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Test Summary
No. of Sections: 6
No. of Questions: 79
Total Duration: 108 min

Section 1 - English

Section Summary
No. of Questions: 18
Duration: 18 min

Additional Instructions:
None

Q1. Find out the antonym for the given words.

ADROIT

labour saving

unskillful

burdened

unemployed

Q2. Identify the one which is opposite in meaning (antonym) to the question word and mark.

cumbersome

senile

gigantic

convenient

unwieldy

Q3. Choose the word opposite in meaning to the given word.

Ostracize

Outrun

Surpass

Transcend

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Embrace

Q4. The study of working conditions, especially the design of equipment and furniture, in order to help people work more
efficiently.

Epistemology

Metaphysics

Ornithology

Ergonomics

Q5. Substitute the given sentence with one word :

Present opposing arguments or evidence

Criticise

Rebuff

Reprimand

Rebut

Q6. In the following question, out of four alternatives choose the one which can be substituted for the given words/sentence.
Give and receive mutually

Present

Reciprocate

Compromise

Co-operate

Q7. Choose the word that is similar in meaning to the given word

SPURIOUS

prudish

immaculate

banal

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forged

Q8. Select most suitable synonym

noxious

destructive

refined

contaminated

murky

Q9. Select most suitable synonym

wearisome

encouraging

pliable

burdensome

toothsome

Q10. Spot the error in the given sentence which is divided into parts. If there is no error mark 'e' as the answer

(a) No one/ (b) practice honesty / (c) these days / (d) especially the youth. / (e) No error.

Q11. The sentence given below, have three parts, indicated by (A), (B) and (C). Read each sentence to find out whether there is an
error. If a sentence has no error, mark the part (D), which stands for ‘No error’. (Ignore the error of punctuation, if any)

The teacher (A)/ scolded him always (B)/ because he rarely reached school on time. (C)/ No error (D)

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(A)

(B)

(C)

(D)

Q12. The sentence is divided into parts , identify the part with error and if there is no error mark E as the answer.

They were (A) / heavily fined (B) / last month (C) / for coming lately.(D) / No error. (E)

Q13. Each question consists of four sentences on a topic. Some sentences are grammatically incorrect or inappropriate. Identify the
incorrect sentence or sentences:

A. Chilika is situated on the eastern sea coast of India.

B. A narrow outer channel connect it to the Bay of Bengal.
C. On an account of its rich biodiversity, chilika was designate as a 'Ramsar site'.
D. Chilika supports some of the larger migratory birds in the country.

Only A

A and B

Only C

C and D

Q14. Each question consists of four sentences on a topic. Some sentences are grammatically incorrect or inappropriate. Identify the
incorrect sentence or sentences:

A. I helped the Polio vaccine for children.

B. I would give a lump of sugar to each child.
C. The health assistant would squeeze drop of the solution to each lump.
D. Then I would check if the children had all swallowed.

Only A

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Only C

B and C

A and D

Common Content:

Read the following passage carefully and answer the respective questions given below each of them.

True, it is the function of the army to maintain law and order in abnormal times. But in normal times there is another force that
compels citizens to obey the laws and to act with due regard to the rights of others. The force also protects the lives and the
properties of law-abiding men. Laws are made to secure the personal safety of its subjects and to prevent murder and crimes of
violence. They are made to secure the property of the citizens against theft and damage to protect the rights of communities and
castes to carry out their customs and ceremonies, so long as they do not conflict with the rights of others. Now the good citizen, of his
own free will obey these laws and he takes care that everything he does is done with due regard to the rights and well-being of others.
But the bad citizen is only restrained from breaking these laws by fear of the consequence of his actions. And the necessary steps to
compel the bad citizen to act as a good citizen are taken by this force. The supreme control of law and order in a State is in the hands
of a Minister who is responsible to the State Assembly and acts through the Inspector General of Police.

Q15. A suitable title for the passage would be

The function of the army

laws and the people's rights

The fear of the law and citizen's security

The functions of the police

Q16. Which of the following is not implied in the passage?

Law protects those who respect it

Law ensures people's religious and social rights

absolutely and unconditionally

A Criminal is dettered from commiting crimes only for

fear of the law

The forces of law help to transform irresponsible

citizens into responsible ones

Q17. The expression 'customs and ceremonies' means

fairs and festivals.

habits and traditions

usual practices and religious rites

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superstitions and formalities

Q18. According to the writer, which one of the following is not the responsibility of the police?

To protect the privileges of all citizens.

To check violent activities of citizens.



Section 2 - Quantitative Ability Tech

Section Summary
No. of Questions: 16
Duration: 20 min

Additional Instructions:
None

Q1. Jessy needs to find a number that has got two digits. This two-digit number is 44 more than double the product of its digits
and 19 more than the sum of squares of its digits. What is the number obtained by Jessy?

72

62

22

12

Q2. A two digit number which is five times digit sum is being interchanged its places and the resultant number obtained by
interchanging the digits is 20% more than the original number. What is the original number?

54

45

63

65

Q3. Remainders 11 and 21 are obtained when two different numbers are divided by the same divisor. Later, their sum was divided
again by the same divisor and the remainder was found to be 4. What is that divisor?

36

28

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14

Q4. The unit digit in the product (121)400 is

Q5. What is the smallest number which when increased (added) by 3 is exactly divisible by 15, 18, 20 and 24?

357

307

363

366

Q6. A shopkeeper faced a scenario where the percentage of profit when he sold a product was exactly equal to the cost price of
the product itself. If he sold the product for Rs. 96, then what would be the cost price of the product?

Rs.80

Rs.60

Rs.90

Rs.100

Q7. A real estate agent sells two lands at the same price. If they sells one of them at a profit of 10% and the other at a loss of 10%,
find the percentage profit/loss.

1% loss

1.5% profit

2% loss

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2% profit

Q8. A sum of rupees 3200 is compounded annually at the rate of 25 paise per rupee per annum. Find the compound interest
payable after 2 years.

1200

1600

1800

2000

Q9. A sum of money becomes 4 times at simple interest in 10 years. What is the rate of interest?

10%

20%

30%

40%

Q10. Two trains of lengths 280 m and 400 m travel at speeds of 24 m/s and 16 m/s respectively in opposite directions to each other.
What is the total time taken by them to cross each other?

17 sec

24 sec

36 sec

52 sec

Q11. If log2 = 0.30103 and log3 = 0.4771, then number of digits in 6485 is ________.

12

13

14

15

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Q12. The value of log343(7) is:

1
3

−1
3

-3

Q13. The number of five digit telephone number having at least one of their digits repeated is

90000

100000

30240

62784

Q14. A palindrome is a number that reads the same left to right as it does from right to left, such as 252. How many six-digit
palindromes are there which are even?

900

500

9 x 105

400

Q15. A bag contains 6 white, 7 red and 5 black balls. If 3 balls are drawn from the bag at random without replacement, then what is
the probability that all of them are white?

