Algebra Play

How would you change this number trick game to make the final answer 3? What about 5?

Answer

Let us first see why the original game always gives 2.

1. Think of a number: x
2. Double it: 2x
3. Add four: 2x + 4
4. Divide by 2: x + 2
5. Subtract the original number: x + 2 - x = 2

The final answer comes from the number added in step 3. After dividing by 2, the added number 4 becomes 2, and this is what survives after the original number is subtracted.

So, to make the final answer n, we must add 2n in step 3.

For the final answer 3:

Add 2 × 3 = 6 in step 3.

1. Think of a number: x
2. Double it: 2x
3. Add six: 2x + 6
4. Divide by 2: x + 3
5. Subtract the original number: x + 3 - x = 3

For the final answer 5:

Add 2 × 5 = 10 in step 3.

1. Think of a number: x
2. Double it: 2x
3. Add ten: 2x + 10
4. Divide by 2: x + 5
5. Subtract the original number: x + 5 - x = 5

Hence, adding 6 in step 3 always gives 3, and adding 10 always gives 5.

Can you come up with more complicated steps that always lead to the same final value?

Answer

Yes. As long as the original number x cancels out at the end, the steps can be made as complicated as we like. Here is one example that always gives 3.

1. Think of a number: x
2. Add 5: x + 5
3. Multiply by 2: 2(x + 5) = 2x + 10
4. Subtract 4: 2x + 6
5. Divide by 2: x + 3
6. Subtract the original number: x + 3 - x = 3

Whatever number we start with, the x cancels in the last step.

Check (x = 7): 7 + 5 = 12 ⇒ 12 × 2 = 24 ⇒ 24 - 4 = 20 ⇒ 20 ÷ 2 = 10 ⇒ 10 - 7 = 3

Hence, this is a more complicated number trick that always leads to the same final value.

Mukta thinks of another date, follows the same steps, and reports her answer as 1390. What date did Mukta start with this time?

Answer

We know that

Final answer = 100M + 165 + D

where M is the month and D is the day.

Given:

Mukta's answer = 1390

⇒ 100M + 165 + D = 1390

⇒ 100M + D = 1390 - 165

⇒ 100M + D = 1225

Since D is a day, it is at most 31 and needs only 2 digits. So the last 2 digits give D and the digits before that give M.

∴ M = 12 and D = 25

This means that the date she thought of was 25th December.

Check: 100 × 12 + 165 + 25 = 1200 + 165 + 25 = 1390

Hence, Mukta started with the 25th of December (25/12).

(a) Find the dates if the final answers are the following:

1. 1269
2. 394
3. 296

(b) Can you change the steps in this trick and still find the original date? Instead of subtracting 165 from the final answer, you might have to subtract some other number.

Answer

(a)

In each case we use:

Final answer = 100M + 165 + D ⇒ 100M + D = Final answer - 165

(i) Final answer = 1269

100M + D = 1269 - 165 = 1104

Last 2 digits ⇒ D = 04, and the digits before ⇒ M = 11

Check: 100 × 11 + 165 + 4 = 1100 + 165 + 4 = 1269

Hence, the date is the 4th of November (04/11).

(ii) Final answer = 394

100M + D = 394 - 165 = 229

Last 2 digits ⇒ D = 29, and the digits before ⇒ M = 2

Check: 100 × 2 + 165 + 29 = 200 + 165 + 29 = 394

Hence, the date is the 29th of February (29/02), which is valid in a leap year.

(iii) Final answer = 296

100M + D = 296 - 165 = 131

Last 2 digits ⇒ D = 31, and the digits before ⇒ M = 1

Check: 100 × 1 + 165 + 31 = 100 + 165 + 31 = 296

Hence, the date is the 31st of January (31/01).

(b)

Yes. We can change some of the numbers used in the trick and still recover the original date.

For example, replace "Add 6" with "Add 10".

