Expressions Using Letter-Numbers

Shabnam is 3 years older than Aftab. When Aftab’s age 10 years, Shabnam’s age will be 13 years. Now Aftab’s age is 18 years, what will Shabnam’s age be?

(i) Use the expression a = s − 3 to find Aftab’s age if Shabnam’s age is 20.

Answer

Let:

a = Aftab's age

s = Shabnam's age

Shabnam is 3 years older than Aftab. Then,

s = a + 3

When Aftab’s age is 18 years, replace a by 18:

⇒ s = 18 + 3

⇒ s = 21

Hence, Shabnam’s age will be 21 years.

(i)

Aftab is 3 years younger than Shabnam.

So, Aftab’s age = Shabnam’s age − 3.

Using 'a' for Aftab’s age and 's' for Shabnam’s age, the algebraic expression is:

a = s − 3

When Shabnam’s age is 20 years, replace s by 20:

⇒ a = 20 − 3

∴ a = 17

Hence, Aftab’s age is 17 years.

Ketaki prepares and supplies coconut-jaggery laddus. The price of a coconut is ₹35 and the price of 1 kg jaggery is ₹60.

(i) How much should she pay if she buys 10 coconuts and 5 kg jaggery?

(ii) How much should she pay if she buys 8 coconuts and 9 kg jaggery?

(iii) Write an algebraic expression to find the total amount to be paid for a given number of coconuts and quantity of jaggery.

(iv) Use this expression (or formula) to find the total amount to be paid for 7 coconuts and 4 kg jaggery.

Answer

Given:

Price of 1 coconut = ₹35

Price of 1 kg of jaggery = ₹60

(i)

Cost of 10 coconuts = 10 × ₹35 = ₹350

Cost of 5 kg of jaggery = 5 × ₹60 = ₹300

∴ Total cost = ₹350 + ₹300 = ₹650

Hence, she should pay ₹650.

(ii)

Cost of 8 coconuts = 8 × ₹35 = ₹280

Cost of 9 kg of jaggery = 9 × ₹60 = ₹540

∴ Total cost = ₹280 + ₹540 = ₹820

Hence, she should pay ₹820.

(iii)

Let c denote the number of coconuts and j denote the number of kilograms of jaggery.

Cost of coconuts = c × 35 = 35c

Cost of jaggery = j × 60 = 60j

∴ Total amount to be paid = 35c + 60j

Hence, the required expression is 35c + 60j (that is, c × 35 + j × 60).

(iv)

Replacing c by 7 and j by 4 in the expression 35c + 60j:

= 35 × 7 + 60 × 4

= 245 + 240

= ₹485

Hence, the total amount to be paid is ₹485.

What is the perimeter of a square with sidelength 7 cm? Use the expression to find out.

Answer

The perimeter of a square is 4 times the length of its side.

Let q denote the sidelength.

Then the expression for the perimeter is:

Perimeter = 4q

When the sidelength is 7 cm, replace q by 7:

⇒ Perimeter = 4 × 7

∴ Perimeter = 28 cm

Hence, the perimeter of the square is 28 cm.

Write formulas for the perimeter of:
(a) triangle with all sides equal
(b) a regular pentagon (as we have learnt last year, we use the word ‘regular’ to say that all sidelengths and angle measures are equal)
(c) a regular hexagon

Answer

Let a denote the length of one side of the figure.

Since all the sides are equal, the perimeter is the number of sides multiplied by a.

(a)

A triangle with all sides equal has 3 sides.

∴ Perimeter = a + a + a = 3a

(b)

A regular pentagon has 5 sides.

∴ Perimeter = 5a

(c)

A regular hexagon has 6 sides.

∴ Perimeter = 6a

Hence, the perimeters are 3a, 5a and 6a respectively, where a is the length of one side.

Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter-number ‘k’ to denote the length in meters of the other pipe.

Answer

The length of the first pipe = 20 m.

The length of the other pipe = k m.

On joining them, the combined length = 20 + k.

Hence, the expression for the combined length of the pipe is (20 + k) metres.

What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the following table:

Answer

The total amount is obtained as:

Total = (No. of ₹100 notes) × 100 + (No. of ₹20 notes) × 20 + (No. of ₹5 notes) × 5.

Hence, the completed table is as shown above.

Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to complete grind ‘y’ kg of grain, assuming the machine is off initially?

(a) 10 + 8 + y
(b) (10 + 8) × y
(c) 10 × 8 × y
(d) 10 + 8 × y
(e) 10 × y + 8

Answer

The machine takes a fixed 10 seconds to start running.

After it starts, each kg of grain takes 8 seconds. So, grinding y kg takes 8 × y = 8y seconds.

∴ Total time = starting time + grinding time

= 10 + 8 × y

Hence, option (d) is the correct option.

Write algebraic expressions using letters of your choice.
(a) 5 more than a number
(b) 4 less than a number
(c) 2 less than 13 times a number
(d) 13 less than 2 times a number

Answer

Let the number be x.

(a) 5 more than a number = x + 5

(b) 4 less than a number = x − 4

(c) 2 less than 13 times a number = 13x − 2

(d) 13 less than 2 times a number = 2x − 13

Describe situations corresponding to the following algebraic expressions:
(a) 8 × x + 3 × y
(b) 15 × j – 2 × k

Answer

(a) 8 × x + 3 × y

A shopkeeper sells pens at ₹8 each and pencils at ₹3 each. If a customer buys x pens and y pencils, the total amount to be paid (in rupees) is 8 × x + 3 × y.

(b) 15 × j – 2 × k

In a quiz, each correct answer earns 15 marks and each wrong answer carries a penalty of 2 marks. If a participant gives j correct answers and k wrong answers, the total score is 15 × j – 2 × k.

In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.

Answer

In a calendar, the dates increase by 1 as we move one cell to the right, and by 7 as we move one cell down (to the same weekday of the next week). So, moving one cell up gives a date that is 7 less.

The bottom middle cell is w.

Bottom row:

• The cell to the left of w is 1 less than w, i.e., (w - 1).

• The cell to the right of w is 1 more than w, i.e., (w + 1).

Top row (each cell is 7 less than the cell directly below it):

• Above (w - 1): (w - 1) - 7 = w - 8

• Above w: w - 7

• Above (w + 1): (w + 1) - 7 = w - 6

So, the dates in the 2 × 3 grid are:

Hence, the blank cells are w - 8, w - 7, w - 6 (top row) and w - 1, w + 1 (bottom row).

Find an algebraic expression to get the nth term of the sequence 4, 8, 12, 16, 20, 24, 28, ....