5/204

3/204

5/256

3/256

Q16.

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Three unbiased coins are tossed. What is the probability of getting at most two heads?

3/4

1/4

3/8


Section 3 - Logical Ability

Section Summary
No. of Questions: 12
Duration: 15 min

Additional Instructions:
None

Q1. In a certain code language, if the word GROUND is coded as HPRQSX, then what is coded as NOURISH ?

OMXNNMO

MQRVDYA

MQRTFXA

OMVNMNO

Q2. In a symbolic language, ‘ish lto inm’ means ‘neat and tidy’, ‘qpr inm sen’ means ‘small but neat’ and ‘hsm sen rso’ means
‘good but erratic’, then in this language, which symbolic code has been used for ‘but’?

ish

sen

inm

hsm

none of these

Q3. AC, DE, FJ, JO, T?

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Q4. In the following question, one term in the number series is wrong. Find out the wrong term.

2, 9, 28, 65, 125, 217

28

125

217

Q5. ACE : HIL :: MOQ : ?

XVT

TVX

VTX

TUX

Q6. In the following question, there is a main statement followed by four statements (i), (ii), (iii) and (iv). From the choices
choose the pair in which the first statement implies the second statement and the two are logically consistent with the main
statement.
Unless you study, you cannot crack CAT
(i) You cracked CAT
(ii) You could not crack CAT
(iii) You did study
(iv) You didn’t study

(i) and (iv)

(ii) and (iii)

(iv) and (i)

None of these

Q7. In the following question, there is a main statement followed by four statements A, B, C and D. From the choices choose the
pair in which the first statement implies the second statement and the two are logically consistent with the main statement.

Players can play, only if weather is good.

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A. Players can play.

B. Weather is not good.
C. Players cannot play.
D. Weather is good.

AB

BC

DA

CD

Q8. Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide
whether the data provided in the statements are sufficient to answer the questions. Read both the statements and give
answer.

A man wants to cover his sitting room (rectangular) with a specially imported carpet. What is the cost of the carpet ?
(A) The labour charge is Rs.95 and each square metre costs Rs.155.
(B) If the room were half as large, the carpet would cost him Rs.320 less at a uniform rate, (of carpet only).

if statement (A) alone is sufficient to solve the question, but statement (B)
alone is not.

if statement (B) alone is sufficient to solve the question, but statement

(A) alone is not.

if neither statement (A) nor statement (B) is individually sufficient to solve

the question, but a combination of both is sufficient to solve the question.

if both the statements taken together are not sufficient and more
information is required to solve the question.

Q9. From his house Salman travelled 8 km straight followed by 8 km towards his right. Further he travelled 6 km to his left
followed by 7 km to his right. Finally if he is facing the setting sun, then he would have started his journey in which direction?

East

West

North

South

Q10. Fact 1: Mary said, "Ann and I both have cats."

Fact 2: Ann said, "I don't have a cat."
Fact 3: Mary always tells the truth, but Ann sometimes lies.
If the first three statements are facts, which of the following statements must also be a fact?
I: Ann has a cat.

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II: Mary has a cat. 

III: Ann is lying.

I only

II only

I and II only

All the statements are facts

Common Content:

Study the following information carefully and answer the given questions.
A, B, C, D, E, F and G are seven friends studying seven different branches of engineering, namely Mechanical, Chemical, Electrical,
Electronics, Civil, Computer and Aeronautical Engineering, not necessarily in this order. Each of them studies in three different
colleges, X, Y and Z. Not less than two study in any college. D studies Electrical engineering in College X. The one who studies
Chemical Engineering does not study in college Z. F studies Aeronautical Engineering in college Y with only B. A does not study in
college X and does not study Civil Engineering. E studies Computer Engineering and does not study in college X. G studies Electronics
Engineering but not in college X. none in college X studies Mechanical or Civil Engineering.

Q11. Which of the following groups represents the persons studying in college Z?

D, B

C, E, G

A, G

G, E, A

None of these

Q12. Who studies Chemical Engineering?

Section 4 - Quantitative Ability

Section Summary
No. of Questions: 14
Duration: 20 min

Additional Instructions:

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None 

Q1. A two-digit number when reversed is 18 more than the actual number itself. Also, the actual number's digit sum is 8. Then
what is the original number?

35

53

32

34

Q2. Vani while adding three consecutive numbers noticed that the resultant was a 61st multiple of 3. Find which three numbers did
Vani add?

70,72,74

71,72,73

72,73,74

none of these

Q3. If ratio between two numbers is 5 : 6 and their LCM is 90. The first number is:

60

45

15

20

Q4. What is the least number which when divided by 12, 14 and 16 leaves a remainder of 7?

343

336

373

383

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Q5. Janavi buys 110Rs. ticket 18 in number. Here, each first class ticket costs Rs 10 and each second class ticket costs Rs 3. What
will another lot of 18 tickets in which the number of first-class and second class tickets have their cost price interchanged ?

112

118

121

124

Q6. A contractor engaged 40 labourers on a job. He was paid Rs.1,050 for the work. He took 20% of it and distributed the
remaining amount amongst the labourers. If the number of men to women labourers is in the ratio 5 : 3 and the wages of men
to the wages of women are in the ratio 3: 2 respectively, then how much will a man get?

Rs.36

Rs.40

Rs.24

Rs.50

Q7. Kavin and Raghu take 9 and 16 more days than if they both were working together to complete the refurbishing of Apple
laptops. Raghu alone can do the refurbishing in __________ days.

12 days

10 days

20 days

28 days

Q8. The rate of interest in two banks DNB and HBI are in the ratio of 7 : 8. If a person invested some amount in both the banks and
received equal amounts from both the banks in two year. The ratio of amount invested in DNB and HBI respectively is :

15 : 1

8:7

7:8

108 : 107

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Q9. Braun invested a certain sum of money at 8% p.a. simple interest for 'n' years. At the end of 'n' years, Braun got back 4 times
his original investment. What is the value of n?

50 years

25 years

12 years 6 months

37 years 6 months

Q10. SAIL has got its own railway track to operate its goods carriage. Two trains move along a circular track of circumference 1.2
km with uniform speeds. When they move in opposite directions, they meet every 15 seconds and when they move in the same
direction, one train overtakes the other every 60 seconds. Find the speed of the slower train.

0.04 km/s

0.03 km/s

0.05 km/s

0.02 km/s

Q11. Nikhil travelled from his house to the market at a speed of 45 kmph to buy some fruits. Having bought the fruits he came back
at a speed of 40 kmph via the same route and hence took an hour longer than the onward journey. How many kilometres did
he drive in this journey?

720 km

360 km

480 km

420 km

Q12. A car moving at a speed of 50 km/hr, returns on the same route with a speed of 40 km/hr. What is the average speed?

