
want to reach airport 20 minute earlier me he want 
to reach airport at 9.05, and he takes 40 minutes to 
reach airport me he should leave office 8.25 me ‘aµ’. 
 
44.  (c); A man leave his home at ‘£8’ me 9.55, and he takes 2 
hours to reach office from his home that me he 
reaches office at 11.55 me ‘ 88’. 
 
Page 3


 
 
 
 
             
 
 
SBI Clerk Mains 2018 (Solutions) 
 
 
 
 
Directions (1-5):  
 
 
1. (d)  2. (a)  3. (c) 
 
4. (b)  5. (e) 
 
Directions (6-10): Logic: In the given Input-Output the 
numbers are arranged from both the ends simultaneously. In 
first step- Lowest number is arranged from the left end and 
highest number is arranged from the right end. In second 
step- 2
nd
 lowest number is arranged from the left end and 2
nd
 
highest number is arranged from the right end and so on... 
Also while arranging the numbers, the numbers which are 
arranged from left end are replaced by the addition of the 
digits of that number whereas the numbers which are 
arranged from the right end are replaced by the difference of 
the digits of that number.   
Input: 75  12  10  94  84  32  63  42  54  22 
Step I:  01  75  12  84  32  63  42  54  22   05 
Step II: 03  01  75  32  63  42  54  22   05  04   
Step III: 04  03  01  32  63  42  54  05  04  02 
Step IV: 05  04  03  01  42  54  05  04  02  03 
Step V:  06   05  04  03  01   05  04  02  03  01 
 
6. (b)  7.   (d)  8. (e) 
 
9. (a)  10. (c) 
 
Directions (11-15): 
 
 
11. (b)  12. (a)  13. (d) 
 
14. (c)  15. (e) 
 
Direction (16-20): Logic- The different number codes for all 
the consonant as per the given condition are, 
 
Step 1: The consonants of the word ‘ NORMAL’ are to be coded 
as the number allotted to the consonant: 
 
Step 2: The numbers immediately preceded and followed by 
the vowels are to be coded as per the given conditions;  
So, the code for consonant for word ‘ NORMAL’ is coded as 
‘4O73(M)A2’, numbers 4 and 7 is immediately followed and 
preceded respectively by ‘ O’ so, ‘4’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. Similarly, ‘3’ and ‘2’ is immediately followed 
and preceded respectively by ‘ A’ so, ‘3’ is coded as ‘#1’ and ‘2’ 
is coded as ‘@#’.       
 
Step 3: Now the vowels are to be coded as per the given 
conditions, as ‘O’ comes after ‘M’ in the alphabetical series so 
‘O’ is coded as ‘$$’ and ‘A’ comes before ‘M’ in the alphabetical 
series so, ‘A’ is to be coded as ‘ **’.  
 
So, the final code for the word ‘ NORMAL’ is ‘#1$$@##1**@#’.   
 
16. (b); Therefore, the code for the word ‘NORMAL’ is 
‘#1$$@##1**@#’.   
 
17. (d); Step 1: The consonants of the word ‘ EMBARKS’ are to 
be coded as the number allotted to the consonant: 
   
Step 2: The numbers immediately preceded and 
followed by the vowels are to be coded as per the 
given conditions;  
So, the code for consonant for word ‘ EMBARKS’ is 
coded as ‘ E3(M)1A711’, number ‘3’ is followed by 
vowel so the code for ‘3’ is ‘@#’ and numbers ‘1’ and 
‘7’ is immediately preceded and followed 
respectively by ‘A’ so, ‘1’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. But the numeric code of ‘K’ and ‘S’ is ‘1’ 
is neither followed by nor preceded by any vowel. 
Hence, there code will remain the same. 
REASONING ABILITY 
 
 
 
 
             
 
      
Step 3: Now the vowels are to be coded as per the 
given conditions, as ‘E’ comes before ‘M’ in the 
alphabetical series so ‘E’ is coded as ‘ **’ and ‘A’ comes 
before ‘M’ in the alphabetical series so, ‘A’ is to be 
coded as ‘ **’.  
   
  So, the final code for the word ‘EMBARKS’ is 
‘**@##1**@#11’.   
 
18. (a); The code for ‘’SMITTLE’ is ‘1#1**@#2#1**’. 
   
 
19. (a); The code for ‘ANNUAL’ is ‘**@##1$$**@#’. 
   
 
20. (a);  
   
Direction (21-25): 
Compartment A Compartment B Compartment C 
R N K 
Q P V 
M Y L 
S X Z 
 
21.  (d)  22. (a)  23. (b) 
 
24.  (b)  25. (d) 
 
26.  (b); Both course of action should be followed. For I, As the 
given statement states that ministry has forecasted 
the situation of flood, so the people of city A should 
move to city B as a preventive action. Also as an 
impact of flood here will be a lack of necessities so the 
people of city A should preserve these things. 
 
Direction (27-30): 
 
27.  (d)  28. (d)  29. (c) 
 
30.  (c)   
 
Direction (31-34):  
Subjects Time slot 
Math 6:30am-9:30am 
English 9:30am-11:00am 
Hindi 11:00am-12:30pm 
Chemistry 12:30pm-2:30pm 
Biology 2:30pm-5:30pm 
Physics 5:30pm-8:30pm 
 
31.  (b)  32. (b)  33. (a) 
 
34.  (b) 
 
Direction (35-37): 
 
35.  (c)   
  
 
36. (d);   
   
 
37. (c);   
   
 
Direction (38-41):  
Months/Dates 8th 17th 25th 
January (31) L P N 
March (31) Q S K 
April (30) O M R 
Q>N>L>S>R>K>M>O>P 
 
38.  (b)  39. (d)  40. (e) 
 
41.  (d) 
 
Directions (42-44): 
 
42. (c); A man leave GIP at ‘®Å’ me 6.10, he takes 20 min to 
reach WOW but he reaches 15 min late me he reach 
Wow at 6.45 me ‘®£’. 
 