Answer

The given sequence is:

4, 8, 12, 16, 20, 24, 28, …

These are the numbers in the multiplication table of 4, i.e., the multiples of 4 in increasing order.

• The 1st term is 4 × 1

• The 2nd term is 4 × 2

• The 3rd term is 4 × 3, and so on.

So, the nth term is 4 times n, which is written as 4 × n.

∴ The nth term of the sequence is 4n.

Some simplifications are shown below where the letter-numbers are replaced by numbers and the value of the expression is obtained.

1. Observe each of them and identify if there is a mistake.
2. If you think there is a mistake, try to explain what might have gone wrong.
3. Then, correct it and give the value of the expression.

(1) If a = – 4, then 10 – a = 6.
(2) If d = 6, then 3d = 36.
(3) If s = 7, then 3s – 2 = 15.
(4) If r = 8, then 2r + 1 = 29.
(5) If j = 5, then 2j = 10.
(6) If m = –6, then 3 (m + 1) = 19.
(7) If f = 3, g = 1 then 2f – 2g = 2.
(8) If t = 4, b = 3 then 2t + b = 24.
(9) If h = 5, n = 6 then h – (3 – n) = 4.

Answer

(1)

If a = – 4, then 10 – a = 6.   There is a mistake.

Here a = – 4, so – a = – (– 4) = + 4.

Subtracting a negative number means adding.

10 – a = 10 – (– 4)

= 10 + 4

= 14

The mistake was subtracting 4 instead of adding it.

∴ The correct value is 14.

(2)

If d = 6, then 3d = 36.   There is a mistake.

3d means 3 × d, not the two digits written together as 36.

3d = 3 × 6 = 18

∴ The correct value is 18.

(3)

If s = 7, then 3s – 2 = 15.   There is a mistake.

3s means 3 × s. We must multiply first and then subtract 2 (not subtract 2 from s before multiplying).

3s – 2 = 3 × 7 – 2

= 21 – 2

= 19

The value 15 comes from wrongly computing 3 × (7 – 2).

∴ The correct value is 19.

(4)

If r = 8, then 2r + 1 = 29.   There is a mistake.

2r means 2 × r, not the digits 2 and 8 written together as 28.

2r + 1 = 2 × 8 + 1

= 16 + 1

= 17

∴ The correct value is 17.

(5)

If j = 5, then 2j = 10.   There is no mistake.

2j = 2 × 5 = 10

∴ The value 10 is correct.

(6)

If m = –6, then 3(m + 1) = 19.   There is a mistake.

The negative sign of m must be kept, and 3 must be multiplied by the whole bracket.

3(m + 1) = 3(−6 + 1)

= 3 × (−5)

= −15

The value 19 comes from ignoring the negative sign and computing 3 × 6 + 1.

∴ The correct value is –15.

(7)

If f = 3, g = 1, then 2f – 2g = 2.   There is a mistake.

Both terms carry the coefficient 2, which must not be dropped.

2f – 2g = 2 × 3 – 2 × 1

= 6 – 2

= 4

The value 2 comes from wrongly computing f – g and forgetting the coefficient 2.

∴ The correct value is 4.

(8)

If t = 4, b = 3, then 2t + b = 24.   There is a mistake.

Here the two terms must be added, not multiplied.

2t + b = 2 × 4 + 3

= 8 + 3

= 11

The value 24 comes from wrongly multiplying, as 2 × 4 × 3.

∴ The correct value is 11.

(9)

If h = 5, n = 6, then h – (3 – n) = 4.   There is a mistake.

The bracket must be solved first (or opened with the correct signs).

h − (3 − n) = 5 − (3 − 6)

= 5 − (−3)

= 5 + 3

= 8

The given value is incorrect because the signs inside the bracket were not handled correctly.

∴ The correct value is 8.

Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days.

Answer

Money earned by selling pencils:

• Day 1: 5 pencils, so 5c

• Day 2: 3 pencils, so 3c

• Day 3: 10 pencils, so 10c

Total from pencils = 5c + 3c + 10c

= (5 + 3 + 10)c

= 18c

Money earned by selling erasers:

• Day 1: 4 erasers, so 4d

• Day 2: 6 erasers, so 6d

• Day 3: 1 eraser, so 1d

Total from erasers = 4d + 6d + 1d

= (4 + 6 + 1)d

= 11d

Total money earned:

Total = 18c + 11d

Hence, the total money earned by the shopkeeper is (18c + 11d) rupees.

If c = ₹50, find the total amount earned by the sale of pencils.

Answer

Given:

c = ₹50

The total amount earned by the sale of pencils is 18c.

Replacing c by 50:

18c = 18 × ₹50 = ₹900

∴ The total amount earned by the sale of pencils is ₹900.

Write the expression for the total money earned by selling erasers. Then, simplify the expression.

Answer

The number of erasers sold on the three days is 4, 6 and 1, and the price per eraser is d.

So, the money earned on the three days is 4d, 6d and 1d.

Total money earned by selling erasers:

4d + 6d + 1d = (4 + 6 + 1)d = 11d

Hence, the total money earned by selling erasers is 11d.

Check that both expressions 18c and 5c + 3c + 10c take the same value when c is replaced by different numbers.

Answer

The two expressions are 5c + 3c + 10c and its simplified form 18c.

Let us replace c by a few different numbers and compare.

When c = 2:

5c + 3c + 10c = 5(2) + 3(2) + 10(2)

= 10 + 6 + 20

= 36

18c = 18 × 2 = 36

When c = 7:

5c + 3c + 10c

= 5(7) + 3(7) + 10(7)

= 35 + 21 + 70

= 126

18c = 18 × 7 = 126

When c = 10:

5c + 3c + 10c = 5(10) + 3(10) + 10(10)

= 50 + 30 + 100

= 180

18c = 18 × 10 = 180

In every case, both expressions take the same value.

Hence, 5c + 3c + 10c and 18c are equal for all values of c.

A shop rents out chairs and tables for a day’s use. To rent them, one has to first pay the following amount per piece. When the furniture is returned, the shopkeeper pays back some amount as follows.

Write an expression for the total number of rupees paid if x chairs and y tables are rented.

Answer

Amount paid at the beginning (for x chairs and y tables):

• Chairs: 40 × x = 40x

• Tables: 75 × y = 75y

So, the amount paid at the beginning = 40x + 75y

Amount returned (when the furniture is returned):

• Chairs: 6 × x = 6x

• Tables: 10 × y = 10y

So, the amount returned = 6x + 10y

Total amount actually paid = Amount paid – Amount returned

∴ The total amount paid = (40x + 75y) – (6x + 10y) rupees.

