Number System PYQs for UPSC (2011-2024) | Solved Questions & PDF Download

Master the Number System for UPSC with solved Previous Year Questions (PYQs) from 2011 to 2024. This page provides detailed solutions, practice questions, and a free downloadable PDF to help you ace your UPSC preparation.

Overview of Number System

The Number System is a fundamental topic in the UPSC exam, especially for the CSAT (Civil Services Aptitude Test) paper. It covers concepts like:

• Divisibility Rules

• Prime and Composite Numbers

• LCM and HCF

• Fractions and Decimals

• Number Series and Patterns

Practicing PYQs helps you understand the exam pattern, difficulty level, and frequently asked questions. Let’s dive into the year-wise questions!

Year-Wise Number System PYQs (2011-2024)

GENERAL QUESTIONS

Q1. A class starts at 11:00 am and lasts till 2:27 pm. Four periods of equal duration are held during this interval. After every period, a rest of 5 minutes is given to the students. The exact duration of each period is [CSAT 2016]

(a) 48 minutes

(b) 50 minutes

(d) 53 minutes

Solution:

Given that,

A class starts at 11:00 am and lasts till 2:27 pm.

Four periods of equal duration are held during this interval.

After every period, a rest of 5 minutes is given to the students.

Now,

Total time = 3 hours 27 minutes = 207 min

Four periods of equal duration are held during, a rest of 5 minutes is given to the students.

Thus total gap is 15 min

Total duration of four classes = (207 - 15) = 192 min

Therefore the exact duration = 192/4 = 48 minutes

Hence, option (a) is correct

Q2. A person is standing on the first step from the bottom of a ladder. If he has to climbs 4 more steps to reach exactly the middle step, how many steps does the ladder have? [CSAT 2016]

(a) 8

(b) 9

(d) 11

Solution:

Given that,

A person is standing on the first step from the bottom of a ladder.

If he has to climbs 4 more steps to reach exactly the middle step

Thus reaches 5th step

Therefore total step = 4 + 1 + 4 = 9

Hence, option (b) is correct

Q3. There are some nectar-filled flowers on a tree and some bees are hovering on it. If one bee lands on each flower, one bee will be left out. If two bees land on each flower, one flower will be left out. The number of flowers and bees respectively are [CSAT 2016]

(a) 2 and 4

(b) 3 and 2

(d) 4 and 3

Solution:

Given that,

If one bee lands on each flower, one bee will be left out.

If two bees land on each flower, one flower will be left out.

Now,

Number of bees > number of flowers

Let bees be A and flowers be B

A = B + 1....(I)

A/2 = B - 1....(II)

Solving equation:

B = 3 and A = 4

Hence, option (c) is correct

Q4. How many numbers are there between 100 and 300 which either begin with or end with 2? [CSAT 2016]

(a) 110

(b) 111

(d) None of the above

Solution:

Numbers between 100 and 300 which either begin with or end with 2

From 100 to 199 = 102, 112, 122, 132, 142, 152......192 total 10 numbers

From 200 to 300 = 100 numbers

Total = 100 + 10 = 110

Hence, option (a) is correct

Q5. How many numbers are there between 99 and 1000 such that the digit 8 occupies the units place? [CSAT 2017]

(a) 64

(b) 80

(d) 104

Solution:

From 99 to 1000 all three digit numbers with 8 as unit's place are

100 to 200 = 10 times

Thus this comes 9 times till 900 to 1000

Therefore 9 x 10 x 1 = 90

Hence, option (c) is correct

Q6. Six boys A, B, C, D, E and F play a game of cards. Each has a pack of 10 cards. F borrows 2 cards from A and gives away 5 to C who in turn gives 3 to B while B gives 6 to D who passes on 1 to E. Then the number of cards possessed by D and E is equal to the number of cards possessed by [CSAT 2017]

(a) A, B and C

(b) B, C and F

(d) A, C and F

Solution:

Given that,

Six boys A, B, C, D, E and F play a game of cards.

Each has a pack of 10 cards.

F borrows 2 cards from A and gives away 5 to C who in turn gives 3 to B while B gives 6 to D who passes on 1 to E.

Now,

A has 8 cards, B has (13 - 6) 7 cards, C has (15 - 3) 12 cards, D has 15 cards, E has 11 cards and F has 7 cards

D + E = 15 + 11 = 26

F + B + C = 26

Hence, option (b) is correct

Q7. There are three pillars X, Y and Z of different heights. Three spiders A, B and C start to climb on these pillars simultaneously. In one chance, A climbs on X by 6 cm but slips down 1 cm. B climbs on Y by 7 cm but slips down 3 cm. C climbs on Z by 6.5 cm but slips down 2 cm. If each of them requires 40 chances to reach the top of the pillars, what is the height of the shortest pillar? [CSAT 2017]

(a) 161 cm

(b) 163 cm

(d) 210 cm

Solution:

Given that,

In one chance, A climbs on X by 6cm but slips down 1cm.

B climbs on Y by 7 cm but slips down 3 cm.

C climbs on Z by 6.5 cm but slips down 2 cm.

Total chance 40

Now,

For A:

Climbs 6cm and slips 1 cm so in one chance it climbs = 6 - 1 = 5cm

So, height of pillar for A = 39 x 5 + 6 (last attempt) = 201 cm = X

For B:

7 - 3 = 4cm

Height of Y = 39 x 4 + 7 = 163 cm

For C:

6.5 - 2 = 4.5 cm

Height of Z = 39 x 4.5 + 6.5 = 182 cm

Therefore the shortest pillar height = 163 cm

Hence option (b) is correct

Q8. X and Y are natural numbers other than 1, and Y is greater than X. Which of the following represents the largest number? [CSAT 2018]

(a) XY

(b) X/Y

(d) (X + Y)/XY

Solution:

Given that,

X and Y are natural numbers other than 1, and Y is greater than X

Considering X = 2 and Y = 3

Thus XY would represent the highest form, rest all are in fraction form.

Hence option (a) is correct

Q9. The number of times the digit 5 will appear while writing the integers from 1 to 1000 is [CSAT 2019]

(a) 269

(b) 300

(d) 302

Solution:

The number of times the digit 5 will appear while writing the integers from 1 to 1000

5 from 1 to 100 = 20 times (55 two times 5)

So, from 1 to 1000 same pattern is repeated = 20 x 10 = 200

500 to 599 = 100 times

Total = 200 + 100 = 300 times

Hence option (b) is correct

Q10. If x is greater than or equal to 25 and y is less than or equal to 40, then which one of the following is always correct? [CSAT 2019]

(a) x is greater than y

(b) (y-x) is greater than 15

(d) (x + y) is greater than or equal to 65

Solution:

Given that,

If x is greater than or equal to 25 and y is less than or equal to 40

Now, y - x is always less than 15 as the difference between x and y would not exceed 15

Option (a) is incorrect as x can be 27 and y can be 29

Option (d) is incorrect as 25 + 27 = 52 is less than 65

Hence option (c) is correct

Q11. If the numerator and denominator of a proper fraction are increased by the same positive quantity which is greater than zero, the resulting fraction is [CSAT 2019]

(a) Always less than the original fraction

(b) Always greater than the original fraction

(d) Such that nothing can be claimed definitely

Solution:

Given that,

If the numerator and denominator of a proper fraction are increased by the same positive quantity

Considering fraction 5/7, if 2 is added = 7/9

7/9 > 5/7

Hence option (b) is correct

Q12. Let p, q, r and s be natural numbers such that p-2016 = q+2017 = r-2018 = s + 2019

Which one of the following is the largest natural number? [CSAT 2020]

(a) p

(b) q

(d) s

Solution:

Given that,

p, q, r and s be natural numbers such that p -2016 = q+2017 = r-2018 = s + 2019

Now,

p - q = 4033 and r - q = 4035

Thus r is greater than p and q.

Also, r - s = 4037

This implies that r is greater than s.

Therefore r is greater than p,q and s

Hence option (c) is correct

Q13. How many integers are there between 1 and 100 which have 4 as a digit but are not divisible by 4?[CSAT 2020]

(a) 5

(b) 11

(d) 13

Solution:

Numbers between 1 and 100 with 4 as digit but not divisible by 4 = 14 34, 41, 42, 43, 45, 46, 47, 49, 54, 74, 94

Therefore total 12 numbers are not divisible by 4

Hence option (c) is correct

Q14. A frog tries to come out of a dried well 4.5 m deep with slippery walls. Every time the frog jumps 30 cm, slides down 15 cm. What is the number of jumps required for the frog to come out of the well? [CSAT 2020]

(a) 28

(b) 29

(d) 31

Solution:

Given that,

A frog tries to come out of a dried well 4.5 m deep with slippery walls.

Every time the frog jumps 30 cm, slides down 15 cm.

Now,

One jump covers = 30 - 15 = 15cm

Total Jumps taken = [(450-30)/15] + 1(last jump) = 29 jumps

Hence option (b) is correct

Q15. The recurring decimal representation 1.272727... is equivalent to [CSAT 2020]

(a)

(b)

(c)

(d)

Solution:

Given that,

recurring decimal representation 1.272727......

Let D = 1.272727

Multiplying 100 both the sides

100D = 127.27....

100D - 126 = 1.27....

100D - D = 126

99D = 126

D = 126/99 = 14/11

Hence option (b) is correct

Q16. For what value of n, the sum of digits in the number (10n+1) is 2? [CSAT 2020]

(a) For n=0 only

(b) For any whole number n

(d) For any real number n

Solution:

(10ⁿ+1) is 2

If n = 0

(1 + 1) = 2

If n = 1

10 + 1 = 11 = 1 + 1 = 2

If n = 2

100 + 1 = 101 = 1 + 0 + 1 = 2

Thus for any whole number value the sum of digits is 2

Hence option (b) is correct

Q17. How many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits?

(a) Zero

(b) One

(d) Ten

Solution:

Given that,

five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits

No prime numbers can be formed as the sum of digits is 15 which is divisible by 3

Hence option (a) is correct

Q18. Which one of the following will have minimum change in its value if 5 is added to both numerator and the denominator of the fractions , , and [CSAT 2020]

(a)

(b)

(c)

(d)

Solution:

Fraction wise change

1st:

2/3 = 0.667

(2 + 5) / (3 + 5) = 7/8 = 0.875

Change= 0.875 - 0.667 = 0.208

2nd:

3/4 = 0.75

(3 + 5) / (4 + 5) = 8/9 = 0.89

Change= 0.89 - 0.75 = 0.14

3rd:

4/5 = 0.8

(4 + 5) / (5 + 5) = 9/10 = 0.9

Change 0.9 - 0.8 = 0.1

4th:

5/6 = 0.84

(5 + 5) / (6 + 5) = 10/11 = 0.9

Change.= 0.9 - 0.84 = 0.06

Hence option (d) is correct

Q19. Integers are listed from 700 to 1000. In how many integers is the sum of the digits 10?[CSAT 2021]

(a) 6

(b) 7

(d) 9

Solution:

From 700 to 1000 the list of integers whose sum of the digits = 10

703, 730, 721, 712, 901, 910, 802, 820, 811

Total 9 combinations

Hence option (d) is correct

Q20. Half of the villagers of a certain village have their own houses. One-fifth of the villagers cultivate paddy. One-third of the villagers are literate. Four-fifth of the villagers are under 25 years of age. Which one of the following statements is certainly correct?

[CSAT 2021]

(a) All the villagers who have their own houses are literate.

(b) Some villagers under 25 years of age are literate.

(d) No villager under 25 years of age has his own house.

Solution:

Given that,

Half of the villagers of a certain village have their own houses.

So, 50% have own house.

One-fifth of the villagers cultivate paddy.

20% cultivate paddy

One-third of the villagers are literate.

33.33% are literate

Four-fifth of the villagers are under 25 years of age.

