Problems on Numbers Questions and Answers

Answer: C

The smallest such number is 123456789.

The digits 0 to 9 add up to 45, so distinct digits summing to 45 must be all ten digits or all of 1 to 9 (leaving out 0, which adds nothing). A nine-digit number is smaller than any ten-digit one, and arranging 1 to 9 in increasing order gives the smallest: 123456789.

123555789 repeats the digit 5, and 99999 repeats 9, so neither has distinct digits. 123457869 uses the right digits but is larger than 123456789, because its sixth digit is 7 instead of 6.

Answer: B

The number is 72.

Write the number as 10a + b. Subtracting 45 reverses the digits:

\[ \begin{aligned} 10a + b - 45 &= 10b + a \\ 9(a - b) &= 45, \ a - b = 5 \\ a + b &= 9 \Rightarrow a = 7,\ b = 2 \end{aligned} \]

Check: 72 - 45 = 27. The other options fail: 63 - 45 = 18, 81 - 45 = 36 and 90 - 45 = 45, none of which reverses the digits.

Answer: C

There are 15 keepers.

Let the number of keepers be k. Hens and keepers have 2 feet, goats and camels have 4, and every member of the caravan has one head.

\[ \begin{aligned} \text{heads} &= 50 + 45 + 8 + k = 103 + k \\ \text{feet} &= 100 + 180 + 32 + 2k = 312 + 2k \end{aligned} \]

\[ \begin{aligned} (312 + 2k) - (103 + k) &= 224 \\ 209 + k &= 224 \\ k &= 15 \end{aligned} \]

Check: with 15 keepers there are 118 heads and 342 feet, and 342 − 118 = 224. The other options give differences of 219, 221 and 227.

Answer: B

The number of new students admitted is 24.

Three new sections were added to make 16, so there were 13 sections before the admissions.

\[ \begin{aligned} \text{before} &= 13 \times 24 = 312 \\ \text{after} &= 16 \times 21 = 336 \\ \text{new} &= 336 - 312 = 24 \end{aligned} \]

The trap is to take 16 sections of 24 as the starting point, which gives 384 students before and a fall rather than a rise. 48 would be two full sections of 24, and 114 and 14 do not come from the given figures. The extra sections exist because the 336 students are now spread more thinly, 21 to a section.

Answer: C

A had 820 coins.

Work backwards from D, who received everything he now has from C.

\[ \begin{aligned} \tfrac{c}{3} + 1 &= 20 \Rightarrow c = 57 \\ \tfrac{b}{4} + 5 &= 57 \Rightarrow b = 208 \\ \tfrac{a}{4} + 3 &= 208 \Rightarrow a = 820 \end{aligned} \]

Here c, b and a are the coins C received, B received and A had at the start. Check forward: A gives \(205 + 3 = 208\) to B, B gives \(52 + 5 = 57\) to C, and C gives \(19 + 1 = 20\) to D. 683 and 799 are not divisible by 4, and 1100 would give B 278 coins, a quarter of which is not a whole number.

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.