Numbers, HCF and LCM, Ratio, Percentage and Average

Arithmetic is the cheapest part of the BCA paper to score in.

About 8–10 of the 50 mathematics questions are pure arithmetic, and this chapter covers the half of it that the paper asks most often: number properties, HCF and LCM, ratio, percentage and average.

None of it needs more than Class 8 mathematics. What the paper tests is speed.

A student who knows the fraction–percentage table and the few shortcuts below answers each question in under 40 seconds, and that saved time pays for the harder coordinate-geometry and calculus questions later in the paper.

1.1 Integers, primes, divisibility, remainders, even and odd results

Divisibility tests.

By	Test
2, 5, 10	last digit even; 0 or 5; 0
3, 9	digit sum divisible by 3, 9
4, 8	last two, last three digits divisible by 4, 8
6	divisible by both 2 and 3
11	(sum of odd-place digits) − (sum of even-place digits) is 0 or a multiple of 11

Primes. A prime has exactly two factors. The primes up to 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. Two is the only even prime and 1 is neither prime nor composite.

To test a number up to 120, divide by 2, 3, 5, 7 only. If none divides it, it is prime. The usual traps are 51=3\times17, 57=3\times19, 87=3\times29, 91=7\times13.

Number of factors. If N=p^a q^b r^c then the number of factors is (a+1)(b+1)(c+1). For 36=2^2\cdot3^2 that is 3\times3=9.

Division statement. N=dq+r with 0\le r<d. If a number leaves remainder 7 on division by 10, it leaves remainder 7-5=2 on division by 5, because 5 divides 10.

Even and odd results.

Expression	Always
n(n+1)	even (one of two consecutive integers is even)
n^2+n, n^2-n	even
n^3-n=(n-1)n(n+1)	divisible by 6
n^2+1	even only if n is odd
odd + odd, even + even	even
odd × odd	odd

(a) Euclid: HCF by repeated division

(b) Eratosthenes: the sieve of primes

1.2 HCF and LCM and their word problems

• Product rule. For two numbers, \text{HCF}\times\text{LCM}=a\times b. Given any three of the four quantities, the fourth follows.
• Fractions. \text{HCF}\left(\frac{a}{b},\frac{c}{d}\right)=\frac{\text{HCF}(a,c)}{\text{LCM}(b,d)} and \text{LCM}=\frac{\text{LCM}(a,c)}{\text{HCF}(b,d)} (fractions in lowest terms).
• Greatest number that divides a, b, c leaving remainders r_1,r_2,r_3 is the HCF of (a-r_1),(b-r_2),(c-r_3).
• Least number that leaves remainder r on division by x,y,z is \text{LCM}(x,y,z)+r.
• Bells, lights, runners that repeat every x,y,z units meet again after \text{LCM}(x,y,z).
• Co-prime numbers have HCF 1, so their LCM is their product.

1.3 Ratio, proportion, variation and ages

• a:b=c:d means ad=bc. The mean proportional of a and b is \sqrt{ab}.
• Chain a ratio. From a:b=2:3 and b:c=4:5, make b equal in both: a:b=8:12, b:c=12:15, so a:b:c=8:12:15.
• Dividing N in the ratio p:q: the shares are \frac{p}{p+q}N and \frac{q}{p+q}N.
• Direct variation: y/x constant. Inverse variation: xy constant (more men, fewer days).
• Age problems: let the present ages be ax and bx if they are in ratio a:b; the difference between ages never changes, so a difference of 8 years equals (b-a) parts.

1.4 Percentage and successive change

• x\% of y equals y\% of x. "Per cent" is "per hundred", so x\% = \frac{x}{100}.
• Learn this table: \frac{1}{2}=50\%, \frac{1}{3}=33\frac{1}{3}\%, \frac{1}{4}=25\%, \frac{1}{5}=20\%, \frac{1}{6}=16\frac{2}{3}\%, \frac{1}{8}=12.5\%, \frac{1}{9}=11\frac{1}{9}\%, \frac{1}{10}=10\%, \frac{1}{12}=8\frac{1}{3}\%.
• Successive change of a\% then b\% is a net a+b+\frac{ab}{100} per cent (use a minus sign for a fall). A 20 % rise followed by a 20 % fall: 20-20-4=-4\%.
• Percent more / less. If A is r\% more than B, then B is \frac{100r}{100+r}\% less than A. (25 % more gives 20 % less.)
• Keeping expenditure fixed after a price rise of r\% needs a cut in consumption of \frac{100r}{100+r}\%.
• Population / depreciation: P\left(1\pm\frac{r}{100}\right)^n.