80% are under 25 years of age

Now, from the given statements

80% are under 25 years of age , so some villagers under age of 25 years are literate

Hence option (b) is correct

Q21. In an objective type test of 90 questions, 5 marks are allotted for every correct answer and 2 marks are deducted for every wrong answer. After attempting all the 90 questions, a student got a total of 387 marks. What is the number of incorrect responses? [CSAT 2021]

(a) 9

(b) 13

(d) 43

Solution:

Given that,

Number of questions in test = 90

Marks for each correct answer = 5

Marks deducted for every incorrect answer = 2

Total marks scored = 387

Let the number of incorrect responses be N

Number of correct responses = 90 - N

So,

5(90 - N) - 2(N) = 387 = 450 - 5N - 2N =387

N = (450 - 387)/ 7 = 9

Hence option (a) is correct

Q22. A person P asks one of his three friends X as to how much money he had. X replied, "If Y gives me ₹40, then Y will have half of as much as Z, but if Z gives me ₹40, then three of us will have equal amount." What is the total amount of money that X, Y and Z have? [CSAT 2021]

(a) ₹420

(b) ₹ 360

(d) ₹270

Solution:

Given that,

If Y gives ₹40, then Y will have half of as much as Z,

if Z gives ₹40, then three of us will have equal amount

Now,

Let X have Rs. a, Y have Rs. b and Z have Rs. c

If Y gives ₹40, then Y will have half of as much as Z,

X = a + 40 and Y = b - 40 = c/2 = Z

b - 40 = c/2......(i)

if Z gives ₹40, then three of us will have equal amount

X = a + 40, Z = c - 40

a + 40 = c - 40.......(ii)

a + 40 = b.......(iii)

b = c - 40 ........(iv)

c/2 + 40 = c - 40.......from (i) and (iv)

c = 160

Then, Y = 120 and X will have 80

Total amount = 120 + 160 + 80 = 360

Hence option (b) is correct

Q23. Jay and Vijay spent an equal amount of money to buy some pens and special pencils of the same quality from the same store. If Jay bought 3 pens and 5 pencils, and Vijay bought 2 pens and 7 pencils, then which one of the following is correct?[CSAT 2021]

(a) A pencil costs more than a pen

(b) The price of a pencil is equal to that of a pen

(d) The price of a pen is three times the price of a pencil

Solution:

Given that,

Jay bought 3 pens and 5 pencils

Vijay bought 2 pens and 7 pencils

Jay and Vijay spent equal amount

Now,

Let the price of pen be a and pencil be b

Jay = 3a + 5b

Vijay = 2a + 7b

They spent equal amount

3a + 5b = 2a + 7b......(i)

a = 2b

Price of pen is twice the price of pencil

Hence option (c) is correct

Q24. A biology class at high school predicted that a local population of animals will double in size every 12 years. The population at the beginning of the year 2021 was estimated to be 50 animals. If P represents the population after n years, then which one of the following equations represents the model of the class for the population? [CSAT 2021]

(a) P = 12 + 50n

(b) P = 50+ 12n

(d) P = 50 (2)^n/12

Solution:

Given that,

Local population of animals will double in size every 12 years.

The population at the beginning of the year 2021 = 50 Animals

P represents the population after n years

Now,

As population of animals will double in size.

So P = 50 x 2ⁿ

It doubles in every 12 years

When n = 0 , P = 50

When n = 12, P should be double

P = 50 x (2)^n/12

Hence option (d) is correct

Q25. On one side of a 1.01 km long road, 101 plants are planted at equal distance from each other. What is the total distance between 5 consecutive plants? [CSAT 2022]

(a) 40 m

(b) 40.4 m

(d) 50.5 m

Solution:

Given that,

Road length = 1.01 km or 1010 m

101 plants are planted at equal distance from each other.

Now,

101 plant planted in 100 parts

Distance = 1010 m

1010/100 = 10.1 m distance each between the plants

For 5 consecutive plants, the total parts would be 4, as distance is measured from the first

So, Total distance between 5 consecutive plants = 4 x 10.1 = 40.4m

Hence option (b) is correct

Q26. The sum of three consecutive integers is equal to their product. How many such possibilities are there? [CSAT 2022]

(a) Only one

(b) Only two

(d) No such possibility is there

Solution:

Given that,

The sum of three consecutive integers is equal to their product.

Now,

let three consecutive integers be 1, 2 and 3

So, 1 + 2 + 3 = 1 x 2 x 3

Thus considering other possibilities

Let three consecutive number be a - 1, a, a + 1

a - 1 + a + a + 1 = (a - 1) x (a) x (a + 1)

3a = a³ - a

4a = a³

a = + 2 or -2

or a = 0

Thus there are three possible values

Hence option (c) is correct

Q27. An Identity Card has the number ABCDEFG, not necessarily in that order, where each letter represents a distinct digit (1, 2, 4, 5, 7, 8, 9 only). The number is divisible by 9. After deleting the first digit from the right, the resulting number is divisible by 6. After deleting two digits from the right of original number, the resulting number is divisible by 5. After deleting three digits from the right of original number, the resulting number is divisible by 4. After deleting four digits from the right of original number, the resulting number is divisible by 3. After deleting five digits from the right of original number, the resulting number is divisible by 2. Which of the following is a possible value for the sum of the middle three digits of the number? [CSAT 2022]

(a) 8

(b) 9

(d) 12

Solution:

Given that,

The number has 7 digits, and has been denoted by: ABCDEFG

The letters can be replaced by 1, 2, 4, 5, 7, 8, 9, not necessarily in the same order.

Now,

Deducing the given information:

The original number (ABCDEFG) is divisible by 9.

So, It has to be as 1 + 2 + 4 + 5 + 7 + 8 + 9 = 36 which is divisible by 9.

After deleting 1 digit from the right i.e. G, the resulting number (ABCDEF) is divisible by 6,

So, It means that, F can be 2, 4 or 8 (i.e, an even number) by divisibility rule of 6. Thus the number should be divisible by 2 and 3 both. So, the remaining number is divisible by 3, then it means G = 9

After deleting 2 digits from the right, the resulting number (ABCDE) is divisible by 5.

So, E = 5

G = 9 and E = 5

After deleting 3 digits from the right, the resulting number (ABCD) is divisible by 4.

So, D can be 2, 4 or 8 (ie. an even number).

After deleting four digits from the right of original number, the resulting number is divisible by 3.

So the sum of first three numbers is divisible by 3

After deleting 5 digits from the right, the resulting number (AB) is divisible by 2.

So, B can be 2, 4 or 8 (e. an even number).

Thus F, D and B are even numbers 2, 4 and 8 and, A, C, E and G are odd numbers (1, 5, 7or 9)

7412589 or 1472589 are the possible values

So the sum of middle digits = 1 + 2 + 5 = 8 or 7 + 2 + 5 = 14

Hence option (a) is correct

Q28. A person X wants to distribute some pens among 6 children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number? [CSAT 2022]

(a) 147

(b) 150

(d) 300

Solution:

Given that,

X distribute some pens among 6 children A, B, C, D, E and F.

A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F.

Now,

Let A gets P pens

B = P/2, C = P/3, D = P/4, E = P/5 and F = P/6

Thus LCM of 2, 3, 4, 5 and 6 = 60

P = 60, 120, 180.....(Possible values)

The pen each should get must be in even numbers

Taking P = 120 minimum possible value

A = 120, B = 60, C = 40, D = 30, E = 24 and F = 20

Total = 120 + 60 + 40 + 30 + 24 + 20 = 294

Hence option (c) is correct

Q29. D is a 3-digit number such that the ratio of the number to the sum of its digits is least. What is the difference between the digit at the hundred's place and the digit at the unit's place of D? [CSAT 2023]

(a) 0

(b) 7

(d) 9

Solution:

Given that,

D is a 3-digit number such that the ratio of the number to the sum of its digits is least.

Now,

To minimize the ratio of number to the sum of the digits, the sum of the digits must be maximum and the number should be minimum

100 is the smallest three digit number, however the sum of digits is not maximum

199, sum of digits = 19 and ratio = 199/19 = 10.47

Thus 199 is the smallest possible value of three digit number

Difference between units place and Hundreds place = 9 - 1 = 8

Hence option (c) is correct

Q30. If p, q, r and s are distinct single digit positive numbers, then what is the greatest value of (p+q) (r + s)? [CSAT 2023]

(a) 230

(b) 225

(d) 221

Solution:

Given that,

p, q, r and s are distinct single digit positive numbers

Now,

Taking the value of p, q, r and s = 9, 8, 7, and 6

(p+q) (r + s) = (9 + 8) x (7 + 6) = 221

interchanging the values

(9 + 6) x (8 + 7) = 225 (maximum possible value)

Hence option (b) is correct

Q31. What is the sum of all 4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4, where none of the digits is repeated?[CSAT 2023]

(a) 7998

(b) 8028

(d) 9238

Solution:

Given that,

4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4

Digits must not be repeated

Now,

From 1000 to 2000

1234, 1243, 1324, 1342, 1423, and 1432.

Sum = 1234 + 1243 + 1324 + 1342 + 1423 + 1432 = 7998

Hence option (a) is correct

Questions based on rule of counting principles ∑N, ∑N², ∑N³

Q32. What is the total number of digits printed, if a book containing 150 pages is to be numbered from 1 to 150? [CSAT 2017]

(a) 262

(b) 342

(d) 450

Solution:

Given that,

A book containing 150 pages is to be numbered from 1 to 150

Now,

One digit print = 9

Two digit print = 2 x 90 = 180

Three digit print = 3 x 51 = 153

Total print = 9 + 180 + 153 = 342

Hence option (b) is correct

Q33. While writing all the numbers from 700 to 1000, how many numbers occur in which the digit at hundred's place is greater than the digit at ten's place, and the digit at ten's place is greater than the digit at unit's place? [CSAT 2018]

(a) 61

(b) 64

(d) 91

Solution:

Given that,

From 700 to 1000, the digit at hundred's place is greater than the digit at ten's place, and the digit at ten's place is greater than the digit at unit's place.

Now,

Considering 700 to 799

710, 720, 721, 730, 731, 732, 740, 741, 742, 743, 750, 751, 752, 753,754, 760, 761, 762, 763, 764, 765 or other possible way

710 = 1, 720 series = 2, 730 series = 3, 740 series = 4, 750 series = 5 and 760 series = 6

1 + 2 + 3 + 4 + 5 + 6 = 21 possible numbers

Considering 800 to 899

The series would repeat in same pattern

1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 possible numbers

900 to 999

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

Total possible numbers = 36 + 28 + 21 = 85

Hence option (c) is correct

Q34. A printer numbers the pages of a book starting with 1 and uses 3089 digits in all. How many pages does the book have? [CSAT 2019]

(a) 1040

(b) 1048

(d) 1050

Solution:

Given that,

Printer numbers the pages of a book starting with 1 and uses 3089 digits in all.

Now,

From 1- 9

The single-digit page =9 digit

Double digit page = 2 x 90 = 180

Three digit page = 3 x 900 = 2700

Total digits = 9 + 180 + 2700 = 2889

Thus, remaining digit = 3089 - 2889 = 200

The number of page with four digits= 200/4 = 50

Total page = 999 + 50=1049

Hence option (c) is correct

Q35. One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 195. The torn page contains which of the following numbers? [CSAT 2020]

(a) 5, 6

(b) 7, 8

(d) 11, 12

Solution:

Given that,

One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1.

The sum of the numbers on the remaining pages is 195.

Now,

Let torn page be P

Total page = P + 195

Sum of first n natural numbers = (n/2) (n + 1)

P + 195 = (n/2) (n + 1)

Hit and trial method

The value of n must be such that the sum should be near 195

n = 20, sum = 210

P + 195 = 210

P = 15

Hence option (b) is correct

Q36. In the series AABABCABCDABCDE..., which letter appears at the 100th place? [CSAT 2022]

(a) G

(b) H

(d) J

Solution:

Given that,

AABABCABCDABCDE...,

Now,

The series continue in format

1 + 2 + 3 + 4 .......

To find 100th position

The sum of first 13 natural numbers = 91

Thus from 92nd position series again start with A

So at 100th position it is "I"

Hence option (c) is correct

Q37. What is the sum of all digits which appear in all the integers from 10 to 100? [CSAT 2023]

(a) 855

(b) 856

(d) 911

Solution:

Given that,

Sum of all digits which appear in all the integers from 10 to 100

Now,

For tens digit

10 to 19 Tens digit = 1

20 - 29 tens digit = 2

30 to 39 = 3....so on till 90 to 99 = 9

This appears ten times

So, (1 + 2 +3....+ 9) x 10 = 45 x 10 = 450

For unit digits

0 to 9 appears 9 times

So, (0 + 1 + 2 + 3 + 4....+ 9) x 9 = 45 x 9 = 405

Total sum of digits = 405 + 450 = 855

Last digit 100 = 1 + 0 + 0 = 1

855 + 1 = 856

Hence option (b) is correct

Q38. 40 children are standing in a circle and one of them (say child-1) has a ring. The ring is passed clockwise. Child-1 passes on to child-2, child-2 passes on to child-4, child-4 passes on to child- 7 and so on. After how many such changes (including child-1) will the ring be in the hands of child-1 again? [CSAT 2023]

(a) 14

(b) 15

(d) 17

Solution:

Given that,

40 children are standing in a circle and one of them (say child-1) has a ring.

The ring is passed clockwise.

Child-1 passes on to child-2, child-2 passes on to child-4, child-4 passes on to child- 7 and so on.

Now,

Ring is passed in following sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, 6, 16, 27, 39, 12, 26, 1.

