Problems on Numbers Questions and Answers

Answer: C

The youngest child got ₹ 16 lakh.

Let the total be T lakh. The eldest got 0.5T + 2 and the youngest 0.25T - 3, so the middle child got the rest:

\[ \begin{aligned} M &= T - (0.5T + 2) - (0.25T - 3) \\ &= 0.25T + 1 \end{aligned} \]

After spending half on a house, he had 6 lakh less than the youngest:

\[ \begin{aligned} 0.125T + 0.5 &= (0.25T - 3) - 6 \\ 0.125T &= 9.5 \Rightarrow T = 76 \end{aligned} \]

The youngest got 0.25 × 76 - 3 = 16 lakh. Check: the eldest got 40 lakh and the middle child 20 lakh, keeping 10 lakh after buying the house, which is 6 lakh less than 16.

Answer: B

The container holds 24 bottles of milk.

Let the capacity be C bottles. Taking out 5 bottles and pouring in 3 is a net loss of 2 bottles, which takes the level from three-quarters to two-thirds:

\[ \begin{aligned} \tfrac{3}{4}C - 2 &= \tfrac{2}{3}C \\ \tfrac{1}{12}C &= 2 \\ C &= 24 \end{aligned} \]

Check: three-quarters of 24 is 18; after losing 2 bottles it holds 16, which is two-thirds of 24.

12 comes from treating the net change as 1 bottle, and 20 and 15 give non-whole quarter or third fills, so they cannot fit.

Answer: B

At most 8 chapters can begin on odd pages.

Chapter 1 starts on page 1. A chapter with an even number of pages keeps the next chapter on the same parity; a chapter with an odd number of pages flips it.

The lengths are ten different numbers from 20 to 30. Only six of those are even (20, 22, 24, 26, 28, 30), so among chapters 1 to 9, whose lengths decide where chapters 2 to 10 start, at least three must have odd lengths.

Best arrangement: chapters 1 to 6 even, so chapters 1 to 7 all start on odd pages; chapter 7 odd (chapter 8 starts even), chapter 8 odd (chapter 9 back to odd), chapter 9 odd (chapter 10 starts even). That is 7 + 1 = 8 odd starts.

9 would allow at most two odd-length chapters among the first nine, which needs seven even lengths, and only six exist. 6 and 5 are possible but not the largest.

Answer: D

The difference is 132.

Let the first number be \(x\). Then the second is \(\frac{x}{2}\) and the third \(\frac{x}{3}\). Their sum is three times the average:

\[ \begin{aligned} x + \tfrac{x}{2} + \tfrac{x}{3} &= \tfrac{11x}{6} = 363 \\ x &= 198,\quad \tfrac{x}{3} = 66 \\ 198 - 66 &= 132 \end{aligned} \]

99 is the second number, and 144 or 77 come from setting up the ratios the wrong way (for example making the second twice the first).

Answer: D

X + 21 = 50.

Let X have tens digit \(a\) and units digit \(b\). Reversing adds 63:

\[ \begin{aligned} (10b + a) - (10a + b) &= 9(b - a) = 63 \\ b - a &= 7,\quad a + b = 11 \\ a &= 2,\ b = 9 \end{aligned} \]

So X = 29 and X + 21 = 50. The option 29 is X itself, not X + 21, and 113 adds 21 to the reversed number 92.

Check: 92 - 29 = 63 and 2 + 9 = 11. In general, reversing a two-digit number changes it by 9 times the difference of its digits, a useful shortcut for such questions.

Answer: B

The fraction is \(\frac{5}{3}\).

Test the condition \(x - \frac{2}{x} = \frac{7}{15}\):

\[ \begin{aligned} \frac{5}{3} - 2 \times \frac{3}{5} &= \frac{5}{3} - \frac{6}{5} \\ &= \frac{25 - 18}{15} = \frac{7}{15} \end{aligned} \]

For \(\frac{3}{5}\) the result is \(\frac{3}{5} - \frac{10}{3}\), negative; for \(\frac{4}{3}\) it is \(\frac{4}{3} - \frac{3}{2} \lt 0\); for \(\frac{3}{4}\) it is also negative. Since \(x\) must exceed twice its reciprocal, it must be greater than \(\sqrt{2}\), and only \(\frac{5}{3}\) is.

Answer: A

Let the smaller be \( y \); the greater is \( 4y + 5 \).

\[ y + 4y + 5 = 80 \quad\Longrightarrow\quad 5y = 75 \]

\[ y = 15, \qquad \text{greater} = 65 \]

Answer: D

\[ x - \frac{3x}{8} = 25 \quad\Longrightarrow\quad \frac{5x}{8} = 25 \]

\[ x = 40 \]

Answer: D

Let the number be \(x\)
Then, \(\dfrac{1}{3}\) of \(\dfrac{1}{4}\) of \(x = 15 \Leftrightarrow x =15 \times 12 = 180\)
So, required number
\(\left( \dfrac{3}{10} \times 180 \right) =54\)

Answer: B

Let the middle digit be \(x.\)
Then, \(2x = 10\) or \(x = 5\). So, the number is either \(253 \text{ or } 352.\)
Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.
Hence, required number \(= 253\)