BODMAS

 

(?)² = √13² + 28 / 4 – (3)³ + 107
= √169 + 7 – 27 + 107
= √256
= √16
= 4

 

The given question is,
(70 ÷ 100)^{? + 3} = (0.49)⁴ x (0.343)⁴ ÷ (0.2401)⁴
As we know that
70 ÷ 100 = 0.7
0.49 = 0.7 ²
0.343 = 0.7 ³
0.2401 = 0.7 ⁴
Put these value in given question , we will get
⇒ (0.7)^{? + 3} = (0.7²)⁴ x ( 0.7³ )⁴ ÷ ( 0.7⁴ )⁴
⇒ (0.7) ^{? + 3} = 0.7 ^{2 x 4} x 0.7 ^{3 x 4} ÷ 0.7 ^{4 x 4}
⇒ (0.7) ^{? + 3} = 0.7 ⁸ x 0.7 ¹² ÷ 0.7 ¹⁶
Apply the law of Fractional Exponents and Laws of Exponents
(a^m)(aⁿ) = a^m+n
a^m÷aⁿ=a^m-n
Or
a^m/aⁿ=a^{m – n}
we will get,
⇒ (0.7)^{? + 3} = (0.7) ^{8 + 12 – 16}
⇒ (0.7)^{? + 3} = (0.7) ^{20 – 16}
⇒ (0.7)^{? + 3} = (0.7) ⁴
On comparing the exponents both sides, we get
? + 3 = 4
∴ ? = 4 – 3 = 1

 

= 5 – [3/4 + {5/2 – (1/2 + 7 – 6/42 )}]/2
= 5 – [3/4 + {5/2 – (1/2 + 1/42)}]/2
= 5 – [3/4 + {5/2 – (21 + 1/42)}]/2
= 5 – [3/4 + {5/2 – 22/42)}]/2
= 5 – [3/4 + {105 – 22/42)}]/2
= 5 – [3/4 + 83/42]/2 = 5 – [63 + 166/84]/2
= 5 – 229/84/2 = 420 – 229/84/2 = 191/84/2
= 191/84 x 2 = 191/168 = 1²³/₁₆₈

 

25. (a)

Numerator: 0.5³ = 0.125,        0.2³ = 0.008,       0.3³ =  0.0270

 3 × 0.5 × 0.3 × 0.2 = 0.09

So, the numerator is: 0.125 + 0.008 + 0.027 − 0.09 = 0.07      

Denominator:

0.52 = 0.25,        0.22 = 0.04,       0.32 = 0.09     

$$0.2\; \times \; 0.3=0.06,0.5\times 0.3=0.15$$

So, the denominator is: 0.25 + 0.04 + 0.09 − 0.1 − 0.06 − 0.15 = 0.07

Thus, the fraction is: 0.07/ 0.07=1

26. (c) 

This is a perfect cube identity:

a³+b³+c³−3abc = (a+b+c) (a²+b²+c²−ab−bc−ac)

Thus, the answer simplifies to:

a+b+c=2.247 +1.730 + 1.023 =5   

27. (c)

This is a sum of cubes formula:

a³+b³/ a²−ab+b² = a+b

Substitute, a=725 and b=371

725 + 371 = 1096

28. (b)

29. (c)

√289 – √625 ÷ √25
= √17 x 17 – √25 x 25 ÷ √5 x 5
= 17 – 25 ÷ 5 = 17 – 5 = 12

30. (c)

We know that, (a + b)² = a² + b² + 2ab
= 234 + 2 x 108 = 450
(a – b)² = a² + b² – 2ab
= 234 – 2 x 108 = 18
∴ (a + b)²/(a – b)² = 450/18 = 25
⇒ [(a + b)/(a – b)]² = 25
∴ (a + b)/(a – b) = √25 = 5

31. (b)

a² + 1 = a
⇒ a + 1/a = 1
On squaring both sides, we get
a² + 1/a² + 2 = 1
On cubing both sides, we get
(a² + 1/a²)³ = (-1)³
⇒ a⁶ + 1/a⁶ + 3a² x 1/a² (a² + 1/a²) = -1
⇒ a⁶ + 1/a⁶ + 3 x (-1) = -1
Now, a⁶ + 1/a⁶ + 1 = 3
As, a¹² + a⁶ + 1 can be written as a⁶ + 1/a⁶ + 1
∴ a¹² + a⁶ + 1 = 3

