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: 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.