The total amount paid = (40x + 75y) – (6x + 10y). Can we simplify this expression? If yes, how? If not, why not?

Answer

Yes, this expression can be simplified.

Recalling how we open brackets, the negative sign before the bracket changes the sign of each term inside it:

(40x + 75y) − (6x + 10y)

= 40x + 75y − 6x − 10y

= 40x − 6x + 75y − 10y [Grouping like terms]

= (40 − 6)x + (75 − 10)y

= 34x + 65y

We could group and combine the like terms 40x and – 6x, and the like terms 75y and – 10y, because each pair contains the same letter-number.

Hence, the expression simplifies to 34x + 65y, which is the total amount paid in rupees.

Could we have written the initial expression as (40x + 75y) + (– 6x – 10y)?

Answer

Yes, we could.

Subtracting a quantity is the same as adding its negative.

So, subtracting (6x + 10y) is the same as adding (– 6x – 10y):

(40x + 75y) – (6x + 10y) = (40x + 75y) + (– 6x – 10y)

On simplifying, this also gives:

40x + 75y – 6x – 10y = 34x + 65y

Both forms lead to the same simplified expression.

Hence, yes — the initial expression can equally be written as (40x + 75y) + (– 6x – 10y).

Charu has been through three rounds of a quiz. Her scores in the three rounds are 7p – 3q, 8p – 4q, and 6p – 2q. Here, p represents the score for a correct answer and q represents the penalty for an incorrect answer.

(i) What do each of the expressions mean?

(ii) If the score for a correct answer is 4 (p = 4) and the penalty for a wrong answer is 1 (q = 1), find Charu’s scores in the second and third rounds.

(iii) What if there is no penalty? What will be the value of q in that situation?

(iv) What is her final score after the three rounds?

(v) Her friend Krishita’s score after three rounds is 23p – 7q. Give some possible scores for Krishita in the three rounds so that they add up to give 23p – 7q.

(vi) Can we say who scored more? Can you explain why?

(vii) Simplify the expression 23p – 7q – (21p – 9q).

Answer

(i)

Here, p is the score gained for each correct answer and q is the penalty (marks lost) for each incorrect answer.

So each expression tells us the round’s score = (number of correct answers) × p – (number of wrong answers) × q.

• 7p – 3q means that in the first round Charu got 7 answers correct and 3 answers wrong.
• 8p – 4q means that in the second round she got 8 answers correct and 4 answers wrong.
• 6p – 2q means that in the third round she got 6 answers correct and 2 answers wrong.

(ii)

Given:

Score for a correct answer = 4

Penalty for a wrong answer = 1

Substituting p = 4 and q = 1:

Second round score = 8p – 4q

= 8 × 4 – 4 × 1

= 32 – 4

∴ Second round score = 28

Third round score = 6p – 2q

= 6 × 4 – 2 × 1

= 24 – 2

∴ Third round score = 22

(iii)

The penalty q is the number of marks deducted for a wrong answer.

If there is no penalty, then no marks are deducted at all.

∴ q = 0

(iv)

Her final score is the sum of the scores of the three rounds:

(7p − 3q) + (8p − 4q) + (6p − 2q)

= 7p − 3q + 8p − 4q + 6p − 2q

= (7 + 8 + 6)p − (3 + 4 + 2)q [Grouping like terms]

= 21p − 9q

Hence, Charu’s final score after the three rounds is 21p – 9q.

(v)

We need three round-scores of the form (correct)p – (wrong)q that add up to 23p – 7q. The three p-coefficients must total 23 and the three q-coefficients must total 7.

One possible set of scores is:

Round 1 : 8p – 3q

Round 2 : 7p – 2q

Round 3 : 8p – 2q

Check: (8p – 3q) + (7p – 2q) + (8p – 2q)

= (8 + 7 + 8)p – (3 + 2 + 2)q

= 23p – 7q.

(vi)

Yes, we can say who scored more.

Charu’s total is 21p – 9q and Krishita’s total is 23p – 7q.

The amount by which Krishita’s score exceeds Charu’s is

23p – 7q – (21p – 9q) = 2p + 2q.

Since p (marks for a correct answer) and q (penalty for a wrong answer) are both positive quantities, 2p + 2q is always positive.

Hence, Krishita scored more than Charu, no matter what the values of p and q are.

(vii)

Given expression: 23p – 7q – (21p – 9q)

Simplifying:

23p − 7q − (21p − 9q)

= 23p − 7q − 21p + 9q $\quad$ [Opening the bracket]

= (23 − 21)p + (−7 + 9)q $\quad$ [Grouping like terms]

= 2p + 2q

Hence, 23p – 7q – (21p – 9q) = 2p + 2q.

Are the expressions 5u and 5 + u equal to each other?

Answer

The expression 5u means 5 times the number u, while 5 + u means 5 more than the number u. These are two different operations, so these two give different values.

Fill the blanks below by replacing the letter-numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take.

Answer

(i) Comparing 5u and 5 + u for different values of u:

Are the expressions 10y – 3 and 10(y – 3) equal?

Answer

Similarly, 10y – 3 means 3 less than 10 times y, while 10(y – 3) means 10 times (3 less than y). Comparing their values:

Again, the two expressions take different values for each y.

So 10y – 3 and 10(y – 3) are not equal.

After filling in the two diagrams, do you think the two expressions are equal?

Answer

No. In both diagrams the paired expressions take different values for the same value of the letter-number.

Hence, the two expressions in each pair are not equal.

Add the numbers in each picture below. Write their corresponding expressions and simplify them. Try adding the numbers in each picture in a couple different ways and see that you get the same thing.

Answer

(i)

The picture contains the terms 5y, x, x, 5y, – 6 and 2.

Method 1 (grouping like terms):

5y + x + x + 5y + (−6) + 2

= (5y + 5y) + (x + x) + (−6 + 2)

= 10y + 2x − 4

Method 2 (adding row by row):

Top row = 5y + (– 6) + x

Bottom row = x + 2 + 5y

Total = (5y + 5y) + (x + x) + (– 6 + 2) = 10y + 2x – 4

∴ The simplified expression is 2x + 10y – 4.

(ii)

The picture contains 2p (four times), 3q (four times) and the numbers – 2, – 2, 3, 3.

Method 1 (grouping by term type):

2p occurs 4 times → 8p

3q occurs 4 times → 12q

Numbers → 3 + (– 2) + (– 2) + 3 = 2

Total = 8p + 12q + 2

∴ The simplified expression is 8p + 12q + 2.