43.  (d); Airplane departure time is ‘£µ’ me 9.25, A person 
want to reach airport 20 minute earlier me he want 
to reach airport at 9.05, and he takes 40 minutes to 
reach airport me he should leave office 8.25 me ‘aµ’. 
 
44.  (c); A man leave his home at ‘£8’ me 9.55, and he takes 2 
hours to reach office from his home that me he 
reaches office at 11.55 me ‘ 88’. 
 
 
 
 
 
             
 
Directions (45-48): 
 
 
45.  (d)  46. (a)  47. (d) 
 
48.  (c) 
49.  (b); The demolition of unauthorized buildings would 
teach a lesson to the unscrupulous builders and also 
serve as a warning for the citizens not to indulge in 
such activities in the future. This is essential, as 
unauthorized constructions impose undue burden on 
the city’s infrastructure. So, only argument II holds 
strong.  
50. (e); Clearly, A wishes to study the degree of effect of pay 
revision on job satisfaction of employees. This me 
that job satisfaction can be measured and A is capable 
of making such a study. So, both I and II are implicit. 
 
 
 
 
 
51.  (c); Rahul runs for 15 minutes at a speed of 5 km/hr and 
25 minutes at a speed of 9 km/hr 
  ? Total distance covered by Rahul on treadmill  
 = 
15
60
× 5 +
25
60
× 9 = 1.25 + 3.75 = 5???? 
  A = 5km 
 
52.  (e); P2 can complete work in = 6 ×
5
4
= 7.5 h???????? 
  P1 and P2 together can complete total work in  
  = 
6×7.5
6+7.5
=
45
13.5
= 3
1
3
h???????? 
   ? P1 and P2 together can complete 75% work in 
   =
10
3
×
75
100
= 2.5 h???????? 
They finish work at 12:30 p.m.  
? They start their work at 12:30 – 2:30 = 10 a.m. 
B = 10 a.m. 
 
53.  (b); P2 can complete work in = 6 ×
5
4
= 7.5 h???????? 
Rahul and P2 can complete same work in 3 hours 
? Rahul can complete same work in 
=
1
1
3
-
1
7.5
=
1
0.2
= 5 h???????? 
Ratio of efficiency of Rahul and P1  is 6 : 5 
C = 
6-5
5
× 100 = 20% 
 
54.  (d); Distance between his house and his office is 45 km 
? ?????? speed = 
45
1.5
= 30???? /h?? 
Speed of stream is 3 km/hr 
? Upstream speed of boat = 30 - 3 = 27 
Time to reach home i.e, D = 
45
27
= 1
2
3
h???????? 
 
55.  (a); Each friend has 2 dices so there are total 36 outcomes 
by one friend. 
  If either Rahul or Aman throw their dices, then there 
are total 36 + 36 outcomes 
  So, E = 36 + 36 = 72 
 
56.  (e); Sum of outcomes of dices should be 8 so it can be 
(4,4), (3,5) and (2,6) 
In (4,4)  
Addition of square of outcomes = 4
2
+ 4
2
= 32 
In (3,5)  
Addition of square of outcomes = 3
2
+ 5
2
= 34 
In (2,6)  
  Addition of square of outcomes = 2
2
+ 6
2
= 40 
Now Raman will win the game if he gets (2,6) and 
remaining two get (3,5) or (4,4) 
So, option (e) is the correct answer 
57.  (d); Let length and breadth of rectangle be l cm and b cm 
respectively 
So, ATQ 
l × (b + 6) – b (l – 6) = 252 
6 (l + b) = 252 
2 (l + b) = 84 cm  
 
58.  (b); Diagonal of square = 2.5v2 × v2 = 5 cm 
Length of rectangle = 5 × 3 = 15 cm 
Breadth = 5 cm 
Area of rectangle = 15 × 5 = 75 cm
2
  
 
59.  (e); Speed of boat in still water = 20 km/hr 
Speed of stream = 
20
7
 km/hr 
Ratio of speed of boat in upstream to that of 
downstream = 6 : 8 ? 3 : 4 
Time taken by boat in upstream to that of 
downstream = 4 : 3 
Required distance = (20 +
20
7
) ×
5×3
7
˜ 50 km 
 
60.  (a); Ratio of profit of   
A : B 
800× 8+ 
900+ 
1000+ 
1100+ 
1200 
: 
1600× 8+ 
1700+ 
1800+ 
1900+ 
2000 
53 : 101 
  Profit of A ? 
7700
154
× 53 = 2650 Rs.  
QUANTITATIVE APTITUDE 
Page 4


 
 
 
 
             
 
 
SBI Clerk Mains 2018 (Solutions) 
 
 
 
 
Directions (1-5):  
 
 
1. (d)  2. (a)  3. (c) 
 
4. (b)  5. (e) 
 
Directions (6-10): Logic: In the given Input-Output the 
numbers are arranged from both the ends simultaneously. In 
first step- Lowest number is arranged from the left end and 
highest number is arranged from the right end. In second 
step- 2
nd
 lowest number is arranged from the left end and 2
nd
 
highest number is arranged from the right end and so on... 
Also while arranging the numbers, the numbers which are 
arranged from left end are replaced by the addition of the 
digits of that number whereas the numbers which are 
arranged from the right end are replaced by the difference of 
the digits of that number.   
Input: 75  12  10  94  84  32  63  42  54  22 
Step I:  01  75  12  84  32  63  42  54  22   05 
Step II: 03  01  75  32  63  42  54  22   05  04   
Step III: 04  03  01  32  63  42  54  05  04  02 
Step IV: 05  04  03  01  42  54  05  04  02  03 
Step V:  06   05  04  03  01   05  04  02  03  01 
 
