Numbers and Calculation 20 Question Learning Review

Attempt this review after the five Numbers and Calculation lessons. Each question has exactly one correct answer and is worth one mark. This original learning review is untimed, with no negative marking and no pass cutoff. The four questions per lesson are a coverage choice, not a prediction of exam weighting. It is not a full RRB CBT simulation. Use natural numbers beginning at 1; factors, HCF and LCM refer to positive integers unless stated otherwise; √ denotes the nonnegative principal square root. Write a short reason or calculation before reading the key. Then revisit the relevant lesson and retry any missed question.

1

In this module, natural numbers begin at 1. Which statement about 0 is correct?

A. It is natural but not an integer B. It is a whole number, an integer and a rational number, but not natural C. It is irrational because it has no sign D. It is neither an integer nor rational

2

Evaluate −10 − (6 − 9).

A. −25 B. −13 C. −7 D. 7

3

Evaluate 48 ÷ 6 × 2 + 5.

A. 21 B. 9 C. 13 D. 56

4

Evaluate (−4)² + [18 − 6 ÷ 3].

A. 0 B. 20 C. 60 D. 32

5

Which number is a factor of 72 but not a factor of 54?

A. 12 B. 18 C. 9 D. 3

6

Which digit d makes the four-digit number 6d25 divisible by 11?

A. 1 B. 3 C. 5 D. 7

7

Which is the prime factorisation of 630?

A. 6 × 3 × 5 × 7 B. 2 × 3 × 5 × 7 C. 2 × 3² × 5 × 7 D. 2² × 3² × 5 × 7

8

A positive integer is divisible by both 8 and 9. Which conclusion is justified?

A. It must be divisible by 144 B. It cannot be divisible by 72 C. Its divisibility by 72 is uncertain D. It must be divisible by 72 because 8 and 9 are co-prime

9

Two positive integers have HCF 8 and LCM 168. One is 24. What is the other?

A. 48 B. 56 C. 64 D. 72

10

Find the LCM of 14, 21 and 30.

A. 105 B. 180 C. 210 D. 420

11

72 red beads and 96 blue beads are put into the greatest possible number of identical packets, with no beads left over. How many packets are made?

A. 24 B. 12 C. 36 D. 288

12

Three signals sound every 9, 12 and 20 seconds at fixed intervals, without skipping. They sound together now. After how many positive seconds will they next sound together?

A. 60 B. 90 C. 120 D. 180

13

Evaluate 1⅔ − 5/6.

A. 1/2 B. 5/6 C. 7/6 D. 4/3

14

Evaluate (3/5) ÷ (9/10).

A. 27/50 B. 3/2 C. 2/3 D. 1/6

15

Evaluate 0.45 × 0.8 + 0.04 exactly.

A. 0.4 B. 0.076 C. 0.36 D. 0.76

16

The block 45 repeats forever in 0.454545…. Which fraction is exactly equal to this number?

A. 9/20 B. 45/1000 C. 11/5 D. 5/11

17

What is √0.81?

A. 0.09 B. −0.9 C. 0.9 D. Both 0.9 and −0.9

18

What is the smallest positive integer by which 450 must be multiplied to obtain a perfect-square integer?

A. 2 B. 3 C. 5 D. 10

19

In finding √6084 by long division, group the digits as 60 | 84. After choosing 7 as the first root digit, the remainder is 11; bringing down 84 gives 1184. Which product selects the next digit correctly?

A. 148 × 78 B. 78 × 78 C. 14 × 8 D. 148 × 8

20

Between which consecutive integers does √200 lie?

A. 12 and 13 B. 14 and 15 C. 15 and 16 D. 19 and 20

Explained answer key

1. B. The stated convention excludes 0 from natural numbers. Whole numbers include it; it is an integer and can be written 0/1, so it is rational. Having no positive/negative sign does not make it irrational.

2. C. First 6 − 9 = −3. Then −10 − (−3) = −10 + 3 = −7. The outside subtraction adds the opposite of the entire inner value.

3. A. 48 ÷ 6 × 2 + 5 = 8 × 2 + 5 = 16 + 5 = 21. Division and multiplication have equal precedence. The distractor 9 comes from incorrectly grouping 6 × 2 as the divisor.