• Multiply month by 5: 5M
• Add 10: 5M + 10
• Multiply by 4: 20M + 40
• Add 9: 20M + 49
• Multiply by 5: 100M + 245
• Add the day: 100M + 245 + D

Now the final answer is 100M + 245 + D.

To recover the date, we subtract 245 from the final answer.

This gives 100M + D

From this, we can identify the month and the day just as before.

Hence, the steps in the trick can be changed, but we must subtract the new constant obtained from the steps in order to recover the original date.

In a number pyramid, each number is the sum of the two numbers directly below it. Use this rule to fill the following pyramids.

Answer

In a number pyramid, each box is the sum of the two boxes directly below it. Since the full bottom row is given in each pyramid, we simply add upwards.

(i) Bottom row: 6, 2

Top = 6 + 2 = 8

(ii) Bottom row: 3, 4, 3

Middle row: (3 + 4), (4 + 3) = 7, 7

Top = 7 + 7 = 14

(iii) Bottom row: 5, 4, 5, 0

Row above bottom: (5 + 4), (4 + 5), (5 + 0) = 9, 9, 5

Next row: (9 + 9), (9 + 5) = 18, 14

Top = 18 + 14 = 32

Hence, the tops of the three pyramids are 8, 14 and 32 respectively.

Fill the following pyramids:

Answer

Here some boxes are given and some are missing. We use letter-numbers for the unknown bottom boxes and form equations from the rule that each box equals the sum of the two below it.

(i) Let the bottom row be 4, x, 6, y, with top 50 and the right box of the second row 22.

Row above bottom: (4 + x), (x + 6), (6 + y)

Second row from top: (4 + 2x + 6), (x + 12 + y) = (10 + 2x), (x + y + 12)

From the second row: x + y + 12 = 22 ⇒ x + y = 10

From the top: (10 + 2x) + (x + y + 12) = 50 ⇒ 22 + 3x + y = 50 ⇒ 3x + y = 28

Subtracting the first relation from the second:

(3x + y) - (x + y) = 28 - 10 ⇒ 2x = 18 ⇒ x = 9, and so y = 1

∴ Bottom row: 4, 9, 6, 1 ⇒ next row: 13, 15, 7 ⇒ next row: 28, 22 ⇒ top: 50

(ii) Let the bottom row be 5, x, 7, y, with the left box of the second row 40 and the right box of the third row 9.

Third row: (5 + x), (x + 7), (7 + y)

Right box: 7 + y = 9 ⇒ y = 2

Second row, left box: (5 + x) + (x + 7) = 12 + 2x = 40 ⇒ 2x = 28 ⇒ x = 14

∴ Bottom row: 5, 14, 7, 2 ⇒ next row: 19, 21, 9 ⇒ next row: 40, 30 ⇒ top: 70

(iii) Let the bottom row be 3, 5, x, y, with top 35 and the right box of the third row 7.

Third row: (3 + 5), (5 + x), (x + y) = 8, (5 + x), (x + y)

Right box: x + y = 7

Second row: 8 + (5 + x), (5 + x) + (x + y) = (13 + x), (5 + 2x + y)

Top: (13 + x) + (5 + 2x + y) = 18 + 3x + y = 35 ⇒ 3x + y = 17

Subtracting x + y = 7:

(3x + y) - (x + y) = 17 - 7 ⇒ 2x = 10 ⇒ x = 5, and so y = 2

∴ Bottom row: 3, 5, 5, 2 ⇒ next row: 8, 10, 7 ⇒ next row: 18, 17 ⇒ top: 35

Hence, the three completed pyramids have topmost numbers 50, 70 and 35 respectively.

Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

Answer

For a pyramid with three rows and bottom row a, b, c, the topmost number is:

a + 2b + c

This is because the middle row is (a + b) and (b + c), and the top is (a + b) + (b + c) = a + 2b + c.

The middle value b is used twice, so it is counted twice.

(i) Bottom row: 4, 13, 8

Topmost = a + 2b + c

= 4 + 2(13) + 8

= 4 + 26 + 8

= 38

Hence, the topmost number is 38.