6. (b)  7.   (d)  8. (e) 
 
9. (a)  10. (c) 
 
Directions (11-15): 
 
 
11. (b)  12. (a)  13. (d) 
 
14. (c)  15. (e) 
 
Direction (16-20): Logic- The different number codes for all 
the consonant as per the given condition are, 
 
Step 1: The consonants of the word ‘ NORMAL’ are to be coded 
as the number allotted to the consonant: 
 
Step 2: The numbers immediately preceded and followed by 
the vowels are to be coded as per the given conditions;  
So, the code for consonant for word ‘ NORMAL’ is coded as 
‘4O73(M)A2’, numbers 4 and 7 is immediately followed and 
preceded respectively by ‘ O’ so, ‘4’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. Similarly, ‘3’ and ‘2’ is immediately followed 
and preceded respectively by ‘ A’ so, ‘3’ is coded as ‘#1’ and ‘2’ 
is coded as ‘@#’.       
 
Step 3: Now the vowels are to be coded as per the given 
conditions, as ‘O’ comes after ‘M’ in the alphabetical series so 
‘O’ is coded as ‘$$’ and ‘A’ comes before ‘M’ in the alphabetical 
series so, ‘A’ is to be coded as ‘ **’.  
 
So, the final code for the word ‘ NORMAL’ is ‘#1$$@##1**@#’.   
 
16. (b); Therefore, the code for the word ‘NORMAL’ is 
‘#1$$@##1**@#’.   
 
17. (d); Step 1: The consonants of the word ‘ EMBARKS’ are to 
be coded as the number allotted to the consonant: 
   
Step 2: The numbers immediately preceded and 
followed by the vowels are to be coded as per the 
given conditions;  
So, the code for consonant for word ‘ EMBARKS’ is 
coded as ‘ E3(M)1A711’, number ‘3’ is followed by 
vowel so the code for ‘3’ is ‘@#’ and numbers ‘1’ and 
‘7’ is immediately preceded and followed 
respectively by ‘A’ so, ‘1’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. But the numeric code of ‘K’ and ‘S’ is ‘1’ 
is neither followed by nor preceded by any vowel. 
Hence, there code will remain the same. 
REASONING ABILITY 
 
 
 
 
             
 
      
Step 3: Now the vowels are to be coded as per the 
given conditions, as ‘E’ comes before ‘M’ in the 
alphabetical series so ‘E’ is coded as ‘ **’ and ‘A’ comes 
before ‘M’ in the alphabetical series so, ‘A’ is to be 
coded as ‘ **’.  
   
  So, the final code for the word ‘EMBARKS’ is 
‘**@##1**@#11’.   
 
18. (a); The code for ‘’SMITTLE’ is ‘1#1**@#2#1**’. 
   
 
19. (a); The code for ‘ANNUAL’ is ‘**@##1$$**@#’. 
   
 
20. (a);  
   
Direction (21-25): 
Compartment A Compartment B Compartment C 
R N K 
Q P V 
M Y L 
S X Z 
 
21.  (d)  22. (a)  23. (b) 
 
24.  (b)  25. (d) 
 
26.  (b); Both course of action should be followed. For I, As the 
given statement states that ministry has forecasted 
the situation of flood, so the people of city A should 
move to city B as a preventive action. Also as an 
impact of flood here will be a lack of necessities so the 
people of city A should preserve these things. 
 
Direction (27-30): 
 
27.  (d)  28. (d)  29. (c) 
 
30.  (c)   
 
Direction (31-34):  
Subjects Time slot 
Math 6:30am-9:30am 
English 9:30am-11:00am 
Hindi 11:00am-12:30pm 
Chemistry 12:30pm-2:30pm 
Biology 2:30pm-5:30pm 
Physics 5:30pm-8:30pm 
 
31.  (b)  32. (b)  33. (a) 
 
34.  (b) 
 
Direction (35-37): 
 
35.  (c)   
  
 
36. (d);   
   
 
37. (c);   
   
 
Direction (38-41):  
Months/Dates 8th 17th 25th 
January (31) L P N 
March (31) Q S K 
April (30) O M R 
Q>N>L>S>R>K>M>O>P 
 
38.  (b)  39. (d)  40. (e) 
 
41.  (d) 
 
Directions (42-44): 
 
42. (c); A man leave GIP at ‘®Å’ me 6.10, he takes 20 min to 
reach WOW but he reaches 15 min late me he reach 
Wow at 6.45 me ‘®£’. 
 
43.  (d); Airplane departure time is ‘£µ’ me 9.25, A person 
want to reach airport 20 minute earlier me he want 
to reach airport at 9.05, and he takes 40 minutes to 
reach airport me he should leave office 8.25 me ‘aµ’. 
 
44.  (c); A man leave his home at ‘£8’ me 9.55, and he takes 2 
hours to reach office from his home that me he 
reaches office at 11.55 me ‘ 88’. 
 
 
 
 
 
             
 
Directions (45-48): 
 
 
45.  (d)  46. (a)  47. (d) 
 
48.  (c) 
49.  (b); The demolition of unauthorized buildings would 
teach a lesson to the unscrupulous builders and also 
serve as a warning for the citizens not to indulge in 
such activities in the future. This is essential, as 
unauthorized constructions impose undue burden on 
the city’s infrastructure. So, only argument II holds 
strong.  
50. (e); Clearly, A wishes to study the degree of effect of pay 
revision on job satisfaction of employees. This me 
that job satisfaction can be measured and A is capable 
of making such a study. So, both I and II are implicit. 
 
