The World of Numbers

For 2 bags of spices, the merchant gets = 15 copper ingots.

For 1 bag of spices, the merchant gets = 152 ingots

Thus, for 12 bags, the number of ingots = 12 × 152 = 6 × 15 = 90.

Therefore, the merchant will leave with 90 copper ingots.

The numbers 11, 13, 17, and 19 are prime numbers (each number is divisible only by 1 and itself).

The next three numbers in this pattern after 19 are 23, 29, 31.

Natural numbers are not closed under subtraction.

Justification:

5 − 3 = 2, which is a natural number.
3 − 5 = −2, which is not a natural number.

Since the subtraction of two natural numbers does not always result in a natural number, natural numbers are not closed under subtraction.

Each hand has 4 fingers (excluding the thumb), and each finger has 3 joints.
So, total joints = 4 × 3 = 12.

We can count up to 12 on one hand.

Since counting on one hand gives 12, this method naturally leads to counting in groups of 12. Hence, it is related to the base-12 system used in ancient times.

Temperature at noon = 4°C
Drop in temperature = 15°C

So, Midnight temperature = 4 − 15 = −11°C

Therefore, the temperature at midnight is −11°C.

Loan (debt) = −850
Profit = +1200
Loss = −450

So, the equation is: −850 + 1200 − 450

Now calculate:
−850 + 1200 – 450 = 350 – 450 = −100

Therefore, his final financial standing is ₹–100, which means he is still in debt of ₹100.

(i) (–12) × 5
Negative × Positive = Negative
(–12) × 5 = –60

(ii) (–8) × (–7)
Negative × Negative = Positive
(–8) × (–7) = 56

(iii) 0 – (–14)
Subtracting a negative = Adding a positive
0 + 14 = 14

(iv) (–20) ÷ 4
Negative ÷ Positive = Negative
(–20) ÷ 4 = –5

Positive numbers represent your fortune, and negative numbers represent debt.

Suppose you have ₹10. Subtracting −₹5 means removing a debt of ₹5.

So, 10 − (−5) = 10 + 5 = 15

Explanation: Removing a debt increases your fortune. Therefore, subtracting a negative number is the same as adding a positive number.

(i) 23 and 46

So, 23=46

(ii) 54 and 108

So, 54=108

(iii) -35 and -610

So, -35=-610

(iv) 93 and 3

So, 93=3

(i) 25+310
LCM of 5 and 10 = 10

(ii) 712+58
LCM of 12 and 8 = 24

(iii) -47+314
LCM of 7 and 14 = 14

(i) 56-14
LCM of 6 and 4 = 12

(ii) 118-34
LCM of 8 and 4 = 8

(iii) -79-(-23)
LCM of 9 and 3 = 9

(i) 23×310=2×33×10
Cancelling 3,

(ii) 711×58=7×511×8=3588

(iii) -47×514=-4×57×14=-2098=-1049

(i) 23÷310=23×103=209=229

(ii) 711÷58=711×85=5655=1155

(iii) -47÷514=-47×145=-5635=-85=-135

LHS =(12+34)×83
LCM of 2 and 4 = 4

RHS =12×83+34×83

Hence, verified LHS = RHS.

67-34
LCM of 7 and 4 = 28

Thus 79(67-34)=79×328=21252=112.

LHS =56(x+35)=56x+56×35=56x+12

RHS =56x+12

Hence, LHS = RHS. This is an identity true for all values of x.

(i) 23

(ii) -54

(iii) 112=32

Converting the given numbers to a common denominator.

So, the rational numbers between -48 and 28 are -34, -24, -14.

(-14)+(512)
LCM of 4 and 12 = 12

So, (-14)+(512)=-312+512=-3+512=212=16.

Total silk = 1534 m = 634 m

Silk needed for 1 kurta = 214 m = 94 m

Therefore, he can make exactly 7 kurtas.

Writing the numbers with more decimal places:
3.1415 = 3.14150
3.1416 = 3.14160

Choose any numbers between them: 3.14151, 3.14152, 3.14153.

Therefore, three rational numbers are 3.14151, 3.14152, 3.14153.

Yes, there are other ways.

Method 1: If a and b are two rational numbers, then a+b2 is a rational number lying between them.

Method 2: Once we find one number between a and b, we can again find the mean between the new numbers to get more rational numbers. In this way, infinitely many rational numbers can be found between any two rational numbers.

A rational number p/q (in lowest terms) has a terminating decimal if and only if the denominator q has no prime factors other than 2 and 5.

(i) 720
20 = 2² × 5
Since the denominator has only the prime factors 2 and 5, the decimal expansion is terminating.

Verification:

720 = 0.35
Hence, the given rational number is a terminating decimal.

(ii) 415
15 = 3 × 5
Since the denominator has a prime factor 3 (other than 2 or 5), the decimal expansion is non-terminating repeating.

Verification:

415 = 0.2666… = 0.26
Hence, the given rational number is a repeating decimal.

(iii) 13250
250 = 2 × 5³
Since the denominator has only the prime factors 2 and 5, the decimal expansion is terminating.