Decoding the sequence: 1, 2, 3 ,4 , .....15

The sequence is formed by adding the given numbers

1 + 1 = 2, 2 + 2 = 4, 4 + 3 = 7, 7 + 4 = 11.......26 + 15 = 41 which will go the child 1 again

Therefore 15 changes are required

Hence option (b) is correct

Q39. What is the middle term of the sequence Z, Z, Y, Y, Y, X, X, X, X, W, W, W, W, W, ..., A? [CSAT 2023]

(a) H

(b) I

(d) M

Solution:

Given that,

The sequence Z, Z, Y, Y, Y, X, X, X, X, W, W, W, W, W, ..., A

Now

Sequence follows: 2 + 3 + 4 .......+ 27

Total sum = [(n)(n + 1)/2 ] - 1 = 377

Middle term = 377/2 = 188.5 or 189th term

A + B + C + D + E + F + G + H = 27 + 26 + 25 + 24 + 23 + 22 + 21 + 20 = 188

So 189th term is I

Hence option (b) is correct

Questions based on 2 digits and 3 digits numbers

Q40. Certain 3-digit numbers have the following characteristics: [CSAT 2017]

1. All the three digits are different.

2. The number is divisible by 7.

3. The number on reversing the digits is also divisible by 7.

How many such 3-digit numbers are there? [CSAT 2017]

(a) 2

(b) 4

(d) 8

Solution:

Given that,

A three digit number

1. All the three digits are different.

2. The number is divisible by 7.

3. The number on reversing the digits is also divisible by 7.

Now,

Let 3 digit number be in the form: 100x + 10y + z

on reversing = 100z + 10y + x

Difference between the digits = (100x + 10y + z) - (100z + 10y + x) = 99 (x - z) = 7a(difference must be 7 or multiple of 7)

So, 99 (x - z) is divisible by 7

As 99 is not divisible, therefore x -z is divisible by 7

(x - z) are single digits thus the difference must be 7

Possibilities: (9,2) or (8,1)

For (9,2)

Numbers are 952 or 259

For (8,1)

Numbers are 861 or 168

Therefore total 4 possibilities.

Hence option (b) is correct

Q41. A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible remainder? [CSAT 2017]

(a) 9

(b) 27

(d) 45

Solution:

Given that,

A 2-digit number is reversed.

The larger of the two numbers is divided by the smaller one.

Now,

Let two digit number be 10a + b

Reversing the digits : 10b + a

10a + b > 10b + a

The number with highest form must have 9

10a + b = 90 + b

10b + a = 10b + 9

Thus number must be 5 or 4

Hence option (d) is correct

Q42. There are certain 2-digit numbers. The difference between the number and the one obtained on reversing it is always 27. How many such maximum 2-digit numbers are there [CSAT 2017]

(a) 3

(b) 4

(d) None of these

Solution:

Given that,

There are certain 2-digit numbers.

The difference between the number and the one obtained on reversing it is always 27.

Now,

Let two digit number be 10a + b

Reversing the digits : 10b + a

Difference = 10a + b - 10b -a = 27

9a - 9b = 27

a - b = 3

Possible numbers = 41, 52, 63, 74, 85 and 96

Total 6 numbers

Hence option (d) is correct

Q43. A number consists of three digits of which the middle one is zero and their sum is 4. If the number formed by interchanging the first and last digits is greater than the number itself by 198, the difference between the first and last digits is [CSAT 2018]

(a) 1

(b) 2

(d) 4

Solution:

Given that,

A number consists of three digits of which the middle one is zero and their sum is 4.

The number formed by interchanging the first and last digits is greater than the number itself by 198.

Now,

Let the number be 100a + 10b + c Then the number of reversed digit will be 100c + 10b + a

100c + a - (100a + c) = 198 .......(i)

99c - 99a = 198

c - a = 2....(ii)

The difference = 2.

Hence option (b) is correct

Q44. The ratio of a 2-digit natural number to a number formed by reversing its digits is 4: 7. The number of such pairs is [CSAT 2019]

(a) 5

(b) 4

(d) 2

Solution:

Given that,

The ratio of a 2-digit natural number to a number formed by reversing its digits is 4: 7.

Now,

Let two digit number be 10a + b

Reversing the digits : 10b + a

(10a + b)/(10b + a) = 4/7

70a + 7b = 40b + 4a

66a = 33b

a/b = 1/2

Possible numbers are: 12, 24, 36, and 48

Hence option (b) is correct

Q45. Let p be a 2-digit number and q be the number consisting of same digits written in reverse order. If px q = 2430, then what is the difference between p and q? [CSAT 2022]

(a) 45

(b) 27

(d) 9

Solution:

Given that,

p be a 2-digit number and q be the number consisting of same digits written in reverse order.

p x q = 2430

So one digit must be 5 as the product ending is with 0

Now,

Let two digit number be 10a + 5 = p.....(i)

Reversing the digits: 50 + a = q......(ii)

(10a + 5) x (50 + a) = 2430

500a + 250 + 10a² + 5a = 2430

10a² + 505a = 2180 ........(iii)

2a² + 101a = 436 (dividing by 5).....(iv)

2a² + 101a - 436 = 0

2 x 436 = 872

a(2a + 109) - 4(2a + 109) =0

(a - 4) (2a + 109) = 0

Thus the possible value of a = 4

Possible number = 45 or 54

So p = 45, q = 54

q - p = 54 - 45 = 9

Hence option (d) is correct

Unit Digit of X^y (Rule of Cyclicity)

Q46. If 3²⁰¹⁹ is divided by 10, then what is the remainder? [CSAT 2021]

(a) 1

(b) 3

(d) 9

Solution:

Given that,

3²⁰¹⁹ is divided by 10

Now,

If Power of 3 is 4, 8, 16....or multiple of 4, then the unit's digit is 1

So, 3²⁰¹⁹ = 3^{4 x 504} x 3³ or 3^{4 x 504} x 27

Now the units place would be 7 as last digit is 27

Thus remainder is 7 when divided by 10

Hence option (c) is correct

Q47. What is the unit digit in the expansion of (57242)^{9×7×5×3×1}? [CSAT 2023]

(a) 2

(b) 4

(d) 8

Solution:

Given that,

Expansion of (57242)^{9×7×5×3×1}

Now,

Considering cyclicity of power of 2

2 repeats in units digit after every 4 cycles

So, (57242)^{9×7×5×3×1}

Power value = 9 x 7 x 5 x 3 x 1 = 945

So, 2⁹⁴⁵, using cyclicity of 2

945/4 = 236 and remainder is 1

Thus 2 will repeat again in unit digit

Therefore unit digit of expansion = 2

Hence option (a) is correct

Exponents and Number of Zero

Q48. How many zeroes are there at the end of the following product? 1 X 5 X 10 X 15 X 20 X 25 X 30 X 35 X 40 X 45 X 50 X 55 X 60 [CSAT 2020]

(a) 10

(b) 12

(d) 15

Solution:

Given that,

1 X 5 X 10 X 15 X 20 X 25 X 30 X 35 X 40 X 45 X 50 X 55 X 60

Now number of 2's and 5's

2⁴ x 5⁶ x 10⁶

So there are only 4 2's

Thus total zeros = 4 + 6 = 10

Hence option (a) is correct

Q49. If 15 X 14 X 13 X … X 3 X 2 X 1 = 3^M X n , where m and n are positive integers, then what is the maximum value of m? [CSAT 2022]

(a) 7

(b) 6

(d) 4

Solution:

Given that,

15 X 14 X 13 X … X 3 X 2 X 1 = 3^M x n

or, 15! = 3^M x n

Now,

Numbers divisible by 3 in 15! = 3, 6, 9, 12 and 15

3 x 1 = 3¹

3 x 2 = 3¹

3 x 3 = 3²

3 x 4 = 3¹

3 x 5 = 3¹

Thus M = 1 + 1 + 2 + 1 + 1 = 6

Hence option (b) is correct

Divisibility

Q50. If ABC × DEED = ABCABC; where A, B, C, D and E are different digits, what are the values of D and E? [CSAT 2015]

(a) D=2, E=0

(b) D=0, E=1

(d) D=1, E=2

Solution:

Given that,

ABC × DEED = ABCABC

A, B, C, D and E are different digits

Now,

When a three digit number ABC is multiplied by 1001

ABC x 1001 = ABCABC

Thus DEED = 1001, Where D = 1 and E = 0

Hence option (c) is correct

Q51. If R and S are different integers both divisible by 5, then which of the following is not necessarily true? [CSAT 2016]

(a) R-S is divisible by 5

(b) R+S is divisible by 10

(d) R2 x S² is divisible by 5

Solution:

Given that,

R and S are different integers both divisible by 5

Now,

Let R = 5a and S = 5b

From the given options

R - S = 5a - 5b = 5 x (a - b), thus it is divisible by 5

R + S = 5a + 5b = 5 x (a + b), thus it is not necessarily true as it can or cannot be divisible by 10

Hence option (b) is correct

Q52. An 8-digit number 4252746B leaves remainder 'O' when divided by 3. How many values of B are possible? [CSAT 2019]

(a) 2

(b) 3

(d) 6

Solution:

Given that,

8-digit number 4252746B leaves remainder '0' when divided by 3

Now,

Divisibility rule of 3

4 + 2 + 5 + 2 + 7 + 4 + 6 + B = 30 + B

B can be 0, 3, 6 or 9

Thus 4 values are possible.

Hence option (c) is correct

Q53. Two Statements are given followed by two Conclusions: [CSAT 2020]

Statements: All numbers are divisible by 2.

All numbers are divisible by 3.

Conclusion-I: All numbers are divisible by 6.

Conclusion-II: All numbers are divisible by 4.

Which of the above Conclusions logically follows/follow from the two given Statements

(a) Only Conclusion-I

(b) Only Conclusion-II

(d) Both Conclusion-I and Conclusion-II

Solution:

Given that,

Statements:

All numbers are divisible by 2.

All numbers are divisible by 3.

Now,

Conclusion-I: All numbers are divisible by 6.

All numbers are divisible by 6 as LCM of 2 and 3 is 6.

Hence conclusion 1 is correct

Conclusion-II: All numbers are divisible by 4.

This is not true as all number divisible by 4 is also divisible by 2 but not all numbers are divisible by 3

Hence conclusion 2 is incorrect

Q54. A digit n > 3 is divisible by 3 but not divisible by 6. Which one of the following is divisible by 4? [CSAT 2020]

(a) 2n

(b) 3n

(d) 3n+1

Solution:

Given that,

A digit n > 3 is divisible by 3 but not divisible by 6.

Now,

Let the value of n = 9

So by hit and trial method

Option (a)

taking 2n, 2 x 9 = 18, which is not divisible by 4

Option (b)

3n = 3 x 9 = 27, which is not divisible by 4

option (c)

2n +4 = 2 x 9 + 4 = 22, which is not divisible by 4

Option(d)

3n+1 = 3 x 9 + 1 = 28, which is divisible by 4

Hence option (d) is correct

Q55. Let XYZ be a 3-digit number, where (X+Y+ Z) is not a multiple of 3. Then (XYZ + YZX + ZXY) is not divisible by [CSAT 2020]

(a) 3

(b) 9

(d) (X + Y + Z)

Solution:

Given that,

XYZ be a 3-digit number

(X+Y+ Z) is not a multiple of 3.

Now,

Three digit format:

XYZ = 100X + 10Y + Z.......(i)

YZX = 100Y + 10Z + X.......(ii)

ZXY = 100Z + 10X + Y........(iii)

Adding all three

XYZ + YZX + ZXY = (100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + y)

XYZ + YZX + ZXY = 111X + 111Y + 111Z

XYZ + YZX + ZXY = 111(X + Y + Z).....(iv)

Thus 111 is the common factor.

So, XYZ + YZX + ZXY is divisible by 111 and it's factors are 3 and 37.

Also, (X + Y + Z) is not multiple of 3

Thus (XYZ + YZX + ZXY) is not divisible by 9.

Hence option (b) is correct

Q56. When a certain number is multiplied by 7, the product entirely comprises ones only (1111...). What is the smallest such number?

[CSAT 2021]

(a) 15713

(b) 15723

(d) 15873

Solution:

Given that,

a certain number is multiplied by 7, the product entirely comprises ones only (1111...).

Now,

By hit and trial on options

11/7 = 1.571

111/7 = 15.857

1111/7 = 158.714

11111/7 = 1587.28

111111/7 = 15873

Hence option (d) is correct

Q57. The number 3798125P369 is divisible by 7. What is the value of the digit P? [CSAT 2021]

(a) 1

(b) 6

(d) 9

Solution:

Given that,

The number 3798125P369 is divisible by 7.

Now,

Three digit number blocks are formed from right to left.

369 ,25P,981,037

Adding and subtracting the numbers formed alternatively

369 - 25P + 981 - 37 = 369 - 250 - P + 981 - 37 = 1063 - p

Thus number 1063 - p must be divisible by 7

151 x 7 + 6 - p

So, 6 - p = 0 or p = 6

Hence option (b) is correct

Q58. For any choices of values of X, Y and Z, the 6-digit number of the form XYZXYZ is divisible by [CSAT 2023]

(a) 7 and 11 only

(b) 11 and 13 only

(d) 7, 11 and 13

Solution:

Given that,

For any choices of values of X, Y and Z, the 6-digit number of the form XYZXYZ

Now,

XYZ x 1001 = XYZXYZ

Where 1001 is divisible by 7, 11 and 13

Hence option (d) is correct

Q59. A number N is formed by writing 9 for 99 times. What is the remainder if N is divided by 13? [CSAT 2023]

(a) 11

(b) 9

(d) 1

Solution:

Given that,

A number N is formed by writing 9 for 99 times.

Now

When 9 is divided by 13 no remainder

9/13 = no remainder

99/13 = 8 remainder

999/13 = 11 remainder

9999/13 = 2 remainder

99999/13 = 3 remainder

999999/13 = 0 remainder

Then cycle repeats

Thus 9 written 99 times

So, 99/6 (as cycle repeats after 6) = remainder 3

Thus when 9 written 99 times and divided by 13 the remainder is 11 as it follows the cycle of 3 times 9

Hence option (a) is correct

Q60. Each digit of a 9-digit number is 1. It is multiplied by itself. What is the sum of the digits of the resulting number? [CSAT 2023]

(a) 64

(b) 80

(d) 100

Solution:

Given that,

Each digit of a 9-digit number is 1.