 
 
 
 
51.  (c); Rahul runs for 15 minutes at a speed of 5 km/hr and 
25 minutes at a speed of 9 km/hr 
  ? Total distance covered by Rahul on treadmill  
 = 
15
60
× 5 +
25
60
× 9 = 1.25 + 3.75 = 5???? 
  A = 5km 
 
52.  (e); P2 can complete work in = 6 ×
5
4
= 7.5 h???????? 
  P1 and P2 together can complete total work in  
  = 
6×7.5
6+7.5
=
45
13.5
= 3
1
3
h???????? 
   ? P1 and P2 together can complete 75% work in 
   =
10
3
×
75
100
= 2.5 h???????? 
They finish work at 12:30 p.m.  
? They start their work at 12:30 – 2:30 = 10 a.m. 
B = 10 a.m. 
 
53.  (b); P2 can complete work in = 6 ×
5
4
= 7.5 h???????? 
Rahul and P2 can complete same work in 3 hours 
? Rahul can complete same work in 
=
1
1
3
-
1
7.5
=
1
0.2
= 5 h???????? 
Ratio of efficiency of Rahul and P1  is 6 : 5 
C = 
6-5
5
× 100 = 20% 
 
54.  (d); Distance between his house and his office is 45 km 
? ?????? speed = 
45
1.5
= 30???? /h?? 
Speed of stream is 3 km/hr 
? Upstream speed of boat = 30 - 3 = 27 
Time to reach home i.e, D = 
45
27
= 1
2
3
h???????? 
 
55.  (a); Each friend has 2 dices so there are total 36 outcomes 
by one friend. 
  If either Rahul or Aman throw their dices, then there 
are total 36 + 36 outcomes 
  So, E = 36 + 36 = 72 
 
56.  (e); Sum of outcomes of dices should be 8 so it can be 
(4,4), (3,5) and (2,6) 
In (4,4)  
Addition of square of outcomes = 4
2
+ 4
2
= 32 
In (3,5)  
Addition of square of outcomes = 3
2
+ 5
2
= 34 
In (2,6)  
  Addition of square of outcomes = 2
2
+ 6
2
= 40 
Now Raman will win the game if he gets (2,6) and 
remaining two get (3,5) or (4,4) 
So, option (e) is the correct answer 
57.  (d); Let length and breadth of rectangle be l cm and b cm 
respectively 
So, ATQ 
l × (b + 6) – b (l – 6) = 252 
6 (l + b) = 252 
2 (l + b) = 84 cm  
 
58.  (b); Diagonal of square = 2.5v2 × v2 = 5 cm 
Length of rectangle = 5 × 3 = 15 cm 
Breadth = 5 cm 
Area of rectangle = 15 × 5 = 75 cm
2
  
 
59.  (e); Speed of boat in still water = 20 km/hr 
Speed of stream = 
20
7
 km/hr 
Ratio of speed of boat in upstream to that of 
downstream = 6 : 8 ? 3 : 4 
Time taken by boat in upstream to that of 
downstream = 4 : 3 
Required distance = (20 +
20
7
) ×
5×3
7
˜ 50 km 
 
60.  (a); Ratio of profit of   
A : B 
800× 8+ 
900+ 
1000+ 
1100+ 
1200 
: 
1600× 8+ 
1700+ 
1800+ 
1900+ 
2000 
53 : 101 
  Profit of A ? 
7700
154
× 53 = 2650 Rs.  
QUANTITATIVE APTITUDE 
 
 
 
 
             
 
61.  (c); Let initial investment of A = x  
  Ratio of profit  
  
A
12 × x
x
  
:
:
:
  
B
6 × 4500
2250
  
:
:
:
  
C
4 × 4500
1500
  
Now ATQ 
x
x+2250+1500
=
49
100
  
x ˜ Rs 3600  
 
62.  (c); S.P. of article D sold by Ravi = Rs.120 
Profit % earned on article D by Ravi = 60% 
Cost price of article D for Ravi =
120
160
× 100 = Rs75 
Profit earned by Shyam = 120×
25
100
= Rs30 
Profit earned by Ravi = 120 – 75 = Rs 45 
Required difference = 45 – 30 = Rs.15 
63.  (b); Cost price of article A =
105
140
× 100 = Rs75 
  Cost price of article C =
150
125
× 100 = Rs120 
  Required % =
120-75
120
× 100 =
45
120
× 100 = 37.5% 
 
64.  (d); Cost price of article B =
60
120
× 100 = Rs 50 
Marked price of article B = 50 × 1.5 = Rs 75 
Required discount % =
75-60
75
× 100 
   =
15
75
× 100 = 20% 
 
65.  (e); Profit earned on selling article E =
90
180
× 80 = ???? 40 
Profit earned on selling article C =
150
125
× 25 = ???? 30 
Required difference = 40 – 30 = Rs 10 
 
66.  (b); Mark price of article A =
105
84
× 100 
= Rs 125 
CP of article A =
105
140
× 100 = Rs75 
Mark up % of article A =
125-75
75
× 100 = 66
2
3
% 
 