1.5 Averages and weighted averages

• \text{Average}=\dfrac{\text{sum}}{\text{number of items}}, so sum = average \times number.
• The average of the first n natural numbers is \frac{n+1}{2}; of any evenly spaced list it is the middle value (or the mean of the first and last).
• A member leaves or joins: work with totals. Removed value = old total - new total.
• Weighted mean of groups with sizes n_1,n_2 and averages A_1,A_2 is \frac{n_1A_1+n_2A_2}{n_1+n_2}.
• Adding a value equal to the current average leaves the average unchanged.

1.6 Eight traps in arithmetic questions

Arithmetic questions are short, so the examiner hides the mark in one small word: "greatest", "least", "more", "less". These eight traps cover the ones that cost the most marks, and each is checked by the figures above.

1. HCF times LCM equals the product of two numbers only. For a and b we have \text{HCF}\times\text{LCM}=ab, which lets you find a missing LCM at once. It fails for three numbers, so do not use it when the stem gives three. In Fig 1.2, 21\times1260=26{,}460=252\times105, which is a quick way to confirm your answer.
2. "Greatest divisor" means HCF; "least multiple" means LCM. A question that asks for the longest rod, the largest tile or the biggest equal groups is an HCF question because the length must divide every given length. A question about bells, lights or runners meeting again is an LCM question because the time must be a multiple of every period. Read the verb before you pick the tool.
3. Remainders change the problem. To find the greatest number that divides 60, 75 and 90 leaving remainders 4, 3 and 6, you take the HCF of 56, 72 and 84, which is 4. For the least number leaving remainder 5 on division by 6, 8 and 12, you take \text{LCM}+5=24+5=29. Subtract for "greatest", add for "least".
4. 1 is not prime, and 2 is the only even prime. An option such as "the smallest prime is 1" is wrong. Fig 1.1 shows 25 primes up to 100, and every prime after 2 is odd. A sum of two primes that is odd must therefore contain 2, which is why 2+11=13 works but 3+11=14 does not.
5. Ratio of parts is not ratio of amounts until you add. If two quantities are in the ratio 3:5 and total 160, one part is 160/8=20, so the amounts are 60 and 100. Students often divide by 5 or by 3 alone. Always divide by the sum of the parts, as in the last row of Fig 1.3.
6. Percent is always taken on a stated base. "A is 20 percent more than B" has base B, while "B is what percent less than A" has base A, so the two answers differ: 20 percent more up, but \frac{100\times20}{120}=16.67 percent less back. The same idea explains why +25 then -25 ends at 93.75 and not at 100 (Fig 1.4).
7. Successive changes multiply; they do not add. Two rises of 10 percent give 10+10+\frac{100}{100}=21 percent, not 20. A rise of 20 followed by a fall of 10 gives 20-10-2=+8 percent. Use the shortcut a+b+\frac{ab}{100} with signs, and always put a minus sign on a fall.
8. Average questions are total questions. When a member joins or leaves, find the old total and the new total, then subtract, as in Fig 1.5. Averaging two averages is only correct when the groups have equal size; otherwise use the weighted mean \frac{n_1A_1+n_2A_2}{n_1+n_2}. If a number equal to the current average is added, the average does not change.

1.6.1 Worked checks

Q. The HCF of two numbers is 12 and their LCM is 180. One number is 36. Find the other.

Answer. Product rule: 12\times180=36\times x, so x=\frac{2160}{36}=60.

Check with the HCF \text{HCF}(36,60)=12 and \text{LCM}(36,60)=180, so both conditions hold. If 60 had given a different HCF, the question would have no consistent answer.

Q. A shopkeeper raises the price of rice by 20 percent, then cuts it by 20 percent. What is the net change?

Answer. Net change =20-20+\frac{20\times(-20)}{100}=-4 percent. Test with Rs 100: it becomes Rs 120, then 120\times0.80=96, a fall of 4 percent. The answer is not zero, because the second change is on the larger base (Fig 1.4).

Q. The average of five numbers is 27. If one number is removed, the average becomes 25. Which number was removed?

Answer. Old total =5\times27=135 and new total =4\times25=100, so the removed number is 135-100=35.

Check the other four add to 100 and average 25. This is the same total method as the teacher problem in Fig 1.5.