It is multiplied by itself.

Now,

111111111 x 111111111 = 12345678987654321

Thus sum of digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 81

Hence option (c) is correct

Remainder Theorem

Q61. What is the remainder when 51 x 27 x 35 x 62 x 75 is divided by 100? [CSAT 2020]

(a) 50

(b) 25

(d) 1

Solution:

Given that,

51 X 27 X 35 X 62 X 75 is divided by 100

Now,

Factorizing

51 = 3 x 17, 27 = 3³, 35 = 5 x 7, 62 = 2 x 31, 75 = 5² x 3

Unit digits = 3, 5, 7, 1 and 10

3 x 5 x 7 x 10 = 1050/100 = remainder 50

Thus remainder = 50

Hence option (a) is correct

Q62. If 3²⁰¹⁹ is divided by 10, then what is the remainder? [CSAT 2021]

(a) 1

(b) 3

(d) 9

Solution:

Given that,

3²⁰¹⁹ is divided by 10

Now,

If Power of 3 is 4, 8, 16....or multiple of 4, then the unit's digit is 1

So, 3²⁰¹⁹ = 3^{4 x 504} x 3³ or 3^{4 x 504} x 27

Now the units place would be 7 as last digit is 27

Thus remainder is 7 when divided by 10

Hence option (c) is correct

Q63. What is the remainder when 91 x 92 x 94 x 95 x 96 x 97 x 98 x 99 is divided by 1261? [CSAT 2022]

(a) 3

(b) 2

(d) 0

Solution:

Given that,

91 X 92 X 94 X 95 X 96 X 97 X 98 X 99 is divided by 1261

Now,

Factors of 1261 = 13 and 97

When 91 X 92 X 94 X 95 X 96 X 97 X 98 X 99/13 x 97

Thus it is completely divisible by1261

Remainder = 0

Hence option (d) is correct

Q64. What is the remainder when 85 x 87 x 89 x 91 x 95 x 96 is divided by 100? [CSAT 2023]

(a) 0

(b) 1

(d) 4

Solution:

Given that,

85 X 87 X 89 X 91 X 95 X 96 is divided by 100

Now,

Number of zeroes = 5² x 2² = 100

Thus it is completely divisible by 100

Hence option (a) is correct

Q65. If today is Sunday, then which day is it exactly on 10¹⁰ th day? [CSAT 2023]

(a) Wednesday

(b) Thursday

(d) Saturday

Solution:

Given that,

Today is Sunday

Now on 10¹⁰th day = 10¹⁰/7 (dividing 10 by 7 the remainder is 3) = 3¹⁰/7 (the remainder becomes the base) = (3²)⁵/7 = 9⁵/7 = 2⁵/7 (the remainder becomes the base) = 32/7 = 4 odd days

So Sunday + 4 odd days = Wednesday as (Sunday is included as 1^st odd day)

Hence option (a) is correct

Q66. What is the remainder if 2¹⁹²is divided by 6? [CSAT 2023]

(a) 0

(b) 1

(d) 4

Solution:

Given that,

2¹⁹²is divided by 6

Now,

Power of 2 repeats after every 4

As 192 is multiple of 4

So, 2⁴/6 = Remainder = 4

Hence option (d) is correct

HCF and LCM

Q67. Three persons start walking together and their steps measure 40 cm, 42 cm and 45 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps? [CSAT 2011]

(a) 25 m 20 cm

(b) 50 m 40 cm

(d) 100 m 80 cm

Solution:

Given that,

Three persons start walking together and their steps measure 40 cm, 42 cm and 45 cm respectively

Now,

LCM of 40 cm, 42 cm and 45 cm = 2520 cm

Thus, 25 m 20 cm

Hence option (a) is correct

Q68. There are five hobby clubs in a college viz, photography, yachting, chess, electronics and gardening. The gardening group meets every second day, the electronics group meets every third day, the chess group meets every fourth day, the yachting group meets every fifth day and the photography group meets every sixth day. How many times do all the five groups meet on the same day within 180 days? [CSAT 2013]

(a) 3

(b) 5

(d) 18

Solution:

Given that,

There are five hobby clubs in a college viz, photography, yachting, chess, electronics and gardening. The gardening group meets every second day, the electronics group meets every third day, the chess group meets every fourth day, the yachting group meets every fifth day and the photography group meets every sixth day.

Now,

LCM of 2, 3, 4, 5 and 6 = 60

Thus number of times all groups meet within 180 days = 180/60 = 3 times

Hence option (a) is correct

Q69. A bell rings every 18 minutes. A second bell rings every 24 minutes. A third bell rings every 32 minutes. If all the three bells ring at the same time at 8 o'clock in the morning, at what other time will they all ring together? [CSAT 2014]

(a) 12:40 hrs

(b) 12:48 hrs

(d) 13:04 hrs

Solution:

Given that,

A bell rings every 18 minutes.

A second bell rings every 24 minutes.

A third bell rings every 32 minutes

all the three bells ring at the same time at 8 o'clock in the morning.

Now,

LCM of 18, 24 and 32 = 288 min or 4 hours 48 min

Thus after 8 they will again ring at = 8 + 4 hr 48 min = 12: 48 PM

Hence option (b) is correct

Q70. Five persons fire bullets at a target at an interval of 6, 7, 8, 9 and 12 seconds respectively. The number of times they would fire the bullets together at the target in an hour is [CSAT 2014]

(a) 6

(b) 7

(d) 9

Solution:

Given that,

Five persons fire bullets at a target at an interval of 6, 7, 8, 9 and 12 seconds respectively

Now,

LCM of 6, 7, 8, 9 and 12 = 504 sec

Number of times bullet fired = 3600/504 = 7.14

Approximately 7

Hence option (b) is correct

Q71. There are five hobby clubs in a college photography, yachting, chess, electronics and gardening. The gardening group meets every second day, the electronics group meets every third day, the chess group meets every fourth day, the yachting group meets every fifth day and the photography group meets every sixth day. How many times do all the five groups meet on the same day within 180 days? [CSAT 2016]

(a) 5

(b) 18

(d) 3

Solution:

Given that,

Now,

LCM of 2, 3, 4, 5 and 6 = 60

Thus number of times all groups meet within 180 days = 180/60 = 3 times

Hence option (d) is correct

Q72. A lift has the capacity of 18 adults or 30 children. How many children can board the lift with 12 adults? [CSAT 2018]

(a) 6

(b) 10

(d) 15

Solution:

Given that,

A lift has the capacity of 18 adults or 30 children.

Now,

LCM of 18 and 30 = 90a

1 adult value = 90a/18 = 5a

1 child value = 90a/30 = 3a

Total adult on lift = 12

Total value = 12 x 5a = 60a

Capacity of lift = 90a

90a - 60a = 30a

Total children can accommodate = 30a/3a = 10

Hence option (b) is correct

Q73. In a school every student is assigned a unique identification number. A student is a football player if and only if the identification number is divisible by 4, whereas a student is a cricketer if and only if the identification number is divisible by 6. If every number from 1 to 100 is assigned to a student, then how many of them play cricket as well as football? [CSAT 2019]

(a) 4

(b) 8

(d) 12

Solution:

Given that,

Every student is assigned a unique identification number.

A student is a football player if and only if the identification number is divisible by 4,

a student is a cricketer if and only if the identification number is divisible by 6.

Every number from 1 to 100 is assigned to a student

Now,

LCM of 4 and 6 = 12

Number divisible by 12 from 1 to 100 = 12, 24, 36, 48, 60, 72, 84 and 96

Thus there are 8 students who will play cricket and football.

Hence option (b) is correct

Q74. Seeta and Geeta go for a swim after a gap of every 2 days and every 3 days respectively. If on 1st January both of them went for a swim together, when will they go together next? [CSAT 2019]

(a) 7th January

(b) 8th January

(d) 13th January

Solution:

Given that,

Seeta and Geeta go for a swim after a gap of every 2 days and every 3 days respectively.

On 1st January both of them went for a swim together

Now,

Seeta goes after gap of two days: 1st Jan, 4th Jan, 7th Jan, 10th Jan, 13th Jan

Geeta goes after gap of three days: 1st Jan, 5th Jan, 9th Jan, 13th Jan

Thus both meet again on 13th Jan

Hence option (d) is correct

Q75. If you have two straight sticks of length 7.5 feet and 3.25 feet, what is the minimum length can you measure? [CSAT 2020]

(a) 0.05 foot

(b) 0.25 foot

(d) 3.25 feet

Solution:

Given that,

Two straight sticks of length 7.5 feet and 3.25 feet

Now,

For minimum length,

HCF of 7.5 feet and 3.25 feet = 0.25

Thus minimum length = 0.25 feet

Hence option (b) is correct

Q76. What is the least 4-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case? [CSAT 2020]

(a) 1012

(b) 1022

(d) 1222

Solution:

Given that,

The least 4-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2

Now,

LCM of 3, 4, 5 and 6 = 60

Least four digit number = 1000

So, 1000/60 = remainder 40

So least four digit number divisible by 3, 4, 5 and 6 = 1000 + (60 - 40) = 1020

Now for remainder 2 = 1020 + 2 = 1022

Alternate method: Use hit and trial from the given options

Hence option (b) is correct

Q77. What is the greatest length x such that 3 and 8 are integral multiples of x? [CSAT 2020]

(a) 1

(b) 1

(d) 1

Solution:

Given that,

3 and 8 are integral multiples of x

Now,

3 = 7/2 and 8 = 35/4

So, HCF of 7/2 and 35/4

Numerator HCF of 7 and 35 = 7

Denominator LCM of 2 and 4 = 4

Thus 7/4 or 1 is the maximum or greatest length

Hence option (d) is correct

Q78. Joseph visits the club on every 5th day, Harsh visits on every 24th day, while Sumit visits on every 9th day. If all three of them met at the club on a Sunday, then on which day will all three of them meet again? [CSAT 2021]

(a) Monday

(b) Wednesday

(d) Sunday

Solution:

Given that,

Joseph visits the club on every 5th day

Harsh visits on every 24th day

Sumit visits on every 9th day.

All three of them met at the club on a Sunday

Now,

LCM of 5, 9 and 24 = 360

Thus all meet on 360th day

1st time is on Sunday, thus Sunday occurs every 7 days

51 x 7 + 3 = Sunday + 3 = Wednesday

They will meet on Wednesday

Hence option (b) is correct

Q79. What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3? [CSAT 2022]

(a) 1063

(b) 1073

(d) 1183

Solution:

Given that,

The smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3

Now,

LCM of 6, 9, 12, 15, 18 = 1080

Therefore to leave remainder 3 the number = 1080 + 3 = 1083

Hence option (c) is correct

Q80. There are three traffic signals. Each signal changes colour from green to red and then from red to green. The first signal takes 25 seconds, the second signal takes 39 seconds and the third signal takes 60 seconds to change the colour from green to red. The durations for green and red colours are same. At 2:00 pm, they together turn green. At what time will they change to green next, simultaneously? [CSAT 2023]

(a) 4:00 pm

(b) 4:10 pm

(d) 4:30 pm

Solution:

Given that,

Each signal changes colour from green to red and then from red to green.

The first signal takes 25 seconds, the second signal takes 39 seconds and the third signal takes 60 seconds to change the colour from green to red.

The durations for green and red colours are same.

At 2:00 pm, they together turn green.

Now,

Time taken by each signal is twice the time taken to change the colour

Thus first signal = 50 sec, second signal = 78 sec, third signal = 120 sec

LCM of 50, 78 and 120 = 7800 sec or 2hr 10min

So next change to green = 2:00pm + 2hr 10min = 4: 10 pm

Hence option (b) is correct

Number of Factors

Q81. Let x be a positive integer such that 7x + 96 is divisible by x. How many values of x are possible? [CSAT 2023]

(a) 10

(b) 11

(d) Infinitely many

Solution:

Given that,

x be a positive integer

7x + 96 is divisible by x

Now,

If x is a positive integer and 7x + 96 is divisible by x

Then, 96 is also divisible by x

So factors of 96 = 1, 2, 3, 4, 6, 8 , 12, 16, 24 , 32, 48, and 96

Total values = 12 possible values

Hence option (c) is correct

Q82. How many natural numbers are there which give a remainder of 31 when 1186 is divided by these natural numbers? [CSAT 2023]

(a) 6

(b) 7

(d) 9

Solution:

Given that,

Natural numbers which give a remainder of 31 when 1186

Now,

The remainder = the difference between the dividend and the largest multiple of the divisor that is less than or equal to the dividend.

Remainder is 31 when 1186 is divided by a certain number.

Thus, (1186 - 31) = 1155

The divisors of 1155 = 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155.

However, the remainder is 31, the divisor must be greater than 31.

possible numbers = 33, 35, 55, 77, 105, 165, 231, 385, 1155.

Therefore total 9 natural numbers are there.

