Arithmetic - Practice Set

📝 Arithmetic — Practice Set 1

Questions 1-10

Q1. The four integers next lower than 81, and the four next higher than 81, are written down and added together, this sum is divisible by,

• (a) 7
• (b) 9
• (c) 11
• (d) 13

Answer: (b) Since 81 is divisible by 9, the sum is divisible by 9.

Q2. How many numbers, between 1 and 300 are divisible by 3 and 5 together?

• (a) 16
• (b) 18
• (c) 20
• (d) 100

Answer: (c) LCM of 3 and 5 = 15. 300/15 = 20 numbers.

Q3. A number when divided by 5 leaves a remainder 3. What is the remainder when the square of the same number is divided by 5?

• (a) 9
• (b) 3
• (c) 0
• (d) 4

Answer: (d) Let number = 5q + 3. (5q+3)² = 5(5q²+6q+1) + 4. Remainder = 4.

Q4. If 5432*7 is divisible by 9, then the digit in place of * is

• (a) 0
• (b) 1
• (c) 6
• (d) 8

Answer: (c) Sum of digits = 5+4+3+2+*+7 = 21+*. For divisible by 9, * = 6.

Q5. At a college football game, 4/5 of the seats in the lower deck of the stadium were sold. If 1/4 of all the seating in the stadium is located in the lower deck, and if 2/3 of all the seats in the stadium were sold, then what fraction of the unsold seats in the stadium was in the lower deck?

• (a) 3/20
• (b) 1/6
• (c) 1/5
• (d) 1/3

Answer: (a) Required fraction = (p/20)/(p/3) = 3/20.

Q6. If a, a + 2 and a + 4 are prime numbers, then the number of possible solutions for a is

• (a) one
• (b) two
• (c) three
• (d) more than three

Answer: (a) Only a = 3 gives 3, 5, 7 as prime numbers.

Q7. The product of two consecutive even numbers is 12768. What is the greater number?

• (a) 110
• (b) 108
• (c) 114
• (d) 112

Answer: (c) 112 × 114 = 12768. Greater number = 114.

Q8. What is 131 times 333?

• (a) 46323
• (b) 43623
• (c) 43290
• (d) 44955

Answer: (b) 333 × 131 = 43623.

Q9. The cost of 3 chairs and 10 tables is 9856. What is the cost of 6 chairs and 20 tables?

• (a) 17227
• (b) 18712
• (c) 19172
• (d) Cannot be determined

Answer: (c) 2 × (3x + 10y) = 2 × 9856 = 19712.

Q10. Three friends A, B and C start running around a circular stadium and complete a single round in 24, 36 and 30 seconds respectively. After how many minutes will they meet again at the starting point?

• (a) 12
• (b) 6
• (c) 8
• (d) 15

Answer: (b) LCM of 24, 36 and 30 = 360 seconds = 6 minutes.

Questions 11-20

Q11. If the fractions 2/5, 3/4, 4/5, 5/7 and 6/11 are arranged in ascending order of their values, which one will be the fourth?

• (a) 2/5
• (b) 3/4
• (c) 4/5
• (d) 6/11

Answer: (b) Order: 2/5 < 6/11 < 5/7 < 3/4 < 4/5. Fourth is 3/4.

Q12. The difference between a two-digit number and the number obtained by interchanging the two digits of the number is 9. The sum of the digits of the number is 15. What is the product of the two digits of the two-digit number?

• (a) 54
• (b) 72
• (c) 56
• (d) Cannot be determined

Answer: (c) Let number = 10x + y. y - x = 1 and x + y = 15. So y = 8, x = 7. Product = 56.

Q13. If (74)² is subtracted from the square of a number, the answer so obtained is 5340. What is the number?

• (a) 98
• (b) 102
• (c) 104
• (d) 110

Answer: (c) x² - 74² = 5340 ⇒ x² = 10816 ⇒ x = 104.

Q14. The H.C.F and L.C.M of two numbers are 21 and 4641 respectively. If one of the numbers lies between 200 and 300, then the two numbers are

• (a) 273, 357
• (b) 273, 361
• (c) 273, 359
• (d) 273, 363

Answer: (a) Product = 21 × 4641 = 21 × 3 × 7 × 13 × 17. Numbers are 273 and 357.

Q15. What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24?

• (a) 264
• (b) 259
• (c) 269
• (d) None of these

Answer: (b) LCM of 8, 11, 24 = 264. Required number = 264 - 5 = 259.

Q16. Which is the least number that must be subtracted from 1856, so that the remainder when divided by 7, 12 and 16 will leave the same remainder 4?

• (a) 137
• (b) 1361
• (c) 140
• (d) 172

Answer: (d) LCM of 7, 12, 16 = 336. 1856 - x = 5 × 336 + 4 = 1684. x = 172.

Q17. Three bells chime at an interval of 18, 24 and 32 minutes respectively. At a certain time they begin to chime together. What length of time will elapse before they chime together again?