(ii) Bottom row: 7, 11, 3

Topmost = a + 2b + c

= 7 + 2(11) + 3

= 7 + 22 + 3

= 32

Hence, the topmost number is 32.

(iii) Bottom row: 10, 14, 25

Topmost = a + 2b + c

= 10 + 2(14) + 25

= 10 + 28 + 25

= 63

Hence, the topmost number is 63.

Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row.

Answer

Let the bottom row of the four-row pyramid be a, b, c, d.

Third row (from bottom): (a + b), (b + c), (c + d)

Second row: (a + b) + (b + c), (b + c) + (c + d) = (a + 2b + c), (b + 2c + d)

Topmost row: (a + 2b + c) + (b + 2c + d)

= a + 2b + c + b + 2c + d $\quad$[Removing brackets]

= a + (2b + b) + (c + 2c) + d

= a + 3b + 3c + d

Hence, the topmost number of a four-row pyramid with bottom row a, b, c, d is a + 3b + 3c + d.

Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.
(i) 8, 19, 21, 13
(ii) 7, 18, 19, 6
(iii) 9, 7, 5, 11

Answer

For a four-row pyramid with bottom row a, b, c, d, the topmost number is

a + 3b + 3c + d

(i) Bottom row: 8, 19, 21, 13

Top = a + 3b + 3c + d

= 8 + 3(19) + 3(21) + 13

= 8 + 57 + 63 + 13

= 141

Hence, the number in the topmost row is 141.

(ii) Bottom row: 7, 18, 19, 6

Top = a + 3b + 3c + d

= 7 + 3(18) + 3(19) + 6

= 7 + 54 + 57 + 6

= 124

Hence, the number in the topmost row is 124.

(iii) Bottom row: 9, 7, 5, 11

Top = a + 3b + 3c + d

= 9 + 3(7) + 3(5) + 11

= 9 + 21 + 15 + 11

= 56

Hence, the number in the topmost row is 56.

Recall the Virahāṅka-Fibonacci number sequence 1, 2, 3, 5, …, where each number is the sum of the two numbers before it.

If the first three Virahāṅka-Fibonacci numbers are written in the bottom row of a number pyramid with three rows, fill in the rest of the pyramid. What numbers appear in the grid? What is the number at the top? Are they all Virahāṅka-Fibonacci numbers?

Answer

The first three Virahāṅka-Fibonacci numbers are 1, 2, 3.

Filling the pyramid by adding the two numbers directly below each box:

Bottom row: 1, 2, 3

Middle row: 1 + 2 = 3, $\quad$ 2 + 3 = 5

Top row: 3 + 5 = 8

The numbers appearing in the grid are 1, 2, 3, 3, 5 and 8, and the number at the top is 8.

Each of these numbers — 1, 2, 3, 5, 8 — belongs to the sequence 1, 2, 3, 5, 8, … . This happens because adding two consecutive Virahāṅka-Fibonacci numbers always gives the next one in the sequence (for example, 2 + 3 = 5 and 3 + 5 = 8).

Hence, every number in the pyramid is a Virahāṅka-Fibonacci number, and the number at the top is 8.

What can you say about the numbers in the pyramid and the number at the top in the following cases?
(i) The first four Virahāṅka-Fibonacci numbers are written in the bottom row of a four row pyramid.
(ii) The first 29 Virahāṅka-Fibonacci numbers are written in the bottom row of a 29 row pyramid.

Answer

(i) The first four Virahāṅka-Fibonacci numbers are 1, 2, 3, 5.

Bottom row: 1, 2, 3, 5

Third row: 1 + 2 = 3, $\quad$ 2 + 3 = 5, $\quad$ 3 + 5 = 8

Second row: 3 + 5 = 8, $\quad$ 5 + 8 = 13

Top row: 8 + 13 = 21

The numbers appearing in the grid are 1, 2, 3, 5, 8, 13 and 21 — all of which are Virahāṅka-Fibonacci numbers. The number at the top is 21, which is the 7^th number of the sequence.