32. (b)

Given, 0.764 x P = 1.236 x Q
⇒ Q / P = 1.236 / 0.764
Now, (Q – P) / (Q + P) = [(Q/P) – 1] / [(Q/P) + 1]
⇒ [(1.236/0.764) – 1] / [(1.236/0.764) + 1]
= (1.236 – 0.764) / (1.236 + 0.764)
= 0.472 / 2.000
= 0.236

33. (c)

∵ a + (1/b) = 1
ab + 1 = b ……(i)

Also, b + (1/c) = 1
⇒ b = 1 – (1/c) ….(ii)

From Eqs. (i) and (ii), we get
ab + 1 = 1 – (1/c)
⇒ ab = -1/c
⇒ abc = -1

34. (a)

Notice that from the given condition $a+b+c=0$, we can use substitutions for the terms involving sums. For example:

• a+b=−c
• b+c=−a
• c+a=−b

Using these in the first term:

(a+b)/ c+(b+c)/ a+ (c+a)/ b =(−c/ c) + (−a/ a) + (−b/ b) ​

This simplifies to:

−1+ (−1) + (−1) = −3

Next, simplify the second part:

a/ (b+c) +b/ (c+a) +c/ (a+b)    ​

Again using the fact that a+b+c=0, we can substitute:

• b+c=−a
• c+a=−b
• a+b=−c

This becomes:

a/ (−a) +b/(−b) + c/(−c) =−1−1−1=−3

Now multiply the results from both parts:

(−3) × (−3) = 9

35. (b)

Since a * b = a + b + a/b
∴ 12 * 4 = 12 + 4 + 12/4
= 12 + 4 + 3 = 19

36. (a)

37. (d)

x + ($\frac{\mathrm{a/}}{\mathrm{x)}}$ = 1

⇒ $x^2$^​​ + a = x….(1)

⇒ $x^2$​​ – x = -a….(2)

Now, ($x^2$^​​ +x+a)/($x^3$^​​ –$x^2$​​ )​ = 2x/[x($x^2$^​​ -x)]

= 2/(-a) [From (1) and (2)]

= -2/a

38. (a)

Given equation is,
√28 – 6√3 = √3a + b
By expanding equation,
√ 1 + 27 – 6√3 = √3a + b
⇒ √(1)² + (3√3)² – 2 x 3√3 = ⇒ √3a + b
⇒ √(1 – 3√3)² = √3a + b
⇒ 1 – 3√3 = √3a + b
On comparing the left and right side equation, we get,
a = -3, b = 1
∴ a + b = -3 + 1 = – 2

39. (b)

GIVEN:
x + y = 1
Formula used:
( x + y)³ = x³ + y³ + 3xy ( x + y)

Calculation:
Here, we have
x + y = 1
On cubing both sides we get
(x + y)³ = 1³
⇒ x³ + y³ + 3xy ( x +y) = 1
⇒ x³ + y³ + 3xy × 1 = 1
⇒ x³ + 3xy + y³
= 1

40. (a)

Given that,
a = (√2 – 1)^1/3
By cubing the both side of the given equation.
a³ = √2 – 1
Then 1/a³ = 1/√2 – 1
Now multiply and divide the above equation by √2 + 1
⇒ 1/a³ = ( √2 + 1 ) / ( √2 + 1 ) x 1/(√2 – 1 )
⇒ 1/a³ = ( √2 + 1 ) x 1 / ( √2 + 1 ) x (√2 – 1 ) [ Use the formula (A + B)(A – B) = A² – B² ]
⇒ 1/a³ = ( √2 + 1 ) / ( 2 – 1 )
⇒ 1/a³ = √2 + 1
According to the question
( a – a^-1 )³ + 3 ( a – a^-1 )
= (a – 1/a)³ + 3(a – 1/a) Use the formula [ ( A – B)³ = A³ – B³ + 3AB(A – B) ]
= a³ – 1/a³ – 3 a x 1/a (a -1/a) + 3(a – 1/a)
= a³ – 1/a³ – 3 (a -1/a) + 3(a – 1/a)
= a³ – 1/a³
Pu the value of a³ and 1/a³
= √2 -1 – ( √2 + 1 )
= √2 -1 – √2 – 1
= -2