40.4 Km/hr

44.4 Km/hr

42.04 Km/hr

46.22 Km/hr

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Q13. An article was sold for Rs. 2770. Had it been sold for Rs. 3000 there would have been an additional gain of 10%. Cost Price of
the article is:

Rs. 2100

Rs. 2200

Rs. 2300

Rs. 2400

Q14. A person X sold an Item to Y at 40% loss, then Y sold it to third person Z at 40% profit and finally Z sold it back to X at 40%
profit. In this whole process what is the percentage loss or profit of X?

70%

62.5%

57.6%

55%

Section 5 - Information Gathering & Synthesis

Section Summary
No. of Questions: 12
Duration: 15 min

Additional Instructions:
None

Q1. The following question consists of five figures marked A, B, C, D and E called the problem figures. Select a figure from among
the Answer Figures which will continue the same series as established by the five Problem Figures.

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Q2. The following question consists of two sets of figures. Figures 1, 2, 3 and 4 constitute the Problem Set while figures (a), (b),
(c) and (d) constitute the Answer Set. There is a definite relationship between figures (1) and (2). Establish a similar
relationship between figures (3) and (4) by selecting a suitable figure from the Answer Set that would replace the question
mark (?) in fig.(4).

Q3. Identify the figure that completes the pattern.

Common Content:

The following pie-chart shows the sales of coffee in a particular country X over the years 1992 to 1996. There are two types of coffee;
Grade A and Grade B. The price for grade A coffee was Rs.50/kg and for grade B was Rs.100/kg. The pie chart shows the combined
sales for grade A and grade B types of coffee.
In each of the given years, sales were 80% of the total production in that particular year.

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Q4. The average annual percentage growth rate of the production of coffee from 1992 to 1996 was

10%

7.5%

5%

2.75%

Q5. If 10000 MT (metric tons) of grade A coffee was sold in the year 1996, then what was the sales of
grade B coffee in the same year (in MT)? (Assume only two types).

7000

5000

4000

8000

Q6. If the sales (kg) of coffee of grade A and grade B in 1992 was in the ratio of 7 : 3, then what was
the average price per kg of coffee sold in 1992?

Rs.60

Rs.65

Rs.70

Rs.75

Q7. Exports = Production - Sales

What were the exports of grade B coffee in 1992 (tons), if only grade B coffee was exported that
year, which brought 25% higher price than the given price?

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1000

2000

3000

4000

Common Content:

Each of these consists of a question and two statements labelled I and II. Decide whether the data provided in the statement(s) is/are
sufficient/necessary to answer the question. Mark your answer as :
(a) If the data given in any one of the statements is sufficient to answer the question.
(b) If the data in either of the statements I or II is sufficient to answer the question.
(c) If the data in both statements I and II together are sufficient to answer the question.
(d) if the data in both statements I and II together are not sufficient to answer the question.

Q8. The population of a country decreased every year by 20% over its previous year. Find its population
in 2002.
1. Population of C was 2 million in 2003.
2. Population of C in 2003 was 0.72 million more than that of 2001.

(a)

(b)

(c)

(d)

Q9. Is A + B – C > 0?
I. A – B + C > 0.
II. A – B – C > 0.

(a)

(b)

(c)

(d)

Q10. If P is a prime number. Find the remainder when P is divided by 6.

1. P + 6 is prime.
2. P + 3 is prime.

(a)

(b)

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(c)

(d)

Q11. When Ramu was as old as Somu is, Somu was one- third as old as he is now. Find the present age of
Somu.
1. The sum of the present ages of Ramu and Somu is 80 years.
2. The present ages of Ramu and Somu are in the ratio 5 : 3.

(a)

(b)

(c)

(d)

Q12. The sum of the three consecutive terms in G.P is S. Find the square of the middle term.
1. The sum of the reciprocals of the terms is 13/9.
2. S = 13.

(a)


Section 6 - Automata fix

Section Summary
No. of Questions: 7
Duration: 20 min

Additional Instructions:
None

Q1. You are required to fix all the syntax errors in the given code.You can use click on run to execute your code.You can use
printf() to debug your code.The submitted code should be logically and syntactically correct and pass all the test cases.Do not
write the main function as it is not required.
Code approach: For this code you will need to write the given implementation.We do not expect you to modify the approach
or incorporate any additional library methods.
Matrix multiplication
Transpose the given matrix, multiply it with the given matrix and print the transpose of the result.
Given matrix
123
456
789
Transpose
147
258
369
Multiplication
14 32 50
32 77 122
50 122 194

Input Format

Input contains rowCount,colCount and the array values

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Output Format

Print the array values

Constraints

1 ≤ array_size ≤ 1000

Sample Input Sample Output


3 3 14 32 50
1 2 3 32 77 122
4 5 6 50 122 194

7 8 9
Time Limit: - ms Memory Limit: - kb Code Size: - kb

Q2. Sort By Frequency:

Design a way to sort of list of positive integers in descending order according to the frequency of elements. The elements with
integers with higher frequency come before lower frequency elements with the same frequency come in the same order as
they appear in the values.

Complete int * frequencySortArray( int *arr, int size) to get the desired output.

For example:
Input: 1,2,2,3,3,3,4,4,5,5,5,5,6,6,6,7,8,9,10)
Output: (5,5,5,5,3,3,3,6,6,6,2,2,4,4,1,7,8,9,10)

Useful Commands:
The length method can help in returning the length of the array arr.
Usage- int len - arr.length;
The following command can be used to declare an array of length len.
int resultnew int[len];

Input Format

The first line of input contains the array size

The next line is the array values

Output Format

The output prints the altered array

Refer to the sample output for specifications.

Constraints

1 <= size <= 1000

Sample Input Sample Output

13 5 5 5 5 5 2 2 2 1 1 4 4 3
1 1 2 2 2 3 4 4 5 5 5 5 5

Sample Input Sample Output

10 4 4 4 2 2 2 3 3 3 5
4 2 3 4 3 5 2 3 2 4

Time Limit: - ms Memory Limit: - kb Code Size: - kb

Q3. You are required to fix all the logical errors in the given code. You can use click on run to execute your code. You can use
printf() to debug your code. The submitted code should be logically and syntactically correct and pass all the test cases. Do
not write the main function as it is not required.
Code approach: For this code, you will need to write the given implementation. We do not expect you to modify the approach
or incorporate any additional library methods.

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Write a C program number, store them in an array and find out the smallest number using a pointer. 

Sample Input Sample Output


3 29
29
45

65
Sample Input Sample Output


5 10
85
96

32
Time Limit: - ms Memory Limit: - kb Code Size: - kb

Q4. You are required to fix the logical error in the code
Mr. Jason has captured your friend and has a collar around his neck. He has locked the collar with a given “locking key". Now it
can only be opened with an “unlocking key”. Your friend sees the locking key but he does not know how to find the unlocking
key. You can calculate the unlocking key if you have the locking key because the unlocking key will be the smallest (in
magnitude) permutation of the digits of the locking key and will never start with zero.
Help your friend write an algorithm that outputs the unlocking key by taking the key as an input
TestCase 1:
Input:
310
Expected Output:
103
TestCase 2:
Input:
918
Expected Output:
189

Input Format

The input to the function/method consists of an argument locking key, and an integer representing the locking key.

