Problems on Numbers Questions and Answers

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.

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.

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.

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.

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 \( 1 - 1 + 1 = 1 \).

Answer: B

There are 10 eucalyptus trees.

Neem trees are one-sixth of the total, and there are 5 of them, so the garden has \(5 \times 6 = 30\) trees. Half, 15, are Ashoka trees. The rest are eucalyptus:

\[ 30 - 5 - 15 = 10 \]

Equivalently, eucalyptus trees are \(1 - \frac{1}{6} - \frac{1}{2} = \frac{1}{3}\) of 30. The option 15 is the number of Ashoka trees and 5 the number of neem trees.

Answer: A

Cheema gets \(\frac{1}{6}\) of the pizza.

Follow the pizza step by step:

\[ \begin{aligned} \text{after Sonu} &= 1 - \tfrac{1}{4} = \tfrac{3}{4} \\ \text{Priya} &= \tfrac{1}{3} \times \tfrac{3}{4} = \tfrac{1}{4} \\ \text{left} &= \tfrac{3}{4} - \tfrac{1}{4} = \tfrac{1}{2} \\ \text{Cheema} &= \tfrac{1}{2} \div 3 = \tfrac{1}{6} \end{aligned} \]

The trap is taking Priya's third of the whole pizza instead of the remainder: that leaves \(1 - \frac{1}{4} - \frac{1}{3} = \frac{5}{12}\), and a third of it is \(\frac{5}{36}\). Dividing Sonu's quarter by three gives \(\frac{1}{12}\), which answers a different question.

Answer: C

Kanta has 25 chocolates left.

Five dozen is 60 chocolates. Work through each gift in turn:

\[ \begin{aligned} \text{to Aruna} &= \tfrac{1}{6} \times 60 = 10 \\ \text{left} &= 60 - 10 = 50 \\ \text{to Radha} &= \tfrac{1}{5} \times 50 = 10 \\ \text{left} &= 50 - 10 = 40 \end{aligned} \]

Each of the three friends gets \(\frac{1}{8}\) of the remaining 40, that is 5 each, all taken from the same 40:

\[ \begin{aligned} 3 \times 5 &= 15 \\ 40 - 15 &= 25 \end{aligned} \]

15 is the number given to the three friends, not the number Kanta keeps, and 20 and 10 do not follow from any correct reading of the three steps.

Answer: B

There are 37 students in the class.

Ruhi is 19th from one end, so 18 students stand before her on that side. She is also 19th from the other end, so 18 students stand on that side too.

\[ \text{Total} = 18 + 1 + 18 = 37 \]

The usual shortcut is (position from the left) + (position from the right) - 1 = 19 + 19 - 1 = 37. The answer 38 is what you get by forgetting to subtract 1, so Ruhi is counted twice; 36 subtracts her as well; 40 has no basis.

Answer: B

The number is 63.

It is more than 6 tens and less than 8 tens, so it lies between 60 and 80, and its tens digit is 6 or 7.

The tens digit is double the ones digit, so it must be even: the tens digit is 6 and the ones digit is 3. Check the digit sum: 6 + 3 = 9, and 63 is larger than half of 100.

The distractors fail one condition each. 72 has digit sum 9 but 7 is not double 2. 54 is less than 60. 81 is more than 8 tens, and 8 is not double 1 either.