Problems on Numbers Questions and Answers

Answer: B

The number is 34.

Let the number be \(10a + b\). Adding 9 reverses the digits:

\[ \begin{aligned} 10a + b + 9 &= 10b + a \\ 9(b - a) &= 9 \Rightarrow b = a + 1 \end{aligned} \]

Two consecutive digits with product 12 are 3 and 4, so the number is 34. Check: \(3 \times 4 = 12\) and \(34 + 9 = 43\).

43 has the right digits but adding 9 gives 52, not 34. 62 has product 12 but \(62 + 9 = 71\). 2 is not a two-digit number.

Answer: D

The third integer is 15.

Three consecutive odd integers can be written \(x\), \(x + 2\), \(x + 4\). The condition is

\[ \begin{aligned} 3x &= 2(x + 4) + 3 \\ 3x &= 2x + 11, \quad x = 11 \end{aligned} \]

So the integers are 11, 13 and 15, and the third is 15. Check: \(3 \times 11 = 33\) and \(2 \times 15 + 3 = 33\). 11 is the first integer, a common slip, and 13 is the middle one.

Answer: D

The sum is always divisible by 11.

Write the number as \(10a + b\), where \(a\) and \(b\) are its digits. Interchanging the digits gives \(10b + a\). Adding the two:

\[ \begin{aligned} (10a + b) + (10b + a) &= 11a + 11b \\ &= 11(a + b) \end{aligned} \]

For example, \(47 + 74 = 121 = 11 \times 11\). The sum is divisible by 3, 5 or 7 only when \(a + b\) happens to be, so none of those holds for every number: \(12 + 21 = 33\) is not divisible by 5 or 7.

Answer: B

6 other two digit numbers increase by 18.

For a number \(10a + b\), reversing gives \(10b + a\), and the increase is

\[ (10b + a) - (10a + b) = 9(b - a) = 18 \]

So the units digit must be 2 more than the tens digit: 13, 24, 35, 46, 57, 68 and 79. That is 7 numbers, and leaving out 13 itself, 6 others. The option 7 counts 13 as well, which the question excludes with the word 'other'.

Answer: B

Raman bought 26 kgs.

Work backwards from the 2 kg left at the end. If x kg was in the store before a buyer took half plus k kg, then what remained was x/2 - k.

\[ \begin{aligned} \text{before Amar} &: x/2 - 3 = 2 \Rightarrow x = 10 \\ \text{before Nazneen} &: y/2 - 2 = 10 \Rightarrow y = 24 \\ \text{before Raman} &: z/2 - 1 = 24 \Rightarrow z = 50 \end{aligned} \]

The store began with 50 kg. Raman bought half of it plus 1 kg: 25 + 1 = 26 kg.

50 kgs is the starting stock, not Raman's share; 24 kgs is what was left after Raman's purchase; and 52 kgs comes from doubling 26.

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.