দেওয়া আছে,
r + q = s ....... (1)
r + q + s = w ........ (2)

(2) সমীকরণে s এর মান বসিয়ে,
r + q + (r + q) = w
⇒ r + q + r + q = w
⇒ 2r + 2q = w

5c + 3 = 3c + 5
বা, 5c - 3c = 5 - 3
বা, 2c = 2
বা, c = 1

x² + 8x + 15 এর
= x² + 3x + 5x + 15 
= x(x + 3) + 5(x + 3) 
= (x + 3) (x + 5)

I = (3/4) ÷ (5/6)
  = (3/4) x (6/5)
  = (9/10)

II = 3 ÷ [(4 ÷ 5) ÷ 6]
   = 3 ÷ {(4/5) x (1/6)}
   = 3 ÷ (4/30)
   = 3 x (30/4)
   = (45/2)

III = [3 ÷ (4 ÷ 5)] ÷ 6
    = [3 ÷ (4/5)] ÷ 6
    = [3 x (5/4)] ÷ 6
    = (15/4) x (1/6)
    = (5/8)

IV = 3 ÷ 4 ÷ (5 ÷ 6)
    = 3 ÷ 4 ÷ (5/6)
    = (3/4) x (6/5)
    = (9/10)

So, I and IV are equal.

a = (4 ÷ 3) ÷ 3 ÷ 4
   = (4/3) x (1/3) x (1/4)
   = (1/9)

b = 4 ÷ (3 ÷ 3) ÷ 4
   = 4 ÷ 1 ÷ 4
   = 4 x (1/1) x (1/4)
   = 1

c = 4 ÷ 3 ÷ (3 ÷ 4)
   = 4 ÷ 3 ÷ (3/4)
   = (4/3) x (4/3)
   = (16/9)

এখানে, (16/9)> 1 > (1/9) ; স্পষ্টত, বৃহত্তম সংখ্যা।

প্রয়োজনীয় সংখ্যা = {37 (1/2)}/ (1/8)
                      = {(75/2)/ (1/8)}
                      = (75/2) x 8
                      = 300

এখানে,

1 (3/16) = (19/16) এর reciprocal = (16/19)

∴ নির্ণেয় পার্থক্য = (19/16) - (16/19)
                    = [{(19)² - (16)²}/304]
                    = [{(19 + 16) x (19 - 16)}/304]
                    = {(35 x 3)/ 304}
                    = (105/304)

প্রদত্ত সমীক্রণ, {(5x + 1)/ x} x {(4y + 3)/ 4} = 20

⇒ (5x + 1) x (4y + 3) = 80x........(i)

Option check করলে দেখা যায়, x = 3 এবং y = 3  (i)নং সমীকরণকে সিদ্ধ করে।

ধরি, ? = x

∴ 5 (2/3) ÷ x (5/6) = 2

⇒ (17/3) ÷ x (5/6) = 2

⇒ [(17/3)/ {x (5/6)}] = 2

⇒ {x (5/6)} = {(17/3) x (1/2)}

⇒ {x (5/6)} = 2 (5/6)

∴  = 2

1 (2/3) ÷ (2/7) x (y/7) = 1 (1/4) x (2/3) ÷ (1/6)

ধরি, 
(5/3) ÷ (2/7) x (y/7) = (5/4) x (2/3) ÷ (1/6)

⇒ (5/3) x (7/2) x (y/7) = (5/4) x (2/3) x 6

⇒ (5/6) y = 5

⇒ y = {(5 x 6)/5}

∴ y = 6

প্রদত্ত রাশি

= (3 + 4 ÷ 2 x 3)/ (4 + 3 x 2 ÷ 3)

প্রদত্ত রাশি

= (5 + 5 + 5)/ (7 + 7 + 7) - (5 + 5)/ (7 + 7) + (5/7)

প্রদত্ত রাশি

= (18 x 14 - 6 x 8) / (488 ÷ 4 - 20)

= (252 - 48) / (122 - 20)