67.  (c);  3
?? +5
. 9
2?? -4
= 9
5?? -14
 
 ? 3
?? +5
. 3
4?? -8
= 3
10?? -28
 
 ? 3
?? +5+4?? -8
= 3
10?? -28
 
 ? 3
5?? -3
= 3
10?? -28
 
 ? 5?? - 3 = 10?? - 28 
 ? 5?? = 25 
 ? ?? = 5 
 And, 2?? 2
- 15?? - 28 = 3?? 2
- 23?? - 13 
 ? ?? 2
- 8?? + 15 = 0 
 ? ?? 2
- 3?? - 5?? + 15 = 0 
 ? ?? (?? - 3) - 5(?? - 3) = 0 
 ? (?? - 5)(?? - 3) = 0 
 ? ?? = 5 , 3 
Quantity I: - Value of x = 5 
Quantity II: - Value of y = 5 , 3 
 ? Quantity I = Quantity II 
 
68.  (b); Quantity I:  
Let C.P. ? Rs 100 
So, S.P. ? Rs 129.6  
ATQ, 
M.P. ?   
129.6
72
× 100 ? Rs 180  
‘x’ ? 
180×(100–30)
100
 -100 ? 26% 
  Quantity II > Quantity I  
 
69.  (a); Let efficiency of 1 man, 1 woman and 1 child is m, w 
and c respectively 
ATQ, 
10 × 12m = 18w × 20  = 27c ×20 
2m = 6w = 9c  
Let total work = 120 m  
Quantity I:  
(9w + 9c) × 16 = (3m + 2m) × 16 = 80 m 
Remaining work = 120 m – 80 m = 40 m  
Number of men required to complete remaining 
work in one day = 40  
Quantity II = 36  
Quantity I > Quantity II 
70.  (a); Quantity I:-  
   
  Let total capacity of tank be 60. 
Units filled in first three minutes = 3 + 4 + 5 = 12 
Hence, total time taken = 5 × 3 = 15 minutes 
Quantity II:- 
Let waste pipe can empty the cistern in x min 
   
1
10
+
1
15
-
1
x
=
1
18
 
  ? 
1
x
=
9+6-5
90
=
10
90
 
? x = 9 minutes 
  Quantity I > Quantity II 
 
Solution (71-75): -  
House A ? 
Units consumed by Other appliances = 120 units 
Let unit consumed by Lights = x 
Then, Units consumed by Fans = ?? - 30 
 ?? + ?? - 30 = 250- 120 
 2?? = 130+ 30 
 ?? = 80 
Units consumed by Lights = 80 units 
Units consumed by Fans = 50 units 
House B ? 
Units consumed by Lights = 80 ?????????? 
Units consumed by Fans = 
160
100
× 50 = 80 ?????????? 
House C ? 
Total units consumed by Lights in all three houses = 200 units 
? Units consumed by Lights in house ‘C’ = 200- 80 - 80 
 = 40 ?????????? 
Units consumed by Fans = 40 ?????????? 
Units consumed by Other appliances = 40 ×
225
100
= 90 ?????????? 
Total units consumed by Other appliances in House ‘B’  
= 320- 90 - 120 = 110 ?????????? 
Units Consumed Fans Lights Other appliances 
House A 50 80 120 
House B 80 80 110 
House C 40 40 90 
Page 5


 
 
 
 
             
 
 
SBI Clerk Mains 2018 (Solutions) 
 
 
 
 
Directions (1-5):  
 
 
1. (d)  2. (a)  3. (c) 
 
4. (b)  5. (e) 
 
Directions (6-10): Logic: In the given Input-Output the 
numbers are arranged from both the ends simultaneously. In 
first step- Lowest number is arranged from the left end and 
highest number is arranged from the right end. In second 
step- 2
nd
 lowest number is arranged from the left end and 2
nd
 
highest number is arranged from the right end and so on... 
Also while arranging the numbers, the numbers which are 
arranged from left end are replaced by the addition of the 
digits of that number whereas the numbers which are 
arranged from the right end are replaced by the difference of 
the digits of that number.   
Input: 75  12  10  94  84  32  63  42  54  22 
Step I:  01  75  12  84  32  63  42  54  22   05 
Step II: 03  01  75  32  63  42  54  22   05  04   
Step III: 04  03  01  32  63  42  54  05  04  02 
Step IV: 05  04  03  01  42  54  05  04  02  03 
Step V:  06   05  04  03  01   05  04  02  03  01 
 
6. (b)  7.   (d)  8. (e) 
 
9. (a)  10. (c) 
 
Directions (11-15): 
 
 
11. (b)  12. (a)  13. (d) 
 
14. (c)  15. (e) 
 
Direction (16-20): Logic- The different number codes for all 
the consonant as per the given condition are, 
 
Step 1: The consonants of the word ‘ NORMAL’ are to be coded 
as the number allotted to the consonant: 
 
Step 2: The numbers immediately preceded and followed by 
the vowels are to be coded as per the given conditions;  
So, the code for consonant for word ‘ NORMAL’ is coded as 
‘4O73(M)A2’, numbers 4 and 7 is immediately followed and 
preceded respectively by ‘ O’ so, ‘4’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. Similarly, ‘3’ and ‘2’ is immediately followed 
and preceded respectively by ‘ A’ so, ‘3’ is coded as ‘#1’ and ‘2’ 
is coded as ‘@#’.       
 
Step 3: Now the vowels are to be coded as per the given 
conditions, as ‘O’ comes after ‘M’ in the alphabetical series so 
‘O’ is coded as ‘$$’ and ‘A’ comes before ‘M’ in the alphabetical 
series so, ‘A’ is to be coded as ‘ **’.  
 
So, the final code for the word ‘ NORMAL’ is ‘#1$$@##1**@#’.   
 
16. (b); Therefore, the code for the word ‘NORMAL’ is 
‘#1$$@##1**@#’.   
 
