EPFO APFC 2023 General Studies · Question 84 of 120

Correct — A, (a) 100111011. Octal is base eight and eight is two cubed, so every octal digit corresponds to exactly three binary digits. That single relationship makes the conversion mechanical: take the octal number one digit at a time, write each as a three-bit group, and join the groups in the same order. Here 4 is 100, 7 is 111 and 3 is 011, so 473 in octal becomes 100 111 011, that is 100111011. Two details decide whether the method is applied correctly. First, the place values inside a group are 4, 2 and 1, so 4 is 100 and not 110, and 3 is 011 and not 111. Second, an inner group must keep its leading zero — the 3 has to be written as three bits, 011, because dropping the zero would shift every bit to its left into the wrong place value. Only the leading group may lose its leading zeros, which is why the answer is written with nine digits and not with a tenth zero in front. The result is worth verifying through decimal, because the check is quick and catches a mis-set group at once. In octal, 473 is 4 × 64 + 7 × 8 + 3, that is 256 + 56 + 3 = 315. Now read 100111011 in binary from the left, with place values 256, 128, 64, 32, 16, 8, 4, 2 and 1: the ones sit at 256, 32, 16, 8, 2 and 1, and 256 + 32 + 16 + 8 + 2 + 1 is 315. The two agree, so the conversion is right. The same check exposes each wrong option immediately, because every one of them is the correct binary form of a different octal number — group the nine bits of any option into threes and read the octal number back out.

• (b)100111111 — Group these nine bits into threes and they read 100, 111, 111, which is octal 477, not 473. The first two groups are right and the last one has been written as 111 instead of 011 — that is, the digit 3 has been converted as though it were a 7. It is the mistake of a candidate who is counting bits rather than reading place values: three bits are needed, so three ones are written. Inside a three-bit group the place values are 4, 2 and 1, so 3 is 2 + 1 and is written 011, while 7 is 4 + 2 + 1 and is written 111. The decimal check confirms the diagnosis: this string is 256 + 32 + 16 + 8 + 4 + 2 + 1 = 319, while 473 in octal is 315, and the difference of four is exactly the value of the extra bit that has been switched on in the last group.
• (c)110111011 — Grouped into threes these bits read 110, 111, 011, which is octal 673. The last two groups are correct and the first has been written as 110 instead of 100, so the leading digit 4 has been converted as though it were a 6. This is the place-value error running the other way from the one in option (b): within a group the bits stand for 4, 2 and 1, so 4 is a single one in the leftmost position, 100, whereas 110 turns on both the 4 and the 2 and therefore means 6. The safeguard is to convert each octal digit from a memorised table rather than by counting — 0 000, 1 001, 2 010, 3 011, 4 100, 5 101, 6 110, 7 111 — which takes a minute to learn and removes every arithmetic step from this kind of question.
• (d)110001011 — These bits group as 110, 001, 011, which is octal 613. Only the last group survives the conversion: the 4 has become 110, which is 6, and the 7 has become 001, which is 1. An option with two of its three groups wrong is not there to be arrived at by a single slip; it is there to be picked by a candidate who is matching the shape of the string against a half-remembered answer instead of decoding it. That is why the reverse check is the habit worth building. Any binary string offered as the answer to an octal conversion can be split into threes from the right and read straight back as an octal number, and if the number that comes out is not the one printed in the stem, the option is wrong and no further arithmetic is needed.

A positional number system fixes a base and gives each digit a place value that is a power of that base. Decimal uses base ten, binary base two, octal base eight and hexadecimal base sixteen. Binary is what a machine actually stores, since a bit has two states, but long binary strings are unreadable and error-prone for a human, so octal and hexadecimal exist as compact shorthands for binary rather than as independent systems. The reason they work as shorthands is arithmetic: eight is two to the power three and sixteen is two to the power four, so one octal digit stands for exactly three bits and one hexadecimal digit for exactly four, with no remainder and no carrying between groups. Conversion between binary and either shorthand is therefore a substitution, not a calculation. Going from octal to binary, replace each digit by its three-bit group; going the other way, group the bits in threes starting from the right, padding the leftmost group with zeros if it is short, and replace each group by its digit. The same procedure in fours converts binary to and from hexadecimal. Conversions that do not involve binary as one of the two bases have no such shortcut and are done by expanding place values, as in 473 octal being 4 × 64 + 7 × 8 + 3 = 315 decimal, or by repeated division of a decimal number by the target base and reading the remainders upwards. Octal survives in practice chiefly in Unix file permissions, where three bits of read, write and execute map onto one octal digit.

