Algebra Play - NCERT Solutions Class 8 Maths (Ganita Prakash)

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Algebra Play – NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 8 Maths (Ganita Prakash).

NCERT Solutions Class 8

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Algebra Play – NCERT Solutions

Q.1: How would you change this game to make the final answer 3? What about 5?

Solution: To make the final answer 3:

Think of a number: {tex}x{/tex}
Double it: 2{tex}x{/tex}
Add 6: {tex}2 x+6{/tex}
Divide by {tex}2:(2 x+6) / 2=x+3{/tex}
Subtract the original number: {tex}\mathrm{x}+3-\mathrm{x}=3{/tex}

To make the final answer 5:

Think of a number: {tex}x{/tex}
Double it: 2 x
Add 10: {tex}2 x+10{/tex}
Divide by 2 : {tex}(2 x+10) / 2=x+5{/tex}
Subtract the original number: {tex}x+5-x=5{/tex}

Q.2:

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

Solution:

Yes, here are some examples:
Example 1 (Final answer = 10):

Think of a number: {tex}x{/tex}
Multiply by 4 : 4x
Add 8: {tex}4 x+8{/tex}
Divide by 2: {tex}(4 x+8) / 2=2 x+4{/tex}
Subtract the original number: {tex}2 x+4-x=x+4{/tex}
Subtract the original number again: {tex}x+4-x=4{/tex}
Add 6:4+6 = 10

Example 2 (Final answer {tex}=7{/tex} ):

Think of a number: {tex}x{/tex}
Add 5: {tex}x+5{/tex}
Multiply by 2 : {tex}2 x+10{/tex}
Subtract 6: {tex}2 x+4{/tex}
Divide by {tex}2: x+2{/tex}
Add 5: {tex}x+7{/tex}
Subtract the original number: {tex}x+7-x=7{/tex}

Q.3: Mukta thinks of another date, follows the same steps, and reports her answer as 1390. What date did Mukta start with this time?

Solution:

Subtracting 165 from 1390, we get: {tex}1390-165=1225{/tex}
The last 2 digits represent the day (D) {tex}=25{/tex}
The remaining digits represent the month {tex}(\mathrm{M})=12{/tex}
Therefore, the date Mukta thought of was 25th December (25/12).

Q.4: Find the dates if the final answers are the following:

1269
394
296

Solution:

- Subtract 165: 1269-165=1104
- Month {tex}(\mathrm{M})=11{/tex} (November)
- Day {tex}(\mathrm{D})=04{/tex}
- Date: 4th November {tex}\left(\frac{04}{11}\right){/tex}
- Subtract 165: {tex}394-165=229{/tex}
- Month (M) = 2 (February)
- Day {tex}(\mathrm{D})=29{/tex}
- Date: 29th February {tex}\left(\frac{29}{02}\right){/tex}
- Subtract 165: 296-165=131
- Month {tex}(\mathrm{M})=1{/tex} (January)
- Day (D) = 31
- Date: 31st January {tex}\left(\frac{31}{01}\right){/tex}

Q.5: Use the same rule to fill these pyramids:

Solution:

Q.6 Fill the following pyramids:

Solution:

Do it yourself.

Q.7: Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

Solution:

Bottom row: 4, 13,8
Middle row:
- {tex}4+13=17{/tex}
- {tex}13+8=21{/tex}
Topmost number: 17+21=38
Bottom row: 7, 11,3
Middle row:
- {tex}7+11=18{/tex}
- {tex}11+3=14{/tex}
Topmost number: {tex}18+14=\mathbf{3 2}{/tex}
Bottom row: 10, 14, 25
Middle row:
- {tex}10+14=24{/tex}
- {tex}14+25=39{/tex}
Topmost number: {tex}24+39=\mathbf{6 3}{/tex}

Q.8: Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row.

Solution:

Let the bottom row be: a, b, c, d
Building up:

Row 2: {tex}(\mathrm{a}+\mathrm{b}){/tex}, {tex}(\mathrm{b}+\mathrm{c}){/tex}, {tex}(\mathrm{c}+\mathrm{d}){/tex}
Row 3: {tex}(a+b)+(b+c),(b+c)+(c+d){/tex} {tex}=(a+2 b+c),(b+2 c+d){/tex}
Row 4: {tex}(a+2 b+c)+(b+2 c+d)=a+3 b+3 c+d{/tex}

Expression: {tex}\mathrm{a}+3 \mathrm{~b}+3 \mathrm{c}+\mathrm{d}{/tex}
This follows the binomial coefficients pattern: {tex}1,3,3,1{/tex} (coefficients from Pascal’s triangle).