17. (d); Step 1: The consonants of the word ‘ EMBARKS’ are to 
be coded as the number allotted to the consonant: 
   
Step 2: The numbers immediately preceded and 
followed by the vowels are to be coded as per the 
given conditions;  
So, the code for consonant for word ‘ EMBARKS’ is 
coded as ‘ E3(M)1A711’, number ‘3’ is followed by 
vowel so the code for ‘3’ is ‘@#’ and numbers ‘1’ and 
‘7’ is immediately preceded and followed 
respectively by ‘A’ so, ‘1’ is coded as ‘#1’ and ‘7’ is 
coded as ‘@#’. But the numeric code of ‘K’ and ‘S’ is ‘1’ 
is neither followed by nor preceded by any vowel. 
Hence, there code will remain the same. 
REASONING ABILITY 
 
 
 
 
             
 
      
Step 3: Now the vowels are to be coded as per the 
given conditions, as ‘E’ comes before ‘M’ in the 
alphabetical series so ‘E’ is coded as ‘ **’ and ‘A’ comes 
before ‘M’ in the alphabetical series so, ‘A’ is to be 
coded as ‘ **’.  
   
  So, the final code for the word ‘EMBARKS’ is 
‘**@##1**@#11’.   
 
18. (a); The code for ‘’SMITTLE’ is ‘1#1**@#2#1**’. 
   
 
19. (a); The code for ‘ANNUAL’ is ‘**@##1$$**@#’. 
   
 
20. (a);  
   
Direction (21-25): 
Compartment A Compartment B Compartment C 
R N K 
Q P V 
M Y L 
S X Z 
 
21.  (d)  22. (a)  23. (b) 
 
24.  (b)  25. (d) 
 
26.  (b); Both course of action should be followed. For I, As the 
given statement states that ministry has forecasted 
the situation of flood, so the people of city A should 
move to city B as a preventive action. Also as an 
impact of flood here will be a lack of necessities so the 
people of city A should preserve these things. 
 
Direction (27-30): 
 
27.  (d)  28. (d)  29. (c) 
 
30.  (c)   
 
Direction (31-34):  
Subjects Time slot 
Math 6:30am-9:30am 
English 9:30am-11:00am 
Hindi 11:00am-12:30pm 
Chemistry 12:30pm-2:30pm 
Biology 2:30pm-5:30pm 
Physics 5:30pm-8:30pm 
 
31.  (b)  32. (b)  33. (a) 
 
34.  (b) 
 
Direction (35-37): 
 
35.  (c)   
  
 
36. (d);   
   
 
37. (c);   
   
 
Direction (38-41):  
Months/Dates 8th 17th 25th 
January (31) L P N 
March (31) Q S K 
April (30) O M R 
Q>N>L>S>R>K>M>O>P 
 
38.  (b)  39. (d)  40. (e) 
 
41.  (d) 
 
Directions (42-44): 
 
42. (c); A man leave GIP at ‘®Å’ me 6.10, he takes 20 min to 
reach WOW but he reaches 15 min late me he reach 
Wow at 6.45 me ‘®£’. 
 
43.  (d); Airplane departure time is ‘£µ’ me 9.25, A person 
want to reach airport 20 minute earlier me he want 
to reach airport at 9.05, and he takes 40 minutes to 
reach airport me he should leave office 8.25 me ‘aµ’. 
 
44.  (c); A man leave his home at ‘£8’ me 9.55, and he takes 2 
hours to reach office from his home that me he 
reaches office at 11.55 me ‘ 88’. 
 
 
 
 
 
             
 
Directions (45-48): 
 
 
45.  (d)  46. (a)  47. (d) 
 
48.  (c) 
49.  (b); The demolition of unauthorized buildings would 
teach a lesson to the unscrupulous builders and also 
serve as a warning for the citizens not to indulge in 
such activities in the future. This is essential, as 
unauthorized constructions impose undue burden on 
the city’s infrastructure. So, only argument II holds 
strong.  
50. (e); Clearly, A wishes to study the degree of effect of pay 
revision on job satisfaction of employees. This me 
that job satisfaction can be measured and A is capable 
of making such a study. So, both I and II are implicit. 
 
 
 
 
 
51.  (c); Rahul runs for 15 minutes at a speed of 5 km/hr and 
25 minutes at a speed of 9 km/hr 
  ? Total distance covered by Rahul on treadmill  
 = 
15
60
× 5 +
25
60
× 9 = 1.25 + 3.75 = 5???? 
  A = 5km 
 
52.  (e); P2 can complete work in = 6 ×
5
4
= 7.5 h???????? 
  P1 and P2 together can complete total work in  
  = 
6×7.5
6+7.5
=
45
13.5
= 3
1
3
h???????? 
   ? P1 and P2 together can complete 75% work in 
   =
10
3
×
75
100
= 2.5 h???????? 
They finish work at 12:30 p.m.  
? They start their work at 12:30 – 2:30 = 10 a.m. 
B = 10 a.m. 
 
53.  (b); P2 can complete work in = 6 ×
5
4
= 7.5 h???????? 
Rahul and P2 can complete same work in 3 hours 
? Rahul can complete same work in 
=
1
1
3
-
1
7.5
=
1
0.2
= 5 h???????? 
Ratio of efficiency of Rahul and P1  is 6 : 5 
C = 
6-5
5
× 100 = 20% 
 
54.  (d); Distance between his house and his office is 45 km 
? ?????? speed = 
45
1.5
= 30???? /h?? 
Speed of stream is 3 km/hr 
? Upstream speed of boat = 30 - 3 = 27 
Time to reach home i.e, D = 
45
27
= 1
2
3
h???????? 
 