Output Format

Return an integer representing the unlocking key

Constraints

-10^7 <= lockingkeys <= 10^7

Sample Input Sample Output

621540 102456

Sample Input Sample Output

75259 25579

Time Limit: - ms Memory Limit: - kb Code Size: - kb

Q5. The following program prints the pattern as given in the below example. Find the logical error in the function/method
display(int n) to get the desired output.
Examples :
Input : 5
Output :
1
1*2
1*2*3

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1*2 

Input : 9
Output :
1
1*2
1*2*3
1*2*3*4
1*2*3*4*5
1*2*3*4
1*2*3
1*2
1

Input Format

A number
 
Output Format

Prints the Pattern

Sample Input Sample Output


5 1
1*2
1*2*3

1*2
Time Limit: - ms Memory Limit: - kb Code Size: - kb

Q6. The function/method rotateList accepts three arguments as inputs - an integer size, an integer k, and a node list_head
representing the size of the list, the rotation index value, and the head node of the linked list, respectively.
It is supposed to rotate the linked list in the counterclockwise direction from the kth node
The function/method compiles successfully but fails to return the desired result for some test cases. Your task is to fix the
code so that it passes all the test cases
Note
0<= k <= size

Sample Input Sample Output

6 2 1 6 5 4 3 2
1 2 3 4 5 6

Sample Input Sample Output

9 3 85 96 62 95 84 35 15 21 30
85 96 30 21 15 35 84 95 62

Time Limit: - ms Memory Limit: - kb Code Size: - kb

Q7. You are required to complete the code. You can use click on run to execute your code. You can use printf() to debug your
code. The submitted code should be logically and syntactically correct and pass all the test cases. Do not write the main
function as it is not required.

Code approach: For this code, you will need to write the given implementation. We do not expect you to modify the approach
or incorporate any additional library methods.

The function/method SameReflection prints all possible right rotations of the given word.

For eg: abc is given, the ouput should be abc,bca,cab

The function/method SameReflection accepts 1 argument- str as character array

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Your task is to use the function/method SameReflection to completed the code so that it passes all the test cases. 

Sample Input Sample Output


sample sample
amples
mplesa

plesam
Sample Input Sample Output


computer computer

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Answer Key & Solution

Section 1 - English
Q1 unskillful

Solution

Adroit means cleverly skilful, resourceful, or ingenious, so unskilful will be the opposite.

Q2 convenient

Solution

'Cumbersome' means unwieldy, bulky or clumsy. 'Senile' means mentally or physically infirm with age. Hence, 'convenient' meaning
comfortable is the right option.

Q3 Embrace

Solution

Explanation:
Ostracize means to exclude from a society or group

Outrun - run or travel faster or further than.

Surpass - exceed; be greater than.
Transcend - be or go beyond the range or limits of (a field of activity or conceptual sphere).
Embrace - hold (someone) closely in one's arms, especially as a sign of affection.

Q4 Ergonomics

Solution

No Solution

Q5 Rebut

Solution

Rebut means to provide some evidence or argument that refutes or opposes.

Q6 Reciprocate

Solution

Present-in a particular place

Reciprocate-to give and take mutually
Compromise-an agreement
Co-operate-work jointly towards same end
the right answer is reciprocate

Q7 forged

Solution

Spurious means lacking authenticity or validity in essence or origin; not genuine

Forged means fake or duplicate, not original.

Q8 contaminated

Solution

The word noxious means harmful or deadly. Contaminated which means impure or polluted, refined refers to advanced and murky
refers to dark. So the best answer is destructive refers to harsh or deadly. Ans.(3)

Q9 burdensome
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Solution

The word wearisome is a negative word which means boring or frustrating. Encouraging refers to hopeful, pliable refers to flexible
and toothsome refers to pleasant. So the best answer is burdensome a negative word means is heavy or troublesome.

Q10 b

Solution

No one takes a singular verb. Hence, the verb should be 'practices'.

Q11 (B)

Solution

The error is in part (B) of the sentence. It is because of the wrong placement of the adverb “always”. Please note that adverbs like
always, seldom, never etc are used before the main verb. So, “always” should be used before the verb “scolded”.

Q12 D

Solution

The use of adverb in the part D of the sentence is incorrect. The correct sentence should be "late" the adjective form and not the
adverb form.

Q13 C and D

Solution

Sentence C should begin with 'On account of ....' (meaning because of) and not with 'on an account of ...'

In sentence D it should be '... the largest of migratory birds,' meaning large in number. As given it suggests large in size.

Sentences A and B are grammatically correct.

Q14 A and D

Solution

You can't 'help' polio vaccine but you can 'help with' polio vaccines. Hence sentence A should read 'I helped with the.....'.
In Sentence D, 'all swallowed' is incorrect usage. It has to be 'swallowed it all'.
So sentences A and D are incorrect.

The functions of the police

Q15.
Solution

"The functions of the police" is a suitable title since in the passage, it depicts the
functions and duties of the force.

Law ensures people's religious and social rights absolutely and unconditionally
Q16.
Solution

All the other options are mentioned in the passage except the option (Law ensures
people's religious and social rights absolutely and unconditionally).

usual practices and religious rites

Q17.
Solution

Here in the passage, "customs and ceremonies" depicts the usual practices and religious
rites of communities and castes.

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To maintain peace during extraordinary circumstances.
Q18.
Solution

From the passage, it is clear that the function of the military is to maintain peace
during extraordinary circumstances and it is not the function of the police.

Section 2 - Quantitative Ability Tech

Q1 72

Solution

10x+y-19=x2+y2 --------(i)
10x+y-44=2xy------(ii)
Subtract eqn (i) and (ii)
x2+y2-2xy = 25
(x-y)2 = 25
x-y =5
Out of the given options, only 72 fits the equation.

Q2 45

Solution

Let the two digit number be “XY”. Thus, we get 10X + Y = 5(X + Y)
=> 5X = 4Y.
Further, we get, (10Y + X) = 1.2 (10X + Y)
=> 8.8Y = 11X.
Solving these equations, we get,
X = 4 and Y = 5. Thus, the number is 45.

Q3 28

Solution

(b) Divisor = [Sum of remainders]-[Remainder when sum is divided] = 11 +21 – 4 = 28

Q4 1

Solution

Since any power of 1 always gives unit digit as 1.

∴ 121400 will also gives unit digit as 1.

