Total distance and average speed of a train, the height of a cuboid from its surface area, relationships in a six-member family, and counting primes between square roots

Understanding the topic

Question 8 mixes a two-stage speed problem, a surface-area equation, a family-relationship puzzle and prime counting between square roots. The speed part is the one to slow down on, because the two halves of the journey are given in different units.

(a) A train in first five successive minutes from its start runs 68 m, 127 m, 208 m, 312 m and 535 m and for next 5 minutes maintains an average speed of 33 km/hr. Find the whole distance covered and the average speed in covering this total distance.

Given: distances of 68 m, 127 m, 208 m, 312 m and 535 m in the first five minutes, then an average speed of 33 km/h for the next five minutes. The fifth figure is the paper's printed value of 535 m.

Method: add the first five distances, convert the second stage to metres using distance = speed × time, then divide total distance by total time.

Working:

68 + 127 + 208 + 312 + 535 = 1,250 m

Time for the second stage = 5/60 hour = 1/12 hour

Distance = 33 × 1/12 = 2.75 km = 2,750 m

Total distance = 1,250 + 2,750 = 4,000 m = 4 km

Total time = 10 minutes = 1/6 hour

Average speed = 4 ÷ (1/6) = 4 × 6 = 24 km/h

Answer: the train covers 4 km in total, at an overall average speed of 24 km/h.

(b) A solid cuboid with base 10 cm by 6 cm is available. The height of the cuboid is 'y' centimetres and the total surface area of the cuboid is given to be 376 cm². Find the height of the cuboid.

Given: length 10 cm, breadth 6 cm, total surface area 376 cm², height y cm.

Method: apply the total surface area formula for a cuboid, 2(lb + lh + bh), substitute the known values and solve the linear equation for y.

Working:

2(lb + lh + bh) = 376

2[(10 × 6) + 10y + 6y] = 376

2(60 + 16y) = 376

120 + 32y = 376

32y = 256

y = 8

Answer: the height of the cuboid is 8 cm.

Check: 2[(10 × 6) + (10 × 8) + (6 × 8)] = 2(60 + 80 + 48) = 376 cm².

(c) In a six-member family (A, B, C, D, E and F), there are two fathers, three brothers and a mother. If C is the sister of F, B is the brother of E's husband, D is the father of A and grandfather of F. Who is E's husband?

Given, as printed:

• D is the father of A.
• D is the grandfather of F, so one of D's children is a parent of F.
• C is the sister of F, so C and F are siblings.
• B is the brother of E's husband.
• The family contains two fathers, three males described as brothers, and one mother.

Method: fix the generations first from the grandfather statement, then place the married pair, then attach the siblings. The wording is compressed, so the assumptions are kept visible rather than folded into the answer.

Working: the arrangement that satisfies all statements is:

• D is the father of A and B, and the grandfather of C and F.
• A is married to E.
• A and E are the parents of C and F.
• B is A's brother.
• F is C's brother.

This gives two fathers, D and A; one mother, E; and three males who are brothers within their own sibling groups, A, B and F.

Answer: E's husband is A.

(d) How many prime numbers are between each of the following pairs of numbers?

Method: approximate each square-root boundary to two decimal places, then list the primes strictly between the two values and count them.

Answers:

1. Pair (a), √3 and √120. √3 is about 1.73 and √120 is about 10.95. Primes: 2, 3, 5, 7. Count = 4.
2. Pair (b), √10 and √410. √10 is about 3.16 and √410 is about 20.25. Primes: 5, 7, 11, 13, 17, 19. Count = 6.
3. Pair (c), √10 and √999. √10 is about 3.16 and √999 is about 31.61. Primes: 5, 7, 11, 13, 17, 19, 23, 29, 31. Count = 9.
4. Pair (d), √28 and √120. √28 is about 5.29 and √120 is about 10.95. Prime: 7. Count = 1.
5. Pair (e), √8 and √400. √8 is about 2.83 and √400 = 20. Primes: 3, 5, 7, 11, 13, 17, 19. Count = 7.

The boundaries are exclusive, so a prime equal to a boundary would not count; here only √400 is exact, and 20 is not prime in any case.

Working a solved paper is different from reading one. How to actually use CSS past papers sets out the difference.