Arithmetic (UPSC) Part- V

Total Questions: 50

11. 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? [2021-II]

(a) ₹420
(b) ₹360
(c) ₹300
(d) ₹270

Correct Answer: (b) ₹360

Solution:y – 40 = 1/2 z
... (1)
z = 2y – 80
z – 40 = x + 40 – y
z = y + 40
... (2)
eqn (i) = eqn (2)
2y – 80 = y + 40
y = 120 put in eqn. (1)

z = 240 – 80 = 160; x = 120 – 40 = 80
Total = 80 + 120 + 160 = 360

12. 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? [2021-II]

(a) 9
(b) 13
(c) 27
(d) 43

Correct Answer: (a) 9

Solution:By using allegation

incorrect responses = 90/10 × 1

13. Consider the following addition problem: [2021-II]

3P + 4P + PP + PP = RQ2; where P, Q and R are different digits.
What is the arithmetic mean of all such possible sums?

(a) 102
(b) 120
(c) 202
(d) 220

Correct Answer: (d) 220

Solution:

14. Consider the following multiplication problem: [2021-II]

(PQ) × 3 = RQQ, where P, Q and R are different digits and R ≠ 0.
What is the value of (P + R) + Q?

(a) 1
(b) 2
(c) 5
(d) insufficient data

Correct Answer: (b) 2

Solution:PQ × 3 = RQQ
Q = 5
P 5
× 3
______
R 55
P = 8
8 5
× 3
______
255
R = 2

(P + R) ÷ Q = (8 + 2) ÷ 5 = 2

15. 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? [2022-II]

(a) 8
(b) 9
(c) 11
(d) 12

Correct Answer: (a) 8

Solution:Sum of the digits = 1 + 2 + 4 + 5 + 7 + 8 + 9 = 36
∴ 36 ÷ 9 = 4, then number is divisible by 9.
After deleting right most digit, number is divisible by 6 when
sum of its digits is divisible by 3 and also lost digit be even
number.
Thus, right most digit is 9 and second digit from the right end
is either 2, 4 or 8.
After deleting, last two digits, the resulting number is divisible
by 5. So, third digit from right is 5.
Again, after deleting last 3 digits resulting number is divisible
by 4. This is only when last two digits of the number formed
by remaining digits is divisible by 4. And after deleting last 5
digits, number is divisible by 2. i.e second digit (from left end)
is an even number.

Based on all statements, we get a number

Where X = 2 or 4 or 8 only Y = 1 or 7

Case–I: When 6th digits = 2
Then, Number formed by first 4 digits are, 1874, 1478, 7418
and 7814. Here all these number is not divisible by 4. Hence,
6th digit is either 4 or 8.

Case–II: When 6th digit is 4. then, number formed by first 4
digits are, 1278, 1872, 7218 and 7812. Here, only 1872 and
7812 are divisible by 4.

Case–III: When 6th digit is 8 then number formed by first 4
digits are 1274, 1472, 7214 and 7412. Here,. only 7412 and
1472 are divisible by 4

Thus, from case II and III, we get that sum of middle three
digits are
1 + 2 + 5 = 8 or 7 + 2 + 5 = 14
Thus option (a) is correct.

16. 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: [2022-II]

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

(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3

Correct Answer: (d) 1, 2 and 3

Solution:Statement–1: 25 notes of 50 = 25 × 50 = 1250
Remaining Amount = 1840 – 1250 = 590
Remaining notes = 50 – 25 = 25
Now, 20X + 10 (25 – X) = 590 [where X = number of 20 notes]
This is not possible even if x = 25 (maximum).
Hence, statement 1 is not correct

Statement–2: 35 notes of ₹20 = 35 × 20 = 700
Remaining Amount = 1840 – 700 = 1140
Remaining notes = 50 – 35 = 15
Now, 50X + 10(15 – X) = 1140
5X + 15 – X = 114
X = 99/4 ≠ not an integer
Hence, statement 2 is not correct

Statement–3: 20 notes of ₹10 = 20 × 10 = 200
Remaining amount = 1840 – 200 = 1640
Remaining notes = 50 – 20 = 30
Now, 50X + 20(30 – X) = 1640
5X + 60 – 2X = 164
X = (164 – 60)/3 = 104/3 ≠ Not an integer
Hence, statement 3 is not correct.

17. Which number amongst 2⁴⁰, 3²¹, 4¹⁸ and 8¹² is the smallest? [2022-II]

(a) 2⁴⁰
(b) 3²¹
(c) 4¹⁸
(d) 8¹²

Correct Answer: (b) 3²¹

Solution:(2)⁴⁰ = (2)⁴⁰
(4)¹⁸ = (2)² × 18 = (2)³⁶
(8)¹² = (2)³ × 12 = (2)³⁶
(3)²¹ = (3)²¹
Hence among the given number, it is obvious that
(3)²¹ is smallest and (2)⁴⁰ is greatest.

18. Consider the question and two statements given below: [2022-II]

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

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

Solution:For X = 2, X/3 = 2/3 (Not an integer).
and, 3X = 3 × 2 = 6 (an Integer)
Again, for X = 2/3 , X/3 = 2/3 ÷ 3 = 2/9 (Not an integer)
and 3X = 3 × 2/3 = 2 (An Integer).
Hence, we can not find X is an integer or fraction.

19. The increase in the price of a certain item was 25%. Then the price was decreased by 20% and then again increased by 10%. What is the resultant increase in the price? [2022-II]

(a) 5%
(b) 10%
(c) 12.5%
(d) 15%

Correct Answer: (b) 10%

Solution:Let price of the item = 100
Price increases by 25%
New price = 100 × (100 + 25)/100 = 125
Price decreases by 20%
⇒ New price = 125 × (100 – 20)/100 = 100
Price increases by 10%
⇒ New price = 100 × (100 + 10)/100 = 110
Resultant increases in price = (110 – 100)/100 × 100 = 10%

Tricks: If three successive percent increase are x%, y% and z%, then single equivalent percent increase
= [ x + y + z + (xy + yz + zx)/100 + xyz/(100)² ]

Here, x = 25%, y = –20% and z = 10%
∴ single percent increase =
[ 25 – 20 + 10 + (25 × (–20) + (–20) × 10 + 25 × 10)/100 + (25 × (–20) × 10)/(100)² ]

= [15 – 4.5 – 0.5] = 10%

20. 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? [2022-II]

(a) 40 m
(b) 40.4 m
(c) 50 m
(d) 50.5 m

Correct Answer: (b) 40.4 m

Solution:Distance between two consecutive plants (in meter)
= (1.01 × 1000)/(101 – 1) = 10.1 m

∴ Distance between 5 consecutive plants
= (5 – 1) × 10.1 = 40.4 m.

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