Q5 357

Solution

LCM of 15, 18, 20 and 24 is 360

3 has to be subtracted from 360,

i.e., 360 - 3 = 357

Q6 Rs.60

Solution

Selling price = 96
Cost price = C
Profit percentage = Cost Price = C
Profit percentage = (Selling Price - Cost Price) / Cost Price
C/ 100 = (96-C)/C
C = 60

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Q7 1% loss

Solution

When S.P is same, it will always loss.

The result will always be a loss of [ x/10]^2%.
Hence, the answer here is [10/10]^2% = 1% loss.

Q8 1800

Solution

P = Rs.3200 R = 25% N = 2 years

CI = P[ (1 + R/100)n] - P
= 3200(125/100)2 - 3200
= 1800

Q9 30%

Solution

Let the principal be x

Now we are given that a sum of money becomes 4 times
Amount = 4x
So, Simple interest = 4x-x = 3x
Time = 10 years
Formula:

Hence the rate of interest​is 30%

Q10 17 sec

Solution

By the concept of relative speed ,speed(relative)= 24+ 16 = 40 m/sec

Time = (280+400) / 40 = 680 / 40 = 17 sec

Q11 15

Solution

log(648)5 = 5 log(81 x 8) = 20 log3 +15 log2

= 20(0.4771) +15(0.30103)
= 14.05745.
Therefore, The number of digits in 6485 is 15.

1
Q12
3

Solution

log3437 = log73 7
= 1/3 log77
= 1/3

Q13 62784

Solution

Total ways of telephone numbers = 90000.

9*10*10*10*10 (digits from 0 to 9)

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Total arrangements of telephone numbers in which no digit is repeated = 9 × 9 × 8 × 7 × 6.

So the telephone numbers having at least one digit repeated

Total ways – ways in which. no digit is repeated. = 90000 – 9 × 9 × 8 × 7 × 6 = 62784.

Q14 400

Solution

In a six-digit palindrome, the first 3 digits should be same as next 3, so we can do our permutations only in first 3.
The first digit should be the same as last digit and it has to be even for the whole number to be even. First digit cannot be zero, so
there are only 4 possibilities for the first digit. The remaining two digits can be any number from 0 to 9, so 10 possibilities for each.
Therefore, a total number of six-digit even palindromes would be 4*10*10.

Q15 5/204

Solution

P(A)=N(A)/N(S)
3 balls can be drawn in 18C3 ways.
Favourable cases = 6C3
Probability = 5/204

Q16 7/8

Solution

Here S = [TTT, TTH, THT, HTT, THH, HTH, HHT, HHH]

Let E = event of getting at most two heads
Then, E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
∴P(E)=n(E) / n(S)
=7 / 8

Section 3 - Logical Ability

Q1 MQRVDYA

Solution

Word : G R O U N D Pattern : +1 -2 +3 -4 +5 -6
Code : H P R Q S X Similarly,
NOURISH
-1 +2 -3 +4 -5 +6 -7
MQ RVDY A

Q2 sen

Solution

Clearly, ‘sen’ stands for ‘but’.

Q3 P

Solution

Logic is sum is a Perfect square

AC=1+3= 4
DE=4+5= 9
FJ=6+10= 16

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JO= 10+15= 25
T?= 20 +16= 36
16th alphabet is P

Q4 125

Solution

This is a number series of sum of a cube and 1

2 = 1*1*1 +1,
9= 2*2*2+2,
28 = 3*3*3+1,
65= 4*4*4+1,
126=5*5*5+1,
217=6*6*6+1.

125 is the correct answer.It should be replaced by 126.

Q5 TUX

Solution

As Similarly
A + 7 ->H M + 7 -> T
C + 6 ->I O + 6 -> U
E + 7 ->L Q + 7 -> X

Q6 (ii) and (iii)

Solution

No Solution

Q7 BC

Solution

No Solution

Q8 if statement (B) alone is sufficient to solve the question, but statement (A) alone is not.

Solution

From statement (A), we can't get an answer because area of the room is unknown.

From statement (B), we can directly get the cost as 320 x 2 = Rs 640.

Q9 South

Solution

The path traversed by Salman is as follows.

Let us assume that Salman started his Journey towards East.

Let A be the initial position and B the final position. Finally he is facing towards south i.e. 90° clockwise direction to his starting
point.
Finally as he is facing the setting sun i.e., West, he started his journey towards South.

Q10 All the statements are facts

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Solution

If Mary always tells the truth, then both Ann and Mary have cats (statements I and II), and Ann is lying (statement III). So all the
statements are facts.

G, E, A
Q11.
Solution

No Solution

C
Q12.
Solution

No Solution

Section 4 - Quantitative Ability

Q1 35

Solution

Unit digit= x ; Ten’s digit = (8-x)[10(8-x)+x] – [10x+(8-x)]

= 1880-10x+x-10x-8+x
= 18-18x = 18-80+818x
= 54X = 54/18
= 38 – x = 8 – 3
=5

Hence the number is 35

Q2 72,73,74

Solution

Let the numbers be x,x+1,x+2 with x>0.

61st multiple of 3 is 183.

So, x+x+1+x+2=183

or, 3x=216or, x=72So the numbers are 72,73,74.

Q3 15

Solution

Let the 2 numbers be 5x and 6x

and HCF be x.
Using formula, a×b = LCM × HCF
5x × 6x = x * 90
=30x2 =90x
x = 90/30 = 3

HCF = x = 3

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So, the first number is 5*3 = 15

Q4 343

Solution

By checking the options.Or the required number = (LCM of 12, 14, 16) + 7 = 336 + 7 = 343.

Q5 124

Solution

Q6 Rs.24

Solution

Number of men labourers = 40 × 5/8 = 25

Number of women labourers = 15
Amount distributed to the labourers = Rs. 1050 × 80/100 = Rs 840
Wages are in the ratio 3 : 2
If a man's wage is Rs.3x, a woman's wage is Rs.2x.
∴ 75x + 30x = 840.
x = 840/105 = 8
∴ A man's wage = Rs. 24.

Q7 28 days

Solution

Let both will take X days

Kavin takes X+9 days
Raghu takes X+16 days
So, (1/X)= (1/(X + 9))+((X + 16))
Solving X= 12,
Raghu takes, 12+16= 28 days

Q8 8:7

Solution

Q9 37 years 6 months

Solution

Let us say Braun invested Rs. 100.