Q.9: Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

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

Solution:

Bottom row: 8, 19, 21, 13
Using formula: {tex}a+3 b+3 c+d=8+3(19)+3(21)+13=8+57+63+13=\mathbf{1 4 1}{/tex}
Bottom row: 7, 18, 19, 6
{tex}7+3(18)+3(19)+6=7+54+57+6=124{/tex}
Bottom row: 9, 7, 5, 11
{tex}9+3(7)+3(5)+11=9+21+15+11=56{/tex}

Q.10: 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?

Solution:

The first three Virahāñka – Fibonacci numbers are: {tex}1,2,3{/tex}
Calculation:

Row 2: {tex}1+2=3,2+3=5{/tex}
Row 3: {tex}3+5=8{/tex}

Numbers in the pyramid: 1, 2, 3, 3, 5, 8
Yes, all numbers in the pyramid are Virahāñka – Fibonacci numbers!
The top number is {tex}\boldsymbol{8}{/tex}, which is also a Virahāṅka – Fibonacci number.

Q.11: What can you say about the numbers in the pyramid and the number at the top in the following cases?

The first four Virahāṅka-Fibonacci numbers are written in the bottom row of a four row pyramid.
The first 29 Virahāṅka-Fibonacci numbers are written in the bottom row of a 29 row pyramid.

Solution:

The first four V-F numbers are: 1, 2, 3, 5
Let me build it properly:
- Bottom row: {tex}1,2,3,5{/tex}
- Row {tex}2:(1+2)=3,(2+3)=5,(3+5)=8{/tex}
- Row 3: {tex}(3+5)=8,(5+8)=13{/tex}
- Row 4: {tex}(8+13)=21{/tex}
All numbers in the pyramid are Virahāṅka-Fibonacci numbers: {tex}1,2,3,5,3,5,8,8,13,21{/tex}
Following the pattern observed above, when we build a pyramid starting with consecutive Virahāṅka – Fibonacci numbers:
- All numbers generated in the pyramid are also V-F numbers
- The top number will be a V-F number
This happens because each number in a pyramid is the sum of two numbers below it, and the VF sequence is defined by the property that each number is the sum of the two preceding numbers.
Therefore, all numbers in the 29-row pyramid will be Virahānka-Fibonacci numbers, and the top will also be a
V-F number.

Q.12: 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?

Solution:

All numbers in the pyramid will be Virahānka-Fibonacci numbers.
This is because:

We start with V-F numbers at the bottom.
Each new number is formed by adding two numbers from below
Since V-F numbers are defined by the sum property, adding any two V-F numbers that appear consecutively in the sequence gives another V-F number

The number at the top will also be a Virahāñka-Fibonacci number.

Q.13: Create your own calendar trick. For instance, choose a grid of a different size and shape.

Solution: Do it yourself.

Q.14: In the following grids, find the values of the shapes and fill in the empty squares:

Solution:

Grid 1
Let:

Blue square {tex}=x{/tex}
Red circle {tex}=y{/tex}

Row 1:
{tex} x+x+y=27 {/tex}
{tex} 2 x+y=27(1) {/tex}
Row 2:
{tex} y+y+x=21 {/tex}
{tex} 2 y+x=21(2) {/tex}
Solving (1) and (2)
From (1):
{tex} y=27-2 x y=27-2 x {/tex}
Substitute in (2):
{tex} 2(27-2 x)+x=212(27-2 {x})+{x}=2154-4 x+x{/tex}{tex}=2154-4 {x}+{x}=2154-3 x=2154-3 {x}=21 {/tex}
{tex} 3 x=33 \Rightarrow x=113 {x}=33 \Rightarrow {x}=11 {/tex}
Now:
{tex} y=27-2(11)=5 y=27-2(11)=5 {/tex}
Values found:

Blue square = 11
Red circle {tex}=5{/tex}

Row 3 (empty total):
{tex} y+x+y=5+11+5=21 {/tex}
Grid 2
Let:

