Number System Questions and Answers

Answer: B

The statement not true for integers is that division is commutative.

Commutativity would need a ÷ b = b ÷ a for all integers, but 6 ÷ 3 = 2 while 3 ÷ 6 = 1/2, which is not even an integer. Division of integers is neither commutative nor closed.

The other three are true. Multiplication of integers is associative: (a × b) × c = a × (b × c). 1 is the multiplicative identity, since a × 1 = a. And subtraction is indeed not commutative: 5 - 3 = 2 but 3 - 5 = -2.

Answer: B

The decreasing order is \(\frac{3}{7}, \frac{12}{63}, \frac{1}{9}, \frac{1}{21}\).

Convert each fraction to the common denominator 63:

\[ \begin{aligned} \tfrac{3}{7} &= \tfrac{27}{63}, \quad \tfrac{12}{63} = \tfrac{12}{63} \\ \tfrac{1}{9} &= \tfrac{7}{63}, \quad \tfrac{1}{21} = \tfrac{3}{63} \end{aligned} \]

So \(27 \gt 12 \gt 7 \gt 3\) gives the order above. Option 1 puts \(\frac{1}{9}\) above \(\frac{12}{63}\), option 3 starts with \(\frac{12}{63}\), and option 4 repeats \(\frac{1}{9}\) and is in increasing order.

Answer: A

One crore is ten million.

One crore is 1,00,00,000 in the Indian system, which is 10,000,000 in the international system, that is, 10 million.

One million is 10 lakh, one billion is 100 crore, and a hundred million is 10 crore.

Answer: A

The difference is 861741.

The greatest six-digit number uses the digits in descending order, 965310. The smallest cannot begin with 0, so it starts with the smallest non-zero digit and puts 0 next: 103569.

\[ 965310 - 103569 = 861741 \]

Putting 0 first (035691) would not give a six-digit number, which is the common trap; the other options come from that or from subtraction slips.

Answer: D

The sum is \(2\frac{39}{40}\).

Compare the four values: \(\frac{6}{7} \approx 0.857\), \(2\frac{3}{5} = 2.6\), \(\frac{3}{8} = 0.375\) and \(2\frac{8}{15} \approx 2.533\). The largest is \(2\frac{3}{5}\) and the smallest \(\frac{3}{8}\):

\[ \begin{aligned} 2\frac{3}{5} + \frac{3}{8} &= 2 + \frac{24}{40} + \frac{15}{40} \\ &= 2\frac{39}{40} \end{aligned} \]

Taking \(2\frac{8}{15}\) as the largest, or \(\frac{6}{7}\) as the smallest, gives the other options.

Answer: B

A prime number is a natural number greater than 1 whose only factors are 1 and itself. The number 1 is not prime, because it has only one factor. The number 2 has just the two factors 1 and 2, so it is prime, and it is the smallest prime number, and also the only even prime. The numbers 4 and 6 are composite, since 4 has the factor 2 and 6 has the factors 2 and 3. The answer is 2, option B.

Answer: A

A composite number has factors other than 1 and itself. The largest three digit number is 999, and it is divisible by 3 (the sum of its digits is 27) and by 9 and 27, since \(999 = 27 \times 37\), so it is composite. As 999 is composite and is the largest three digit number, it is the largest three digit composite number. The options 998, 989 and 979 are smaller, even though they are composite as well (998 is even, 989 = 23 x 43, 979 = 11 x 89). The answer is 999, option A.

Answer: B

The value of B is 4.

The units digit of A × 6 must be 8, so A is 3 or 8. A = 3 gives 18, and then 6B + 1 must end in 8, which is impossible. A = 8 gives 48 with a carry of 4, so 6B + 4 must end in 8, which needs 6B to end in 4, so B = 4 (24 + 4 = 28, carry 2, C = 2).

\[ 48\times6=288 \]

2, 6 and 1 do not give a product of the form C88.

You can check the result by multiplying: 48 × 6 = 288, which has the form C88 with C = 2. A trial with A = 3 gives 43 × 6 = 258 and 63 × 6 = 378, which end in 8 but do not have 8 in the tens place. So only A = 8 with B = 4 works.

Answer: C

The value of B is 5.

Write AB as 10A + B. The sum AB + 37 = 6A means 10A + B + 37 = 60 + A, so 9A + B = 23.

\[ 9A+B=23 \Rightarrow A=2,\ B=5 \]

Check: 25 + 37 = 62. B = 3, 4 or 6 would give no digit A that fits.

You can check by putting in the values: with A = 2 and B = 5 the sum is 25 + 37 = 62, which has the form 6A with A = 2. Trying B = 3 gives A = 20/9, which is not a digit, and B = 4 and B = 6 give no whole digit either, so B must be 5.

Answer: D

The highest value of A is 6.

The units column is 2 + 8 + 2 = 12, so 2 is written and 1 is carried. In the tens column A + B + C + 1 must give 7 with no carry, because the hundreds column 4 + 7 + 3 = 14 needs no extra carry to make 14. So A + B + C = 6.

\[ A+B+C=6 \Rightarrow A_{max}=6\ (B=C=0) \]

A larger digit is not possible, so 3, 4 and 5 are not the highest.

Since A, B and C are digits and the tens column sum is 6 plus the carry from the units, the largest A is found by making B and C as small as possible, namely 0. Then A = 6 and the sum 462 + 708 + 302 = 1472 checks out.