101.
One-fifth of a number added to 8 gives 20. The number is:
Answer: B
\[ \frac{x}{5} + 8 = 20 \quad\Longrightarrow\quad \frac{x}{5} = 12 \]
\[ x = 60 \]
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Problems on Numbers Questions and Answers
Important Instructions
102.
The least number that must be added to 1056 to make it exactly divisible by 23 is:
Answer: C
Dividing 1056 by 23 gives 45 with a remainder of 21, since \(23 \times 45 = 1035\). To reach the next multiple, \(23 \times 46 = 1058\), we must add \(1058 - 1056 = 2\). The remainder 21 is what must be added to reach the multiple only if we were subtracting instead.
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103.
The greatest number that divides 43, 91 and 183 leaving the same remainder in each case is:
Answer: D
When a divisor leaves the same remainder, it divides the differences exactly. The differences are \(91 - 43 = 48\), \(183 - 91 = 92\) and \(183 - 43 = 140\). The highest common factor of 48, 92 and 140 is 4. Check: each of 43, 91 and 183 leaves remainder 3 on division by 4.
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104.
A number leaves remainder 4 when divided by 7. The remainder when twice that number is divided by 7 is:
Answer: A
Write the number as \(7k + 4\). Twice it is \(14k + 8 = 7(2k + 1) + 1\), so the remainder is 1. Working with remainders directly, \(2 \times 4 = 8\) and \(8\) leaves 1 on division by 7 - remainders can be multiplied and then reduced.
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105.
The least number which when divided by 6, 7 and 8 leaves remainder 3 in each case is:
Answer: B
A number leaving the same remainder with each divisor is the lowest common multiple plus that remainder. The LCM of 6, 7 and 8 is 168, so the number is \(168 + 3 = 171\). Check: 171 divided by 6, 7 and 8 leaves 3 each time.
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106.
The smallest three-digit number exactly divisible by 11 is:
Answer: C
The first multiple of 11 at or above 100:
\[ 11 \times 9 = 99, \qquad 11 \times 10 = 110 \]
So 110. 111 is not a multiple of 11 - its alternating digit sum is \( 1 - 1 + 1 = 1 \).
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