Hence option (d) is correct

Surds, Indices and exponents

Q83. What is the largest number among the following? [CSAT 2020]

(a)
(b)
(c)
(d)
Solution:

= 2⁶ = 64
= 4³ = 64
= 3⁴ = 81
= 6² = 36

Hence option (c) is correct

Q84. Which number amongst 2⁴⁰, 3²¹, 4¹⁸ and 8¹² is the smallest? [CSAT 2022]

(a) 2⁴⁰

(b) 3²¹

(d) 8¹²

Solution:

Given that,

2⁴⁰, 3²¹, 4¹⁸ and 8¹²

Now,

8¹² = 2³⁶, 4¹⁸= 2³⁶

So among 2³⁶ and 3²¹, 3²¹ is smallest

Hence option (b) is correct

Square, Cube and Square root

Q85. A gardener has 1000 plants. He wants to plant them in such a way that the number of rows and the number of columns remains the same. What is the minimum number of plants that he needs more for this purpose? [CSAT 2013]

(a) 14

(b) 24

(d) 34

Solution:

Given that,

Total plants = 1000

The number of rows and the number of columns remains the same.

Now,

As 1000 is not a perfect square, thus the nearest square = 31² or 32²

For 32² = 1024

Thus extra plant needed = 1024 - 1000 = 24

Hence option (b) is correct

Q86. In aid of charity, every student in a class contributes as many rupees as the number of students in that class. With the additional contribution of Rs. 2 by one student only, the total collection is Rs. 443. Then how many students are there in the class? [CSAT 2016]

(a) 12

(b) 21

(d) 45

Solution:

Given that,

Every student in a class contributes as many rupees as the number of students in that class.

Additional contribution of Rs. 2 by one student only, the total collection is Rs. 443.

Now,

Let the number of student be S

Each student will give Rs. S

Total collection = Number of student x each contribution

S x S = S²

Now, one student's contribution is increased by 2

So, total collection now = S² + 2 = 443

S² = 443 - 2

S = 21 or - 21

Thus S = 21

Hence option (b) is correct

Q87. The age of Mr. X last year was the square of a number and it would be the cube of a number next year. What is the least number of years he must wait for his age to become the cube of a number again? [CSAT 2017]

(a) 42

(b) 38

(d) 16

Solution:

Given that,

The age of Mr. X last year was the square of a number and it would be the cube of a number next year.

Now,

Assuming the age of last year which is a square can be 1, 4, 9, 16, 25, 36.....

After 2 years the age is cube

After 2 years age will be 3, 6, 11, 18, 27, 38....

Thus age last year was 25 and next year will be 27 and present age is 26

Again the age will be cube = 64

Difference 64 - 26 = 38

Thus it will take 38 years

Hence option (b) is correct

Q88. If X is between -3 and -1, and Y is between -1 and 1, then X² - Y² is in between which of the following? [CSAT 2018]

(a) -9 and 1

(b) -9 and -1

(d) 0 and 9

Solution:

Given that,

If X is between -3 and -1

Y is between -1 and 1

Now,

X lies between = 0 < X² < 9

Y lies between = 0 < Y² < 1

So, 0 < X² - Y²< 9

Hence option (d) is correct

Q89. How many pairs of natural numbers are there such that the difference of whose squares is 63? [CSAT 2020]

(a) 3

(b) 4

(d) 2

Solution:

Given that,

Pairs of natural numbers , difference of whose squares is 63

Now,

X² - Y² = 63

(X + Y) x (X - Y) = 63

Pairs of 63 = (63, 1), (21, 3) and (9, 7)

Thus 3 such pairs are possible

Hence option (a) is correct

Number System (Statement Based Questions)

Q90. If x-y = 8, then which of the following must be true? [CSAT 2018]

1. Both x and y must be positive for any value of x and y.

2. If x is positive, y must be negative for any value of x and y.

3. If x is negative, y must be positive for any value of x and y.

Select the correct answer using the code given below.

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2 nor 3

Solution:

Given that,

X - Y = 8

Now,

1st option: x - y = 8 at x = 4 and y= - 4

so option does not satisfy the given condition

2nd option = x - y = 8 at x = 12 and y = 4

so option does not satisfy the given condition

3rd option = x - y = 8 at x = - 1 and y = - 9

so option does not satisfy the given condition

So no option follows.

Hence option (d) is correct

Q91. Consider two statements S1 and S2 followed by a question: [CSAT 2019]

S1: p and q both are prime numbers.

S2: p + q is an odd integer.

Question: Is pq an odd integer?

Which one of the following is correct?

(a) S1 alone is sufficient to answer the question

(b) S2 alone is sufficient to answer the question

(d) Both S1 and S2 are necessary to answer the question

Solution:

Given that,

S1: p and q both are prime numbers.

S2: p + q is an odd integer.

Now

p x q =Even, If p and q are even or odd and even numbers

p x q = Odd if p and q both are odd.

S1: As in a set of prime numbers two numbers can be both even and odd.

Hence statement 1 alone is not sufficient

S2: p + q is an odd integer:

Thus one of the numbers is odd and another is even.

Therefore pq will always be even.

Hence statement 2 alone is sufficient

Q92. Two Statements S1 and S2 are given below with regard to four numbers P, Q, R and S followed by a Question:

S1: R is greater than P as well as Q.

S2: S is not the largest one.

Question:

Among four numbers P, Q, R and S, which one is the largest?

Which one of the following is correct in respect of the above Statements and the Question? [CSAT 2020]

(a) S1 alone is sufficient to answer the Question.

(b) S2 alone is sufficient to answer the Question.

(c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.

(d) S1 and S2 together are not sufficient to answer the Question.

Solution:

Given that,

S1: R is greater than P as well as Q.

Possibility 1: R > P > Q

Possibility 2: R > Q > P

Hence statement 1 alone is insufficient

S2: S is not the largest one.

Hence statement 2 alone is insufficient

Combining both the statements 1 and 2

R is the largest number.

Hence, S1 and S2 together are sufficient to answer the Question, but neither $1 alone nor $2 alone is sufficient to answer the Question.

Q93. Two Statements S1 and S2 are given below followed by a Question: [CSAT 2020]

S1: There are not more than two figures on any page of a 51-page book.

S2: There is at least one figure on every page. Question: Are there more than 100 figures in that book?

Which one of the following is correct in respect of the above Statements and the Question?

(a)Both S1 and S2 are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.

(b) S1 alone is sufficient to answer the Question.

(d) S2 alone is sufficient to answer the Question.

Solution:

Given that,

S1: There are not more than two figures on any page of a 51-page book.

S2: There is at least one figure on every page.

Now,

S1: There are not more than two figures on any page of a 51-page book.

So, there are 0, 1 or 2 figures on each page.

Maximum possible figures = 51 x 2 = 102

Minimum possible figures: = 51 x 0 = 0

Hence statement 1 alone is insufficient

S2: There is at least one figure on every page.

Thus information is not sufficient

Hence statement 2 alone is insufficient

Combining both the statements then

Maximum possible figures = 51 x 2 = 102

Minimum possible figures: = 51 x 1 = 51

Thus both combined statements cannot answer the question

Hence S1 and S2 together are not sufficient to answer the Question.

Q94. Two Statements S1 and S2 are given below followed by a Question: [CSAT 2020]

S1: n is a prime number.

S2: n leaves a remainder of 1 when divided by 4.

Question: If n is a unique natural number between 10 and 20, then what is n?

Which one of the following is correct in respect of the above Statements and the Question?

(a) S1 alone is sufficient to answer the Question.

(b) S2 alone is sufficient to answer the Question.

(c)S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.

(d) S1 and S2 together are not sufficient to answer the Question.

Solution:

Given that,

S1: n is a prime number.

S2: n leaves a remainder of 1 when divided by 4.

Now,

Combining both the statements:

Between 10 to 20 prime numbers are 11, 13, 17, 19,

Also, 11/4, 13/4 17/4, 19/4

Remainder = 3, 1, 1, 3

Thus there is no unique solution to the given statements

Hence, S1 and S2 together are not sufficient to answer the Question.

Q95. Two Statements S1 and S2 are given below with regard to two numbers followed by a Question: [CSAT 2020]

S1: Their product is 21.

S2: Their sum is 10.

Question: What are the two numbers?

Which one of the following is correct in respect of the above Statements and the Question?

(a) S1 alone is sufficient to answer the Question.

(b) S2 alone is sufficient to answer the Question.

(c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.

(d) S1 and S2 together are not sufficient to answer the Question.

Solution:

Given that,

S1: Their product is 21.

S2: Their sum is 10.

Now,

Combining both the statements:

Let the numbers be a and b

So, a x b = 21.....(i)

a + b = 10....(ii)

From (i) a = 21/b....(iii)

Putting (iii) in (ii)

By solving, b = 3 or 7

Hence S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question

Q96. Consider all 3-digit number (without repetition of digits) obtained using three non-zero digits, which are multiples of 3. Let S be their sum. Which of the following is/are correct? [CSAT 2021]

1. S is always divisible by 74.

2. S is always divisible by 9.

Select the correct answer using the code given below:

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Given that,

all 3-digit number (without repetition of digits) obtained using three non-zero digits, which are multiples of 3.

Now,

Possible numbers are: 369, 396, 936, 963, 639, 693

Sum of the possible numbers S = 3996

1. S is always divisible by 74.

3996/74 = 54

Hence statement 1 is correct

2. S is always divisible by 9.

3996/9 = 444

Hence statement 2 is correct

Q97. The difference between a 2-digit number and the number obtained by interchanging the positions of the digits is 54. [CSAT 2021]

Consider the following statements:

1. The sum of the two-digits of the number can be determined only if the product of the two digits is known.

2. The difference between the two digits of the number can be determined.

Which of the above statements is/are correct?

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Given that,

The difference between a 2-digit number and the number obtained by interchanging the positions of the digits is 54

Now,

Let two digit number = 10a + b.....(i)

Reverse digit = 10b + a....(ii)

10a + b - 10b - a = 54

a - b = 6......(iii)

Thus possible values are 71 or 17, 82 or 28 and 93 or 39

Statement 1: The sum of the two-digits of the number can be determined only if the product of the two digits is known.

Hence statement 1 is incorrect

Statement 2: The difference between the two digits of the number can be determined.

a - b = 6

Hence statement 2 is correct

Q98. A bill for₹ 1,840 is paid in the denominations of ₹50, ₹20 and₹ 10 notes. 50 notes in all are used. Consider the following statements:

1. 25 notes of₹ 50 are used and the remaining are in the denominations of₹ 20 and ₹10.

2. 35 notes of ₹20 are used and the remaining are in the denominations of ₹ 50 and₹ 10.

3. 20 notes of ₹10 are used and the remaining are in the denominations of ₹ 50 and ₹20.

Which of the above statements are not correct? [CSAT 2022]

(a) 1 and 2 only

(b) 2 and 3 only

(d) 1, 2 and 3

Solution:

Given that,

Bill for₹ 1,840 is paid in the denominations of ₹50, ₹20 and₹ 10 notes.

50 notes in all are used.

Now,

Statement 1: 25 notes of ₹50 are used and the remaining are in the denominations of 20 and 10.

25 notes of ₹ 50 = 25 x 50 = Rs 1250

Remaining amount = 1840 - 1250 = 590

As, Rs. 590 cannot be paid in the combination of ₹20 and ₹10 notes

Hence statement 1 is incorrect

Statement 2: 35 notes of 20 are used and the remaining are in the denominations of ₹50 and 10.

So, 35 notes of Rs. 20 = Rs 700

Remaining amount = 1840 - 700 = 1140

As, Rs 1140 cannot be paid in the combination of 50 and 10.

Hence statement 2 is incorrect

Statement 3: 20 notes of 10 are used and the remaining are in the denominations of ₹50 and ₹20

20 notes of Rs. 10 = Rs 200

Remaining amount= 1840 - 200 = Rs. 1640

As Rs 1640 cannot be paid in the combination of ₹50 and 20.

Hence statement 3 is incorrect

Q99. Five friends P, Q, X, Y and Z purchased some notebooks. The information is given below: relevant

1. Z purchased 8 notebooks more than X did.

2. P and Q together purchased 21 notebooks.

3. Q purchased 5 notebooks less than P did.

4. X and Y together purchased 28 notebooks.

5. P purchased 5 notebooks more than X did.

If each notebook is priced ₹40, then what is the total cost of all the notebooks? [CSAT 2022]

(a) ₹2,600

(b) ₹2,400

(d) ₹2,320

Solution:

Given that,

Five friends P, Q, X, Y and 2 purchased some notebooks

1. Z purchased 8 notebooks more than X did.

2. P and Q together purchased 21 notebooks.

3. Q purchased 5 notebooks less than P did.

4. X and Y together purchased 28 notebooks.

5. P purchased 5 notebooks more than X did.

Each notebook is priced ₹40

Now,

P + Q = 21......(i)

P - Q = 5.......(ii)

Q purchased 5 notebooks less than P did.

From (i) and (ii)

P = 13 and Q = 8

P purchased 5 notebooks more than X did.

X = p - 5 = 13 - 5 = 8

X and Y together purchased 28 notebooks.