41. (a)

42. (b)

Given:
p + q = 10 and pq = 5

Concept used:
a² + b² = (a + b)² – 2ab

Calculation:
Here,
p + q = 10 and pq = 5
⇒ p/q + q/p = (p² + q²)/ pq
⇒ [(p + q)² – 2 × pq]/pq
⇒ [102 – 2 × 5]/5
⇒ (100 – 10)/5 = 90/5 = 18

43. (c)

Given in question,
x + y = 18
By using algebraic formula
∵ x² + y² = (x + y)² – 2xy
Put the value of x + y and xy as per given question,
⇒ x² + y² = (18)² – 2 x 72
⇒ x² + y² = 324 – 144
⇒ x² + y² = 180

44. (a)

We can utilize the identity for the sum of cubes. The sum of cubes of three variables a, b, and $c$ is:

a³+b³+c³=3abc if a+b+c = 0 

In our case, we have:  (m−n)  + (n−r) + (r−m)

Simplifying this:

(m−n) + (n−r) + (r−m) = m − n + n − r + r − m = 0 

Since the sum is zero, we can apply the identity for the sum of cubes:

(m−n)³ + (n−r)³ + (r−m)³ = 3(m−n)(n−r)(r−m)

Now substitute 3(m−n)(n−r)(r−m) into the original expression:

3(m−n)(n−r)(r−m)/ 6(m−n)(n−r)(r−m)​

Cancel out (m−n)(n−r)(r−m) from the numerator and denominator:

3​/ 6 = 2/ 1

45. (c)​

We know that,
(a + b + c)² = (a² + b² + c²) + 2(ab + bc + ca)
⇒ 196 = 96 + 2(ab + bc + ca)
⇒ 2(ab + bc + ca) = 196 – 96 = 100
∴ (ab + bc + ca) = 100/2 = 50

46. (a)

Given expression can be written as
= 1/(1 x 4) + 1/(4 x 7) + 1/(7 x 10) + 1/(10 x 13) + 1/(13 x 16)
= 1/3{ 3/(1 x 4) + 3/(4 x 7) + 3/(7 x 10) + 3/(10 x 13) + 3/(13 x 16) }
= 1/3{ (4 – 1)/(1 x 4) + (7 – 4)/(4 x 7) + (10 – 7 )/(7 x 10) + (13 – 10)/(10 x 13) + (16 – 13/(13 x 16) }
= 1/3[(1 – 1/4) + (1/4 – 1/7) + (1/7 – 1/10) + (1/10 – 1/13) + (1/13 – 1/16)]
= 1/3[1 – 1/4 + 1/4 – 1/7 + 1/7 – 1/10 + 1/10 – 1/13 + 1/13 – 1/16]
= 1/3[1 – 1/16]
= 1/3[ (16 – 1)/16]
= 1/3[ (15)/16]
= 1 x 15/ 3 x 16
= 5/16

47. (a)

Let’s simplify it step by step:

(1−1/2)=1/2, (1−1/3)=2/3, (1−1/4) =3/4, …, (1−1/m)=(m−1)/m​

The entire product becomes:

1/ 2 × 2/ 3 × 3/ 4 ×4/ 5⋯(m−1)/m​

Notice that most terms in the numerator and denominator cancel out, leaving: 1/m

48. (c)​

We can simplify each term inside the parentheses: 2 – (1/3) = 5/3, 2 – (3/5) = 7/5, 2 – (5/7) = 9/7

In general, the term for $n$-th factor is:

2−(2n−1)/(2n+1) = 2(2n+1)−(2n−1)/ 2n+1

= 4n+2−2n+1/ 2n+1

=2n+3/ 2n+1​

Thus, the product becomes:

5/ 3 × 7/ 5 × 9/ 7⋅⋯⋅1001/ 999​

Notice that most terms in the numerator and denominator cancel out, leaving: 1001/ 3

49. (d)​

(1 + 1/2) (1 + 1/3) (1 + 1/4) …. (1 + 1/150)
= 3/2 x 4/3 x 5/4 x … 151/150
= (1/2) x 151
= 151/2 = 75.5

50. (b)