Then, at the end of "n" years he would have got back Rs. 400.
Therefore, the simple interest earned =400−100= Rs. 300
Simple interest = PRT/100
300= 100xnx8/100

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n = 37.5 years

Q10 0.03 km/s

Solution

Let the speed of the train be x & y m/s

∴(x + y) 15 = 1200 ;
(x -y) 60 = 1200
x+ y = 80
x- y = 20
2x = 100
x= 50 m/s,
y = 30 m/s
Speed of the slower train= 30/1000= 0.03 km/s

Q11 720 km

Solution

X is the Distance covered

x/40 – x/45 = 1
9x - 8x / 360 = 1
x/360 = 1
x = 360 km
Whole distance covered = 2x = 720 km

Q12 44.4 Km/hr

Solution

Avg speed = 2 xy / (x + y)
= 2(50 * 40) /90 = 44 .44 Km/hr

Q13 Rs. 2300

Solution

Let the cost price be Rs. x

Original Selling Price = Rs. 2770
Original Profit = (2770 - x) Rs.
Profit % = (2770 - x)*100/CP
Now, new Selling Price = Rs. 3000
Original CP = Rs. x
New profit = (3000 - x) Rs.
New profit % = (3000 - x)*100/CP
(3000 - x)*100/CP - (2770 - x)*100/CP = 10
(300000 - 100x)/CP - (277000 - 100x)/CP = 10
23000/CP = 10
10CP = 23000
CP = 23000/10
CP = 2300

Q14 57.6%

Solution

Let the CP = Rs.100. for X.

Y’s CP = Rs.60.
Z’s CP = Rs.84.
Finally, X’s CP = Rs.117.6.
:. X’s loss = 117.6 – 60 = Rs.57.6
:. X’s loss percent = 57.6%

Section 5 - Information Gathering & Synthesis

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Q1

Solution

A has 1 shaded square.

B has the same square shaded + 2 others
C has the squares shaded in B - 1 of them
D has the squares shaded in C + 2 others
E has the squares shaded in D - 1 of them
The next figure will have the squares shaded in E + 2 other squares shaded. The only options which fits this criteria is 2.

Q2 d

Solution

From (1) to (2) the changes are :

The diamond shape in right diagnol moves to left end of the diagnol.

The shape in bottom right end of the diagnol, moves to the top right.

Q3 2

Solution

The diagonal part of the square has inverse shapes (i.e., Horizontal lines becomes vertical and vice versa and Left slash becomes
right slash and vice versa)

2.75%
Q4.
Solution

There is no increase as is evident (Both the sales values are 100 for '92 & for
'96).

% change from 1992 to 1993 = [(21-17)/17) * 100] = 23.52%

% change from 1993 to 1994 = [(24 - 21)/ 21 * 100] = 14.285%

% change form 1994 to 1995 = [(21-24)/24 * 100] = -12.5%

% change form 1995 to 1996 = [(17-21)/21 * 100] = -14.285%

Average percentage change from 1992 to 1996 = [23.52 + 14.258 - 12.5 - 14.2580] /
4

= 2.75%

5000
Q5.
Solution

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tal sales = Sales (A) + Sales (B)

100 x 107 = 50 x 10000 x 1000 + 100X x 1000 (X = Quantity of B sold)
X = 5000
ice of A & B are taken as 50 & 100 Rs./kg Respectively as sales for '92 = sales for
).

Rs.65
Q6.
Solution

(7X * 50 + 3X * 100)/10X = Rs. 65/Kg

2000
Q7.
Solution

Production = Sales / 0.8

= 1.25 x Sales.
Sales in 1992 = 100 crore
=> Production in 1992 = 125 crore
=> Exports = 25 crore

Required answer = (25 x 107)/(100 x 1.25 x

1000)
= 2000 tons.

(b)
Q8.
Solution

Let the population of C in 2003 be x.

Using statement I, C’s population in 2002 = x = 0.8x x = 2

million

∴ 0.8x = 1.6 million.

I is sufficient.

Using statement II, C’s population in 2002 = 0.8x.

Its population in 2001 = 0.8x = 0.64x

x – 0.64x = 0.72 million

=> 0.36x = 0.72 million

=> X = 2 million

=> ∴0.8x = 1.6 million

II is sufficient. Either of the statement is sufficient. Ans(2)

(d)
Q9.
Solution

From statement I, A – (B – C) > 0 But A + (B – C) may be positive or negative.

Example: A = 10 B – C = –8
And A = 10 B – C = –12
From statement II, (A – C) – B > 0 but (A – C) + B may be positive or negative.
Example: A – C = 10 B = –8
And A – C =10 B = –12

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mbining both the statements we cannot conclude whether it is positive or
gative.

(a)
Q10.
Solution

Statement I : if P + 6 = 13 => P = 7 when (P = 7) divided by 6 the remainder is 1.

P + 6 = 23 => P = 17

When (P =17) divided by 6 it leaves a remainder 5 ∴ Statement I is not sufficient

Statement II : All the prime numbers more than 3 are odd numbers.

When P + 3 = odd and this is only possible only when P = 2 (∴ P is prime) ∴ Statement
II alone sufficient. Ans(2)

(a)
Q11.
Solution

Let the present ages of Ramu and Somu be r years and s years respectively. When Ramu
was s years old, Somu was Image not present years old. But Somu was 5/3 years old
(2/3)s years ago.

Then Ramu was s years old r=5s/3

Using statement I, r + s = 80

5s/3+s=80

s = 30

I is sufficient. From Statement II, we cannot find the present age of Somu, as we only
know about the ratio of their ages.

∴ II is not sufficient. Ans (1)

(c)
Q12.
Solution

Let the first term and the common ratio be a and r respectively.Middle term = 2nd term
= ar.Its square = (ar)2 = a2r2.The terms are a, ar and ar2.Using statement I,

→ (1)Or, Since r is unknown,

cannot be found∴ ar cannot be found.∴ (ar)2 cannot be found.I is not sufficient.

Using Statement II, S = a + ar + ar2 = 13 → (2)a + ar2 is not known.∴ ar cannot be

found∴ a2r2 cannot be found.II is not sufficient.

Using both statements,(2) => a(1 + r + r2) = 13 (1) => (13/a)/ar2 =

13/9So,

∴ Both the statements taken together are sufficient to answer the question. Ans (3)