X + Y = 28 so Y = 20

Z purchased 8 notebooks more than X did.

Z = X + 8 = 8 + 8 = 16

Total number of notebooks P + Q + X + Z + Y = 65

Cost of one notebook Rs. 40

Thus cost of total notebooks = 40 x 65 = 2600

Hence option (a) is correct

Q100. Consider the Question and two Statements given below: [CSAT 2022]

Question: Is x an integer?

Statement-1: x/3 is not an integer.

Statement-2: 3x is an integer.

Which one of the following is correct in respect of the Question and the Statements?

(a) Statement-1 alone is sufficient to answer the Question.

(b) Statement-2 alone is sufficient to answer the Question.

(d) Both Statement-1 and Statement-2 are not sufficient to answer the Question.

Solution:

Given that,

Statement-1: x/3 is not an integer.

Statement-2: 3x is an integer.

Now,

Statement-1: x/3 is not an integer.

If x = 9, then x/3 = 3

Hence statement 1 is not sufficient

Statement-2: 3x is an integer

If 3x = 2, then x = 3/2

Hence statement 2 is not sufficient

Combining both statement

Hence Both Statement-1 and Statement-2 are not sufficient to answer the Question.

Q101. Let A, B and C represent distinct non-zero digits. Suppose x is the sum of all possible 3-digit numbers formed by A, B and C without repetition. [CSAT 2022]

Consider the following statements:

1. The 4-digit least value of x is 1332.

2. The 3-digit greatest value of x is 888.

Which of the above statements is/are correct?

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Given that,

A, B and C represent distinct non-zero digits.

x is the sum of all possible 3-digit numbers formed by A, B and C without repetition.

Now,

The minimum possible value of A, B and C = 1,2, and 3.

All possible 3-digit numbers formed by A, B, and C = 123, 132, 213, 231, 312, and 321.

Thus, x = 123 + 132 + 213 + 231 + 312 + 321 = 1332

Least value of x is 1332,

1. The 4-digit least value of x is 1332.

Hence statement 1 is correct

2. The 3-digit greatest value of x is 888.

Hence statement 2 is incorrect

Q102. Consider the following statements in respect of two natural number p and q such that p is a prime number and q is a composite number: [CSAT 2022]

1. p x q can be an odd number.

2. q/p can be a prime number.

3. p+q can be a prime number.

Which of the above statements are correct?

(a) 1 and 2 only

(b) 2 and 3 only

(d) 1, 2 and 3

Solution:

Given that,

Two natural number p and q such that p is a prime number and q is a composite number

Now,

Statement 1: p x q can be an odd number.

Let p = 7 and q = 15, so p x q = 105

Thus 105 an odd number

Hence statement 1 is correct

Statement 2: q/p can be a prime number.

If p = 5 and q = 10, so q/p = 2, which is a prime number

Hence statement 2 is correct

Statement 3: p + q can be a prime number.

p = 3 and q = 14, so p + q = 17, which is a prime number

Hence statement 3 is correct

Q103. Three of the five positive integers p, q, r, s, t are even and two of them are odd (not necessarily in order). Consider the following: [CSAT 2023]

1. p+q+r-s-t is definitely even.

2. 2p+q+2r-2s+t is definitely odd.

Which of the above statements is/are correct?

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Given that,

Three of the five positive integers p, q, r, s, t are even and two of them are odd (not necessarily in order)

Now,

The sum of p, q, r, s and t is always even as 3 even and 2 odds are even

Then

Statement 1: p+q+r-s-t is definitely even

This is even, hence statement 1 is correct

Statement 2: 2p+q+2r-2s+t is definitely odd.

The sum can be both even and odd

Hence statement 2 is incorrect

Q104. Consider the following in respect of prime number p and composite number c. [CSAT 2023]

1. can be even.

2. 2p + c can be odd.

3. pc can be odd.

Which of the statements given above are correct?

(a) 1 and 2 only

(b) 2 and 3 only

(d) 1, 2 and 3

Solution:

Given that,

Prime number p and composite number c.

Now,

1. can be even.

If p is odd and c is even, then:

then,

p + c = odd + even = odd

p - c = odd - even = odd

Thus odd/odd can be even

Hence statement 1 is correct

2. 2p + c can be odd.

If p is odd and c is odd

So, 2 x odd + odd = even + odd = odd

Hence statement 2 is correct

3. pc can be odd.

If p is odd and c is odd then p x c can be odd

as odd x odd = odd

Hence statement 3 is correct

Q105. Question: Is p greater than q? [CSAT 2023]

Statement-1: pxq is greater than zero.

Statement-2: p² is greater than q².

Which one of the following is correct in respect of the above Question and the Statements?

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.

(b) The Question can be answered by using either Statement alone.

(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.

(d) The Question cannot be answered even by using both the Statements together.

Solution:

Given that,

Statement-1: p x q is greater than zero.

Statement-2: p² is greater than q².

Now,

Statement-1: p x q is greater than zero.

No particular information is given. Thus data is not sufficent

Hence statement 1 is not sufficient

Statement-2: p² is greater than q².

p² > q²

However, p and q can both be negative

Hence statement 2 is not sufficient

Combining both statements

There is no unique solution for the given question as p can be greater or smaller depends on the sign of p and q

Hence the Question cannot be answered even by using both the Statements together.

Q106. Question: Is (p + q - r) greater than (p – q + r), where p, q and r are integers? [CSAT 2023]

Statement-1: (p-q) is positive.

Statement-2: (p-r) is negative.

Which one of the following is correct in respect of the above Question and the Statements?

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.

(b) The Question can be answered by using either Statement alone.

(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.

(d) The Question cannot be answered even by using both the Statements together.

Solution:

Given that,

Statement-1: (p-q) is positive.

Statement-2: (p-r) is negative

Now,

Statement-1: (p-q) is positive.

Thus p > q, however there is no information about r

Hence statement 1 is not sufficient

Statement-2: (p-r) is negative

Thus p < r, but there is no information of q

Hence statement 2 is not sufficient

Combining both the statements

r > p > q

So, (p + q - r) < (p – q + r) as r > p > q

Hence the Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.

Q107. Consider a 3-digit number. [CSAT 2023]

Question: What is the number?

Statement-1: The sum of the digits of the number is equal to the product of the digits.

Statement-2: The number is divisible by the sum of the digits of the number.

Which one of the following is correct in respect of the above Question and the Statements?

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.

(b) The Question can be answered by using either Statement alone.

(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.

(d) The Question cannot be answered even by using both the Statements together.

Solution:

Given that,

Statement-1: The sum of the digits of the number is equal to the product of the digits.

Statement-2: The number is divisible by the sum of the digits of the number.

Now,

Let three digit number = ABC

Statement-1: The sum of the digits of the number is equal to the product of the digits.

ABC = 123

1 + 2 + 3 = 1 x 2 x 3

Thus ABC can be 123, 132, 213, 231, 321, 312

Therefore there is no unique solution.

Hence statement 1 is not sufficient

Statement-2: The number is divisible by the sum of the digits of the number.

ABC can be 111, 222, 333.....

Therefore there is no unique solution.

Hence statement 2 is not sufficient

Combining both the statements

There is no unique solution for the question

Hence the Question cannot be answered even by using both the statements together.

Q108. Consider the following statements: [CSAT 2021]

1. The sum of 5 consecutive integers can be 100.

2. The product of three consecutive natural numbers can be equal to their sum.

Which of the above statements is/are correct?

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Given that,

1. The sum of 5 consecutive integers can be 100.

Let 5 consecutive numbers be x - 2, x - 1, x, x + 1, x + 2

Sum = x - 2 + x - 1 + x + x + 1 + x + 2 = 100

5x = 100, so x = 20

Thus numbers are: 18, 19, 20, 21 and 22

Hence statement 1 is correct

2. The product of three consecutive natural numbers can be equal to their sum.

Let the number be 123

1 x 2 x 3 = 3 + 2 + 1

Hence statement 2 is correct

Q109. There are large numbers of silver coins weighing 2 gm, 5 gm, 10 gm, 25 gm, 50 gm each. Consider the following statements:

1. To buy 78 gm of coins one must buy at least 7 coins.

2. To weigh 78 gm using these coins one can use less than 7 coins.

Which of the statements given above is/are correct? ? [CSAT 2023]

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Given that,

There are large numbers of silver coins weighing 2 gm, 5 gm, 10 gm, 25 gm, 50 gm each.

Now,

1. To buy 78 gm of coins one must buy at least 7 coins.

50 + 10 + 10 + 2 + 2 + 2 + 2 = 78

Thus at least 7 coins one must buy

Hence statement 1 is correct

2. To weigh 78 gm using these coins one can use less than 7 coins

To weigh 78 gm = 50 + 25 + 5 - 2

One can use 4 coins

Hence statement 2 is correct

Q110. Consider the following:

I. A + B means A is neither smaller nor equal to B.

II. A - B means A is not greater than B.

III. A×B means A is not smaller than B.

IV. A ÷ B means A is neither greater nor equal to B.

V. A ± B means A is neither smaller nor greater than B.

Statement: P×Q, P – T, T ÷ R, R± S

Conclusion-1: Q T

Conclusion-2: S + Q

Which one of the following is correct in respect of the above Statement and the Conclusions? [CSAT 2023]

(a) Only Conclusion-1 follows from the Statement.

(b) Only Conclusion-2 follows from the Statement.

(d) Neither Conclusion-1 nor Conclusion-2 follows from the Statement.

Solution:

Given that,

I. A + B means A is neither smaller nor equal to B.

II. A - B means A is not greater than B.

III. A×B means A is not smaller than B.

IV. A ÷ B means A is neither greater nor equal to B.

V. A ± B means A is neither smaller nor greater than B.

Statement: P × Q, P – T, T ÷ R, R ± S

Conclusion-1: Q T

Conclusion-2: S + Q

Now,

P × Q = P is not smaller than Q.

P – T = P is not greater than T.

T ÷ R = T is neither greater nor equal to R.

R ± S = R is neither smaller nor greater than S.

Conclusion-1: Q ± T

Q is neither smaller nor greater than T.

This conclusion does not follow as P is not smaller than Q.

So it can be equal or greater.

Conclusion-2: S + Q

S is neither smaller nor equal to Q.

From the given statement it can be inferred that

S > or = Q

Therefore conclusion 2 follows the statement

BASIC NUMERACY

Q111. Consider the following sum: *+1 *+2 *+*3+*1=21* [CSAT 2018]

In the above sum, stands for

(a) 4

(b) 5

(d) 8

Solution:

Given that,

*+1 *+2 *+*3+*1=21*

Now,

Using the given options:

Let's take the value 8

8 + 18 + 28 + 83 + 81 = 218

Thus satisfies the value 8 satisfies the equation

hence option (d) is correct

Q112. Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3.

The only possible value of B is [CSAT 2019]

(a) 2

(b) 5

(d) 8

Solution:

Given that,

Number 136 is added to 5B7 and the sum obtained is 7A3

A and B are integers.

7A3 is exactly divisible by 3.

Now,

7 + A + 3 = 10 + A

Possible values are 2, 5, 8

136 + 5B7 = 7A3

taking A = 2 and B = 8

723 = 7 + 2 + 3 = 12 (divisible by 3)

Thus B = 8

hence option (d) is correct

Q113. In the sum ®+1®+5®+®®+®1] = 1®®, for which digit does the symbol ® stand? [CSAT 2020]

(a) 2

(b) 3

(d) 5

Solution:

Given that,

®+1®+5®+®®+®1 = 1®®

Now,

From the given options taking the value 3

3 + 13 + 53 + 33 + 31 = 133

Hence option (b) is correct

Q114. Let A3BC and DE2F be four-digit numbers where each letter represents a different digit greater than 3. If the sum of the numbers is 15902, then what is the difference between the values of A and D? [CSAT 2020]

(a) 1

(b) 2

(d) 4

Solution:

Given that,

A3BC and DE2F be four-digit numbers where each letter represents a different digit greater than 3.

The sum of the numbers is 15902

Now,

A 3BC + DE2F = 15902

So, C + F = 2

Thus C + F must be 12

Possible values of C and F: (5, 7) or (4, 8)

Then, B + 2 = 10

So B must be 7 as the sum of C and F will give a carry of 1.

Then, 3 + E = 9

So, E must be 5 as it will get 1 as carry from the sum of B + 7

Then, A + D = 15

So, A and D possible values: (7, 8) or (9, 6)

As all digits are different , thus A and D are (9, 6)

Difference of A and D = 9 - 6 = 3

Hence option (c) is correct

Q115. Consider the following addition problem: 3P+4P+PP + PP = RQ2; where P, Q and R are different digits.

What is the arithmetic mean of all such possible sums? [CSAT 2021]

(a) 102

(b) 120

(d) 220

Solution:

Given that,

3P+4P+PP + PP = RQ2

P, Q and R are different digits.

Now,

As, P is added 4 times in the units place and resultant value of units place is 2

so possible values for P = 3 or 8

If P = 3

33 + 43 + 33 + 33 = 142

R = 1 and Q = 4

If P = 8

38 + 48 + 88 + 88 = 262

R = 2 and Q = 6

Arithmetic mean of possible values = (262 + 142)/ 2 = 202

Hence option (c) is correct

Q116. Consider the following multiplication problem: (PQ) × 3 = RQQ, where P, Q and R are different digits and R ≠ 0. [CSAT 2021]

What is the value of (P+R) ÷ Q?