The computers block of this paper sets one base-conversion item in each of its two runs — a hexadecimal-to-decimal conversion earlier and this octal-to-binary one — and the pair is a good illustration of how differently two conversions of the same family behave. Hexadecimal to decimal has to be computed, digit by digit against powers of sixteen, and the wrong options in that item sit one or two away from the right one so that an arithmetic slip is punished. Octal to binary does not have to be computed at all: it is a substitution of three-bit groups, and the wrong options are built by substituting the wrong group. That difference tells you what to practise. For the substitution conversions, memorise the eight three-bit groups and the sixteen four-bit groups until they are automatic, and the question becomes a reading exercise. For the computed conversions, practise the place-value expansion and the repeated-division method until neither needs thought. The habit that makes this particular item safe is the reverse check. Every wrong option here is a correct answer to a different question — 477, 673 and 613 in octal — so regrouping the offered bits into threes and reading the octal number back out identifies the right option without doing the original conversion twice. That check costs a few seconds and catches every error this option set is designed to induce.

• Eight is two to the power three, so one octal digit corresponds to exactly three binary digits and the conversion between octal and binary is a substitution rather than a calculation. Similarly sixteen is two to the power four, so one hexadecimal digit corresponds to exactly four bits.
• The eight groups to memorise are 0 = 000, 1 = 001, 2 = 010, 3 = 011, 4 = 100, 5 = 101, 6 = 110 and 7 = 111. Within a group the three places carry the values 4, 2 and 1, which is why 3 is 011 and 4 is 100, and why writing either as 110 or 111 changes the digit.
• Every inner group must be written with its full three bits, leading zeros included, because dropping a zero shifts every bit to its left into a wrong place value. Only the leading group may shed leading zeros, which is why 473 octal is written as the nine-bit 100111011.
• The decimal cross-check on this item: 473 in octal is 4 × 64 + 7 × 8 + 3 = 315, and 100111011 in binary is 256 + 32 + 16 + 8 + 2 + 1 = 315. Agreement between the two routes is the standard verification for any base conversion.
• Converting the other way, group the binary digits in threes from the right, padding the leftmost group with zeros if it is short, and read each group as an octal digit. Applied to the three wrong options here, this yields 477, 673 and 613, which is a faster disqualification than redoing the forward conversion.

• Dropping the leading zero of an inner group. The digit 3 must be written 011; writing it 11 shortens the string and pushes every bit to its left into a wrong place value.
• Counting bits rather than reading place values. A digit needing three bits is not written as three ones: inside a group the places are worth 4, 2 and 1, so 3 is 011 and 7 is 111.
• Reversing a group. The digit 4 is 100 and 6 is 110; these differ by one bit and a hurried candidate can substitute either for the other, which converts 473 into 673 without any obvious sign of error.
• Trying to convert octal to binary through decimal. It works, but it turns a substitution into two calculations and introduces two chances of arithmetic error where there was none.
• Grouping from the left when converting binary back to octal. The grouping must start from the right, because it is the low-order places that must stay together; a short group belongs at the left, padded with zeros.

Number-system items in this exam are set in four shapes and each paper usually carries one or two. The first is the direct substitution asked here — octal or hexadecimal into binary, or binary back into one of them — where the wrong options are built by substituting a neighbouring group and can all be eliminated by decoding the offered string. The second is a computed conversion into or out of decimal, where the wrong options sit close to the right one so that an arithmetic slip lands on a printed answer. The third asks a structural fact rather than a conversion: how many bits an octal or hexadecimal digit represents, how many values a given number of bits can hold, or what the largest number in a fixed width is. The fourth embeds the idea in an application, most often Unix file permissions, where read, write and execute are three bits and therefore one octal digit. Preparation for all four is the same small table of three-bit and four-bit groups, plus the discipline of checking a converted answer by converting it back.

No directly related past PYQ was found.

• practice — not a real PYQ

  The binary number 110101110 is to be written in octal. Which one of the following is its octal equivalent?

  • (a)556
  • (b)646
  • (c)656
  • (d)665

  Answer(c) 656 — grouping the bits in threes from the right gives 110, 101 and 110, which read as 6, 5 and 6. The decimal check agrees: the binary string is 256 + 128 + 32 + 8 + 4 + 2 = 430, and 656 in octal is 6 × 64 + 5 × 8 + 6 = 384 + 40 + 6 = 430. Grouping must begin at the right-hand end, since it is the low-order places that have to stay together.

• practice — not a real PYQ

  The hexadecimal number 2F is to be expressed in binary using eight bits. Which one of the following is correct?

  • (a)00101111
  • (b)00111111
  • (c)00101110
  • (d)11110010

  Answer(a) 00101111 — one hexadecimal digit corresponds to exactly four bits, so 2 becomes 0010 and F, which is fifteen, becomes 1111. Joining them in order gives 00101111. The decimal check agrees: 2F is 2 × 16 + 15 = 47, and 00101111 is 32 + 8 + 4 + 2 + 1 = 47. The last option is the same two groups written in the wrong order, which is the commonest slip in this conversion.