= 204/120

= 2

(113 x 4 - x x 2)/ (13 x 9 - 5 x 7) = 5

⇒ (452 - 2x)/ (117 - 35) = 5

⇒ (452 - 2x)/ 82 = 5

⇒ 452 - 2x = 410

⇒ 2x = 452 - 410 

⇒ 2x = 42

⇒ x 21

প্রদত্ত রাশি
=(343 x 49)/ (216 x 16 x 81)

= (7³ x 7²)/(6³ x 2⁴ x 3⁴)

= {7⁽³ ⁺ ²⁾} / {6³ x (2 x 3)⁴}

= 7⁵ / (6³  x 6⁴)

=  7⁵ / 6⁽³ ⁺ ⁴⁾

= 7⁵ / 6⁷

ধরি, ? = x

∴ 45 - [28 - {37 - (15 - x)}] = 58

তাহলে, 45 - [28 - {37 - (15 - x)}] = 58

⇒ 45 - [28 - {22 + x}] = 58

⇒ 45 - [28 - 22 - x] = 58

⇒ 45 - [6 - x] = 58

⇒ 45 - 6 + x = 58

⇒ 39 + x = 58

⇒ x = 58 - 39

∴ x = 19

ধরি, ? এর স্থলে x (গুণ) চিহ্ন

∴ 2 x 6 - 12 ÷ 4 + 2 = 11

তাহলে, 2 x 6 - 3 + 2 = 11

⇒ 2 x 6  = 11 + 3 - 2 = 12

সুতরাং, ? এর স্থলে x (গুণ) চিহ্ন বসবে।

সমাধানঃ (x - a) (x - b) (x - c)....(x - y) (x- z)

= (x - a) (x - b) (x - c)....(x - x) (x - y) (x- z)

= {(x - a) (x - b) (x - c)....(x - w) x 0 x [(x - y) (x -z)]

= 0

প্রদত্ত রাশি

= 999 x 99 x 9 ÷ 99 ÷ 9 ÷ 3

= 999 x 99 x (99/9) ÷ 9 ÷ 3

= 999 x 99 x (99/9) x (1/9) ÷ 3

= 999 x 99 x (99/9) x (1/9) x (1/3)

= 333

প্রদত্ত রাশি

= [(125)² ÷ 50 x 20] ÷ 25

= [{(125 x 125)/50} x 20] ÷ 25

= 6250 ÷ 25

= 250

ধরি, ? = x

∴ (x - 968) ÷ 79 x 4 = 512.

তাহলে, {(x - 968)/ 79} x 4 = 512

⇒ x - 968 = (512 x 79)/ 4

⇒ x - 968 = 10112

⇒ x = 10112 + 968

⇒ x = 11080 

Follow BODMAS rule to solve this question, as per the order given below,

Step-1- Parts of an equation enclosed in 'Brackets' must be solved first,

Step-2- Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3- Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4- Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression:

\(\begin{array}{l} 2\frac{1}{8}\% \;of\;320\;of\left( {1\frac{1}{4}\% \;of\;280\; \div \left( {\sqrt {441} - 11} \right)\% \;of\;140} \right) = ?\\ \Rightarrow \frac{{8 \times 2 + 1}}{{8 \times 100}}\;of\;320\;of\left( {\frac{{1 \times 4 + 1}}{{4 \times 100}}\;of\;280\; \div \frac{{\left( {21 - 11} \right)}}{{100}}\;of\;140} \right) = ?\\ \Rightarrow \frac{{17}}{{800}}\;of\;320\;of\left( {\frac{7}{2}\; \div \frac{{10}}{{100}}\;of\;140} \right) = ?\\ \Rightarrow \frac{{34}}{5}\;of\left( {\frac{7}{2}\; \div 14} \right) = ?\\ \Rightarrow \frac{{34}}{5}\;of\left( {\frac{7}{2} \times \frac{1}{{14}}} \right) = ?\\ \Rightarrow \frac{{34}}{5}\;of\frac{1}{4}\; = ?\\ \Rightarrow ? = \frac{{17}}{{10}} = 1\frac{7}{{10}} \end{array}\)

In this type of question, we are expected to calculate Approximate value (not exact value), so we can replace the given numbers by their nearest perfect places which makes the calculation easy.