Method 2 (Group by columns):

Column 1: 2p + 3q

Column 2: 3q + 2p

Column 3: (– 2) + 3 + 2p + 3q = 1 + 2p + 3q

Column 4: 3 + (– 2) + 3q + 2p = 1 + 3q + 2p

Total: 8p + 12q + 2

(iii)

The picture contains – 5g at the four corners and 5k in the remaining twelve places.

Method 1 (grouping by term type):

– 5g occurs 4 times → – 20g

5k occurs 12 times → 60k

Total = – 20g + 60k

Method 2 (Group by columns):

First column: –5g + 5k + 5k + (–5g) = –10g + 10k

Middle columns 2 and 3 all 5k = 8 × 5k = 40k

Last column: –5g + 5k + 5k + (–5g) = –10g + 10k

Total: = –10g + 10k + 40k –10g + 10k = – 20g + 60k

∴ The simplified expression is – 20g + 60k.

Simplify each of the following expressions:
(a) p + p + p + p, p + p + p + q
(b) p + q + p – q
(c) p – q + p – q
(d) p + q – p + q
(e) p + q – (p + q)
(f) p – q – p – q
(g) 2d – d – d – d
(h) 2d – d – d – c
(i) 2d – d – (d – c)
(j) 2d – (d – d) – c
(k) 2d – d – c – c

Answer

(a)

p + p + p + p

= (1 + 1 + 1 + 1)p

= 4p

p + p + p + q

= (1 + 1 + 1)p + q

= 3p + q

(b)

p + q + p – q

= (p + p) + (q – q)

= 2p

(c)

p – q + p – q

= (p + p) + (– q – q)

= 2p – 2q

(d) p + q – p + q

= (p – p) + (q + q)

= 2q

(e)

p + q – (p + q)

= p + q – p – q

= 0

(f)

p – q – p – q

= (p – p) + (– q – q)

= – 2q

(g)

2d – d – d – d

= (2 – 1 – 1 – 1)d

= – d

(h)

2d – d – d – c

= (2 – 1 – 1)d – c

= – c

(i)

2d – d – (d – c)

= 2d – d – d + c

= (2 – 1 – 1)d + c

= c

(j)

2d – (d – d) – c

= 2d – 0 – c

= 2d – c

(k)

2d – d – c – c

= (2 – 1)d – 2c

= d – 2c

Some simplifications of algebraic expressions are done below. The expression on the right-hand side should be in its simplest form.

• Observe each of them and see if there is a mistake.
• If you think there is a mistake, try to explain what might have gone wrong.
• Then, simplify it correctly.

Answer

The completed table, with the correct simplest forms, is:

Explanation of the mistakes:

1. Mistake. 3a and 2b are unlike terms (different letter-numbers), so they cannot be added into a single number. The expression is already in its simplest form: 3a + 2b.

2. No mistake. 3b – 2b – b = (3 – 2 – 1)b = 0b = 0. Correct.

3. Mistake. By the distributive property, both terms inside the bracket must be multiplied by 6. Here 6 × 2 = 12 was wrongly written as 8. Correct: 6(p + 2) = 6p + 12.

4. Mistake. When the bracket with a minus sign is opened, the sign of every term inside must change. The sign of 4y was not changed. Correct: 4x + 3y – 3x – 4y = x – y.

5. Mistake. Opening – (2 – 6z) changes the sign of – 6z to + 6z. Correct: 5 – 2 + 6z = 3 + 6z.

6. Mistake. The result is completely wrong; there is no term to double and nothing to subtract. Correct: 2 + x + 3 = x + 5.

7. Mistake. 2y and 3y are like terms and must be added, not subtracted. Correct: 2y + 3y – 6 = 5y – 6.

8. Mistake. 7p – p = 6p, not 7p. Correct: 6p + 3q.

9. Mistake. The right-hand side is not in simplest form — the like terms 10w and 20w have not been added. Correct: 30w + 15x.

10. Mistake. The right-hand side is written in a factored form, but simplest form requires the brackets to be removed. Correct: 3j + 6k + 9h + 12.

11. Mistake. The signs are wrong and 4 × 5 = 20 (not – 20). Correct: 4(2r + 3s + 5) = 8r + 12s + 20.

Take a look at all the corrected simplest forms (i.e. brackets are removed, like terms are added, and terms with only numbers are also added). Is there any relation between the number of terms and the number of letter-numbers these expressions have?

Answer

Let us count the number of terms and the number of distinct letter-numbers in each corrected simplest form:

In every case, the number of letter-numbers is less than or equal to the number of terms.

Hence, the number of letter-numbers in a simplest form is never more than the number of terms in it.

Look at the picture given. In each case, the number machine takes in the 2 numbers at the top of the ‘Y’ as inputs, performs some operations and produces the result at the bottom. The machine performs the same operations on its inputs in each case. Find out the formula of this number machine.

Answer

Looking at the first machine, the inputs are 5 and 2 and the result is 8. We notice that 2 × 5 – 2 = 8, i.e. two times the first number minus the second number.

If the first input is a and the second input is b, the formula is:

2a - b

Checking the formula for each set of inputs:

• Inputs 5, 2 : 2 × 5 – 2 = 8
• Inputs 8, 1 : 2 × 8 – 1 = 15
• Inputs 9, 11 : 2 × 9 – 11 = 7
• Inputs 10, 10 : 2 × 10 – 10 = 10
• Inputs 6, 4 : 2 × 6 – 4 = 8

The formula holds true for every set of inputs.

Hence, the formula of the number machine is 2a – b.

Find the formulas of the number machines below and write the expression for each set of inputs.

Answer

Machine (i): Taking the first inputs 5 and 2 with result 5, we see 5 + 2 – 2 = 5, i.e. two is subtracted from the sum of the two numbers.

Formula: a + b - 2

• Inputs 5, 2 : 5 + 2 – 2 = 5
• Inputs 8, 1 : 8 + 1 – 2 = 7
• Inputs 9, 11 : 9 + 11 – 2 = 18
• Inputs 10, 10 : 10 + 10 – 2 = 18
• Inputs a, b : a + b – 2

Machine (ii): Taking the inputs 4 and 1 with result 5, we see 4 × 1 + 1 = 5, i.e. one is added to the product of the two numbers.

Formula: a × b + 1 = ab + 1

• Inputs 4, 1 : 4 × 1 + 1 = 5
• Inputs 6, 0 : 6 × 0 + 1 = 1
• Inputs 3, 2 : 3 × 2 + 1 = 7
• Inputs 10, 3 : 10 × 3 + 1 = 31
• Inputs a, b : ab + 1

Hence, the formulas are a + b – 2 for machine (i) and ab + 1 for machine (ii).

Somjit noticed a repeating pattern along the border of a saree.