(a) 1

(b) 2

(d) Cannot be determined due to insufficient data.

Solution:

Given that,

(PQ) × 3 = RQQ,

P, Q and R are different digits and R ≠ 0

Now,

If Q = 5

P5 x 3 = R55

Now, P = 8

85 x 3 = 255, R = 2

(P+R) ÷ Q

10/5 = 2

Hence option (b) is correct

Q117. If ABC and DEF are both 3-digit numbers such that A, B, C, D, E and F are distinct non-zero digits that ABC + DEF = 1111, then what is the value of A+B+C+D+E+F[CSAT 2023]

(a) 28

(b) 29

(d) 31

Solution:

Given that,

If ABC and DEF are both 3-digit numbers such that A, B, C, D, E and F are distinct digits that ABC + DEF = 1111

Now, Taking values ABC = 785 and DEF = 326, 785 + 326 = 1111

So, 7 + 8 + 5 + 3 + 2 + 6 = 31

Hence option (d) is correct

Q118. A 3-digit number ABC, on multiplication with D gives 37DD where A, B, C and D are different non-zero digits. What is the value of A+B+C?[CSAT 2023]

(a) 18

(b) 16

(d) Cannot be determined due to insufficient data

Solution:

Given that,

A 3-digit number ABC, on multiplication with D gives 37DD

A, B, C and D are different non-zero digits.

Now,

Taking values by hit and trial method

If D = 4, then four digit number = 3744 which is divisible by 4

So, ABC = 3744/4 = 936

A = 9, B = 3, C = 6

9 + 3 + 6 = 18

Hence option (a) is correct

Q119. AB and CD are 2-digit numbers. Multiplying AB with CD results in a 3-digit number DEF. Adding DEF to another 3-digit number GHI results in 975. Further A, B, C, D, E, F, G, H, I are distinct digits. If E = 0, F = 8, then what is A + B + C equal to? [CSAT 2023]

(a) 6

(b) 7

(d) 9

Solution:

Given that,

AB and CD are 2-digit numbers.

Multiplying AB with CD results in a 3-digit number DEF.

Adding DEF to another 3-digit number GHI results in 975.

A, B, C, D, E, F, G, H, I are distinct digits.

E = 0, F = 8

Now,

To obtain a product of 08X

If (AB or CD) is a multiple of 4 and the other number ends with 2 or 7.

The possible pairs of numbers = (17, 24) (34, 12) (12, 34) (24, 17)

The sum of DEF + GHI = 975,

The sum of the digits D, G, H, I must be 9 (As E = 0 and F = 8).

The only pair satisfying this condition = (12, 34)

Product = 408.

Thus, A = 1 B = 2 C = 3 D = 4

A + B + C = 1 + 2 + 3 = 6

Hence option (a) is correct

Q120. Let pp, qq and rr be 2-digit numbers, where p

1. The number of possible values of p is 5.

2. The number of possible values of q is 6.

Which of the above statements is/are correct?

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

Solution:

Given that,

pp, qq and rr be 2-digit numbers,

pp + qq + rr = tt0

tt0 is a 3-digit number ending with zero

Now,

The sum of the digits p, q and r is either 10 or 20

Since the sum of their double-digit = three-digit number ending in zero

So it can be either 110 or 220.

Statement 1: The number of possible values of p is 5.

p is the smallest digit

p + q + r is either 10 or 20,

The possible values of p = 1 or 2 for the sum 10,

1, 2, 3, 4 or 5 for the sum 20.

5 possible values for p

Hence statement 1 is correct

Statement 2: The number of possible values of q is 6.

q is the middle digit

The sum of p + q + r is either 10 or 20,

the possible values of q = 2, 3 or 4 for the sum 10,

6, 7 or 8 for the sum 20.

So, total possible values = 6

Hence statement 2 is correct

BODMAS

Q121. If $ means 'divided by'; @ means 'multiplied by'; # means 'minus', then the value of 10#5@1$5 is [CSAT 2019]

(a) 0

(b) 1

(d) 9

Solution:

Given that,

If $ means 'divided by'; @ means 'multiplied by'; # means 'minus',

Now,

10#5@1$5 = 10 - 5 x 1/5 = 9

Hence option (d) is correct

CSAT 2024

Q122. How many consecutive zeros are there at the end of the integer obtained in the product 1² x 2⁴ x 3⁶ x 4⁸ x... x25⁵⁰? [CSAT 2024]

(a) 50

(b) 55

(d) 200

Solution:

Given that,

1² x 2⁴ x 3⁶ x 4⁸ x..x 25⁵⁰

Now,

Number of 5's = 5¹⁰ x 10²⁰ x 15³⁰ x 20⁴⁰ x 25⁵⁰

= 5¹⁰ x 5²⁰ x 5³⁰ x 5⁴⁰ x 5¹⁰⁰ = 5²⁰⁰

Number of 2's = 2⁴ x 4⁸ x 6¹² x 8¹⁶ x 10²⁰ x 12²⁴ x 14²⁸ x 16³² x 18³⁶ x 20⁴⁰ x 22⁴⁴ x 24⁴⁸

= 2⁴ x 2¹⁶ x 2¹² x 2⁴⁸ x 2²⁰ x 2⁴⁸ x 2²⁸ x 2¹²⁸ x 2³⁶ x 2⁸⁰ x 2⁴⁴ x 2¹⁴⁴ = No. of 2's is definitely more than 200

For number of zeros = no. of 5's x no. of 2's

no. of 5's < no. of 2's

Required no. of 2's to form zeros = 200

= 5²⁰⁰ x 2²⁰⁰

Consecutive zeros = 200

Hence option (d) is correct

Q123. 222³³³ +333²²² is divisible by which of the following numbers? [CSAT 2024]

(a) 2 and 3 but not 37

(b) 3 and 37 but not 2

(d) 2, 3 and 37

Solution:

Given that,

222³³³ + 333²²²

Now,

222³³³ + 333²²² = 111²²²(2 x 111¹¹¹ + 3)

111²²² = (37 x 3)¹¹¹

Thus the sum is divisible by 37 and 3

Hence option (b) is correct

Q124. What is the rightmost digit preceding the zeros in the value of 30³⁰? [CSAT 2024]

(a) 1

(b) 3

(d) 9

Solution:

Given that,

The value = 30³⁰

Now,

The value of 30³⁰ = (3 x 10)³⁰ = 3³⁰ x 10³⁰

For non-zero numbers the value is determined by the cyclicity of 3

Cyclicity of 3 is 4

So, 30/4 = remainder 2

Thus, 3²⁸ has unit digit as 1

Remaining value = 3² = 9

Therefore, the rightmost digit preceding the zeros = 9

Hence option (d) is correct

Q125. 421 and 427, when divided by the same number, leave the same remainder 1. How many numbers can be used as the divisor in order to get the same remainder 1? [CSAT 2024]

(a) 1

(b) 2

(d) 4

Solution:

Given that,

421 and 427 when divided by the same number the remainder is same i.e. 1

Now,

Let the number be X

421/X = Remainder 1

427/X = Remainder 1

So,

421 - 1 = 420

427 - 1 = 426

Prime factors of 420 = 2² x 3 x 5 x 7

Prime factors of 426 = 2 x 3 x 71

Common factors = 2, 3 and 6

Thus when 421 and 427 is divided by 2 or 3 or 6 then the remainder is 1

Therefore there are 3 divisors

Hence option (c) is correct

Q126. A can X contains 399 litres of petrol and a can Y contains 532 litres of diesel. They are to be bottled in bottles of equal size so that whole of petrol and diesel would be separately bottled. The bottle capacity in terms of litres is an integer. How many different bottle sizes are possible? [CSAT 2024]

(a) 3

(b) 4

(d) 6

Solution:

Given that,

Petrol = 399 L

Diesel = 532 L

Now,

Factors of 399 = 1 x 3 x 7 x 19

Factors of 532 = 1 x 2² x 7 x 19

Common factors = 1, 7, 19 and 133

Total different sizes of bottle possible are 4

Hence option (b) is correct

Q127. Consider the following statements in respect of the sum S=x+y+z, where x, y and z are distinct prime numbers each less than 10: [CSAT 2024]

1. The unit digit of S can be 0.

2. The unit digit of S can be 9.

3. The unit digit of S can be 5.

Which of the statements given above are correct?

(a) 1 and 2 only

(b) 2 and 3 only

(d) 1, 2 and 3

Solution:

Given that,

S = x + y + z

x, y and z are distinct prime numbers less than 10

Now,

1. The unit digit of S can be 0

If x = 2, y = 3 and z = 5

Sum = 2 + 3 + 5 = 10

Thus unit digit can be 0

Hence statement 1 is correct

2. The unit digit of S can be 9

Considering the value of x, y and z

2 + 3 + 5 = 10

3 + 5 + 7 = 15

5 + 7 + 2 = 14

No other combinations are possible

Hence statement 2 is incorrect

3. The unit digit of S can be 5

Considering the value of x, y and z

3 + 5 + 7 = 15

Hence statement 3 is correct

Q128. Consider the following: [CSAT 2024]

1. 1000 litres = 1 m³

2. 1 metric ton = 1000 kg

3. 1 hectare = 10000 m²

Which of the above are correct?

(a) 1 and 2 only

(b) 2 and 3 only

(d) 1, 2 and 3

Solution:

Given that,

1. 1000 litres = 1m³

2. 1 metric ton = 1000 kg

3. 1 hectare = 10000 m²

Now,

The given relations are correct

1. 1000 litres = 1m³

2. 1 metric ton = 1000 kg

3. 1 hectare = 10000 m²

Hence option (d) is correct

Q129. Let X be a two-digit number and Y be another two-digit number formed by interchanging the digits of X. If (X+Y) is the greatest two-digit number, then what is the number of possible values of X? [CSAT 2024]

(a) 2

(b) 4

(d) 8

Solution:

Given that,

X and Y are two digit numbers

Y is formed by interchanging the digits

(X + Y) = Greatest two digit number

Now,

Let X be 10a + b

Y = 10b + a

X + Y = 10a + b + 10b + a = 11a + 11b = 11(a + b)

Greatest two digit value = 99

11(a + b) = 99

(a + b) = 9

Possible values of a and b = (1, 8), (8,1), (2, 7), (7, 2), (3, 6), (6, 3), (4, 5), (5, 4) = total 8 values

Hence option (d) is correct

Q130. Let p, q, r and s be distinct positive integers. Let p, q be odd and r, s be even. Consider the following statements: [CSAT 2024]

1. (p-r)²(qs) is even.

2. (q-s)q²s is even.

3. (q+r)²(p+s) is odd.

Which of the statements given above are correct?

(a) 1 and 2 only

(b) 2 and 3 only

(d) 1, 2 and 3

Solution:

Given that,

p, q, r and s are distinct positive integers

p and q are odd and r and s are even

Now,

1. (p -r)²(qs) is even

(odd - even)² x (odd x even) = (odd)² x (even) = even

Hence statement 1 is correct

2. (q - s)(q)²s is even

(odd - even) x (odd)² x (even) = even

Hence statement 2 is correct

3. (q + r)² x (p + s) is odd

(odd + even)² x (odd + even) = (odd)² x (odd) = odd

Hence statement 3 is correct

Q131. What is the number of fives used in numbering a 260-page book? [CSAT 2024]

(a) 55

(b) 56

(d) 60

Solution:

Given that,

Number of pages in book = 260

Now,

From 1 to 100 number of 5's

Unit place = 10 times

Tens place = 10 times

Total = 20 times

From 101 to 200 number of 5's

Unit place = 10 times

Tens place = 10 times

Total = 20 times

From 201 to 260 number of 5's

Unit place = 6 times (205, 215, 225,235, 245 and 255)

Tens place = 10 times (250, 251, 252, 253, 254, 255, 256, 257, 258, 259)

Total = 16 times

Total number of times 5 comes from 1 to 260 = 20 + 20 + 16 = 56 times

Hence option (b) is correct

Q132. If the sum of the two-digit numbers AB and CD is the three-digit number 1CE, where the letters A, B, C, D, E denote distinct digits, then what is the value of A? [CSAT 2024]

(α) 9

(b) 8

(d) Cannot be determined due to insufficient data

Solution:

Given that,

AB + CD = 1CE

Where A, B, C, D and E are distinct numbers

AB and CD are two digit numbers

Now,

AB + CD = 1CE

Concept used:

The answer depends on Value of A

A + C = C

It is possible in two cases 0 and 9

0 is not possible as AB is a two digit number

9 + C = C is possible when the addition has a carry forward

This means that B + D must be more than 10, where 1 is carry forwarded

9 + 1 + C = C

Thus, the value of AB + CD = 1CE is only possible when value of A is 9

Hence option (a) is correct

Q133. Three numbers x, y, z are selected from the set of the first seven natural numbers such that x > 2y > 3z. How many such distinct triplets (x, y, z) are possible? [CSAT 2024]

(a) One triplet

(b) Two triplets

(d) Four triplets

Solution:

Given that,

The value of x, y and z will be selected from first seven natural numbers

x > 2y > 3z

Now,

Possible values of x, y and z

X	2y	3z	Triplets
5	y = 2, 2y = 4	Z = 1, 3z = 3	5, 4, 3
6	y = 2, 2y = 4	Z = 1, 3z = 3	6, 4, 3
7	y = 2, 2y = 4	Z = 1, 3z = 3	7, 4, 3
7	y = 3, 2y = 6	Z = 1, 3z = 3	7, 6, 3

Thus there are 4 triplets

Hence option (d) is correct

Q134. 32⁵ +2²⁷ is divisible by [CSAT 2024]

(a) 3

(b) 7

(d) 11

Solution:

Given that,

32⁵ + 2²⁷

Now,

32⁵ + 2²⁷ = 2²⁵ + 2²⁷ = 2²⁵(1 + 2²) = 2²⁵ x (5) = 2²⁴ x 10

Thus the given number is exactly divisible by 10

Hence option (c) is correct

Q135. Let p and q be positive integers satisfying p

(a) 3

(b) 4

(d) 6

Solution:

Given that,

p and q are positive integers

p < q

p + q = k

Now,

P	Q	p + q = k
1	2	1 + 2 = 3
1	3	1 + 3 = 4
1	4	1 + 4 = 5
2	3	2 + 3 = 5

Thus the value of K = 5 is not unique where p < q

The value of p can be 1 or 2 and the value of q can be 4 or 3

Hence option (c) is correct

Q136. A Question is given followed by two Statements I and II. Consider the Question and the Statements.