Blue circle {tex}=a{/tex}
Purple diamond {tex}=b{/tex}

Row 1:
{tex} a+b+b=18 {/tex}
{tex} a+2 b=18(1) {/tex}
Row 2:
{tex} b+a+a=15 {/tex}
{tex} 2 a+b=15(2) {/tex}
Solving (1) and (2):
Multiply (1) by 2:
{tex} 2 a+4 b=36 {/tex}
Subtract (2):
{tex} (2 a+4 b)-(2 a+b)=36-15 {/tex}
{tex} 3 b=21 \Rightarrow b=7 {/tex}
Substitute into (1):
{tex} a+2(7)=18 {/tex}
{tex} a=4 {/tex}
Values found:

Blue circle = 4
Purple diamond {tex}=7{/tex}

Row 3 (empty total):
{tex} b+a+a=7+4+4=15 {/tex}

Q.15: Fill the digits 1, 3, and 7 in __ __ {tex}\times{/tex} __ to make the largest product possible.

Solution:

Following the rule: largest digit as multiplier, others in decreasing order.
Digits: 1, 3, 7

Largest digit: 7 (multiplier)
Remaining digits in decreasing order: 3, 1

Answer: {tex}\mathbf{3 1} \boldsymbol{\times} \mathbf{7}=\mathbf{2 1 7}{/tex}
Verification by checking all possibilities:

{tex}13 \times 7=91{/tex}
{tex}17 \times 3=51{/tex}
{tex}31 \times 7=217 {/tex} (largest)
{tex}37 \times 1=37{/tex}
{tex}71 \times 3=213{/tex}
{tex}73 \times 1=73{/tex}
{tex}31 \times 7=217{/tex} is the largest product.

Q.16: Fill the digits 3, 5, and 9 in __ __ {tex}\times{/tex} __ to make the largest product possible.

Solution:

Following the rule:

Largest digit: 9 (multiplier)
Remaining digits in decreasing order: 5, 3

Answer: {tex}\mathbf{5 3} \boldsymbol{\times} \mathbf{9 = 4 7 7}{/tex}
Verification:

{tex}35 \times 9=315{/tex}
{tex}39 \times 5=195{/tex}
{tex}53 \times 9=477 {/tex} (largest)
{tex}59 \times 3=177{/tex}
{tex}93 \times 5=465{/tex}
{tex}95 \times 3=285{/tex}

{tex}53 \times 9=477{/tex} is the largest product.

Q.17: 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?

Solution:

Using the example from the textbook:

Original number: 47
Reversed number: 74
Difference: {tex}74-47=27{/tex}
Quotient when divided by {tex}9: 27 \div 9=3{/tex}

The quotient is the difference between the two digits!
Let’s verify with algebra:

Original number: {tex}10 \mathrm{a}+\mathrm{b}{/tex}
Reversed number: {tex}10 \mathrm{~b}+\mathrm{a}{/tex}
Difference (if {tex}b>a):(10 b+a)-(10 a+b)=9 b-9 a=9(b-a){/tex}
Quotient: {tex}9(b-a) \div 9=b-a{/tex}

Yes, the quotient equals the difference between the two digits of the original number.
Examples:

47: difference of digits {tex}=7-4=3 {/tex}
{tex}28{/tex}: reversed {tex}=82{/tex}, difference {tex}=82-28=54{/tex}, quotient {tex}=54 \div 9=6=(8-2){/tex}
63: reversed {tex}=36{/tex}, difference {tex}=63-36=27{/tex}, quotient {tex}=27 \div 9=3=(6-3) {/tex}

Q.18: In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? Can we justify this claim using algebra?

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?

Solution:

Let’s examine the given examples:

{tex}31+13=44{/tex} (divisible by 11)
{tex}28+82=110{/tex} (divisible by 11)
{tex}12+21=33{/tex} (divisible by 11)

Using algebra:

Original number: {tex}10 \mathrm{a}+\mathrm{b}{/tex}
Reversed number: {tex}10 \mathrm{~b}+\mathrm{a}{/tex}
Sum: {tex}(10 a+b)+(10 b+a)=11 a+11 b=11(a+b){/tex}

Therefore, the sum is always divisible by 11!
This is because the sum can be written as {tex}11(a+b){/tex}, which is clearly a multiple of 11.
The quotient when dividing by 11 is {tex}(a+b){/tex}, which is the sum of the two digits.
Examples:

{tex}31+13=44=11 \times 4{/tex}, where {tex}4=3+1 {/tex}
{tex}28+82=110=11 \times 10{/tex}, where {tex}10=2+8 {/tex}
{tex}57+75=132=11 \times 12{/tex}, where {tex}12=5+7 {/tex}

Q.19: Consider any 3-digit number, say abc {tex}(100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}){/tex}. 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?