Somjit wonders if there is a way to describe all the positions where the (i) Design A occurs, (ii) Design B occurs, and (iii) Design C occurs.

(a) Where would design C appear for the nth time?

(b) Similarly, find the formula that gives the position where the other Designs appear for the nth time.

Answer

The designs repeat in the order A, B, C, A, B, C, … so the pattern repeats after every 3 positions.

(a) Design C. It appears at positions 3, 6, 9, 12, … — these are the multiples of 3. The first time at 3 (= 3 × 1), the second time at 6 (= 3 × 2), and so on.

Hence, Design C appears for the nth time at position 3n.

(b) Design B. It appears at positions 2, 5, 8, 11, … Each of these is one less than the corresponding position of Design C (3n – 1).

∴ Design B appears for the nth time at position 3n – 1.

Design A. It appears at positions 1, 4, 7, 10, … Each of these is two less than the corresponding position of Design C (3n – 2).

∴ Design A appears for the nth time at position 3n – 2.

Given a position number can we find out the design that appears there? Which Design appears at Position 122? Can the remainder obtained by dividing the position number by 3 be used for this? Observe the table below. Use this to find what design appears at positions 99, 122, and 148.

Answer

The designs occur in a repeating cycle A, B, C, and their positions are:

Design A → 3n – 2, $\quad$ Design B → 3n – 1, $\quad$ Design C → 3n

So the remainder left on dividing the position number by 3 tells us the design:

• Remainder 0 (the position is a multiple of 3) → Design C

• Remainder 2 (the position is one less than a multiple of 3, i.e. 3n – 1) → Design B

• Remainder 1 (the position is two less than a multiple of 3, i.e. 3n – 2) → Design A

Applying this to the given positions:

Position 99:

99 = 3 × 33, remainder = 0.

∴ Design C appears at position 99.

Position 122:

122 = 3 × 40 + 2, remainder = 2.

∴ Design B appears at position 122.

Position 148:

148 = 3 × 49 + 1, remainder = 1.

∴ Design A appears at position 148.

Hence, positions 99, 122 and 148 show Designs C, B and A respectively.

Will the diagonal sums be equal in every 2 × 2 square in this endless grid? How can we be sure?

Answer

Yes, the diagonal sums are equal in every 2 × 2 square. We cannot check this for all squares, since there are unlimited many of them, so we use algebra to be sure.

Let the top left number of any 2 × 2 square be ‘a’.

In the calendar grid the number to the right of a number is 1 more, and the number just below it is 7 more. So the square is:

$\begin{array}{|c|c|} \hline a & a + 1 \\ \hline a + 7 & a + 8 \\ \hline \end{array}$

Now find the two diagonal sums:

a + (a + 8) = 2a + 8

(a + 1) + (a + 7) = 2a + 8

Both diagonal sums equal 2a + 8 for every value of ‘a’.

Hence, the diagonal sums are equal in every 2 × 2 square of the grid.

Given that we know the top left number, how do we find the other numbers in this 2 × 2 square?

Answer

Let the top left number be ‘a’. Using the way dates are arranged in the calendar grid:

• the number to the right of ‘a’ is 1 more than it → a + 1

• the number below ‘a’ is 7 more than it (next row) → a + 7

• the number diagonal to ‘a’ is 8 more than it → a + 8

So the 2 × 2 square is:

$\begin{array}{|c|c|} \hline a & a + 1 \\ \hline a + 7 & a + 8 \\ \hline \end{array}$

Hence, once the top left number ‘a’ is known, the other three numbers are a + 1, a + 7 and a + 8.

Verify this expression for diagonal sums by considering any 2 × 2 square and taking its top left number to be ‘a’.

Answer

Take any 2 × 2 square and let its top left number be ‘a’. The four numbers are:

$\begin{array}{|c|c|} \hline a & a + 1 \\ \hline a + 7 & a + 8 \\ \hline \end{array}$

First diagonal (top left and bottom right):

a + (a + 8)

= a + a + 8

= 2a + 8

Second diagonal (top right and bottom left):

(a + 1) + (a + 7)

= a + 1 + a + 7

= 2a + 8

Both diagonal sums are 2a + 8, which is the same for any value of ‘a’.

∴ The diagonal sums of every 2 × 2 square are equal, and each equals 2a + 8.

Consider a set of numbers from the calendar (having endless rows) forming under the following shape:

Find the sum of all the numbers. Compare it with the number in the centre: 15. Repeat this for another set of numbers that forms this shape. What do you observe?

Answer

The shape is a ‘plus’ (cross) with the centre number 15. The five numbers are 8, 14, 15, 16 and 22.

Sum of the numbers:

8 + 14 + 15 + 16 + 22

= 75

= 5 × 15

So the sum is 5 times the centre number.

Repeating for another such set (centre number 9): the five numbers are 2, 8, 9, 10 and 16.

2 + 8 + 9 + 10 + 16

= 45

= 5 × 9

Again the sum is 5 times the centre number.

Hence, we observe that the sum of all the numbers in this shape is always 5 times the number in the centre.

Will this always happen? How do you show this?
[Hint: Consider a general set of numbers that forms this shape. Take the number at the centre to be ‘a’. Express the other numbers in terms of ‘a’.]

Answer

Yes, it will always happen. To show this, take the centre number to be ‘a’ and describe the other numbers in terms of ‘a’:

• the number above the centre is 7 less → a – 7

• the number below the centre is 7 more → a + 7

• the number to the left is 1 less → a – 1

• the number to the right is 1 more → a + 1

So the numbers forming the shape are:

Now find the sum:

(a - 7) + (a - 1) + a + (a + 1) + (a + 7)

= a + a + a + a + a + (- 7 - 1 + 1 + 7)

= 5a + 0

= 5a

The sum is 5a, i.e. 5 times the centre number, for every value of ‘a’.

∴ The sum of the numbers in this shape is always 5 times the centre number.

Find other shapes for which the sum of the numbers within the figure is always a multiple of one of the numbers.

Answer

Such shapes can be built by choosing cells that are placed symmetrically about one central number ‘a’, so that they pair up as (a – k) and (a + k). Each such pair adds up to 2a, and the middle number ‘a’ is left over, making the total a multiple of ‘a’.

A few examples are shown below.

(i) A horizontal row of three dates (with middle number ‘a’): a – 1, a, a + 1

(a - 1) + a + (a + 1) = 3a

The sum is 3 times the middle number.

(ii) A vertical column of three dates (with middle number ‘a’): a – 7, a, a + 7

(a - 7) + a + (a + 7) = 3a

The sum is 3 times the middle number.