This is because adding two consecutive Virahāṅka-Fibonacci numbers gives the next one, so every row of the pyramid is again a run of consecutive Virahāṅka-Fibonacci numbers.

Hence, all the numbers in the pyramid are Virahāṅka-Fibonacci numbers, and the number at the top is 21.

(ii) Each time we move up a row, every box is the sum of two consecutive Virahāṅka-Fibonacci numbers, which is again a Virahāṅka-Fibonacci number. So each higher row is once more a block of consecutive Virahāṅka-Fibonacci numbers, shifted two places forward.

Starting from the first 29 numbers in the bottom row and moving up 28 rows, the single number at the top is the (2 × 29 − 1) = 57^th Virahāṅka-Fibonacci number.

Hence, every number in this pyramid is a Virahāṅka-Fibonacci number, and the number at the top is the 57^th Virahāṅka-Fibonacci number.

If the bottom row of an n row pyramid contains the first n Virahāṅka-Fibonacci numbers, what can we say about the numbers in the pyramid? What can we say about the number at the top?

Answer

Let the Virahāṅka-Fibonacci sequence be V₁, V₂, V₃, … = 1, 2, 3, 5, 8, … , where each number is the sum of the two before it, i.e. V_k + V_k+1 = V_k+2.

The bottom row holds V₁, V₂, …, V_n, which are consecutive Virahāṅka-Fibonacci numbers. When we add each adjacent pair to form the row above, the rule V_k + V_k+1 = V_k+2 means every entry is again a Virahāṅka-Fibonacci number, and the new row V₃, V₄, …, V_n+1 is once more a run of consecutive Virahāṅka-Fibonacci numbers.

So the smallest number in each row increases by two positions as we move up. After climbing all the way to the top (n − 1 rows up), the single top number is

V_1+2(n-1) = V_2n-1

Hence, every number in the pyramid is a Virahāṅka-Fibonacci number, and the number at the top is the (2n − 1)^th Virahāṅka-Fibonacci number.

A friend selects a 2 × 2 grid from a calendar and tells you only the sum of the four numbers in the grid.

Can we find the 4 numbers in the grid from just knowing this sum?

Answer

Yes, we can.

In any 2 × 2 grid taken from a calendar, once the top-left number is fixed, the other three are completely determined by the calendar layout. Let the top-left number be a. Then the grid is:

So the four numbers are a, a + 1, a + 7 and a + 8, and their sum is:

a + (a + 1) + (a + 7) + (a + 8) = 4a + 16

Since the sum depends only on a, knowing the sum lets us solve for a, and then the remaining three numbers a + 1, a + 7 and a + 8 follow at once.

Hence, from just the sum we can recover all four numbers of the grid.

Suppose you are told that the sum is 36. Can you find the 4 numbers in the grid?

Answer

We have:

4a + 16 = 36

⇒ 4a = 36 - 16 $\quad$[Subtracting 16 from both sides]

⇒ 4a = 20

⇒ a = 5 $\quad$[Dividing both sides by 4]

The other three numbers are a + 1 = 6, a + 7 = 12 and a + 8 = 13. So the grid is:

Hence, the four numbers in the grid are 5, 6, 12 and 13.

In the following grids, find the values of the shapes and fill in the empty squares:

Answer

(i) Let the blue square = s and the red circle = c.

From the first two rows:

Row 1: s + s + c = 27

⇒ 2s + c = 27 $\quad$…(1)

Row 2: c + c + s = 21

⇒ s + 2c = 21 $\quad$…(2)

Multiplying (2) by 2:

2s + 4c = 42 $\quad$…(3)

(2s + 4c) - (2s + c) = 42 - 27 $\quad$[(3) - (1)]

⇒ 2s + 4c - 2s - c = 15

⇒ 3c = 15

⇒ c = $\dfrac{15}{3}$

⇒ c = 5

Substituting c = 5 in (1):

2s + 5 = 27

⇒ 2s = 27 - 5

⇒ 2s = 22

⇒ s = 11

So, blue square = 11 and red circle = 5.

