Problems on Numbers Questions and Answers

Answer: A

The number is 87649.

Fill in the digits from the clues, starting with the ones that are fixed directly:

\[ \begin{aligned} \text{tens} &= 2^{2} = 4 \\ \text{hundreds} &: \tfrac{2}{3} \times h = 4,\ h = 6 \\ \text{ten thousands} &: \tfrac{3}{4} \times t = 6,\ t = 8 \\ \text{thousands} &= 7,\quad \text{units} = 9 \end{aligned} \]

The smallest prime is 2, so the tens digit is 4. The largest single-digit prime is 7 and the largest single-digit odd number is 9. Reading from the ten thousands place: 8, 7, 6, 4, 9, giving 87649.

The other numbers fail at once: 49327 and 42937 have the wrong tens digit, and 83419 has 4 in the hundreds place, which is not three-fourths of 8.

Answer: B

The sum of the first three numbers is 9.

Five consecutive numbers have their middle number equal to their average:

\[ \begin{aligned} \text{middle} &= 20 \div 5 = 4 \\ \text{numbers} &= 2, 3, 4, 5, 6 \\ 2 + 3 + 4 &= 9 \end{aligned} \]

Check: 2 + 3 + 4 + 5 + 6 = 20. The answer 12 is 3 + 4 + 5, the middle three, and 5 is just 2 + 3, the first two.

Answer: B

The number is 8.

Let the number be \(x\). Increasing it by 8 doubles it:

\[ \begin{aligned} x + 8 &= 2x \\ x &= 8 \end{aligned} \]

Check: 8 + 8 = 16, which is double 8. For the other options, 6 + 8 = 14 (not 12), 12 + 8 = 20 (not 24) and 16 + 8 = 24 (not 32), so none of them doubles.

Answer: A

The original number is 39.

Let the tens digit be \(a\) and the units digit \(b\). The number is \(10a + b\) and its reverse is \(10b + a\).

\[ \begin{aligned} a + b &= 12 \\ (10b + a) - (10a + b) &= 9(b - a) = 54 \\ b - a &= 6 \\ a = 3,\ b &= 9 \end{aligned} \]

Check: 93 - 39 = 54. 93 has the right digits but reversing it gives a smaller number. 57 reversed is 75, larger by only 18, and 66 reversed is unchanged.

Answer: A

Raju has 5 notes of Rs. 10.

Let t be the number of Rs. 10 notes and w the number of Rs. 20 notes:

\[ \begin{aligned} t + w &= 14 \\ 10t + 20w &= 230 \\ 10t + 20(14 - t) &= 230 \\ 280 - 10t &= 230 \Rightarrow t = 5 \end{aligned} \]

So he has 5 notes of Rs. 10 and 9 of Rs. 20: 50 + 180 = 230. The option 9 is the number of Rs. 20 notes, the usual slip of solving for the wrong variable. 6 and 10 give totals of 220 and 180.

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.