Solution:

Let’s use algebra:

First number (abc): {tex}100 a+10 b+c{/tex}
Second number (bca): {tex}100 \mathrm{~b}+10 \mathrm{c}+\mathrm{a}{/tex}
Third number {tex}(\mathrm{cab}): 100 \mathrm{c}+10 \mathrm{a}+\mathrm{b}{/tex}

{tex} \text { Sum: }=(100 a+10 b+c){/tex}{tex}+(100 b+10 c+a)+(100 c+10 a+b){/tex} {tex}=100 a+a+10 a+10 b+100 b+b+c {/tex}
{tex} +10 c+100 c=111 a+111 b+111 c=111(a+b+c) {/tex}
Now, {tex}111=3 \times 37{/tex}
Therefore, the sum {tex}=3 \times 37 \times(a+b+c){/tex}
This proves that:

The sum is always divisible by {tex}\mathbf{3 7}{/tex} (since it equals {tex}37 \times 3 \times(a+b+c){/tex})
The sum is always divisible by 3 (since it equals {tex}3 \times 37 \times(a+b+c){/tex})
The sum is also divisible by {tex}\mathbf{1 1 1}{/tex} (since it equals {tex}111 \times(a+b+c){/tex})

Example:

{tex}\mathrm{abc}=123: 123+231+312=666{/tex}
{tex}666 \div 37=18 {/tex}
{tex}666 \div 3=222{/tex}
{tex}666 \div 111=6{/tex}

Q.20: 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.

Solution:

Let’s try with an example first:

3-digit number: 123
6-digit number: 123123
{tex}123123 \div 7=17589{/tex}
{tex}17589 \div 11=1599{/tex}
{tex}1599 \div 13=123{/tex}

We get back the original 3-digit number!
Let’s try another:

3-digit number: 456
6-digit number: 456456
{tex}456456 \div 7=65208{/tex}
{tex}65208 \div 11=5928{/tex}
{tex}5928 \div 13=456{/tex}

Using algebra to understand why:
The 6 -digit number abcabc can be written as: {tex}=100000 \mathrm{a}+10000 \mathrm{~b}+1000 \mathrm{c}+100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}={/tex}
{tex} 100000 a+100 a+10000 b+10 b+1000 c+c{/tex}{tex}=100 a(1000+1)+10 b(1000+1)+c(1000+1){/tex}
{tex}= (100 a+10 b+c) \times 1001 {/tex}
Now, let’s find what 1001 equals: {tex}\mathbf{1 0 0 1}=\mathbf{7} \times \mathbf{1 1} \times \mathbf{1 3}{/tex} (given in hint)
Therefore: {tex}a b c a b c=(100 a+10 b+c) \times 7 \times 11 \times 13{/tex}
When we divide by 7 , then 11 , then 13: {tex}=(100 a+10 b+c) \times 7 \times 11 \times 13 \div 7 \div 11 \div 13=100 a+{/tex}10b {tex}+c=a b c{/tex} (the original 3-digit number)
This works because {tex}1001=7 \times 11 \times 13{/tex}, and repeating a 3-digit number creates a multiple of 1001.

Q.21: 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?

Solution:

Let the number of flowers placed in each shrine be {tex}k {/tex}..
Let the number of flowers he started with be {tex}\mathbf{x}{/tex}.
Step 1: First pond

Flowers double: {tex}2 x{/tex}
He places {tex}k{/tex} flowers in shrine 1
Flowers left: {tex}2 x-k{/tex}

Step 2: Second pond

Flowers double: {tex}(2 x-k)=4 x-2 k{/tex}

He places {tex}k{/tex} flowers in shrine 2

Flowers left:( {tex}4 x-2 k)-k=4 x-3 k{/tex}

Step 3: Third pond

Flowers double: {tex}2(4 x-3 k)=8 x-6 k{/tex}
He places all remaining flowers in shrine 3

