Playing with Numbers Quiz - ICSE Class 8 Mathematics

Free practice quiz: Playing with Numbers (ICSE Class 8 Mathematics). 10 multiple-choice questions with answers and explanations. Play online, no sign-up needed.

This ICSE Class 8 Mathematics practice quiz has 10 multiple-choice questions on Playing with Numbers, and 10 of them come with an explanation. Try the 5 sample questions below (tap an option to check yourself), then play the full quiz online. No sign-up needed.

Sample questions

1. A number when divided by 13 leaves a remainder of 5. What is the remainder when the same number is divided by 17?

   1. 5
   2. 12
   3. Cannot be determined
   4. 0
   Show answer

   Answer: C. Cannot be determined

   Let the number be N. We are given that N = 13k + 5 for some integer k. We are asked to find the remainder when N is divided by 17. N can be 5, 18, 31, 44, 57, 70, 83, 96, 109, 122, 135, 148, 161, 174, etc. When these numbers are divided by 17, the remainders are 5, 1, 14, 10, 6, 2, 14, 11, 7, 3, 15, 12, 8, 4, etc. Since the remainders are not constant, it cannot be determined.

2. The sum of the digits of a two-digit number is 12. If the digits are reversed, the new number is 36 greater than the original number. Find the original number.

   1. 39
   2. 48
   3. 57
   4. 66
   Show answer

   Answer: B. 48

   Let the two-digit number be represented as $10x + y$, where x is the tens digit and y is the units digit. We are given that $x + y = 12$. When the digits are reversed, the new number is $10y + x$. We are also given that $10y + x = (10x + y) + 36$. Simplifying this equation, we get $9y - 9x = 36$, which means $y - x = 4$. Now we have a system of two linear equations: $x + y = 12$ and $y - x = 4$. Adding these two equations, we get $2y = 16$, so $y = 8$. Substituting y = 8 into $x + y = 12$, we get $x + 8 = 12$, so $x = 4$. The original number is $10x + y = 10(4) + 8 = 48$.

3. Find the smallest number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, but is divisible by 7.

   1. 101
   2. 121
   3. 301
   4. 421
   Show answer

   Answer: C. 301

   A number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6 must be of the form LCM(2, 3, 4, 5, 6) * k + 1. The LCM of 2, 3, 4, 5, and 6 is 60. So, the number is of the form $60k + 1$. We need to find the smallest such number that is also divisible by 7. We test values of k: k=1: $60(1) + 1 = 61$. $61 \div 7 = 8$ remainder 5. k=2: $60(2) + 1 = 121$. $121 \div 7 = 17$ remainder 2. k=3: $60(3) + 1 = 181$. $181 \div 7 = 25$ remainder 6. k=4: $60(4) + 1 = 241$. $241 \div 7 = 34$ remainder 3. k=5: $60(5) + 1 = 301$. $301 \div 7 = 43$ remainder 0. So, the smallest number is 301.

4. If $557 imes 97 = 54029$ and $557 imes 98 = 54586$, what is $557 imes 97.5$?

   1. 54307.5
   2. 54307
   3. 54308
   4. 54308.5
   Show answer

   Answer: A. 54307.5

   We are given two multiplications. We need to find $557 imes 97.5$. Notice that $97.5$ is exactly halfway between 97 and 98. Therefore, $557 imes 97.5$ will be exactly halfway between $557 imes 97$ and $557 imes 98$. Average = $(54029 + 54586) / 2 = 108615 / 2 = 54307.5$. Alternatively, $557 imes 97.5 = 557 imes (97 + 0.5) = (557 imes 97) + (557 imes 0.5) = 54029 + 278.5 = 54307.5$.

5. What is the least number which when divided by 2, 3, 4, 5, 6, 7 and 8 leaves no remainder?

   1. 840
   2. 1680
   3. 2520
   4. 5040
   Show answer

   Answer: A. 840

   The least number which when divided by several numbers leaves no remainder is their Least Common Multiple (LCM). We need to find the LCM of 2, 3, 4, 5, 6, 7, and 8. Prime factorization of each number: $2 = 2$ $3 = 3$ $4 = 2^2$ $5 = 5$ $6 = 2 imes 3$ $7 = 7$ $8 = 2^3$ To find the LCM, we take the highest power of each prime factor present: Highest power of 2 is $2^3 = 8$. Highest power of 3 is $3^1 = 3$. Highest power of 5 is $5^1 = 5$. Highest power of 7 is $7^1 = 7$. LCM = $2^3 imes 3 imes 5 imes 7 = 8 imes 3 imes 5 imes 7 = 24 imes 35$. $24 imes 35 = 24 imes (30 + 5) = 720 + 120 = 840$. So, the least number is 840.

More questions in this quiz

Answers and explanations are in the full quiz.

1. If a number is divisible by both 15 and 18, then it must be divisible by which of the following?
2. What is the smallest 6-digit number that is divisible by 3, 4, and 5?
3. What is the value of a positive integer N if $N \div 3$ has remainder 2, $N \div 5$ has remainder 3, and $N \div 7$ has remainder 2?
4. The sum of three consecutive multiples of 11 is 297. What is the largest of these multiples?
5. If $123456789 imes 9 = 1111111101$, then $123456789 imes 18$ is:

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