(iii) A diagonal of three dates (with middle number ‘a’): a – 8, a, a + 8

(a - 8) + a + (a + 8) = 3a

The sum is again 3 times the middle number.

(iv) An ‘X’ shape of five dates (with centre ‘a’): a – 8, a – 6, a, a + 6, a + 8

(a - 8) + (a - 6) + a + (a + 6) + (a + 8) = 5a

The sum is 5 times the centre number.

Hence, any shape whose cells are symmetric about a central number gives a sum that is a multiple of that central number.

Look at the picture below. It is a pattern using matchsticks.

How many matchsticks will there be in Step 33, Step 84, and Step 108? Of course, we can draw and count, but is there a quicker way to find the answers using the pattern present here?

Answer

Yes, there is a quicker way.

The number of matchsticks increases by 2 at each step, and the number of matchsticks at Step y is given by the expression:

3 + 2 × (y – 1)

We just replace ‘y’ by the step number.

Step 33:

3 + 2 × (33 - 1)

= 3 + 2 × 32

= 3 + 64

= 67

Hence, Step 33 has 67 matchsticks.

Step 84:

3 + 2 × (84 - 1)

= 3 + 2 × 83

= 3 + 166

= 169

Hence, Step 84 has 169 matchsticks.

Step 108:

3 + 2 × (108 - 1)

= 3 + 2 × 107

= 3 + 214

= 217

Hence, Step 108 has 217 matchsticks.

You might have already noticed that there is a 2 in the first step also, 3 = 1 + 2. Using this, the expression we get is 2y + 1. Does the above expression also give the number of matchsticks at each step correctly? Are these expressions the same?

Answer

Yes, the expression 2y + 1 also gives the number of matchsticks correctly at each step. We can check this by simplifying the earlier expression 3 + 2 × (y – 1):

3 + 2 × (y - 1)

= 3 + 2y - 2

= 2y + (3 - 2)

= 2y + 1

On simplifying, 3 + 2 × (y – 1) becomes exactly 2y + 1.

Hence, both expressions are the same and give the same number of matchsticks at every step.

Matchsticks are placed in two orientations — (a) horizontal ones at the top and bottom, and (b) the ones placed diagonally in the middle. For example, in step 2 there are 2 matchsticks placed horizontally and 3 matchsticks placed diagonally.

What are these numbers in Step 3 and Step 4?

Answer

Looking at the pattern, the horizontal matchsticks follow 1, 2, 3, 4, … and the diagonal matchsticks follow 2, 3, 4, 5, … So at each step the number of diagonal matchsticks is one more than the number of horizontal matchsticks.

Step 3:

No. of matchsticks placed horizontally = 3

No. of matchsticks placed diagonally = 4

Step 4:

No. of matchsticks placed horizontally = 4

No. of matchsticks placed diagonally = 5

Hence, in Step 3 there are 3 horizontal and 4 diagonal matchsticks, and in Step 4 there are 4 horizontal and 5 diagonal matchsticks.

How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step ‘y’ in each orientation. Do the two expressions add up to 2y + 1?

Answer

In the triangle pattern, the matchsticks are placed in two orientations — the horizontal ones (at the top and bottom) and the ones placed diagonally (in the middle).

Horizontal matchsticks:

The number of horizontal matchsticks increases by 1 at each step, and equals the step number.

∴ At Step ‘y’, the number of horizontal matchsticks = y.

Diagonal matchsticks:

The number of diagonal matchsticks also increases by 1 at each step, and is always one more than the step number.

∴ At Step ‘y’, the number of diagonal matchsticks = y + 1.

Do they add up to 2y + 1?

Total matchsticks = (horizontal) + (diagonal)

= y + (y + 1)

= 2y + 1

Hence, the number of horizontal matchsticks at Step ‘y’ is y and the number of diagonal matchsticks is y + 1; the two expressions add up to 2y + 1.

One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day?
(a) 30x + 20y
(b) (30 + 20) × (x + y)
(c) 20x + 30y
(d) (30 + 20) × x + y
(e) 30x – 20y

Answer

The amount earned from Jowar roti = ₹30 per plate × x plates = 30x

The amount earned from Pulao = ₹20 per plate × y plates = 20y

Total amount earned = 30x + 20y

Hence, option (a) is the correct option.

Pushpita sells two types of flowers on Independence day: champak and marigold. ‘p’ customers only bought champak, ‘q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day?
(a) p + q + r
(b) p + q + 2r
(c) 2 × (p + q + r)
(d) p + q + r + 2
(e) p + q + r + 1
(f) 2 × (p + q)

Answer

The three groups of customers are all different people:

p customers bought only champak, q customers bought only marigold, and r customers bought both.

So the total number of customers = p + q + r.

Since each customer (including those who bought both flowers) received exactly one flag, the number of flags given away equals the number of customers.

∴ Number of flags = p + q + r

Hence, option (a) is the correct option.

A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights.
(a) Write an expression describing how far away the snail is from its starting position.
(b) What can we say about the snail’s movement if d > u?

Answer

(a)

In one full day-and-night cycle, the snail climbs u cm and slips down d cm.

Net distance covered in one day and one night = (u – d) cm.

For 10 days and 10 nights, this happens 10 times.

∴ Distance from the starting position = 10(u – d) cm.

Hence, the snail is 10(u – d) cm away from its starting position.

(b)

If d > u, then (u – d) is negative, so 10(u – d) is also negative.

This means the snail slips down more than it climbs in each cycle, so it keeps moving below its starting position.

Hence, if d > u, the snail moves downward overall and will never reach the top of the well.

Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks?

Answer

There are 7 days in a week.

Week 1: She cycles 5 km each day.

Distance in Week 1 = 5 × 7 = 35 km

Week 2: The daily distance increases by z km, so she cycles (5 + z) km each day.

Distance in Week 2 = (5 + z) × 7 = 35 + 7z km

Week 3: The daily distance increases by z km again, so she cycles (5 + 2z) km each day.

Distance in Week 3 = (5 + 2z) × 7 = 35 + 14z km

Total distance after 3 weeks:

= (35) + (35 + 7z) + (35 + 14z)

= (35 + 35 + 35) + (7z + 14z)

= 105 + 21z

Hence, Radha would have cycled (105 + 21z) km after 3 weeks.

In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.

