Explore a comprehensive collection of UPSC Prelims Topic Wise Questions focusing on Basic Numeracy, covering the Number System, Fractions, LCM and HCF, Sequence, and Series. Delve into a diverse range of questions designed to enhance your understanding and preparation for the UPSC Preliminary examination in these fundamental mathematical concepts. This resource offers a structured approach to mastering key concepts and techniques necessary for success in basic numeracy. With a strategic emphasis on topic-wise questions, aspirants can refine their numerical reasoning skills and sharpen their problem-solving abilities across various domains. Whether you’re a beginner or an advanced candidate, these meticulously curated questions provide valuable insights and practice opportunities to excel in the competitive UPSC environment. Prepare diligently, grasp essential insights, and boost your confidence to ace the UPSC Prelims with proficiency in the foundational aspects of basic numeracy.
Q1. The missing fraction in the series given below is (1998)
4/9,9/20,…,39/86
(a) 17/40
(b) 19/42
(c) 20/45
(d) 29/53
Ans. (b)
Given pattern,
4/9, (4 X 2 + 1)/(9 X 2 + 2) = 9/20, (9 X 2 + 1)/(20 X 2 + 1) = (19/12, 19 X 2 + 1)/(42 X 2 + 2) = 39/86
Hence, the missing fraction is 19/42.
Q2. In the sequence of numbers 5, 8, 13, X, 34, 55, 89,… the value of X is (1999)
(a) 20
(b) 21
(c) 23
(d) 29
Ans. (b)
In the provided sequence, each number is generated by adding the two preceding numbers.
i.e., 13 = 8 + 5, X = 8 + 13 = 21
13 + 21 = 34, 34 + 21 = 55 and so on,
Therefore, The value of X is 21
Q3. A club has 108 members. Two-third of them are men and the rest are women. All members are married except 9 women members. How many married women are there in the club? (2000)
(a) 20
(b) 24
(c) 27
(d) 30
Ans. (c)
According to the question,
Number of women = ((1/3) X 108 = 36)
∴ Number of married women = Number of women – Number of unmarried women = 36 – 9 = 27
Q4. In a class, there are 18 boys who are over 160 cm tall If these boys constitute three-fourths of the boys and the total number of boys is two-third of the number of students in the class, then what is the number of girls in the class? (2000)
(a) 6
(b) 12
(c) 18
(d) 24
Ans. (b)
Let’s denote the total number of boys as ( n ). As per the question, the number of boys above 160 cm in height is 18.
(3/4)n = 18
⇒ n = 24
Also, let total number of students be N.
Then, (2/3)N = 24
N = (3/2) X 24 = 36
∴ Number of girls = N – n = 36 – 24 = 12
Q5. A conveyer belt delivers baggage at the rate of 3 tonnes in 5 min and a second conveyer belt delivers baggage at the rate of 1 tonnes in 2 min. How much time will it take to get 33 tonnes of baggage delivered using both the conveyer belts? (2001)
(a) 25 min and 30 sec
(b) 30 min
(c) 35 min
(d) 40 min and 45 sec
Ans. (b)
For first belt,
∵ Baggage deliver 3 tonnes in 5 min.
∴ Baggage deliver in 1 min = 3/6 tonnes.
For second belt,
∵ Baggage deliver 1 tonne in 2 min.
∴ Baggage deliver in 1 min = 1/2 tonne
∴ Total baggage deliver in 1 min = 3/5 + 1/2 = 11/10 tonnes.
∴ 11/10 tonnes delivered in 10/11 min.
Hence, time taken to deliver 33 tonnes = 10/11 X 33 = 30 min.
Q6. Three bells toll at intervals of 9, 12 and 15 min respectively. All the three begin to toll at 8 am. At what time will they toll together again? (2003)
(a) 8:45 am
(b) 10:30 am
(c) 11:00 am
(d) 1:30 pm
Ans. (c)
The bells will ring simultaneously again after a duration equal to the least common multiple (LCM) of their individual tolling intervals, which is the LCM of 9, 12, and 15, resulting in 180 minutes. Thus, they will ring together again after 180 minutes, or 3 hours. Adding 3 hours to the initial time of 8:00 am gives us 11:00 am.
Q7. There are 6 persons; A, B, C, D, E and F. (2005)
A has 3 items more than C.
D has 4 items less than B.
E has 6 items less than F.
C has 2 items more than E.
F has 3 items more than D.
Which one of the following figure cannot be equal to the total number of items possessed by all the 6 persons?