The empty square is Row 3 = c + s + c

= 5 + 11 + 5

= 21.

Hence, the blue square is 11, the red circle is 5, and the missing sum is 21.

(ii) Let the blue circle = m and the purple diamond = d.

From the first two rows:

Row 1: m + d + d = 18

⇒ m + 2d = 18 $\quad$…(1)

Row 2: d + m + m = 15

⇒ 2m + d = 15 $\quad$…(2)

Multiplying (1) by 2:

2m + 4d = 36 $\quad$…(3)

(2m + 4d) - (2m + d) = 36 - 15 $\quad$[(3) - (2)]

⇒ 2m + 4d - 2m - d = 21

⇒ 3d = 21

⇒ d = $\dfrac{21}{3}$

⇒ d = 7

Substituting d = 7 in (1):

m + 14 = 18

⇒ m = 4

So, blue circle = 4 and purple diamond = 7.

The empty square is Row 3 = d + m + m

= 7 + 4 + 4

= 15.

Filling the Last Row:

Each number in the last row is the sum of the shapes in that column.

Column 1: m + d + d = 4 + 7 + 7 = 18

Column 2: d + m + m = 7 + 4 + 4 = 15

Column 3: d + m + m = 7 + 4 + 4 = 15

Bottom-right Square

The bottom-right square is the sum of the three row totals (or the three column totals):

18 + 15 + 15 = 48

Hence, the blue circle is 4, the purple diamond is 7, and the missing sum is 15.

Fill the digits 1, 3, and 7 in $\square\square \times \square$ to make the largest product possible.

Answer

The digits are 1, 3 and 7.

The largest digit is 7. Using it as the multiplier and arranging the other two digits in decreasing order gives:

Largest digit (multiplier) = 7

Remaining digits in decreasing order (multiplicand) = 31

31 × 7 = 217

Hence, the largest possible product is 31 × 7 = 217.

Fill the digits 3, 5, and 9 in $\square\square \times \square$ to make the largest product possible.

Answer

The digits are 3, 5 and 9.

The largest digit is 9. Using it as the multiplier and arranging the other two digits in decreasing order gives:

Largest digit (multiplier) = 9

Remaining digits in decreasing order (multiplicand) = 53

53 × 9 = 477

Hence, the largest possible product is 53 × 9 = 477.

If we choose other 2-digit numbers, and follow the steps, will there always be no remainder? Can you work out what happens if a > b?

Answer

Yes, there will always be no remainder.

Let the two-digit number be ab, that is 10a + b. Reversing the digits gives ba, that is 10b + a.

Taking the difference (with b > a):

(10b + a) - (10a + b)

= 10b - b - 10a + a

= 9b - 9a

= 9(b - a)

Taking the difference (with a > b):

(10a + b) - (10b + a)

= 10a - a - 10b + b

= 9a - 9b

= 9(a - b)

In both cases, the difference is a multiple of 9.

∴ Dividing the difference by 9 always leaves no remainder.

Hence, for any two-digit number with different digits, the result is always divisible by 9.

In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient?

Answer

Let the chosen 2-digit number have tens digit a and units digit b, so the number is (10a + b). On reversing, the new number is (10b + a).

When b > a:

Difference = (10b + a) - (10a + b)

= 9b - 9a

= 9(b - a)

⇒ Quotient on dividing by 9 = (b − a)

When a > b:

Difference = (10a + b) - (10b + a)

= 9a - 9b

= 9(a - b)

⇒ Quotient on dividing by 9 = (a − b)

So in both cases, the quotient is simply the difference between the two digits.

For example, with 74 and 47: $\dfrac{74 - 47}{9} = \dfrac{27}{9} = 3$, which is exactly 7 − 4.

Hence, the quotient equals the difference between the two digits of the chosen number.

In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? For example:

• We start with 31. After reversing we get 13. Adding 31 and 13, we get 44.
• We start with 28. After reversing we get 82. Adding 28 and 82, we get 110.
• We start with 12. After reversing we get 21. Adding 12 and 21, we get 33.

Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra?

Answer

Let the chosen 2-digit number have tens digit a and units digit b, so the number is (10a + b).

On reversing, the new number is (10b + a).

Sum = (10a + b) + (10b + a)

= 10a + a + 10b + b

= 11a + 11b

= 11(a + b)

Since the sum is 11 × (a + b), it is a multiple of 11 for every choice of digits.

Therefore, the sum of a 2-digit number and its reverse is always divisible by 11. The quotient obtained on dividing by 11 is a + b, the sum of the two digits.

Hence, the sum of a 2-digit number and its reverse is always divisible by 11.

Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3? [Hint: Look at some multiples of 37.]

Answer

The three numbers formed by cycling the digits are:

abc = 100a + 10b + c

bca = 100b + 10c + a

cab = 100c + 10a + b

Adding the three numbers:

Sum = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)

= (100a + a + 10a) + (10b + 100b + b) + (c + 10c + 100c)

= 111a + 111b + 111c

= 111(a + b + c)

Since 111 = 3 × 37,

we have

111(a + b + c) = 3 × 37 × (a + b + c)

Since the factor 37 is always present, the sum is always divisible by 37. Since the factor 3 is also always present, the sum is always divisible by 3 as well.

Hence, the sum abc + bca + cab = 111(a + b + c) is always divisible by both 37 and 3.

Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the digits, that is abcabc. Divide this number by 7, then by 11, and finally by 13. What do you get? Try this with other numbers. Figure out why it works. [Hint: Multiply 7, 11 and 13.]

Answer

Writing the 6-digit number abcabc in terms of the 3-digit number abc:

abcabc = (abc × 1000) + abc

= abc × (1000 + 1)

= abc × 1001

Now, multiplying 7, 11 and 13 as suggested in the hint:

7 × 11 × 13 = 1001

Therefore:

abcabc = abc × 7 × 11 × 13

So dividing abcabc successively by 7, by 11 and by 13 cancels out the factor 1001, leaving the original 3-digit number abc.

For example,

532532 ÷ 7 = 76076
76076 ÷ 11 = 6916
6916 ÷ 13 = 532

Thus, we obtain the original 3-digit number.

Hence, dividing abcabc by 7, then 11, then 13 always gives back the original 3-digit number abc, because abcabc = abc × 7 × 11 × 13.

There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person has some flowers. He dips them all in the first pond and then places some flowers in shrine 1. Next, he dips the remaining flowers in the second pond and places some flowers in shrine 2. Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine?

Answer

Let the number of flowers he started with be x, and let the equal number placed in each shrine be a.

Shrine 1: After the first pond, the flowers double to 2x. He places a flowers in shrine 1.

Remaining = 2x − a

Shrine 2: After the second pond, the flowers double to 2(2x − a) = 4x − 2a. He places a flowers in shrine 2.

Remaining = (4x − 2a) − a = 4x − 3a

Shrine 3: After the third pond, the flowers double to 2(4x − 3a) = 8x − 6a. He places all of these, and this must equal a.

8x - 6a = a

⇒ 8x = 7a

For positive integers, this gives x = 7k and a = 8k, where k is a positive integer. Therefore, there are infinitely many positive integer solutions. The least positive solution is obtained when k = 1, giving x = 7 and a = 8.

Check: Start with 7 → double to 14, place 8, leaving 6 → double to 12, place 8, leaving 4 → double to 8, place all 8. He places 8 in each shrine.

Hence, the least possible positive number of flowers is 7, and the person places 8 flowers in each shrine.

A farm has some horses and hens. The total number of heads of these animals is 55 and the total number of legs is 150. How many horses and how many hens are on the farm? Can you solve this without letter-numbers? [Hint: If all the 55 animals were hens, then how many legs would there be? Using the difference between this number and 150, can you find the number of horses?]

Answer

Using the hint:

If all 55 animals were hens, the number of legs would be 55 × 2 = 110.