Question: What are the values of m and n, where m and n are natural numbers? [CSAT 2024]

Statement-I: m+n > mn and m > n.

Statement-II: The product of m and n is 24.

Which one of the following is correct in respect of the above Question and the Statements?

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone

(b) The Question can be answered by using either Statement alone

(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

(d) The Question cannot be answered even by using both the Statements together

Solution:

Given that,

Statement I: m + n > mn and m > n

Statement II: The product of m and n is 24

Also, m and n are natural numbers

Now,

From statement I:

m + n > mn and m > n

The values of m and n can be many and thus cannot have a unique solution

Hence statement I is insufficient to answer the question

From statement II:

m x n = 24

Cases are: 1 x 24, 2 x 12, 3 x 8, 4 x 6 or 24 x 1, 12 x 2, 8 x 3, 6 x 4

Thus there are 8 cases and no unique solution

Hence statement II is insufficient to answer the question

Combining the statements

The cases reduce to 2 where the values can be

24 x 1 or 1 x 24

However, m > n

So, the value of m = 24 and n = 1

Hence option (c) is correct

Q137. A Question is given followed by two Statements I and II. Consider the Question and the Statements.

Question: What are the unique values of x and y, where x, y are distinct natural numbers? [CSAT 2024]

Statement-I: x/y is odd.

Statement-II: xy = 12

Which one of the following is correct in respect of the above Question and the Statements?

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone

(b) The Question can be answered by using either Statement alone

(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

(d) The Question cannot be answered even by using both the Statements together

Solution:

Given that,

Statement I: x/y is odd

Statement II: xy = 12

x and y are distinct natural numbers

Now,

From statement I:

x / y is odd, the values of x and y can be many and does not have any unique solutions

Hence statement I is insufficient to answer the question

From statement II:

xy = 12

The values are: 12 x 1, 6 x 2, 3 x 4 or 1 x 12, 2 x 6, 4 x 3

Thus there are 6 values and does not have any unique solutions

Hence statement II is insufficient to answer the question

Combining the statements:

The values are: 12 x 1, 6 x 2, 3 x 4 or 1 x 12, 2 x 6, 4 x 3

and x/y = odd

There is only one possible case

x = 6 and y = 2

6/2 = 3 which is odd

Note: Only integers can be odd or even

Hence option (c) is correct

Q138. A Question is given followed by two Statements I and II. Consider the Question and the Statements. There are three distinct prime numbers whose sum is a prime number. [CSAT 2024]

Question: What are those three numbers?

Statement-I: Their sum is less than 23.

Statement-II: One of the numbers is 5.

Which one of the following is correct in respect of the above Question and the Statements?

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone

(b) The Question can be answered by using either Statement alone

(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

(d) The Question cannot be answered even by using both the Statements together

Solution:

Given that,

There are three distinct prime numbers and sum is also a prime number

Statement I: Their Sum is less than 23.

Statement II: One of the number is 5.

Now,

Let prime numbers be P, Q and R

From statement I:

P + Q + R < 23

Possible case

3 + 5 + 7 = 15 (sum is not a prime number)

5 + 7 + 11 = 23 (sum should be less than 23)

3 + 5 + 11 = 19 (possible)

2 + 3 + 5 = 10 (as even + odd + odd = even so 2 cannot be used as prime numbers)

The only possible case is 3 + 5 + 11 = 19 (possible)

Hence statement I is sufficient to answer the question

From statement II:

One number is 5

Possible cases:

3 + 5 + 7 = 15 (sum is not a prime number)

5 + 7 + 11 = 23 (possible)

3 + 5 + 11 = 19 (possible)

2 + 3 + 5 = 10 (as even + odd + odd = even so 2 cannot be used as prime numbers)

There are two possible cases

Hence statement II is insufficient to answer the question

Hence option (a) is correct

Q139. A Question is given followed by two Statements I and II. Consider the Question and the Statements. [CSAT 2024]

Question: Is (x + y) an integer?

Statement-I: (2x+y) is an integer.

Statement-II: (x+2y) is an integer.

Which one of the following is correct in respect of the above Question and the Statements?

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone

(b) The Question can be answered by using either Statement alone

(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

(d) The Question cannot be answered even by using both the Statements together

Solution:

Given that,

Statement I: (2x + y) is an integer.

Statement II: (x + 2y) is an integer.

Now,

From statement I:

Possible cases

x = 1 and y = 2

(2x + y) = 4

x = 1/3 and y = 1/3

(2x + y) = 1

As the value of x and y are not unique it can be both integers or fraction

Hence statement I is insufficient to answer the question

From statement II:

Possible cases

x = 1 and y = 2

(x + 2y) = 5 (integer)

x = 1/3 and y = 1/3

(x + 2y) = 1(Integer)

As the value of x and y are not unique it can be both integers or fraction

Hence statement II is insufficient to answer the question

Combining both the statements:

The value of x and y is still not unique

Hence option (d) is correct

Q140. In some code, letters P, Q, R, S, T represent numbers 4, 5, 10, 12, 15. It is not known which letter represents which number. If Q – S = 2S and

T = R + S + 3, then what is the value of P+R-T? [CSAT 2024]

(a) 1

(b) 2

(d) Cannot be determined due to insufficient data

Solution:

Given that,

P, Q, R, S , T represents 4, 5, 10, 12, 15

Q - S = 2S

T = R + S + 3

Now,

Q = 3S.....(i)

Two cases possible

either S = 4 or 5 and Q = 12 or 15

T = R + S + 3

If R = 10

Then T = 10 + 4 + 3 = 17

or T = 10 + 5 + 3 = 18

Both the cases are not possible as the numbers are 4, 5, 10, 12, 15

So R < 10

So either R = 4 or 5

If R = 4

T = 4 + 5 + 3 = 12 (Possible)

If R = 5

T = 5 + 4 + 3 = 12 (Not possible as S = 4 then Q = 12 so, T cannot be 12)

The values are

P = 10, Q = 15, R = 4, S = 5, T = 12

P + R - T = 10 + 4 - 12 = 2

Hence option (b) is correct

Q141. In the expression 5 * 4 * 3 * 2 * 1, * is chosen from +, -, x each at most two times. Which is the smallest non-negative value of the expression? [CSAT 2024]

(a) 3

(b) 2

(d) 0

Solution:

Given that,

The expression : 5 * 4* 3 * 2 * 1

* is chosen from +, - , x each at most two times

Now,

Smallest non negative value that can be achieved = 5 - 4 - 3 + 2 x 1 = 0

Hence option (d) is correct

Q142. On January 1^st, 2023, a person saved Rs. 1.

On January 2^nd, 2023, he saved Rs. 2 more than that on the previous day. On January 3^rd, 2023, he saved Rs. 2 more than that on the previous day and so on. At the end of which date was his total savings a perfect square as well a perfect cube? [CSAT 2024]

(a) 7^th January 2023

(b) 8^th January, 2023

(d) Not possible

Solution:

Given that,

On 1st Jan person saves Rs.1

On 2nd Rs.2 more than previous

Now,

Date	Total savings in Rs.
1st Jan	1
2nd Jan	3
3rd Jan	5
4th Jan	7
5th Jan	9
6th Jan	11
7th Jan	13
8th Jan	15
Total savings = 64

Thus, 64 is a perfect square of 8 and perfect cube of 4

Therefore, On 8th Jan total savings will become a perfect square and a perfect cube

Hence option (b) is correct

Q143. A Question is given followed by two Statements I and II. Consider the Question and the Statements. [CSAT 2024]

A person buys three articles p, q and r for 50. The price of the article q is 16 which is the least.

Question: What is the price of the article p?

Statement-I: The cost of p is not more than that of r.

Statement-II: The cost of r is not more than that of p.

Which one of the following is correct in respect of the above Question and the Statements?

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone

(b) The Question can be answered by using either Statement alone

(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

(d) The Question cannot be answered even by using both the Statements together

Solution:

Given that,

p + q + r = Rs. 50

q = Rs. 16

Statement I: The cost of p is not more than that of r.

Statement II: The cost of r is not more than that of p.

Now,

p + r = 50 - 16 = Rs. 34

From statement I:

p ≤ r

p + r = 34

Two cases possible

p < r

p = r

Thus there is no unique solution possible

Hence statement I is insufficient to answer the question

From statement II:

Possible cases

p ≥ r

p + r = 34

Two cases possible

p > r

p = r

Thus there is no unique solution possible

Hence statement II is insufficient to answer the question

Combining both the statements:

The conclusion is p = r

So,

p + p = 34 or p = 17 and r = 17

Hence option (c) is correct

Q144.If a + b means a - b; a - b means a x b; a x b means a ÷ b; a ÷ b means a + b, then what is the value of 10 + 30 - 100 x 50 ÷ 25 (Operations are to be replaced simultaneously) [CSAT 2024]

(a) 15

(b) 0

(d) -25

Solution:

Given that,

a + b means a - b; a - b means a x b; a x b means a÷b and a÷b means a+ b

10 + 30 - 100 x 50 ÷25

Now,

10 - 30 x 100 ÷ 50 + 25

Applying BODMAS

10 - 30 x 2 + 25 = 10 - 60 + 25 = - 25

Hence option (d) is correct

Q. No.	Answer	Q. No.	Answer	Q. No.	Answer	Q. No.	Answer
1	a	41	d	81	c	121	d
2	b	42	d	82	d	122	d
3	c	43	b	83	c	123	b
4	a	44	b	84	b	124	d
5	c	45	d	85	b	125	c
6	b	46	c	86	b	126	b
7	b	47	a	87	b	127	c
8	a	48	a	88	d	128	d
9	b	49	b	89	a	129	d
10	c	50	c	90	d	130	d
11	b	51	b	91	b	131	b
12	c	52	c	92	c	132	a
13	c	53	a	93	c	133	d
14	b	54	d	94	d	134	c
15	b	55	b	95	c	135	c
16	b	56	d	96	c	136	c
17	a	57	b	97	b	137	c
18	d	58	d	98	d	138	a
19	d	59	a	99	a	139	d
20	b	60	c	100	d	140	b
21	a	61	a	101	a	141	d
22	b	62	c	102	d	142	b
23	c	63	d	103	a	143	c
24	d	64	a	104	d	144	d
25	b	65	a	105	d
26	c	66	d	106	c
27	a	67	a	107	d
28	c	68	a	108	c
29	c	69	b	109	c
30	b	70	b	110	b
31	a	71	d	111	d
32	b	72	b	112	d
33	c	73	b	113	b
34	c	74	d	114	c
35	b	75	b	115	c
36	c	76	b	116	b
37	b	77	d	117	d
38	b	78	b	118	a
39	b	79	c	119	a
40	b	80	b	120	c

Download Number System PYQs PDF

Download the complete set of Number System PYQs (2011-2024) with solutions in PDF format.

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FAQs on Number System for UPSC

1. Is Number System important for UPSC?

Yes, the Number System is a crucial topic for the UPSC CSAT paper. It tests your quantitative aptitude and problem-solving skills.

2. How to prepare for Number System in UPSC?

• Understand basic concepts like divisibility, LCM, HCF, and number series.

• Practice PYQs regularly.

• Use reliable resources like NCERT books and online platforms like iassetu.com.

3. What are the most frequently asked Number System questions in UPSC?

• LCM and HCF problems

• Divisibility rules

• Unit digit and remainder-based questions

Conclusion

Practicing Number System PYQs is essential for cracking the UPSC CSAT paper. This page provides year-wise solved questions, practice problems, and a free PDF download to help you prepare effectively. Bookmark this page and revisit it regularly for updates.

For more UPSC preparation resources, explore iassetu.com. Happy learning!

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