Number System Questions and Answers

Answer: C

She can make five different rectangles.

Each rectangle made from 36 unit squares has sides whose product is 36. List the factor pairs, counting each shape once whichever way round it is placed:

\[ \begin{aligned} 36 &= 1 \times 36 = 2 \times 18 = 3 \times 12 \\ &= 4 \times 9 = 6 \times 6 \end{aligned} \]

That is five arrangements (the 6 by 6 one is a square, which is a special rectangle). Eight comes from counting 36 x 1 and 1 x 36 as different, and four from leaving out the square.

Answer: C

One hundred lakh is one crore.

In the Indian system, 1 lakh = 100,000 and 1 crore = 100 lakh:

\[ 100 \times 100000 = 10000000 = 1\ \text{crore} \]

In the international system that is ten million. One million is only 10 lakh, one billion is 100 crore, and ten crore is 1,000 lakh, so none of these equals 100 lakh.

Answer: B

The value of (p + q - 2r) is 7.

Factorise 313632 into primes:

\[ \begin{aligned} 313632 &= 2^{5} \times 9801 \\ 9801 &= 99^{2} = 3^{4} \times 11^{2} \\ 313632 &= 11^{2} \times 2^{5} \times 3^{4} \end{aligned} \]

Matching the powers, p = 11, q = 2 and r = 3, so p + q - 2r = 11 + 2 - 6 = 7.

The key step is matching each prime to its exponent: 2 is the prime raised to the fifth power, 3 to the fourth and 11 to the second. Swapping any two of them changes the result, so 8, 9 and 6 do not fit.

Answer: A

The sum is 280.

List the primes between 31 and 60. Check each odd number for divisibility by 3, 5 and 7 (enough, since \(7^{2} = 49\) and \(8^{2} = 64\)):

37, 41, 43, 47, 53, 59

(33, 39, 45, 51 and 57 are multiples of 3; 35 and 55 of 5; 49 is \(7^{2}\).)

\[ 37 + 41 + 43 + 47 + 53 + 59 = 280 \]

31 itself is prime but is not between 31 and 60; including it gives 311, which is the trap in option B.

Answer: B

The sum is 1192209.

The greatest six-digit number from 5, 9, 0, 2, 8, 6 puts the digits in descending order: 986520.

The smallest puts them in ascending order, but a six-digit number cannot start with 0, so the smallest non-zero digit, 2, goes first and 0 second: 205689.

\[ 986520 + 205689 = 1192209 \]

Option A, 1192290, comes from a slip in the units digits, and the options starting 1292 come from adding 100000 too much. Starting the smallest number with 0 would give only a five-digit number, which is the trap the question sets.

Answer: C

The difference of the place values is 49930.

In 352876, the digit 5 is in the ten-thousands place and the digit 7 is in the tens place:

\[ \begin{aligned} \text{place value of } 5 &= 50000 \\ \text{place value of } 7 &= 70 \\ 50000 - 70 &= 49930 \end{aligned} \]

Option D, 49993, is 50000 − 7, which wrongly uses the face value of 7, and 49997 and 50007 match no pairing of the two place values. The place value of a digit is the digit times the value of its position, so the 7 here is worth 70.

Answer: A

The numbers \(\frac{5}{-8}, \frac{-2}{3}, \frac{-3}{5}\) lie between the two.

As decimals, \(\frac{-7}{10} = -0.7\) and \(\frac{-1}{4} = -0.25\). The three numbers are -0.625, about -0.667 and -0.6, all greater than -0.7 and less than -0.25.

The other sets each contain a number outside the range: 0 and \(\frac{-1}{5} = -0.2\) are greater than -0.25, \(\frac{-3}{4} = -0.75\) is less than -0.7, and \(\frac{2}{3}\) is positive.

Answer: B

The smallest positive integer that is not a factor is 13.

Factorise each number and combine:

\[ \begin{aligned} 264 &= 2^{3} \times 3 \times 11 \\ 90 &= 2 \times 3^{2} \times 5 \\ 1680 &= 2^{4} \times 3 \times 5 \times 7 \\ \text{product} &= 2^{8} \times 3^{4} \times 5^{2} \times 7 \times 11 \end{aligned} \]

Every integer from 1 to 12 is built from these primes within these powers, so all divide the product (12 = 2² × 3). 13 is a prime that does not occur, so it is not a factor. 14 = 2 × 7 and 15 = 3 × 5 are factors.

Answer: D

The answer is A = 8, B = 9.

The product ends in 0, so A × 5 ends in 0 and A is even. The product 44B0 is at least 4400, so ABA is at least 880, which makes A = 8 (A = 9 is odd). Now write 8B8 × 5 out:

\[ \begin{aligned} (808 + 10B) \times 5 &= 4400 + 10B \\ 4040 + 50B &= 4400 + 10B \\ 40B &= 360,\ B = 9 \end{aligned} \]

Check: 898 × 5 = 4490. Option B (A = 9) fails because 9 × 5 ends in 5, and options A and C give products far below 4400.

Answer: C

The statement that is not true is Every factor of a given number is always greater than the given number.

A factor divides the number exactly, so for natural numbers it can never exceed the number; the largest factor is the number itself. For example, the factors of 12 are 1, 2, 3, 4, 6 and 12.

The other statements are true. Every number has finitely many factors (A). Every factor is less than or equal to the number (B). Every multiple, such as 12, 24, 36 for 12, is greater than or equal to the number (D).