Section 6 - Automata fix

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Q1
Hidden Test Case

Input Output


3 3 66 78 90
1 4 7 78 93 108
2 5 8 90 108 126

3 6 9

Weightage - 25

Input Output

2 3 29 44
2 3 4 44 77
5 6 4

Weightage - 25

Input Output

1 5 55
1 2 3 4 5

Weightage - 25

Input Output

 
4 4 4 8 12 16
1 1 1 1 8 16 24 32
2 2 2 2 12 24 36 48
 
3 3 3 3 16 32 48 64

Weightage - 25

Sample Input Sample Output


3 3 14 32 50
1 2 3 32 77 122
4 5 6 50 122 194

7 8 9

Solution

Header
#include<stdio.h>
//#include<conio.h>
#include<stdio.h> int main()
//#include<conio.h> {
int main() int a[10][10],b[10][10],mul[10][10],m,n,i,j,k;
{ //clrscr();
int a[10][10],b[10][10],mul[10][10],m,n,i,j,k; //printf("Enter order of matrix A: ");
//clrscr(); scanf("%d%d",&m,&n);
//printf("Enter order of matrix A: "); for(i=1;i<=m;i++)
scanf("%d%d",&m,&n); {
for(j=1;j<=n;j++)
{
//printf("Enter value of a[%d][%d]: ",i,j);
scanf("%d",&a[i][j]);
for(i=1;i<=m;i++)
b[j][i]=a[i][j];
{
}
for(j=1;j<=n;j++)
}
{
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//printf("Enter value of a[%d][%d]: ",i,j); //printf("\nAnswer:\n");
scanf("%d",&a[i][j]); for(i=1;i<=m;i++)
b[j][i]=a[i][j]; {
} for(j=1;j<=m;j++)
} {
//printf("\nAnswer:\n"); mul[i][j]=0;
for(i=1;i<=m;i++) for(k=1;k<=n;k++)
{ mul[i][j]=mul[i][j]+a[i][k]*b[k][j];
for(j=1;j<=m;j++) printf("%4d",mul[i][j]);
{ }
mul[i][j]=0; printf("\n");
for(k=1;k<=n;k++) }
mul[i][j]=mul[i][j]+a[i][k]*b[k][j]; // getch();
printf("%d ",mul[i][j]); }
}

Footer

printf("\n");
}
// getch();
}

Q2
Hidden Test Case

Input Output

38 96 96 29 29 78 78 54 54 73 73 61 61 83 83 55 55
20 96 85 29 78 25 54 89 73 51 54 43 69 61 4 73
   

Weightage - 50

Input Output

100 14 14 14 14 14 83 83 83 83 44 44 44 87 87 87 77
44 64 15 45 99 69 7 16 43 26 83 6 66 4 33 48 4

   

Weightage - 25

Input Output

138 83 83 83 83 83 83 14 14 14 14 14 95 95 95 95 77
5 3 50 42 1 5 2 71 83 75 77 81 65 15 14 30 20

   

Weightage - 25

Sample Input Sample Output

13 5 5 5 5 5 2 2 2 1 1 4 4 3
1 1 2 2 2 3 4 4 5 5 5 5 5

Sample Input Sample Output

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10 4 4 4 2 2 2 3 3 3 5
4 2 3 4 3 5 2 3 2 4

Solution

Header

#include <stdio.h>
#include<limits.h>
#include<malloc.h>
int * frequencySortArray( int *arr,int size)
{

int** occurence = NULL;

int maxpos, count = 0, count_flag, o_row, index,max;
int newindex = 0, num, ctr;

occurence = (int**)calloc(1,sizeof(int*));
occurence[0] = (int*)calloc(2,sizeof(int));
occurence[0][0] = arr[0];
occurence[0][1]++;
count++;
for(index = 1 ; index < size ; index++)
{
//search in occurence array
for(o_row = 0, count_flag = 0 ; o_row < count ; o_row++)
{
if(occurence[o_row][0] == arr[index])
{
occurence[o_row][1]++;
count_flag = 1;
}
} //search completed
if(count_flag == 0)
{
occurence = (int**)realloc(occurence,(count+1)*sizeof(int*));
occurence[count] = (int*)calloc(2,sizeof(int));
occurence[count][0] = arr[index];
occurence[count][1]++;
count++;
}
}
for(index = 0 ; index < count ; index++)
{
//find maximum in occurence
for(o_row = 0, max = INT_MIN ; o_row < count ; o_row++)
{
if(occurence[o_row][1] > max)
{
max = occurence[o_row][1];
maxpos = o_row;
}
}
num = occurence[maxpos][0];

for(ctr = 1 ; ctr <= max ; ctr++)

arr[newindex++] = num;
occurence[maxpos][1] = -1;
}

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// for(index=0 ; index < count ; index++)

// free(occurence[index]);
// free(occurence);
return arr;

Footer

}
int main()
{
int arr[1000];
int size,index;

scanf("%d",&size);
for(index =0 ;index < size; index++)
scanf("%d",&arr[index]);
frequencySortArray(arr,size);
for(index = 0 ; index < size ; index++)
printf("%d ",arr[index]);
return 0;
}

Header

#include <stdio.h>
#include<limits.h>
#include<malloc.h>

int * frequencySortArray( int *arr,int size)

{
int** occurence = NULL;
int maxpos, count = 0, count_flag, o_row, index,max;
int newindex = 0, num, ctr;

occurence = (int**)calloc(1,sizeof(int*));
occurence[0] = (int*)calloc(2,sizeof(int));
occurence[0][0] = arr[0];
occurence[0][1]++;
count++;
for(index = 1 ; index < size ; index++)
{
//search in occurence array
for(o_row = 0, count_flag = 0 ; o_row < count ; o_row++)
{
if(occurence[o_row][0] == arr[index])
{
occurence[o_row][1]++;
count_flag = 1;
}
} //search completed
if(count_flag == 0)
{
occurence = (int**)realloc(occurence,(count+1)*sizeof(int*));
occurence[count] = (int*)calloc(2,sizeof(int));
occurence[count][0] = arr[index];
occurence[count][1]++;
count++;
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}
}
for(index = 0 ; index < count ; index++)
{
//find maximum in occurence
for(o_row = 0, max = INT_MIN ; o_row < count ; o_row++)
{
if(occurence[o_row][1] > max)
{
max = occurence[o_row][1];
maxpos = o_row;
}
}
num = occurence[maxpos][0];

for(ctr = 1 ; ctr <= max ; ctr++)

arr[newindex++] = num;
occurence[maxpos][1] = -1;
}

// for(index=0 ; index < count ; index++)

// free(occurence[index]);
// free(occurence);
return arr;
}

Footer

int main()
{
int arr[1000];
int size,index;

scanf("%d",&size);
for(index =0 ;index < size; index++)
scanf("%d",&arr[index]);
frequencySortArray(arr,size);
for(index = 0 ; index < size ; index++)
printf("%d ",arr[index]);
return 0;
}

Q3
Hidden Test Case

Input Output


5 1
3
4

2

Weightage - 25

Input Output


9 2
2
3

5
Weightage - 25

https://admin.srm615.examly.io/course 42/51
7/16/25, 10:31 PM SRM PAT

Input Output


4 17
85
96

32

Weightage - 25

Input Output


6 23
56
23

95

Weightage - 25

Sample Input Sample Output


3 29
29
45

65

Sample Input Sample Output


5 10
85
96

32

Solution

Header

#include<stdio.h>
int main()
{
int *s,i,small,m;
scanf("%d",&m);
int a[m];
s=&a[0];
for(i=0;i<m;i++,s++)
scanf("%d",s);

s=&a[0];
small=*s;
for(i=0;i<m;i++,s++)
if(*s<small)
small=*s;

