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.
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A number when divided by 13 leaves a remainder of 5. What is the remainder when the same number is divided by 17?
- 5
- 12
- Cannot be determined
- 0
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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.
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.
- 39
- 48
- 57
- 66
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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 $10x + y = 10(4) + 8 = 48$.
Find the smallest number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, but is divisible by 7.
- 101
- 121
- 301
- 421
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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.
If $557 imes 97 = 54029$ and $557 imes 98 = 54586$, what is $557 imes 97.5$?
- 54307.5
- 54307
- 54308
- 54308.5
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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, $557 imes 97.5 = 557 imes (97 + 0.5) = (557 imes 97) + (557 imes 0.5) = 54029 + 278.5 = 54307.5$.
What is the least number which when divided by 2, 3, 4, 5, 6, 7 and 8 leaves no remainder?
- 840
- 1680
- 2520
- 5040
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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.