(a) 41
(b) 47
(c) 53
(d) 59
Ans. (d)
According to the given equations:
[ A = C + 3 ]
[ D = B – 4 ]
[ E = F – 6 ]
[ C = E + 2 ]
[ F = D + 3 ]
Adding all the equations together, we get:
[ A = B – 2 ]
Substituting this into the expression for the total number of items, we have:
[ Total number of items= A + B + C + D + E + F = A + (A + 2) + (A – 3) + (A – 2) + (A – 5) + (A + 1) = 6A – 7 ]
If ( A = 8 ), then the total number of items is ( 8 \times 6 – 7 = 41 ).
If ( A = 9 ), then the total number of items is ( 9 \times 6 – 7 = 47 ).
If ( A = 10 ), then the total number of items is ( 10 \times 6 – 7 = 53 ).
If ( A = 11 ), then the total number of items is ( 11 \times 6 – 7 = 59 ).
Thus, option (d) is the correct answer.
Q8. Each of the five persons A, B, C, D and E possesses unequal number of similar items. A, B and C possesses twenty-one items in all, while C, D and E possess seven items in all. How many items do A and B possess in all? (2006)
(a) 15
(b) 17
(c) 18
(d) Data is insufficient
Ans. (b)
According to the question,
A + B + C = 21 …(i)
C + D + E = 7 ….(ii)
for eq. (ii) C can take values 1, 2 and 4 as 1 + 2 + 4 = 7
For C, = 1, A + B + 1 = 21, A + B = 20
Similarly, for C = 2, A + B = 21 – 2 = 19
and for C = 4, A + B = 21 – 4 = 17
Q9. If all the numbers from 501 to 700 are written, what is the total number of times does the digit 6 appear? (2007)
(a) 138
(b) 139
(c) 140
(d) 141
Ans. (c)
For numbers between 600 and 700:
Number of occurrences of 6 at the unit’s place = 10
Number of occurrences of 6 at the ten’s place = 10
Number of occurrences of 6 at the hundredth place = 100
For numbers between 501 and 599:
Number of occurrences of 6 at the unit’s place = 10
Number of occurrences of 6 at the ten’s place = 10
Hence, the total number of occurrences of 6 between 501 and 700 is:
10 + 10 + 100 + 10 + 10 = 140
Q10. A person has to completely put each of three liquids: 403 L of petrol, 465 L of diesel and 496 L of mobile oil in bottles of equal size without mixing any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required? (2007)
(a) 34
(b) 44
(c) 46
(d) None of these
Ans. (b)
Maximum capacity of each bottle can be found by taking the HCF of the three given liquids.
Maximum capacity of each bottle = HCF of 403, 465 and 496 = 31
Number of bottles for 403 L petrol = 403/32 = 13
Number of bottles for 465 L of diesel = 465 / 31 = 15
Number of bottles for 496 L of mobile oil = 496 / 31 = 16
Hence, total number of bottles = 13 + 15 + 16 = 44.
Q11. What is the number of terms in the series 117, 120, 123, 126, …, 333? (2008)
(a) 72
(b) 73
(c) 76
(d) 79
Ans. (b)
Given series is 117, 120, 123, 126, …. 333.
Given series is an AP series with first term, a = 117 last term, 1333 and common difference, d = 3
Last term, l = a + (n – 1) * d
where, n = number of terms.
⇒ 117 + (n – 1) * 3 = 333
⇒ (n – 1) * 3 = 216
⇒ n – 1 = 72
∴ n = 72 + 1 = 73
Q12. Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut? (2009)
(a) 27
(b) 36
(c) 43
(d) 400
Ans. (b)
Since each rod must be cut into parts of equal length, and each part must be as long as possible, the highest common factor (HCF) should be considered.
HCF of 78, 104, 117 and 169=13
Number of parts from 78 cm rod = 78/13 = 6
Number of parts from 104 cm rod = 104/13 = 8
Number of parts from 117 cm rod = 117/13 = 9
Number of parts from 169 cm rod = 169/13 = 13
∴ Maximum number of pieces=6+8+9+13=36
Q13. While adding the first few continuous natural numbers, a candidate missed one of the numbers and wrote the answer as 177. What was the number missed? (2009)
(a) 11
(b) 12
(c) 13
(d) 14
Ans. (c)
Let there be n natural numbers and ‘x’ be the number missed out.
Now, (n(n + 1)/2) – x = 177
⇒ n(n+1)- 2x = 354
⇒ n(n+1)=354+2x
Out of the given options only x 13 satisfies it, as n = 19 and x = 13
19(19+1)=354 + 2(13) = 380.
⇒ 380 = 380
hence, x = 13
Q14. 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? (2013)
(a) 3
(b) 5
(c) 10
(d) 18
Ans. (a)
The gardening group meets once every 2 days, the electronics group meets once every 3 days, the chess group meets once every 4 days, the yachting group meets once every 5 days, and the photography group meets once every 6 days.
