Problems on Numbers Questions and Answers

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.

Answer: A

The number is 87649.

Fill in the digits from the clues, starting with the ones that are fixed directly:

\[ \begin{aligned} \text{tens} &= 2^{2} = 4 \\ \text{hundreds} &: \tfrac{2}{3} \times h = 4,\ h = 6 \\ \text{ten thousands} &: \tfrac{3}{4} \times t = 6,\ t = 8 \\ \text{thousands} &= 7,\quad \text{units} = 9 \end{aligned} \]

The smallest prime is 2, so the tens digit is 4. The largest single-digit prime is 7 and the largest single-digit odd number is 9. Reading from the ten thousands place: 8, 7, 6, 4, 9, giving 87649.

The other numbers fail at once: 49327 and 42937 have the wrong tens digit, and 83419 has 4 in the hundreds place, which is not three-fourths of 8.

Answer: B

The sum of the first three numbers is 9.

Five consecutive numbers have their middle number equal to their average:

\[ \begin{aligned} \text{middle} &= 20 \div 5 = 4 \\ \text{numbers} &= 2, 3, 4, 5, 6 \\ 2 + 3 + 4 &= 9 \end{aligned} \]

Check: 2 + 3 + 4 + 5 + 6 = 20. The answer 12 is 3 + 4 + 5, the middle three, and 5 is just 2 + 3, the first two.

Answer: B

The number is 8.

Let the number be \(x\). Increasing it by 8 doubles it:

\[ \begin{aligned} x + 8 &= 2x \\ x &= 8 \end{aligned} \]

Check: 8 + 8 = 16, which is double 8. For the other options, 6 + 8 = 14 (not 12), 12 + 8 = 20 (not 24) and 16 + 8 = 24 (not 32), so none of them doubles.

Answer: A

The original number is 39.

Let the tens digit be \(a\) and the units digit \(b\). The number is \(10a + b\) and its reverse is \(10b + a\).

\[ \begin{aligned} a + b &= 12 \\ (10b + a) - (10a + b) &= 9(b - a) = 54 \\ b - a &= 6 \\ a = 3,\ b &= 9 \end{aligned} \]

Check: 93 - 39 = 54. 93 has the right digits but reversing it gives a smaller number. 57 reversed is 75, larger by only 18, and 66 reversed is unchanged.

Answer: A

Raju has 5 notes of Rs. 10.

Let t be the number of Rs. 10 notes and w the number of Rs. 20 notes:

\[ \begin{aligned} t + w &= 14 \\ 10t + 20w &= 230 \\ 10t + 20(14 - t) &= 230 \\ 280 - 10t &= 230 \Rightarrow t = 5 \end{aligned} \]

So he has 5 notes of Rs. 10 and 9 of Rs. 20: 50 + 180 = 230. The option 9 is the number of Rs. 20 notes, the usual slip of solving for the wrong variable. 6 and 10 give totals of 220 and 180.