4. D. (−4)² = 16. Inside the brackets, divide before subtracting: 18 − 2 = 16. The total is 16 + 16 = 32. Treating the parenthesised base as −4² would give the incorrect option 0.

5. A. 72 = 12 × 6, while 54 = 12 × 4 + 6, so 12 satisfies both conditions. The other three numbers divide both 72 and 54.

6. B. The alternating difference is (6 + 2) − (d + 5) = 3 − d. For digits 0 through 9, it ranges from 3 to −6, so it must equal 0. Thus d = 3; 6325 = 11 × 575.

7. C. 630 = 63 × 10 = 3² × 7 × 2 × 5. A has the correct product but leaves a composite factor 6, so it is not a completed prime factorisation. B and D give 210 and 1260.

8. D. 8 = 2³ and 9 = 3² have no common prime factor. Both requirements together imply divisibility by 2³ × 3² = 72. They do not guarantee 144; 72 itself is a counterexample.

9. B. For two positive integers, their product equals HCF × LCM. The other number is (8 × 168) ÷ 24 = 56. Verify: HCF(24,56) = 8 and LCM(24,56) = 168.

10. C. 14 = 2 × 7; 21 = 3 × 7; 30 = 2 × 3 × 5. Taking each needed prime at its greatest required exponent gives 2 × 3 × 5 × 7 = 210. It is divisible by all three.

11. A. The packet count must divide both totals, so take HCF(72,96) = 24. Each packet has 3 red and 4 blue beads. The LCM 288 describes a common multiple, not the possible packet count.

12. D. Use the least positive common multiple: 9 = 3², 12 = 2² × 3, 20 = 2² × 5. LCM = 2² × 3² × 5 = 180 seconds. This is 20, 15 and 9 full intervals, respectively; time 0 is the present coincidence.

13. B. 1⅔ = 5/3 = 10/6. Therefore 10/6 − 5/6 = 5/6. A mixed numeral means the whole number plus the fraction, not multiplication.

14. C. Multiply by the reciprocal of the divisor: (3/5) × (10/9) = 30/45 = 2/3. Check: (2/3) × (9/10) = 3/5. Invert the divisor, not the dividend.

15. A. First 0.45 × 0.8 = 0.36; then 0.36 + 0.04 = 0.40 = 0.4. Multiplication precedes addition and the trailing zero does not change the value. No rounding is involved.

16. D. Let x = 0.454545…. Then 100x = 45.454545…. Subtracting gives 99x = 45, so x = 45/99 = 5/11. The finite decimal 0.45 equals 9/20 and is a different number.

17. C. 0.9² = 0.81. The radical symbol selects the nonnegative principal root, so √0.81 = 0.9. Both signs solve x² = 0.81, but that is a different question.

18. A. 450 = 2 × 3² × 5². Only the factor 2 is unpaired, so multiply by 2. The result is 900 = 30². No smaller positive integer than 2 except 1 is available, and 450 is not already a square.

19. D. Double the root prefix 7 and shift one place: 20 × 7 = 140. For next digit 8, (140 + 8) × 8 = 148 × 8 = 1184. The new root is 78. The trial divisor multiplies the next single digit, not the whole new root.

20. B. 14² = 196 < 200 < 225 = 15², so 14 < √200 < 15. These are bounds; they do not say that the exact root is 14.5.

Quick reference

Revision checklist

• Use the stated natural-number convention. Distinguish a number’s sign from the subtraction operation
• Evaluate innermost groups and powers first. Then perform multiplication/division from left to right, followed by addition/subtraction from left to right. Division by zero is undefined
• A factor divides exactly. Prime and co-prime mean different things; 1 is neither prime nor composite. The product-divisibility guarantee requires co-prime divisors
• Choose HCF for a greatest shared divisor and LCM for a least positive common multiple. State the quantity and conditions in a grouping or repeating-event problem
• Use equivalent common-sized parts for fraction addition. In division invert the nonzero divisor. Written finite-place decimal methods use terminating decimals; recurring decimals can be handled as exact fractions
• The radical √ selects the nonnegative principal root. In root long division, multiply the trial divisor by the next single digit. A permitted last digit does not prove a perfect square; bounds are not an exact midpoint

Use the explained key to identify the missed method and return to its numbered lesson. There is no pass cutoff.