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(e) √? = 2296/√1681= 2296/41= 56
? = (56)² = 3136

Formula used∶

(a^m)ⁿ = a^{m + n}, a^-m = 1/a^m, Basic Knowledge of addition of fraction.

Calculation∶

[(2401 a¹⁶)/(1296 b⁴)]^-3/4

= [(7⁴ a¹⁶)/(6⁴ b⁴)]^-3/4

= [7^{4 × (-3/4)}. a^{16 × (-3/4)}]/[6^{4 × (-3/4)}. b^{4 ×(-3/4)}]

= [7^-3. a^-12]/[6^-3. b^-3]

= [6³. b³]/[7³. a¹²]

= 216 b³/ 343 a¹²

= (216 a^-12 b³)/343

∴ The required value is 216 a^-12 b³/343.

Given:

\(6{2\over3} + {6\over7} + {7\over3} = ?\)

Calculations:

Solving the given fraction,

⇒ {[(6 × 3) + 2]/3} + (6/7) + (7/3)

⇒ [(18 + 2)/3] + (18/21) + (49/21)

⇒ [(20 × 7)/21] + (67/21)

⇒ (140/21) + (67/21)

⇒ 207/21

⇒ 69/7

∴ 69/7 will replace the question mark in the question

(a) Let the larger and smaller numbers be x and y
respectively.
Then, x – y = 3   ...(i)
and, x² – y² = 63
(x + y) (x – y) = 63
(x y) =63/3 = 21   ...(ii)
From equation (i) and (ii), x = 12

Given:

\((\frac{2}{9})^3 \times (\frac{4}{81})^{-6} = (\frac{2}{9})^{2m-1}\)

Formula required:

\(\sqrt[2]{a} = a^{1/2}\)

\(\sqrt[3]{a} = a^{1/n}\)

\(a^m \times a^n = a^{m + n}\)

a-m = 1/am

Calculations:

\(\frac{4}{81} = (\frac{2}{9})^2\)

[(2/9)²]^-6 = (2/9)^-12

\((\frac{2}{9})^3 \times (\frac{2}{9})^{-12} = (\frac{2}{9})^{2m-1}\)

⇒ 3 - 12 = 2m - 1

⇒ 2m = 3 - 12 + 1

⇒ 2m = - 8

⇒ m = - 8/2

Hence, m = - 4

∴ The value of m is -4

Part of salary left=1-(2/5+3/10+1/8) 

Let the monthly salary be Rs.x 

Then, 7/40 of x=1400 

X=(1400*40/7) =8600 

Expenditure on food=Rs.(3/10*800)=Rs.2400 

Expenditure on conveyence=Rs.(1/8*8000)=Rs.1000

Concept used:

(a + b)² = a² + b² + 2ab

Calculation:

(106 + 94)² = 106² + 94² + 2 × 106 × 94

⇒ 200² = 106² + 94² + 19928

⇒ 40000 – 19928 = 106² + 94²

⇒ 20072 = 106² + 94²

∴ The value of A is 20072

Given:

\(\frac{{27 × {{\left( {0.25} \right)}^3} + 125{{\left( {0.05} \right)}^3}}}{{{{\left( {0.75} \right)}^2} - 0.25 × 0.5}}\)

Calculation:

\(\frac{{27 × {{\left( {0.25} \right)}^3} + 125{{\left( {0.05} \right)}^3}}}{{{{\left( {0.75} \right)}^2} - 0.25 × 0.5}}\)

\( ⇒\frac{{27 × {{\left( {0.25} \right)}^3} + {{\left( {0.25} \right)}^3}}}{{{{\left( {0.75} \right)}^2} - 0.25 × 0.5}}\)

\( ⇒ \frac{{{{\left( {0.25} \right)}^3}×(27 + 1)}}{{{{\left( {0.25} \right)}^2}×9 - 0.25 × 0.5}}\)

\( ⇒ \frac{{{{\left( {0.25} \right)}^3}×(28)}}{{{{\left( {0.25} \right)}}×[(9 × 0.25) - 0.5]}}\)

⇒ (0.25)² × 28]/(1.75)

⇒ [(0.25) × 28]/7

⇒ 1

∴ The value is 1

Concept used∶

Follow the 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, and following the BODMAS rule in the bracket -

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.

Since we need to find out the approximate value, we can write these values to their nearest integers.

Calculation∶

39.99% of (1/399.99) + 50.14% of (1/249.99) = ? - 14.99% of (1/149.98)

⇒ 40% of (1/400) + 50% of (1/250) = ? - 15% of (1/150)

⇒ (1/1000) + (1/500) = ? - (1/1000)

⇒ (3/1000) = ? - (1/1000)

⇒ ? = 1/250

∴ The value of ? is 1/250

Calculation:

1.42 × 0.005 = ?

⇒ (142/100) × (5/1000) = ?

⇒ 710/100000 = ?

⇒ 0.0071 = ?

∴ The value of ? is 0.0071

Suppose Arun has Rs. X and Sjal has Rs. Y. then,

2(x-30)= y+30 => 2x-y =90 …(i) 

and x +10 =3(y-10) => x-3y = - 40 …(ii) 

Solving (i) and (ii), we get x =62 and y =34. 

Arun has Rs. 62 and Sajal has Rs. 34.