Answer

Starting from the oval w + 2, we apply the operation in each box to get the expression in the next oval. Doing this along each path:

Path 1 (given):

w + 2 $\xrightarrow{+3}$ w + 5 $\xrightarrow{×4}$ 4w + 20

Path 2:

w + 2 $\xrightarrow{-5}$ w – 3 $\xrightarrow{×3}$ 3(w – 3) = 3w – 9

Path 3:

w + 2 $\xrightarrow{-4}$ w – 2 $\xrightarrow{×3}$ 3(w – 2) = 3w – 6

Path 4:

w + 2 $\xrightarrow{-8}$ w – 6 $\xrightarrow{-4}$ w – 10

So the missing blanks are:

Hence, the missing expressions are 3w – 9, w – 2, w – 6 and w – 10.

A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations.
(a) If t = 4, what is the time taken to travel from Yahapur to Vahapur?
(b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur? [Hint: Draw a rough diagram to visualise the situation]

Answer

The three stations lie at equal distances between Yahapur and Vahapur, dividing the whole journey into 4 equal parts:

Yahapur — Station 1 — Station 2 — Station 3 — Vahapur

Time for travelling each part = t minutes, and there are 4 such parts.

∴ Total travelling time = 4 × t = 4t minutes.

The train halts for 2 minutes at each of the 3 stations.

∴ Total halting time = 3 × 2 = 6 minutes.

(a) When t = 4:

Total time = 4t + 6

= 4 × 4 + 6

= 16 + 6

= 22 minutes

Hence, when t = 4, the time taken to travel from Yahapur to Vahapur is 22 minutes.

(b) Total time = travelling time + halting time

= 4t + 6 minutes

Hence, the algebraic expression for the time taken is (4t + 6) minutes.

Simplify the following expressions:
(a) 3a + 9b – 6 + 8a – 4b – 7a + 16
(b) 3 (3a – 3b) – 8a – 4b – 16
(c) 2 (2x – 3) + 8x + 12
(d) 8x – (2x – 3) + 12
(e) 8h – (5 + 7h) + 9
(f) 23 + 4(6m – 3n) – 8n – 3m – 18

Answer

(a)

Given expression:

3a + 9b – 6 + 8a – 4b – 7a + 16

= (3a + 8a – 7a) + (9b – 4b) + (– 6 + 16)

= 4a + 5b + 10

∴ 3a + 9b – 6 + 8a – 4b – 7a + 16 = 4a + 5b + 10

(b)

Given expression:

3 (3a – 3b) – 8a – 4b – 16

= 9a – 9b – 8a – 4b – 16

= (9a – 8a) + (– 9b – 4b) – 16

= a – 13b – 16

∴ 3 (3a – 3b) – 8a – 4b – 16 = a – 13b – 16

(c)

Given expression:

2 (2x – 3) + 8x + 12

= 4x – 6 + 8x + 12

= (4x + 8x) + (– 6 + 12)

= 12x + 6

∴ 2 (2x – 3) + 8x + 12 = 12x + 6

(d)

Given expression:

8x – (2x – 3) + 12

= 8x – 2x + 3 + 12

= (8x – 2x) + (3 + 12)

= 6x + 15

∴ 8x – (2x – 3) + 12 = 6x + 15

(e)

Given expression:

8h – (5 + 7h) + 9

= 8h – 5 – 7h + 9

= (8h – 7h) + (– 5 + 9)

= h + 4

∴ 8h – (5 + 7h) + 9 = h + 4

(f)

Given expression:

23 + 4(6m – 3n) – 8n – 3m – 18

= 23 + 24m – 12n – 8n – 3m – 18

= (24m – 3m) + (– 12n – 8n) + (23 – 18)

= 21m – 20n + 5

∴ 23 + 4(6m – 3n) – 8n – 3m – 18 = 21m – 20n + 5

Add the expressions given below:
(a) 4d – 7c + 9 and 8c – 11 + 9d
(b) – 6f + 19 – 8s and – 23 + 13f + 12s
(c) 8d – 14c + 9 and 16c – (11 + 9d)
(d) 6f – 20 + 8s and 23 – 13f – 12s
(e) 13m – 12n and 12n – 13m
(f) – 26m + 24n and 26m – 24n

Answer

(a)

(4d – 7c + 9) + (8c – 11 + 9d)

= (4d + 9d) + (– 7c + 8c) + (9 – 11)

= 13d + c – 2

∴ The sum is 13d + c – 2.

(b)

(– 6f + 19 – 8s) + (– 23 + 13f + 12s)

= (– 6f + 13f) + (– 8s + 12s) + (19 – 23)

= 7f + 4s – 4

∴ The sum is 7f + 4s – 4.

(c)

(8d – 14c + 9) + [16c – (11 + 9d)]

= 8d – 14c + 9 + 16c – 11 – 9d

= (8d – 9d) + (– 14c + 16c) + (9 – 11)

= – d + 2c – 2

= 2c – d – 2

∴ The sum is 2c – d – 2.

(d)

(6f – 20 + 8s) + (23 – 13f – 12s)

= (6f – 13f) + (8s – 12s) + (– 20 + 23)

= – 7f – 4s + 3

∴ The sum is – 7f – 4s + 3.

(e)

(13m – 12n) + (12n – 13m)

= (13m – 13m) + (– 12n + 12n)

= 0

∴ The sum is 0.

(f)

(– 26m + 24n) + (26m – 24n)

= (– 26m + 26m) + (24n – 24n)

= 0

∴ The sum is 0.

Subtract the expressions given below:
(a) 9a – 6b + 14 from 6a + 9b – 18
(b) – 15x + 13 – 9y from 7y – 10 + 3x
(c) 17g + 9 – 7h from 11 – 10g + 3h
(d) 9a – 6b + 14 from 6a – (9b + 18)
(e) 10x + 2 + 10y from –3y + 8 – 3x
(f) 8g + 4h – 10 from 7h – 8g + 20

Answer

To subtract the first expression from the second, we change the sign of every term of the first expression and then add.

(a)

(6a + 9b – 18) – (9a – 6b + 14)

= 6a + 9b – 18 – 9a + 6b – 14

= (6a – 9a) + (9b + 6b) + (– 18 – 14)

= – 3a + 15b – 32

∴ The result is – 3a + 15b – 32.

(b)

(7y – 10 + 3x) – (– 15x + 13 – 9y)

= 7y – 10 + 3x + 15x – 13 + 9y

= (3x + 15x) + (7y + 9y) + (– 10 – 13)

= 18x + 16y – 23

∴ The result is 18x + 16y – 23.

(c)

(11 – 10g + 3h) – (17g + 9 – 7h)

= 11 – 10g + 3h – 17g – 9 + 7h

= (– 10g – 17g) + (3h + 7h) + (11 – 9)

= – 27g + 10h + 2

∴ The result is – 27g + 10h + 2.