55.  (a); Each friend has 2 dices so there are total 36 outcomes 
by one friend. 
  If either Rahul or Aman throw their dices, then there 
are total 36 + 36 outcomes 
  So, E = 36 + 36 = 72 
 
56.  (e); Sum of outcomes of dices should be 8 so it can be 
(4,4), (3,5) and (2,6) 
In (4,4)  
Addition of square of outcomes = 4
2
+ 4
2
= 32 
In (3,5)  
Addition of square of outcomes = 3
2
+ 5
2
= 34 
In (2,6)  
  Addition of square of outcomes = 2
2
+ 6
2
= 40 
Now Raman will win the game if he gets (2,6) and 
remaining two get (3,5) or (4,4) 
So, option (e) is the correct answer 
57.  (d); Let length and breadth of rectangle be l cm and b cm 
respectively 
So, ATQ 
l × (b + 6) – b (l – 6) = 252 
6 (l + b) = 252 
2 (l + b) = 84 cm  
 
58.  (b); Diagonal of square = 2.5v2 × v2 = 5 cm 
Length of rectangle = 5 × 3 = 15 cm 
Breadth = 5 cm 
Area of rectangle = 15 × 5 = 75 cm
2
  
 
59.  (e); Speed of boat in still water = 20 km/hr 
Speed of stream = 
20
7
 km/hr 
Ratio of speed of boat in upstream to that of 
downstream = 6 : 8 ? 3 : 4 
Time taken by boat in upstream to that of 
downstream = 4 : 3 
Required distance = (20 +
20
7
) ×
5×3
7
˜ 50 km 
 
60.  (a); Ratio of profit of   
A : B 
800× 8+ 
900+ 
1000+ 
1100+ 
1200 
: 
1600× 8+ 
1700+ 
1800+ 
1900+ 
2000 
53 : 101 
  Profit of A ? 
7700
154
× 53 = 2650 Rs.  
QUANTITATIVE APTITUDE 
 
 
 
 
             
 
61.  (c); Let initial investment of A = x  
  Ratio of profit  
  
A
12 × x
x
  
:
:
:
  
B
6 × 4500
2250
  
:
:
:
  
C
4 × 4500
1500
  
Now ATQ 
x
x+2250+1500
=
49
100
  
x ˜ Rs 3600  
 
62.  (c); S.P. of article D sold by Ravi = Rs.120 
Profit % earned on article D by Ravi = 60% 
Cost price of article D for Ravi =
120
160
× 100 = Rs75 
Profit earned by Shyam = 120×
25
100
= Rs30 
Profit earned by Ravi = 120 – 75 = Rs 45 
Required difference = 45 – 30 = Rs.15 
63.  (b); Cost price of article A =
105
140
× 100 = Rs75 
  Cost price of article C =
150
125
× 100 = Rs120 
  Required % =
120-75
120
× 100 =
45
120
× 100 = 37.5% 
 
64.  (d); Cost price of article B =
60
120
× 100 = Rs 50 
Marked price of article B = 50 × 1.5 = Rs 75 
Required discount % =
75-60
75
× 100 
   =
15
75
× 100 = 20% 
 
65.  (e); Profit earned on selling article E =
90
180
× 80 = ???? 40 
Profit earned on selling article C =
150
125
× 25 = ???? 30 
Required difference = 40 – 30 = Rs 10 
 
66.  (b); Mark price of article A =
105
84
× 100 
= Rs 125 
CP of article A =
105
140
× 100 = Rs75 
Mark up % of article A =
125-75
75
× 100 = 66
2
3
% 
 
67.  (c);  3
?? +5
. 9
2?? -4
= 9
5?? -14
 
 ? 3
?? +5
. 3
4?? -8
= 3
10?? -28
 
 ? 3
?? +5+4?? -8
= 3
10?? -28
 
 ? 3
5?? -3
= 3
10?? -28
 
 ? 5?? - 3 = 10?? - 28 
 ? 5?? = 25 
 ? ?? = 5 
 And, 2?? 2
- 15?? - 28 = 3?? 2
- 23?? - 13 
 ? ?? 2
- 8?? + 15 = 0 
 ? ?? 2
- 3?? - 5?? + 15 = 0 
 ? ?? (?? - 3) - 5(?? - 3) = 0 
 ? (?? - 5)(?? - 3) = 0 
 ? ?? = 5 , 3 
Quantity I: - Value of x = 5 
Quantity II: - Value of y = 5 , 3 
 ? Quantity I = Quantity II 
 
68.  (b); Quantity I:  
Let C.P. ? Rs 100 
So, S.P. ? Rs 129.6  
ATQ, 
M.P. ?   
129.6
72
× 100 ? Rs 180  
‘x’ ? 
180×(100–30)
100
 -100 ? 26% 
  Quantity II > Quantity I  
 
69.  (a); Let efficiency of 1 man, 1 woman and 1 child is m, w 
and c respectively 
ATQ, 
10 × 12m = 18w × 20  = 27c ×20 
2m = 6w = 9c  
Let total work = 120 m  
Quantity I:  
(9w + 9c) × 16 = (3m + 2m) × 16 = 80 m 
Remaining work = 120 m – 80 m = 40 m  
Number of men required to complete remaining 
work in one day = 40  
Quantity II = 36  
Quantity I > Quantity II 
70.  (a); Quantity I:-  
   
  Let total capacity of tank be 60. 
Units filled in first three minutes = 3 + 4 + 5 = 12 
Hence, total time taken = 5 × 3 = 15 minutes 
Quantity II:- 
Let waste pipe can empty the cistern in x min 
   