Since the number of flowers placed in each shrine is equal:
{tex} 8 x-6 k=k {/tex}
Step 4: Solve
{tex} 8 x=7 k {/tex}
So,
{tex} k=8 {x} / 7 {/tex}
To get whole numbers, {tex}x{/tex} must be a multiple of 7.
Taking the smallest possible value:
{tex} x=7 {/tex}
Then,
{tex} k=(8 \times 7) / 7=8 {/tex}
Verification:

Start with 7 flowers
First pond {tex}\boldsymbol{\rightarrow} \mathbf{1 4} \boldsymbol{\rightarrow}{/tex} place {tex}\mathbf{8}{/tex}, left {tex}\mathbf{6}{/tex}
Second pond {tex}\boldsymbol{\rightarrow} \mathbf{1 2} \boldsymbol{\rightarrow}{/tex} place {tex}\mathbf{8}{/tex}, left {tex}\mathbf{4}{/tex}
Third pond {tex}\rightarrow 8 \rightarrow{/tex} place 8

All shrines receive {tex}\mathbf{8}{/tex} flowers each
Answer:

Number of flowers he started with: 7
Number of flowers placed in each shrine: 8

Q.22: 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?

Solution:

Method 1: Using algebra
Let {tex}\mathrm{h}={/tex} number of horses, {tex}\mathrm{c}={/tex} number of hens

Each horse has 1 head, each hen has 1 head: {tex}\mathrm{h}+\mathrm{c}=55{/tex}… (1)
Each horse has 4 legs, each hen has 2 legs: {tex}4 \mathrm{~h}+2 \mathrm{c}=150{/tex}… (2)

From equation (1): {tex}\mathrm{c}=55-\mathrm{h}{/tex}
Substituting in equation (2):

{tex}4 h+2(55-h)=150{/tex}
{tex}4 \mathrm{~h}+110-2 \mathrm{~h}=150{/tex}
{tex}2 \mathrm{~h}=40{/tex}
{tex}\mathrm{h}=20{/tex}

Therefore: {tex}c=55-20=35{/tex}
Method 2: Without letter-numbers (as suggested in hint)
If all 55 animals were hens:

Total legs would be: {tex}55 \times 2=110{/tex} legs
But actual legs = 150
Difference {tex}=150-110=40 \mathrm{legs}{/tex}

Each time we replace a hen with a horse:

We remove 2 legs (hen) and add 4 legs (horse)
Net increase {tex}=2{/tex} legs

Number of horses needed {tex}=40 \div 2=20{/tex} horses
Number of hens {tex}=55-20=35{/tex} hens
Answer: 20 horses and 35 hens
Verification:

Heads: {tex}20+35=55{/tex}
Legs: {tex}20(4)+35(2)=80+70=150{/tex}

Q.23: 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?

Solution:

Let d = daughter’s current age
Then mother’s current age = 5d
After 6 years:

Daughter’s age {tex}=\mathrm{d}+6{/tex}
Mother’s age {tex}=5 d+6{/tex}
Mother will be 3 times daughter’s age: {tex}5 d+6=3(d+6){/tex}

Solving:

{tex}5 d+6=3 d+18{/tex}
{tex}5 d-3 d=18-6{/tex}
{tex}2 \mathrm{~d}=12{/tex}
{tex}d=6{/tex}

Answer: The daughter is 6 years old now.
Verification:

Current: Daughter {tex}=6{/tex}, Mother {tex}=30{/tex} (5 times)
After 6 years: Daughter {tex}=12{/tex}, Mother {tex}=36{/tex} (3 times)

Q.24: 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?

Solution:

Let {tex}\mathrm{g}={/tex} number of Gauri’s cows
Then Naina has 2 g cows (twice as many)
After Naina gives 3 cows to Gauri:

Gauri will have: {tex}g+3{/tex}
Naina will have: {tex}2 g-3{/tex}
They’ll be equal: {tex}\mathrm{g}+3=2 \mathrm{~g}-3{/tex}

Solving:

{tex}\mathrm{g}+3=2 \mathrm{~g}-3{/tex}
{tex}3+3=2 \mathrm{~g}-\mathrm{g}{/tex}
{tex}6=\mathrm{g}{/tex}

Answer:

Gauri has 6 cows
Naina has {tex}\mathbf{1 2}{/tex} cows

Verification:

Currently: Naina {tex}(12)=2 \times{/tex} Gauri {tex}(6) {/tex}
After transfer: Gauri (9) = Naina (9)

Q.25: 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.