Chapter summary

• Check divisibility by digit sums and last digits before dividing; primes up to 120 need testing by 2, 3, 5, 7 only.
• n(n+1) is always even; n^3-n is always a multiple of 6.
• HCF × LCM equals the product of the two numbers, and of no other count of numbers.
• "Greatest divisor" means HCF of the differences from the remainders; "least number" means LCM plus the remainder.
• Chain ratios by matching the common term, then divide in parts.
• Percentage change is always on the base the question names; successive changes multiply.
• A 25 % rise is undone by a 20 % fall, not a 25 % fall.
• For averages, deal in totals, not in averages.

Practice questions

Set A — Recall (20 questions)

1. The sum of the first five prime numbers isa18b25c26d28
2. Which number is divisible by 9?a4,321b5,238c7,415d6,073
3. The HCF of 36 and 48 isa12b6c18d24
4. The LCM of 12, 15 and 20 isa30b120c60d180
5. The product of two numbers is 2,160 and their HCF is 12. Their LCM isa144b180c240d360
6. 12.5 % of 640 isa64b72c90d80
7. If a:b=2:3 and b:c=4:5, then a:c isa6:15b2:5c8:15d4:5
8. The average of the first ten natural numbers isa5.5b5c6d6.5
9. A number leaves remainder 7 when divided by 10. The remainder on division by 5 isa0b1c3d2
10. Which of the following is a prime number?a51b53c57d91
11. The number of factors of 36 isa6b8c9d12
12. 0.375 expressed as a percentage isa37.5 %b3.75 %c33.3 %d0.375 %
13. If n is an integer, which expression must be even?an+1bn^2+1cn(n+2)dn(n+1)
14. The only even prime number isa0b1c2d4
15. 200 increased by 15 % isa215b230c245d300
16. When Rs 56 is divided in the ratio 3 : 4, the larger share isaRs 32bRs 24cRs 28dRs 36
17. The average of five numbers is 20. Their sum isa25b80c120d100
18. \frac{1}{8} expressed as a percentage isa8 %b12.5 %c15 %d18 %
19. Which number is divisible by 4?a3,214b5,102c7,316d9,018
20. The LCM of the co-prime numbers 7 and 9 isa63b1c16d126

Answers and explanations — Set A

1. (d) — 2 + 3 + 5 + 7 + 11 = 28. Option (b) 25 is what you get after dropping 2.

2. (b) — digit sum of 5,238 is 18, a multiple of 9. The others give 10, 17 and 16.

3. (a) — 36 = 2²·3², 48 = 2⁴·3, so the HCF is 2²·3 = 12. 6 is a common divisor but not the greatest.

4. (c) — 12 = 2²·3, 15 = 3·5, 20 = 2²·5, so LCM = 2²·3·5 = 60.

5. (b) — LCM = 2160 ÷ 12 = 180.

6. (d) — 12.5 % = 1/8, and 640 ÷ 8 = 80.

7. (c) — make b equal: 2:3 = 8:12 and 4:5 = 12:15, so a:c = 8:15.

8. (a) — (1 + 10)/2 = 5.5.

9. (d) — N = 10q + 7 = 5(2q + 1) + 2, remainder 2.

10. (b) — 53 has no divisor among 2, 3, 5, 7. 51 = 3×17, 57 = 3×19, 91 = 7×13.

11. (c) — 36 = 2²·3², so (2+1)(2+1) = 9.

12. (a) — 0.375 = 3/8 = 37.5 %. Multiply a decimal by 100.

13. (d) — n(n+1) is a product of consecutive integers. n(n+2) is odd when n is odd.

14. (c) — every other even number has 2 as a factor besides 1 and itself.

15. (b) — 200 × 1.15 = 230.

16. (a) — 56 ÷ 7 = 8 per part, 4 × 8 = 32.

17. (d) — sum = average × number = 20 × 5 = 100.

18. (b) — 100 ÷ 8 = 12.5.

19. (c) — the last two digits 16 are divisible by 4. 14, 02 and 18 are not.

20. (a) — co-prime numbers have LCM equal to their product, 63.