1
10
+
1
15
-
1
x
=
1
18
 
  ? 
1
x
=
9+6-5
90
=
10
90
 
? x = 9 minutes 
  Quantity I > Quantity II 
 
Solution (71-75): -  
House A ? 
Units consumed by Other appliances = 120 units 
Let unit consumed by Lights = x 
Then, Units consumed by Fans = ?? - 30 
 ?? + ?? - 30 = 250- 120 
 2?? = 130+ 30 
 ?? = 80 
Units consumed by Lights = 80 units 
Units consumed by Fans = 50 units 
House B ? 
Units consumed by Lights = 80 ?????????? 
Units consumed by Fans = 
160
100
× 50 = 80 ?????????? 
House C ? 
Total units consumed by Lights in all three houses = 200 units 
? Units consumed by Lights in house ‘C’ = 200- 80 - 80 
 = 40 ?????????? 
Units consumed by Fans = 40 ?????????? 
Units consumed by Other appliances = 40 ×
225
100
= 90 ?????????? 
Total units consumed by Other appliances in House ‘B’  
= 320- 90 - 120 = 110 ?????????? 
Units Consumed Fans Lights Other appliances 
House A 50 80 120 
House B 80 80 110 
House C 40 40 90 
 
 
 
 
             
 
71.  (a); Required % = 
80-40
40
× 100 = 100% 
 
72.  (c); Total number of units consumed by Other appliances 
in House ‘B’, ‘C’ and ‘D’ together  
  = 110× 3 = 330 ?????????? 
  Units consumed by Other appliances in House ‘D’ 
   = 330- 110- 90 = 130 ?????????? 
 
73.  (e); Total units consumed in House ‘ A’ and ‘C’ together  
  = 50 + 80 + 120+ 40 + 40 + 90 = 420 ?????????? 
 
74.  (b); Required difference = 110- 90 = 20 ?????????? 
 
75.  (d); Total units consumed by Fans and Lights in House ‘C’ 
= 40 + 40 = 80 ?????????? 
  Total units consumed By Lights and Other appliances 
in House ‘A’ = 80 + 120 = 200 ?????????? 
  Required % = 
200-80
200
× 100 =
120
200
× 100 = 60% 
 
76.  (c); C.P. of 10 note books ? 140 × 10 = 1400 Rs.  
  Profit on selling one pen ? 
50×200
100
= Rs 100  
  Number of pen required ? 
1400
100
 = 14 
 
77.  (d); Let speed of slower train = 2x 
  ? speed of faster train = 5x 
  ATQ, 
150+200
2x+5x
= 15   
x =
10
3
  
Time required = 
350
50
3
–
20
3
 = 35 second 
 
78.  (b); Ratio of profit share of B and E is  
35% × 80,000 × 9 : 15% of 80,000 × 12 
= 7 : 4 
Required difference =
(7-4)
11
× 15400 
   =
3
11
× 15400= ???? 4200 
 
79.  (d); Ratio of profit share of A, C and D is 
  
  ATQ, 
   
30
30+16+3?? =
6750
13050
 
? 
30
46+3?? =
15
29
 
? 46 + 3x = 58 
x = 4 months 
 
80.  (a); Amount invested by F =
15
100
× 80,000+ 4000 
= 12000 + 4000 = Rs 16,000 
Amount invested by A = 
25
100
× 80,000 = Rs 20,000 
Ratio of profit share of F, C and A 
   
ATQ, 
5 + 2 ? 8750 
Then total annual profit = 9 ?
8750
7
× 9 = Rs 11,250 
 
???? . (?? ); I. (x - 2)² = 9 
? (x - 2) = ± 3 
? x = 5, -1 
II. (2y + 8)² = 16 
(2y + 8) = ± 4 
? y = -2, -6 
x > y 
 
???? . (?? ); I. x² - 16x + 64 = 0 
x² - 8x – 8x + 64 = 0 
x(x - 8) - 8(x - 8) = 0 
(x - 8) (x - 8) = 0 
x = 8, 8 
II. y² - 16y + 63 = 0 
y² - 7y – 9y + 63 = 0 
y(y - 7) - 9(y - 7) = 0 
(y - 9) (y - 7) = 0 
y = 9, 7 
No relation can be established between x & y 
 
???? . (?? ); I.
25
x²
-
15
x
+ 2 = 0 
? 2x
2
- 15x+ 25 = 0 
? 2x
2
- 10x- 5x + 25 = 0 
2x (x - 5) - 5(x - 5) = 0 
(2x - 5)(x - 5) = 0 
x =
5
2
, 5 
II.
40
y²
+ 1 =
13
y
 
? y
2
- 13y + 40 = 0 
? y
2
- 8y - 5y + 40 = 0 
? y(y - 8) - 5(y - 8) = 0 
(y - 5)(y - 8) = 0 
y = 5, 8 
y = x 
 
 
???? . (?? ); I.
48
x²
-
14
x
+ 1 = 0 
? x
2
- 14x+ 48 = 0 
? x
2
- 8x - 6x + 48 = 0 
? x(x - 8) - 6(x - 8) = 0 
? (x - 8)(x - 6) = 0 
x = 8, 6 
II.  
45
y²
+
1
y
= 2 
? 2y
2
- y - 45 = 0 
? 2y
2
- 10y + 9y - 45 = 0 
? 2y(y - 5) + 9(y - 5) = 0 
? (2y + 9)(y - 5) = 0 
y = 5, -
9
2
 
x > y 
A : C : D
25% 80,000 6 : 15% 80,000 x : 10% 80,000 8
150 : 15x : 80
30 : 3x : 16
? ? ? ? ? ?
F : C : A
16000 6 : 12000 8 : 20,000 12
2 : 2 : 5
? ? ?
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