But the actual number of legs is 150, which is 150 − 110 = 40.

Each horse has 2 legs more than a hen, so the number of horses = $\dfrac{40}{2}$ = 20.

Number of hens = 55 − 20 = 35.

Using algebra:

Let the number of horses be h and the number of hens be n.

Each animal has 1 head, so:

h + n = 55 $\quad$… (1)

Each horse has 4 legs and each hen has 2 legs, so:

4h + 2n = 150 $\quad$… (2)

From equation (1), n = 55 − h.

Substituting in equation (2):

4h + 2(55 - h) = 150

⇒ 4h + 110 - 2h = 150

⇒ 2h = 150 - 110

⇒ 2h = 40

⇒ h = 20

So, n = 55 − 20 = 35.

Hence, the farm has 20 horses and 35 hens.

A mother is 5 times her daughter's age. In 6 years' time, the mother will be 3 times her daughter's age. How old is the daughter now?

Answer

Let the daughter's present age be d years.

Then the mother's present age = 5d years.

In 6 years' time:

Daughter's age = (d + 6) years

Mother's age = (5d + 6) years

At that time, the mother will be 3 times the daughter's age:

5d + 6 = 3(d + 6)

⇒ 5d + 6 = 3d + 18

⇒ 5d - 3d = 18 - 6

⇒ 2d = 12

⇒ d = 6

Therefore, the daughter is 6 years old.

Check: Daughter is 6, mother is 30. In 6 years, daughter is 12 and mother is 36, and 36 = 3 × 12.

Hence, the daughter is 6 years old now.

Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. Gauri says to Naina, "You have twice as many cows as I do". Naina says, "That's true, but if I gave you three of my cows, we would each have the same number of cows". How many cows do Gauri and Naina have?

Answer

Let the number of cows Gauri has be g.

Since Naina has twice as many, Naina has 2g cows.

If Naina gives 3 cows to Gauri:

Gauri would have (g + 3) cows.

Naina would have (2g − 3) cows.

These two amounts are equal:

2g - 3 = g + 3

⇒ 2g - g = 3 + 3

⇒ g = 6

So Gauri has 6 cows, and Naina has 2 × 6 = 12 cows.

Check: After Naina gives 3 cows, Gauri has 6 + 3 = 9 and Naina has 12 − 3 = 9. They are equal.

Hence, Gauri has 6 cows and Naina has 12 cows.

I run a small dosa cart and my expenses are as follows:

• Rent for the dosa cart is ₹5000 per day.
• The cost of making one dosa (including all the ingredients and fuel) is ₹10.

(i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹2000?
(ii) If my customers are willing to pay only ₹50 for a dosa, how many dosas should I aim to sell in a day to make a profit of ₹2000?

Answer

(i) Let the selling price of one dosa be ₹p.

Total cost for the day = Rent + (cost of making 100 dosas)

= 5000 + (100 × 10) = ₹6000

Total income from selling 100 dosas = ₹100p

Profit = Income − Cost, and the required profit is ₹2000:

100p - 6000 = 2000

⇒ 100p = 2000 + 6000

⇒ 100p = 8000

⇒ p = $\dfrac{8000}{100}$

⇒ p = 80

Hence, the dosa should be sold for ₹80 each.

(ii) Let the number of dosas sold in a day be n.

Selling price per dosa = ₹50, so income = ₹50n

Total cost for the day = Rent + (cost of making n dosas)

= 5000 + 10n

Profit = Income − Cost, and the required profit is ₹2000:

50n - (5000 + 10n) = 2000

⇒ 50n - 10n - 5000 = 2000

⇒ 40n = 2000 + 5000

⇒ 40n = 7000

⇒ n = $\dfrac{7000}{40}$

⇒ n = 175

Hence, 175 dosas should be sold in a day.

Evaluate the following sequence of fractions:

$\dfrac{1}{3}, \quad \dfrac{1+3}{5+7}, \quad \dfrac{1+3+5}{7+9+11}$

What do you observe? Can you explain why this happens? [Hint: Recall what you know about the sum of the first n odd numbers.]