If they meet on the same day once, the next time they will meet on the same day again will be the least common multiple (LCM) of 2, 3, 4, 5, and 6, which equals 60 days.
Hence, within 180 days, all five groups will meet on the same day 180/60 = 3 times.
Q15. The letters L, M, N, O, P, Q, R, S and T in their order are substituted by nine integers 1 to 9 but not in that order. 4 is assigned to P. The difference between P and Tis 5. The difference between Nand Tis 3. What is the integer assigned to N? (2014)
(a) 7
(b) 5
(c) 4
(d) 6
Ans. (d)
According to the question,
P=4
Now, difference between P and T = 5
⇒ T-P=5
⇒ T-4=5
⇒ T=9
Now, again difference between N and T = 3
⇒ T-N=3
⇒ 9-N=3
⇒ N=6
Hence, option (d) is correct.
Q16. For a charity show, the total tickets sold were 420 Half of these tickets were sold at the rate of ₹5 each, one-third at the rate of ₹3 each and the rest for ₹ 2 each. What was the total amount received? (2014)
(a) ₹900
(b) ₹1540
(c) ₹1610
(d) ₹2000
Ans. (c)
Given, total number of tickets = 420
Number of tickets sold at ₹5= 420/2 = 210
Number of tickets sold at ₹3 = 420/3 = 140
Remaining number of tickets sold at ₹2=420-(210+140)
=420-350=70 tickets
(oros) Now, total amount received
= (210 x5)+(140 x 3) + (70×2)
=1050+420 + 140
= ₹1610
Hence, option (c) is correct.
Q17. There are 240 balls and n number of boxes B₁, B,, B3,…, B. The balls are to be placed in the boxes such that B, should contain 4 balls more than B₂, B should contain 4 balls more than B, and so on. 2 Which one of the following cannot be the possible value of n? (2009)
(a) 4
(b) 5
(c) 6
(d) 7
Ans. (d)
Since, common difference between number of balls in box is 4, so it can be taken as an AP series.
B1-B2 = B2-B3… =4
Total number of balls =240
n/2[2a + (n-1)d] = 240
(since, common difference, d = B2 – B1 =4)
n[2a(n-1)4] = 480
2a-(n-1)4 = 480/n
As we know, n i.e. must be an integer and divide 480 completely. Only option (d), i.e. 7 can not divide 480 completely.
Q18. Each person’s performance compared with all other persons is to be done to rank them subjectively. How many comparisons are needed in total, if there are 11 persons? (2010)
(a) 66
(b) 55
(c) 54
(d) 45
Ans. (b)
According to the question, each person is to be compared with the other 10 persons, resulting in 10 comparisons for the first person. Similarly, the second person is to be compared with the remaining 9 persons, leading to 9 comparisons. This pattern continues until the last person, who does not need to be compared with anyone since all comparisons have already been made.
Hence, total number of comparison = 10+9+…+1
= 10(10+1)/2 [since, sum of n terms = n(n+1)/2]
= 55
Q19. A contract on construction job specifies a penalty for delay in completion of the work beyond a certain date is as follows. ₹200 for the first day, ₹250 for the second day, ₹300 for the third day etc., the penalty for each succeeding day being ₹50 more than that of the preceding day. How much penalty should the contractor pay if he delays the work by 10 days? (2011)
(a) ₹4950
(b) ₹4250
(c) ₹3600
(d) ₹650
Ans. (b)
The penalty charges increase in an arithmetic progression, with the first term being 200 and an increasing common difference of 50. There are a total of 10 terms in this progression.
a = 200, d = 50 and n = 10
since, Required sum to be paid as penalty
= n/2[2a+(n-1) x d]
= 10/2[2 x 200 + (10-1) x 50]
= [400 + 450] x 5 = ₹4250
Q20. There are some balls of red, green and yellow colour lying on a table. There are as many red balls as there are yellow balls. There are twice as many yellow balls as there are green ones. The number of red balls (2013)
(a) is equal to the sum of yellow and green balls
(b) is double the number of green balls
(c) is equal to yellow balls minus green balls
(d) Cannot be ascertained
Ans. (b)
Let, the number of yellow balls = y
Then, number of red balls = y
and number of green balls = y/2
y = 2 x number of green balls
Hence, the number of red balls is double the number of green balls.
In case you still have your doubts, contact us on 9811333901.
For UPSC Prelims Resources, Click here
For Daily Updates and Study Material:
Join our Telegram Channel – Edukemy for IAS
- 1. Learn through Videos – here
- 2. Be Exam Ready by Practicing Daily MCQs – here
- 3. Daily Newsletter – Get all your Current Affairs Covered – here
- 4. Mains Answer Writing Practice – here