Verification:

13250 = 0.052
Hence, the given rational number is a terminating decimal.

113 = 0.076923

Repeating block: 076923 (6 digits)

Now,
213 = 0.153846
313 = 0.230769
413 = 0.307692
513 = 0.384615
613 = 0.461538

Observations: These decimal expansions show a cyclic property. The repeating digits are the same (0, 7, 6, 9, 2, 3) but appear in a different order in each case.

(i) √81=√9²=9
Since 9 is rational number.
∴ √81 is rational.

(ii) √12=√3×4=√3×2²=2√3
Since √3 is irrational, 2√3 is also irrational.
∴ √12 is irrational.

(iii) 0.33333 …
This is a non-terminating repeating decimal.
∴ it is rational.

(iv) 0.123451234512345 … = 0.12345
This is a non-terminating repeating decimal.
∴ it is rational.

(v) 1.01001000100001 …
This is a non-terminating, non-repeating decimal.
∴ 1.01001000100001… is an irrational number.

Let x = 0.9999… ……… (1)

Multiply both sides by 10:
10x = 9.9999… ……… (2)

Now subtract equation (1) from equation (2):
10x − x = 9.9999… – 0.9999…
9x = 9
x = 1

Since x = 0.9999…, we get: 0.9999… = 1

We know that 17=0.142857

It has a repeating block whose digits appear in cyclic order.

Similarly, we find other values of n such that 1n gives a cyclic repeating decimal.

113=0.076923

117=0.0588235294117647

119=0.0526315789417368421

123=0.0434782608695652173913

In each case, the digits in the repeating block are the same, and the decimal expansions of 2n, 3n, … are obtained by cyclic shifts of these digits.

(i) 350

350 = 0.06
This is a terminating decimal.

(ii) 29

29 = 0.2222….
This is a non-terminating and repeating decimal.

Assume that √5 is rational.

Then it can be written in the form √5=ab where a and b are integers having no common factor other than 1.

Squaring both sides:
5 = a²b²
5b² = a²

This shows that a² is divisible by 5, so a is divisible by 5.
Let a = 5k, where k is an integer.

Substituting:
5b² = (5k)²
5b² = 25k²
b² = 5k²

This shows that b² is divisible by 5, so b is also divisible by 5.

Thus, both a and b are divisible by 5, which contradicts the fact that a and b have no common factor other than 1.

Therefore, our assumption is wrong. √5 is irrational.

(i) 12.6
12.6 = 12610=635

(ii) 0.0120
0.0120 = 12010000=3250

(iii) 3.052
Let x = 3.052 ……… (1)
Multiply both sides by 100:
100x = 305.252 ……… (2)
Now subtract equation (1) from equation (2):
100x − x = 305.252 – 3.052
99x = 302.2
99x = 302210
x = 3022990=1511495

(iv) 1.235
Let x = 1.235 ……… (1)
Multiply both sides by 100:
100x = 123.535 ……… (2)
Now subtract equation (1) from equation (2):
100x − x = 123.535 – 1.235
99x = 122.3
99x = 122310
x = 1223990

(v) 0.23
Let x = 0.23 ……… (1)
Multiply both sides by 100:
100x = 23.23 ……… (2)
Now subtract equation (1) from equation (2):
100x − x = 23.23 – 0.23
99x = 23
x = 2399

(vi) 2.05
Let x = 2.05 ……… (1)
Multiply both sides by 10:
10x = 20.55 ……… (2)
Now subtract equation (1) from equation (2):
10x − x = 20.55 – 2.05
9x = 18.5
9x = 18510
x = 18590=3718

(vii) 2.125
Let x = 2.125 ……… (1)
Multiply both sides by 10:
10x = 21.255 ……… (2)
Now subtract equation (1) from equation (2):
10x − x = 21.255 – 2.125
9x = 19.13
9x = 1913100
x = 1913900

(viii) 3.125
Let x = 3.125 ……… (1)
Multiply both sides by 10:
10x = 31.255 ……… (2)
Now subtract equation (1) from equation (2):
10x − x = 31.255 – 3.125
9x = 28.13
9x = 2813100
x = 2813900

(ix) 2.1625
Let x = 2.1625 ……… (1)
Multiply both sides by 10000:
10000x = 21625.1625 ……… (2)
Now subtract equation (1) from equation (2):
10000x – x = 21625.1625 – 2.1625
9999x = 21623
x = 216239999

(i) 0.532
Here, 0.532 = 5321000

To locate on the number line:
Divide the segment from 0 to 1 into 10 equal parts. Each part represents 0.1.
Take the point 0.5.
Now divide the segment from 0.5 to 0.6 into 10 equal parts. Each part represents 0.01.
Take the point 0.53.
Now divide the segment from 0.53 to 0.54 into 10 equal parts. Each part represents 0.001.
Take the second division after 0.53.
So, we get the point 0.532 on the number line.