We can write the given values as:

89898 ≈ 9 × 10000

2.486 ≈ 2.5

Now the given expression will be transformed as:

\(\begin{array}{l} \sqrt {89898} \times 2.486 = ?\\ \Rightarrow ? \approx \sqrt {9 \times 10000} \times 2.486 \end{array}\)

⇒ ? ≈ 3 × 100 × 2.5

⇒ ? ≈ 750

Given expression is,

\(\begin{array}{l} \frac{{3\; + \;\sqrt 5 }}{{3\; - \;\sqrt 5 }} - \;\frac{{3\; - \;\sqrt 5 }}{{3\; + \;\sqrt 5 }}\\ = \;\frac{{{{\left( {3\; + \;\sqrt 5 } \right)}^2} - \;{{\left( {3\; - \;\surd 5} \right)}^2}}}{{\left( {3\; - \;\sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}}\\ = \;\frac{{\left( {9\; + \;5\; + \;6\sqrt 5 } \right) - \;\left( {9\; + \;5\; - 6\;\sqrt 5 } \right)}}{{{{\left( 3 \right)}^2} - \;{{\left( {\sqrt 5 } \right)}^2}}}\\ = \frac{{12\;\sqrt 5 }}{{9 - 5}} \end{array}\)

= 12√5/4

= 3√5

79.97 ≈ 80

84.95 ≈ 85

107.93 ≈ 108

4.89 ≈ 5

4.99 ≈ 5

So, final value by using BODMAS rule,

(80 + 85 – 108 + 5) × 5 = ?

? = 62 × 5

? = 310

56 + 12 × 0.45 – 3

= 56 + (12 × 0.45) – 3         → by using BODMAS rule

= 56 + 5.4 – 3

= 58.4

x = √6 + √5

And \(\frac{1}{x} = \frac{1}{{\sqrt 6 {\rm{\;}} + {\rm{\;}}\sqrt 5 }}\)

\(= \;\frac{{\sqrt 6 \; - \;\sqrt 5 }}{{\left( {\sqrt 6 + \;\sqrt 5 } \right)\left( {\sqrt 6 \; - \;\sqrt 5 } \right)}}\) (Multiplying by \(\frac{{\sqrt 6 - \sqrt 5 }}{{\sqrt 6 - \sqrt 5 }}\) )

\(= \frac{{\sqrt 6 \; - \;\sqrt 5 }}{{6\; - \;5}}\)

= √6 - √5

∴ x – (1/x) = √6 + √5 - √6 + √5 = 2√5

∴ x³ – (1/x)³ = [x – (1/x)]³ + 3[x – (1/x)]

= (2√5)³ + 3 × 2√5

= 40√5 + 6√5

= 46√5

\(\begin{array}{l} \sqrt[2]{{2230}}\\ = \;\sqrt[2]{{2209 + 11}}\\ = \;\sqrt[2]{{{{47}^2} + 11}}\\ \approx \;\sqrt[2]{{{{47}^2}}} \end{array}\)

≈ 47

Given expression is

\(999\frac{1}{7} + 999\frac{2}{7} + 999\frac{3}{7} + 999\frac{4}{7} + 999\frac{5}{7} + 999\frac{6}{7}.\)

Now, this expression can also be written as

⇒ 999 + 1/7 + 999 + 2/7 + 999 + 3/7 + 999 + 4/7 + 999 + 5/7 + 999 + 6/7

⇒ (999 + 999 + 999 + 999 + 999 + 999) + (1/7 + 2/7 + 3/7 + 4/7 + 5/7 + 6/7)

⇒ 5994 + (21/7)

⇒ 5994 + 3 = 5997

Hence, the required answer is 5997,

8100 ÷ 15 ÷ 5 = ?

When this type of situation arises where same type of signs are present then we solve the given expression from left to right.

⇒ (8100/15) ÷ 5 = ?

⇒ 540 ÷ 5 = ?

⇒ ? = 108