Problems on Numbers Questions and Answers

Answer: B

2 only is correct.

Statement 1: two digits with sum 10 and product 24 are 6 and 4, so the number is 64 or 46. The digits of 63 add to 9 and multiply to 18, so it does not qualify.

Statement 2: two digits with difference 4 and product 45 are 9 and 5 (\(9 - 5 = 4\), \(9 \times 5 = 45\)), so the number can be 95 (or 59). Correct.

A quick way to solve such pairs: list the factor pairs of the product (24: 3 × 8, 4 × 6; 45: 5 × 9) and test the other condition.

Answer: A

There are 110 pigeons.

Let there be \(r\) rabbits and \(p\) pigeons. Each has one head; rabbits have 4 legs and pigeons 2:

\[ \begin{aligned} r + p &= 200 \\ 4r + 2p &= 580 \\ 2r &= 580 - 400 = 180 \end{aligned} \]

So \(r = 90\) and \(p = 110\). Check: \(4 \times 90 + 2 \times 110 = 580\). 105, 115 and 120 pigeons give 590, 570 and 560 legs.

Answer: C

The expenditure on donations is ₹ 300.

He spends 2/5 of 5,000, that is 2,000. Food, rent and travel take 1/4 + 1/5 + 2/5 of the expenditure:

\[ \begin{aligned} \tfrac{5}{20} + \tfrac{4}{20} + \tfrac{8}{20} &= \tfrac{17}{20} \\ \text{donations} &= \tfrac{3}{20} \times 2000 = 300 \end{aligned} \]

The trap is to take the fractions of the total earnings rather than of the expenditure, which gives 750; 100, 200 and 400 do not follow from the fractions.

Answer: B

The number is 27.

Let the number be \(10a + b\). Adding 45 reverses the digits:

\[ \begin{aligned} 10a + b + 45 &= 10b + a \\ 9(b - a) &= 45 \\ b - a &= 5 \end{aligned} \]

With the product of the digits 14, the digits are 2 and 7 (\(2 \times 7 = 14\), and \(7 - 2 = 5\)), so the number is 27. Check: \(27 + 45 = 72\).

72 has the right digits but adding 45 gives 117, not 27. 54 and 36 have digit products of 20 and 18, not 14.

Answer: B

The missed number is 20.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). The full sum must be just above 800, because one number was left out:

\[ \begin{aligned} n = 39 &: \frac{39 \times 40}{2} = 780 \\ n = 40 &: \frac{40 \times 41}{2} = 820 \end{aligned} \]

780 is below 800, so Shantanu must have been adding 1 to 40, whose total is 820. The missing number is \(820 - 800 = 20\).

A larger \(n\) does not work: for \(n = 41\) the total is 861, and the missing number would be 61, which is larger than 41 and so cannot be one of the numbers. 15, 25 and 10 would leave sums of 805, 795 and 810, not 800.

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: 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 \(= 253\)