Footer

printf("%d",small);
return 0;
}

Q4
Hidden Test Case

https://admin.srm615.examly.io/course 43/51
7/16/25, 10:31 PM SRM PAT

Input Output

4856 4568

Weightage - 20

Input Output

6489 4689

Weightage - 20

Input Output

321691 112369

Weightage - 30

Input Output

312465697 123456679

Weightage - 30

Sample Input Sample Output

621540 102456

Sample Input Sample Output

75259 25579

Solution

Header

#include<stdio.h>
int unlock(int num)
{
int arr[10],index=0,n,result=0,temp;
int itr;
while(num)
{
arr[index]=num%10;
index++;
num=num/10;
}

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7/16/25, 10:31 PM SRM PAT

n=index;
for(itr=1;itr<=n-1;itr++)
{
for(index=1;index<n;index++)
{
if(arr[index-1] > arr[index])
{
int temp=arr[index];
arr[index]=arr[index-1];
arr[index-1]=temp;
}
}
}
index=0;
while(arr[index]==0)
index++;
temp=arr[index];
arr[index]=arr[0];
arr[0]=temp;

Footer

for(index=0;index<n;index++)
{
result=result*10+arr[index];
}
return result;
}
int main()
{
int num,res;
scanf("%d",&num);
res=unlock(num);
printf("%d",res);
return 0;
}

Q5
Hidden Test Case

Input Output


9 1
1*2
1*2*3

1*2*3*4

Weightage - 25

Input Output


11 1
1*2
1*2*3

1*2*3*4

Weightage - 25

https://admin.srm615.examly.io/course 45/51
7/16/25, 10:31 PM SRM PAT

Input Output


7 1
1*2
1*2*3

1*2*3*4

Weightage - 25

Input Output


15 1
1*2
1*2*3

1*2*3*4

Weightage - 25

Sample Input Sample Output


5 1
1*2
1*2*3

1*2

Solution

Header

#include<stdio.h>
void display(int n)
{

int sp = n / 2, st = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= sp; j++) {
printf(" ");
}

int count = 1;
for (int k = 1; k <= st; k++) {
if (k % 2 == 0)
printf("*");
else
printf("%d",count++);
}
printf("\n");
if (i <= n / 2) {
sp = sp - 1;
st = st + 2;
}
else {
sp = sp + 1;
st = st - 2;
}
}

Footer

https://admin.srm615.examly.io/course 46/51
7/16/25, 10:31 PM SRM PAT

int main()
{
int n;
scanf("%d",&n);
display(n);
return 0;
}

Q6
Hidden Test Case

Input Output

7 4 17 649 393 7 7 2 9
17 649 393 9 2 7

Weightage - 25

Input Output

6 5 1 7 9 45 45 5
1 7 9 45 5 45

Weightage - 25

Input Output

10 2 1 4 8 32 6 8 9 5 4 5
1 5 4 5 9 8 6 32 8 4

Weightage - 25

Input Output

5 4 95 62 51 86 42
95 62 51 42 86

Weightage - 25

Sample Input Sample Output

6 2 1 6 5 4 3 2
1 2 3 4 5 6

Sample Input Sample Output

9 3 85 96 62 95 84 35 15 21 30
85 96 30 21 15 35 84 95 62

Solution

https://admin.srm615.examly.io/course 47/51
7/16/25, 10:31 PM SRM PAT

Header

#include<stdio.h>
#include<stdlib.h>
typedef struct node
{
int data;
struct node *next;
}NODE;
NODE *start;
void displaySLL()
{
NODE *tptr;
for(tptr=start;tptr!=NULL;tptr=tptr->next)
printf("%d ",tptr->data);
printf("\n");
}
void insertData(int givenData)
{
struct node *newnode;
NODE *tptr,*prev;
newnode=(struct node *)malloc(sizeof(struct node));
newnode->data=givenData;
newnode->next=NULL;
if(start==NULL)
{
start=newnode;
}
else
{
for(tptr=start;tptr!=NULL;prev=tptr,tptr=tptr->next);
prev->next=newnode;
}
}
void rotate(int n,int k,NODE *head)
{

NODE *tptr,*prev,*safeNext,*safePrev;
int count;
tptr=head;
for(count=1;count<k;count++)
{
prev=tptr;
tptr=tptr->next;
}
safePrev=NULL;
while(tptr!=NULL)
{
safeNext=tptr->next;
tptr->next=safePrev;
safePrev=tptr;
tptr=safeNext;
}
prev->next=safePrev;

Footer

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}
int main()
{
int n,k,index,num;
scanf("%d %d",&n,&k);
for(index=0;index<n;index++)
{
scanf("%d",&num);
insertData(num);
}
rotate(n,k,start);
displaySLL();
return 0;
}

Q7
Hidden Test Case

Input Output


action action
ctiona
tionac

ionact

Weightage - 20

Input Output


welcome welcome
elcomew
lcomewe

comewel

Weightage - 20

Input Output


examly examly
xamlye
amlyex

mlyexa

Weightage - 25

Input Output


Digitalera Digitalera
igitaleraD
gitaleraDi

italeraDig

Weightage - 25

Input Output


Memory Memory
emoryM
moryMe

oryMem

Weightage - 5

Input Output

https://admin.srm615.examly.io/course 49/51
7/16/25, 10:31 PM SRM PAT


Power Power
owerP
werPo

erPow

Weightage - 5

Sample Input Sample Output


sample sample
amples
mplesa

plesam

Sample Input Sample Output


computer computer
omputerc
mputerco

putercom

Solution

Header Header

#include<bits/stdc++.h> #include<stdio.h>
using namespace std; #include<string.h>

void SameReflection(char str[])

void SameReflection(char str[]) {
{ int len = strlen(str);
int len = strlen(str); char temp[len];
char temp[len]; for (int i = 0; i < len; i++)
for (int i = 0; i < len; i++) {
{ int j = i;
int j = i; int k = 0;
int k = 0; while (str[j] != '\0')
while (str[j] != '\0') {
{ temp[k] = str[j];
temp[k] = str[j]; k++;
k++; j++;
j++; }
} j = 0;
j = 0; while (j < i)
while (j < i) {
{ temp[k] = str[j];
temp[k] = str[j]; j++;
j++; k++;
k++; }
} temp[k]='\0';
temp[k]='\0'; printf("%s\n", temp);
printf("%s\n", temp); }
} }
}

Footer
Footer

int main()
int main() {
{ char str[50];
char str[50]; scanf("%s",str);
cin>>str; SameReflection(str);
SameReflection(str);

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return 0; return 0;
} }

Header

import java.util.*;
class Main
{

public static void SameReflection(char str[])

{
int len = str.length;
char[] temp=new char[len];
for (int i = 0; i < len; i++)
{
int j = i;
int k = 0;
while (j< str.length)
{
temp[k] = str[j];
k++;
j++;
}
j = 0;
while (j < i)
{
temp[k] = str[j];
j++;
k++;
}
System.out.println(temp);
}
}

Footer

public static void main(String[] args)

{
int n,i;
Scanner s=new Scanner(System.in);
String str=s.nextLine();
char [] ch = str.toCharArray();
SameReflection(ch);
}
}

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