If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹2000?
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?

Solution:

Let p = selling price per dosa
Total cost:
1. Fixed cost (rent) = ₹5000
2. Variable cost for 100 dosas {tex}=100 \times ₹ 10=₹ 1000{/tex}
3. Total cost {tex}={/tex} ₹5000 + ₹1000 = ₹6000
Total revenue needed for ₹2000 profit:
1. Revenue {tex}={/tex} Total cost + Profit
2. Revenue {tex}=₹ 6000+₹ 2000=₹ 8000{/tex}
Selling price per dosa:
1. {tex}100 \times \mathrm{p}=₹ 8000{/tex}
2. {tex}\mathrm{p}=₹8000\div 100{/tex}
3. p = ₹80 per dosa
Verification:
1. Revenue: {tex}100 \times ₹ 80=₹ 8000{/tex}
2. Cost: ₹6000
3. Profit: ₹8000 – ₹6000 = ₹2000
Let {tex}\mathrm{n}={/tex} number of dosas to sell
Revenue: 50n
Total cost:
1. Fixed cost {tex}=₹ 5000{/tex}
2. Variable cost {tex}=10 \mathrm{n}{/tex}
3. Total cost {tex}=₹ 5000+10 n{/tex}
For ₹2000 profit:
1. Revenue – Cost = Profit
2. {tex}50 n-(5000+10 n)=2000{/tex}
3. {tex}50 n-5000-10 n=2000{/tex}
4. {tex}40 n=7000{/tex}
5. {tex}\mathrm{n}=175{/tex}
Answer: Need to sell 175 dosas per day
Verification:
1. Revenue: {tex}175 \times{/tex} ₹50 = ₹8750
2. Cost: ₹5000 + (175 {tex}\times{/tex} ₹10) {tex}={/tex} ₹5000 + ₹1750 = ₹6750
3. Profit: ₹8750 – ₹6750 = ₹2000

Q.26: Evaluate the following sequence of fractions:
{tex} \frac{1}{3}, \frac{(1+3)}{(5+7)}, \frac{(1+3+5)}{(7+9+11)} {/tex}
What do you observe? Can you explain why this happens?

Solution:

Let’s evaluate each fraction:
First fraction: 1/3
Second fraction:

Numerator: {tex}1+3=4{/tex}
Denominator: {tex}5+7=12{/tex}
Fraction: {tex}4 / 12=1 / 3{/tex}

Third fraction:

Numerator: {tex}1+3+5=9{/tex}
Denominator: {tex}7+9+11=27{/tex}
Fraction: {tex}9 / 27=1 / 3{/tex}

Observation: All fractions equal {tex}\mathbf{1} \boldsymbol{/} \mathbf{3}{/tex} !
Explanation using algebra:
Recall that the sum of the first {tex}n{/tex} odd numbers {tex}=n^2{/tex}
Pattern in numerator:

First fraction: 1 odd number starting from 1
Second fraction: 2 odd numbers starting from {tex}1(1,3){/tex}
Third fraction: 3 odd numbers starting from {tex}1(1,3,5){/tex}
nth fraction: n odd numbers starting from 1

Sum of first {tex}n{/tex} odd numbers {tex}=n^2{/tex}
Pattern in denominator:

First fraction: 1 odd number starting from 3 (which is the 2 nd odd number)
Second fraction: 2 odd numbers starting from 5 (3rd odd number)
Third fraction: 3 odd numbers starting from 7 (4th odd number)
nth fraction: {tex}n{/tex} odd numbers starting from ({tex}n+1{/tex})th odd number

For the denominator, we’re summing {tex}n{/tex} consecutive odd numbers starting from the ({tex}n+1{/tex})th odd number.
General formula:

Numerator {tex}=1+3+5+\ldots{/tex} ( n terms) {tex}=\mathrm{n}^2{/tex}
Denominator {tex}={/tex} sum of {tex}n{/tex} odd numbers starting from {tex}(2 n+1){/tex}