(d)

[6a – (9b + 18)] – (9a – 6b + 14)

= 6a – 9b – 18 – 9a + 6b – 14 $\quad$[Removing the brackets]

= (6a – 9a) + (– 9b + 6b) + (– 18 – 14)

= – 3a – 3b – 32

= – (3a + 3b + 32)

∴ The result is – (3a + 3b + 32).

(e)

(– 3y + 8 – 3x) – (10x + 2 + 10y)

= – 3y + 8 – 3x – 10x – 2 – 10y

= (– 3x – 10x) + (– 3y – 10y) + (8 – 2)

= – 13x – 13y + 6

∴ The result is – 13x – 13y + 6.

(f)

(7h – 8g + 20) – (8g + 4h – 10)

= 7h – 8g + 20 – 8g – 4h + 10

= (– 8g – 8g) + (7h – 4h) + (20 + 10)

= – 16g + 3h + 30

∴ The result is – 16g + 3h + 30.

Describe situations corresponding to the following algebraic expressions:
(a) 8x + 3y
(b) 15x – 2x

Answer

More than one situation can correspond to each expression. One example for each is given below.

(a) 8x + 3y

Situation: A fruit seller sells mangoes at ₹8 each and bananas at ₹3 each. If a customer buys x mangoes and y bananas, the costs are calculated as follows.

Cost of x mangoes = ₹8x

Cost of y bananas = ₹3y

∴ Total amount to be paid = ₹(8x + 3y)

(b) 15x – 2x

Situation: A packet contains 15 identical pens, and the cost of one pen is ₹x, so the value of a full packet is ₹15x. Two pens in the packet are found damaged and cannot be sold; their value is ₹2x.

∴ Amount received on selling the packet = ₹(15x – 2x) = ₹13x

Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?

Answer

Let us observe the pattern:

Each extra fold increases the number of pieces by 1. So the number of pieces is always 2 more than the number of folds.

∴ Number of pieces = Number of folds + 2

When the rope is folded 10 times and cut:

Number of pieces = 10 + 2 = 12

When the rope is folded r times and cut:

Number of pieces = r + 2

Hence, folding 10 times gives 12 pieces, and folding r times gives (r + 2) pieces.

Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?

Answer

Observe the pattern:

• The 1^st square is made with 4 matchsticks.
• Every next square shares one side with the previous square, so it needs only 3 more matchsticks.

So, for any number of squares, we start with 4 and add 3 for each square after the first.

For 10 squares:

Number of matchsticks = 4 + 3 × 9 = 4 + 27 = 31

For w squares:

Number of matchsticks = 4 + 3(w – 1) = 3w + 1

Hence, 10 squares need 31 matchsticks, and w squares need (3w + 1) matchsticks.

Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour.

Answer

The colours repeat in a cycle of 4 in the order: Red, Yellow, Green, Yellow.

From this pattern:

• Red appears at positions 1, 5, 9, … , described by 4n – 3.
• Green appears at positions 3, 7, 11, … , described by 4n – 1.
• Yellow appears at every even position 2, 4, 6, 8, … , described by 2n.

To find the colour at a position, we check its remainder on division by 4.

Position 90:

90 ÷ 4 = 22 remainder 2

A remainder of 2 corresponds to an even position.

∴ The colour is Yellow.

Position 190:

190 ÷ 4 = 47 remainder 2

∴ The colour is Yellow.

Position 343:

343 ÷ 4 = 85 remainder 3

Positions with remainder 3 are 3, 7, 11, 15,…

These are the positions of Green.

∴ The colour is Green.

Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?

Answer

Count the number of squares at each step:

Each step adds 4 more squares to the previous step. Starting from 5 squares at Step 1, the number of squares at Step n is:

Number of squares = 5 + (n – 1) × 4 = 4n + 1

Step 4:

4n + 1

= 4 × 4 + 1

= 17 squares

Hence, there are 17 squares in Step 4.

Step 10:

4n + 1

= 4 × 10 + 1

= 41 squares

Hence, there are 41 squares in Step 10.

Step 50:

4n + 1

= 4 × 50 + 1

= 201 squares

Hence, there are 201 squares in Step 50.

∴ General formula for the number of squares = 4n + 1

Number of vertices: Each square has 4 vertices. Counting the vertices of all the squares:

Number of vertices = 4 × (Number of squares)

= 4 × (4n + 1)

= 16n + 4

Hence, the number of squares is 4n + 1 and the number of vertices is 16n + 4.

Numbers are written in a particular sequence in this endless 4-column grid.

(a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4).
(b) In which row and column will the following numbers appear:
    (i) 124
    (ii) 147
    (iii) 201
(c) What number appears in row r and column c?
(d) Observe the positions of multiples of 3. Do you see any pattern in it? List other patterns that you see.

Answer

Let r be the row number and c be the column number.

(a)

In any row r, the four entries increase by 1 from column 1 to column 4, and the first entry of row r is 4(r – 1) + 1. So the numbers in each column are:

Column 1: 4(r – 1) + 1

Column 2: 4(r – 1) + 2

Column 3: 4(r – 1) + 3

Column 4: 4(r – 1) + 4

In general, the number in row r and column c is 4(r – 1) + c.

(b)

To locate a number N, we write N = 4(r – 1) + c, where c is 1, 2, 3 or 4.

Dividing N by 4, the quotient decides the row and the remainder decides the column.

(i) 124

124 ÷ 4 gives quotient 31 and remainder 0.

A remainder of 0 corresponds to column 4:

124 = 4 × 30 + 4 = 4(31 – 1) + 4

⇒ r – 1 = 30 and c = 4

∴ 124 appears in row 31, column 4.

(ii) 147

147 ÷ 4 gives quotient 36 and remainder 3.

147 = 4 × 36 + 3 = 4(37 – 1) + 3

⇒ r – 1 = 36 and c = 3

∴ 147 appears in row 37, column 3.

(iii) 201

201 ÷ 4 gives quotient 50 and remainder 1.

201 = 4 × 50 + 1 = 4(51 – 1) + 1

⇒ r – 1 = 50 and c = 1

∴ 201 appears in row 51, column 1.

(c)

The number that appears in row r and column c is:

Number = 4(r – 1) + c

(d)

Positions of multiples of 3:

The columns of the multiples of 3 follow a repeating cycle: 3, 2, 1, 4, 3, 2, 1, 4, …

Some other patterns in the grid:

1. All numbers in Column 4 are multiples of 4.
2. Even numbers always appear in Column 2 and Column 4.
3. Odd numbers always appear in Column 1 and Column 3.

Hence, the number in row r and column c is 4(r – 1) + c, and the multiples of 3 repeat through the columns in the order 3, 2, 1, 4.