• (a) 2 hours 24 minutes
• (b) 4 hours 48 minutes
• (c) 1 hour 36 minutes
• (d) 5 hours

Answer: (b) LCM of 18, 24, 32 = 288 minutes = 4 hours 48 minutes.

Q18. The L.C.M. of two number is 630 and their H.C.F. is 9. If the sum of numbers is 153, their difference is

• (a) 17
• (b) 23
• (c) 27
• (d) 33

Answer: (c) xy = 630 × 9 = 5670. x+y = 153. (x-y)² = 23409 - 22680 = 729. x-y = 27.

Q19. Suppose you have 108 green marbles and 144 red marbles. You decide to separate them into packages of equal number of marbles. Find the maximum possible number of marbles in each package.

• (a) 4
• (b) 36
• (c) 9
• (d) 12

Answer: (b) HCF of 108 and 144 = 36.

Q20. I have a certain number of beads which lie between 600 and 900. If 2 beads are taken away the remainder can be equally divided among 3, 4, 5, 6, 7 or 12 boys. The number of beads I have

• (a) 729
• (b) 842
• (c) 576
• (d) 961

Answer: (b) LCM of 3, 4, 5, 6, 7, 12 = 420. Number = 420 × 2 + 2 = 842.

Questions 21-30

Q21. Three men start together to travel the same way around a circular track of 11 kms. Their speeds are 4, 5.5 and 8 kms per hour respectively. When will they meet at the starting point?

• (a) 22 hrs
• (b) 12 hrs
• (c) 11 hrs
• (d) 44 hrs

Answer: (a) LCM of 11/4, 2, 11/8 = 22 hrs.

Q22. A man was engaged for a certain number of days for Rs. 404.30 but being absent for some days he was paid only Rs. 279.90. His daily wages cannot exceed by:

• (a) Rs. 29.10
• (b) Rs. 31.30
• (c) Rs. 31.10
• (d) Rs. 31.41

Answer: (c) HCF of 404.30 and 279.90 = 31.10.

Q23. The L.C.M. of two numbers is 45 times their H.C.F. If one of the numbers is 125 and the sum of H.C.F. and L.C.M. is 1150, the other number is:

• (a) 215
• (b) 220
• (c) 225
• (d) 235

Answer: (c) l = 45h, l + h = 1150 ⇒ h = 25, l = 1125. Other number = (25 × 1125)/125 = 225.

Q24. There are 4 numbers. The H.C.F. of each pair is 3 and the L.C.M. of all the 4 numbers is 116. What is the product of 4 numbers?

• (a) 9782
• (b) 9396
• (c) 9224
• (d) 9100

Answer: (b) Product = (HCF for each pair)⁴ × LCM = 3⁴ × 116 = 9396.

Q25. Three wheels can complete respectively 60, 36, 24 revolutions per minute. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

• (a) 5/2 seconds
• (b) 5/3 seconds
• (c) 5 seconds
• (d) 7.5 seconds

Answer: (c) LCM of 1, 5/3, 5/2 = 5 seconds.

Q26. In a school there are 391 boys and 323 girls. These are to be divided into the largest possible equal classes, so that there are equal number of boys and girls in each class. How many classes are possible?

• (a) 32
• (b) 37
• (c) 42
• (d) 49

Answer: (c) HCF of 391 and 323 = 17. Number of classes = (391+323)/17 = 42.

Q27. A shopkeeper has three kinds of sugar 184 kg; 230 kg and 276kg. He wants to store it into minimum number of bags to equal size without mixing. Find the size of the bag and the number of bags required.

• (a) 23 kg; 30
• (b) 38 kg; 23
• (c) 46 kg; 15
• (d) 46 kg; 25

Answer: (c) HCF of 184, 230, 276 = 46. Bags = 184/46 + 230/46 + 276/46 = 4+5+6 = 15.

Q28. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

• (a) 37
• (b) 13
• (c) 23
• (d) 33

Answer: (c) LCM of 5, 6, 4, 3 = 60. 2497 ÷ 60 gives remainder 37. Number to add = 60 - 37 = 23.

Q29. The H.C.F. of two numbers is 8. Which one of the following can never be their L.C.M.?

• (a) 24
• (b) 48
• (c) 56
• (d) 60

Answer: (d) HCF must divide LCM. 8 does not divide 60.

Q30. The sum of two numbers is 462 and their highest common factor is 22. What is the maximum number of pairs that satisfy these conditions?

• (a) 1
• (b) 3
• (c) 5
• (d) 6

Answer: (d) Pairs: (22,440), (44,418), (88,374), (110,352), (176,286), (220,242). Total 6 pairs.

Complete Answer Key

1. (b)	2. (c)	3. (d)	4. (c)	5. (a)
6. (a)	7. (c)	8. (b)	9. (c)	10. (b)
11. (b)	12. (c)	13. (c)	14. (a)	15. (b)
16. (d)	17. (b)	18. (c)	19. (b)	20. (b)
21. (a)	22. (c)	23. (c)	24. (b)	25. (c)
26. (c)	27. (c)	28. (c)	29. (d)	30. (d)