(ii) 1.15
Let x = 1.15 ……… (1)
Multiply both sides by 10:
10x = 11.55 ……… (2)
Now subtract equation (1) from equation (2):
10x − x = 11.55 – 1.15
9x = 10.4
9x = 10410
x = 10490=5245

To locate on the number line:
5245=45+745=1+745

So it lies between 1 and 2.
Divide the segment from 1 to 2 into 45 equal parts.
Mark the point 7th division to the right of 1.
This point represents 1.15 on the number line.

Multiply the numerator and denominator by 7,
3 = 3×71×7=217, 4 = 4×71×7=287

Now six rational numbers between 217 and 287: 227, 237, 247, 257, 267, 277.

So, six rational numbers between 3 and 4 are 227, 237, 247, 257, 267, 277.

Convert both fractions to a common denominator:
25=2×75×7=1435 and 35=3×75×7=2135

Now pick five numbers between 1435 and 2135: 1535, 1635, 1735, 1835, 1935

So, five rational numbers between 25 and 35 are 1535, 1635, 1735, 1835, 1935.

Convert both fractions to a common denominator:
16=1×56×5=530 and 25=2×65×6=1230

Now pick five numbers between 530 and 1230: 630, 730, 830, 930, 1030

So, five rational numbers between 16 and 25 are 630, 730, 830, 930, 1030.

x3+x5=1615
LCM of 3 and 5 = 15

x = 16×1515×8
x = 2.

a +1b = 0

a = -1b

ab = a·b = -1b· b = –1

ab = –1

So, ab is negative.

Let the rational number be x. Since its decimal expansion terminates and the last non-zero digit is at the 4th decimal place, we can write

x = 0.a₁a₂a₃a₄, where a₄ ≠ 0

Multiplying by 10⁴, we get
10⁴x = a₁a₂a₃a₄ = p (an integer)

x = a₁a₂a₃a₄10⁴

Since a₄ ≠ 0, p is not divisible by 10.

Now, 10⁴ = (2 × 5)⁴ = 2⁴ × 5⁴

When p10⁴ is written in lowest terms, common factors of 2 and/or 5 may cancel.

Therefore, it is necessary that the denominator in the lowest form is divisible by 2⁴ or 5⁴.

Prime factorisation of the denominator:
125 = 5 × 5 × 5 = 5³

Since the denominator has only the prime factor 5, the decimal expansion is terminating.

Further, 18125=18×8125×8=1441000

So, the decimal expansion terminates and has 3 decimal places.

Given denominator = 2³ × 5 = 8 × 5 = 40.

Since the denominator has the prime factors 2 and 5, the decimal expansion is terminating.

To find the number of decimal places, make the denominator a power of 10.
So multiply by 5²:
2³ × 5 × 5² = 2³ × 5³ = 10³

Thus, the denominator becomes 10³.

Hence, the decimal expansion will terminate and have 3 decimal places.

Given, a = 712 and b = 56

Convert to a common denominator:
a = 712=7×212×2=1424, b = 56=5×46×4=2024

Here k₁ = 14, k₂ = 20, m = 24
k₂ – k₁ = 20 – 14 = 6

So we need a larger common denominator. Scaling up to m = 48

a = 712=1424=14×224×2=2848, b = 56=2024=20×224×2=4048

Now k₁ = 28, k₂ = 40, m = 48
k₂ – k₁ = 40 – 28 = 12 > 6

Numbers between 28 and 40 are 29, 30, 31, 32, 33.

So, the required rational numbers are 2948, 3048, 3148, 3248, 3348.

Explanation: With a shared denominator, the only candidates are fractions with integer numerators strictly between k₁ and k₂. There are exactly k₂ – k₁ – 1 such integers. To fit at least n of them:

Given that
x + y + z = 0 ……… (1)
xy + yz + zx = 0 ……… (2)

Squaring equation (1),
(x + y + z)² = 0
x² + y² + z² + 2(xy + yz + zx) = 0

Using equation (2)
x² + y² + z² + 2(0) = 0
x² + y² + z² = 0

Since x, y, z are rational numbers, each of x², y², z² is non-negative.

A sum of non-negative terms equals zero if and only if every term is zero:
x² = 0, y² = 0, z² = 0

∴ x = 0, y = 0, z = 0

Let us assume a < b.

Since a < b, adding a to both sides:
a + a < a + b
2a < (a + b)
a < (a+b)2

Again, since a < b, adding b to both sides:
a + b < b + b
(a + b) < 2b
(a+b)2 < b

Therefore, a < (a+b)2 < b

Hence, (a+b)2 lies between a and b.

This is the square root spiral. Each triangle in the spiral is formed by taking one given leg and the previous hypotenuse as the other leg.

1st triangle: Hypotenuse = √1²+0² = 1

2nd triangle: Hypotenuse = √1²+1²=√2

3rd triangle: Hypotenuse = √(√2)²+1²=√3

4th triangle: Hypotenuse = √(√3)²+1²=√3+1=2=√4

Continuing this pattern, each time: New hypotenuse = √n

Therefore, the hypotenuses are: √1, √2, √3, √4, √5, √6, √7, √8, √9, …