Answer

Evaluating each fraction:

$\phantom{=} \dfrac{1}{3} = \dfrac{1}{3} \\[1em] \dfrac{1 + 3}{5 + 7} = \dfrac{4}{12} = \dfrac{1}{3} \\[1em] \dfrac{1 + 3 + 5}{7 + 9 + 11} = \dfrac{9}{27} = \dfrac{1}{3}$

Observation: Every fraction in the sequence equals $\dfrac{1}{3}$.

Why this happens:

Recall that the sum of the first n odd numbers is n².

For the n-th fraction:

The numerator is the sum of the first n odd numbers = n².

The denominator is the sum of the next n odd numbers. This equals (sum of the first 2n odd numbers) − (sum of the first n odd numbers):

Denominator = (2n)² - n²

= 4n² - n²

= 3n²

Therefore each fraction is:

$\dfrac{n^2}{3n^2} = \dfrac{1}{3}$

Hence, every fraction in the sequence equals $\mathbf{\dfrac{1}{3}}$, because the numerator is n² and the denominator is 3n².

Karim and the Genie

Karim was taking a nap under a tree. He had a dream about a magical lamp and a genie. He heard a voice saying, "I have come to serve you, Oh master". He woke up and to his surprise, it was a genie!

"Do you want to make money?", asked the genie. Karim nodded dumbly in bewilderment. The genie continued, "Do you see the banyan tree over there? All you have to do is go around it once. The money in your pocket will double".

Karim immediately started towards the tree, only to be stopped by the genie. "One moment!", said the genie. "Since I am bringing you great riches, you should share some of your gains with me. You must give me 8 coins each time you go around the tree."

Thinking that was a trifling amount, Karim readily agreed.

He went around the tree once. Just as the genie had said, the number of coins in his pocket doubled! He gave 8 coins to the genie. He made another round. Again the number of coins doubled. He gave 8 more coins to the genie. He went around the tree for the third time. The number of coins doubled again, but to his horror, he was left with only 8 coins, exactly the number of coins he owed the genie!

As Karim began to wonder how the genie tricked him, the genie let out a loud laugh and disappeared.

(i) How many coins did Karim initially have?
(ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?
(iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim's coins?

Answer

(i) Let the number of coins Karim started with be x.

In each round, the coins double and then 8 are given to the genie.

Round 1:

2x - 8

Round 2:

2(2x - 8) - 8

= 4x - 16 - 8

= 4x - 24

Round 3 (after doubling):

2(4x - 24)

= 8x - 48

After the third doubling, he was left with exactly 8 coins:

8x - 48 = 8

⇒ 8x = 8 + 48

⇒ 8x = 56

⇒ x = 7

Check: Start with 7 → double to 14, give 8, left with 6 → double to 12, give 8, left with 4 → double to 8, which is exactly the 8 coins owed.

Hence, Karim initially had 7 coins.

(ii) Let the cost per round be c.

Starting with x coins:

After the first round:

2x - c

After the second round:

2(2x − c) − c = 4x − 3c

After the third round:

2(4x − 3c) − c = 8x − 7c

For Karim to end up with more coins than he started with,

8x − 7c > x

7x > 7c

x > c

Since Karim started with 7 coins,

c < 7

Karim should agree only if the cost per round is less than his initial number of coins, i.e. less than 7 coins

(iii) Suppose Karim starts with x coins and the cost per round is c.

After three rounds, he has 8x − 7c coins

For the genie to get all of Karim's coins, Karim must be left with no coins:

8x − 7c = 0

7c = 8x

c = $\dfrac{8x}{7}$

Therefore, the genie should set the cost per round to $\dfrac{8x}{7}$ coins.

For Karim, x = 7, so

c = $\dfrac{8 \times 7}{7} = 8$

which is exactly the fee charged in the story.

Hence, the genie should charge $\mathbf{\dfrac{8x}{7}}$ coins per round to obtain all of Karim's coins