Set B — Understanding (12 questions)

1. The HCF of two numbers is 8 and their LCM is 240. If one number is 48, the other isa30b36c40d80
2. The greatest number that divides 62, 132 and 237 leaving remainders 2, 6 and 5 respectively isa2b4c6d12
3. The least number which leaves remainder 5 when divided by 6, 8 and 12 isa24b25c48d29
4. Rs 1,750 is shared among A, B, C so that A:B = 2:3 and B:C = 4:5. C receivesaRs 700bRs 750cRs 800dRs 600
5. A price is raised by 20 % and then reduced by 20 %. The net effect isano changeb4 % increasec4 % decreased2 % decrease
6. A town of 20,000 people grows by 10 % every year. After two years its population isa24,200b24,000c22,000d24,400
7. The average of five numbers is 27. When one is removed the average of the rest is 25. The removed number isa30b32c37d35
8. The average age of 20 students is 15 years. Including the teacher, it is 16. The teacher isa31 yearsb36 yearsc41 yearsd46 years
9. A father is three times as old as his son. After 10 years he will be twice as old. The son's present age isa10 yearsb12 yearsc15 yearsd20 years
10. A is 25 % more than B. B is less than A bya25 %b12.5 %c33⅓ %d20 %
11. If 35 % of a number is 84, the number isa210b235c240d294
12. Three numbers are in the ratio 1:2:3 and the sum of their squares is 504. The largest number isa12b18c24d36

Answers and explanations — Set B

1. (c) — other = (8 × 240) ÷ 48 = 40.

2. (a) — HCF of 60, 126, 232: 60 = 2²·3·5, 126 = 2·3²·7, 232 = 2³·29, so only 2 is common.

3. (d) — LCM(6, 8, 12) = 24, and 24 + 5 = 29. The LCM alone (24) leaves remainder 0.

4. (b) — A:B:C = 8:12:15, total 35 parts, 1750 ÷ 35 = 50, C = 15 × 50 = 750.

5. (c) — 20 − 20 − (20×20)/100 = −4 %. The common error is "no change".

6. (a) — 20,000 × 1.1 × 1.1 = 24,200. Simple growth would give 24,000.

7. (d) — removed = 5×27 − 4×25 = 135 − 100 = 35.

8. (b) — 21 × 16 − 20 × 15 = 336 − 300 = 36.

9. (a) — 3S + 10 = 2(S + 10) gives S = 10, father 30.

10. (d) — A = 125, B = 100, so B is 25/125 = 20 % less than A.

11. (c) — 84 ÷ 0.35 = 240.

12. (b) — numbers k, 2k, 3k: 14k² = 504, k = 6, largest 18.

Set C — Application (10 questions)

1. Three bells ring at intervals of 12, 18 and 30 seconds and ring together at 9:00:00. They next ring together ata9:02b9:03c9:04d9:06
2. The price of rice rises by 25 %. To keep the monthly spending unchanged, consumption must fall bya25 %b15 %c12.5 %d20 %
3. In a two-candidate election the winner gets 55 % of the votes and wins by 2,000 votes. The total number of votes isa20,000b22,000c18,000d25,000
4. The mean mark of 30 students is 60 and of 20 other students is 75. The mean of all 50 isa65b67.5c66d68
5. Two ages are in the ratio 3:5 and the elder is 8 years older. The younger isa9 yearsb12 yearsc15 yearsd18 years
6. How many numbers from 1 to 100 are divisible by both 4 and 6?a6b7c9d8
7. Two numbers have HCF 6 and LCM 180, and their sum is 66. The larger number isa30b32c36d40
8. A number is increased by 20 % and the result is decreased by 25 %, giving 270. The number isa300b320c250d280
9. The remainder when 2^{10} is divided by 7 isa1b4c3d2
10. The average of seven consecutive integers is 20. The largest isa22b23c24d26

Answers and explanations — Set C

1. (b) — LCM(12, 18, 30) = 180 s = 3 min.

2. (d) — 100 × 25/125 = 20 %. Cutting by 25 % would leave spending below the old value.

3. (a) — winner − loser = 55 % − 45 % = 10 % of the total = 2,000, so total = 20,000.

4. (c) — (30×60 + 20×75) ÷ 50 = 3300 ÷ 50 = 66. The simple mean 67.5 ignores the group sizes.

5. (b) — 5 − 3 = 2 parts = 8, one part = 4, younger = 12.

6. (d) — LCM(4, 6) = 12 and 12, 24, …, 96 gives 8 multiples.

7. (c) — numbers are 6a, 6b with ab = 30 and a + b = 11, so a, b = 5, 6 and the numbers are 30, 36.

8. (a) — x × 1.2 × 0.75 = 0.9x = 270, so x = 300.

9. (d) — 2³ = 8 ≡ 1 (mod 7), so 2¹⁰ = (2³)³ · 2 ≡ 2.

10. (b) — the middle (fourth) number is 20, so the seven are 17 to 23.