The {tex}(n+1){/tex} th odd number {tex}=2(n+1)-1=2 n+1{/tex}
Sum of {tex}n{/tex} consecutive odd numbers starting from {tex}(2 n+1):=(2 n+1)+(2 n+3)+\ldots+[2 n+1+2(n-1)]{/tex}
{tex}= (2 n+1)+(2 n+3)+\ldots+(4 n-1) {/tex}
This is an arithmetic series with:

First term {tex}\mathrm{a}=2 \mathrm{n}+1{/tex}
Last term {tex}\mathrm{l}=4 n-1{/tex}
Number of terms {tex}=n{/tex}

{tex} \text { Sum }=n / 2 \times(\text { first }+ \text { last }){/tex} {tex}=n / 2 \times[(2 n+1)+(4 n-1)]=n / 2 \times(6 n)=3 n^2 {/tex}
Therefore: Fraction {tex}=\mathrm{n}^2 /\left(3 \mathrm{n}^2\right)=1 / 3{/tex}
This proves that all fractions in this sequence equal {tex}\frac 13{/tex}!

Q.27: 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.

How many coins did Karim initially have?
For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?
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?

Solution:

(i) Let {tex}\mathrm{x}={/tex} initial number of coins
After 1st round:

Coins double: 2 x
Give 8 to genie: {tex}2 x-8{/tex}

After 2nd round:

Coins double: {tex}2(2 x-8)=4 x-16{/tex}
Give 8 to genie: {tex}4 x-16-8=4 x-24{/tex}

After 3rd round:

Coins double: {tex}2(4 x-24)=8 x-48{/tex}
Give 8 to genie: {tex}8 x-48-8=8 x-56{/tex}

The problem states that after the 3rd round, Karim has exactly 8 coins left:

{tex}8 x-56=8{/tex}
{tex}8 x=64{/tex}
{tex}\mathrm{x}=8{/tex}

Wait, but then he gives these 8 to the genie and has 0 left. Let me re-read.
“he was left with only 8 coins, exactly the number of coins he owed the genie!”
So after doubling the third time, he had 8 coins, which he gave to the genie.

{tex}8 x-48=8{/tex}
{tex}8 x=56{/tex}
{tex}x=7{/tex}

Answer: Karim initially had 7 coins
Verification:

Start: 7
Round {tex}1: 7 \times 2=14{/tex}, give 8 , left with 6
Round 2: {tex}6 \times 2=12{/tex}, give 8 , left with 4
Round 3: {tex}4 \times 2=8{/tex}, give 8 , left with 0

(ii) Let c = cost per round (coins to give genie)
Starting with 7 coins:

After round {tex}1: 2(7)-\mathrm{c}=14-\mathrm{c}{/tex}
After round 2: {tex}2(14-c)-c=28-3 c{/tex}
After round {tex}3: 2(28-3 c)-c=56-7 c{/tex}

For Karim to increase his coins: {tex}56-7 \mathrm{c}>7{/tex}

{tex}56-7>7 \mathrm{c}{/tex}
{tex}49>7 c{/tex}
{tex}\mathrm{c}<7{/tex}

Answer: The cost per round should be less than 7 coins.
For example, if {tex}\mathrm{c}=6{/tex} :

After 3 rounds: {tex}56-7(6)=56-42=14{/tex} coins (doubled his money!)

(iii) Let Karim start with n coins, and let {tex}\mathrm{c}={/tex} cost per round.
After 3 rounds, Karim has: 8n-7c coins
For the genie to get all coins: {tex}8 n-7 c=0{/tex}

{tex}7 \mathrm{c}=8 \mathrm{n}{/tex}
{tex}\mathrm{c}=8 \mathrm{n} / 7{/tex}

Answer: The genie should charge {tex}(\mathbf{8 n}) / \mathbf{7}{/tex} coins per round, where {tex}\mathbf{n}{/tex} is Karim’s starting amount.
For this to be a whole number, n must be a multiple of 7.
General formula: If Karim has {tex}n{/tex} coins initially, the genie should charge {tex}8 n / 7{/tex} coins per round.
For {tex}\mathrm{n}=7: \mathrm{c}=8(7) / 7=8{/tex} coins per round (as in the original story)
For {tex}\mathrm{n}=14: \mathrm{c}=8(14) / 7=16{/tex} coins per round
For {tex}\mathrm{n}=21: \mathrm{c}=8(21) / 7=24{/tex} coins per round