MCQ's Ratio and Proportion, Indices, Logarithms - 2 - Quantitative Aptitude for CA

Q1: If (25)¹⁵⁰ = (25x) ⁵⁰, then the value of x will be
(a) 5³
(b) 5⁴
(c) 5² 
(d) 5
Ans: (b)

Sol:
Given, (25)¹⁵⁰ = (25x) ⁵⁰
⇒ (5²)¹⁵⁰ = (5²x)⁵⁰
⇒ 5³⁰⁰ = 5¹⁰⁰x⁵⁰ 
⇒ x = 5 ^200/50
⇒ x = 5⁴

Q2: Find the value of a from
(a) 2/21
(b) 21/2
(c) -21/2
(d) -2/21
Ans: (c)

Sol: 
Given,

 

Q3: If  then the value of a is
(a) 4, –1
(b) –4, 1
(c) –4, –1
(d) 4, 1
Ans: (d)

Sol:
Given,

⇒ a² + 4 = 5a 
⇒ a² - 5a + 4 = 0
⇒ a² - 4a - a + 4 = 0 
⇒ a(a - 4) - 1(a - 4) = 0 
⇒ (a - 1)(a - 4) = 0 
⇒ a = 1, 4

Q4: (18)^3.5 ÷ (27)^3.5 × 6^3.5 = 2^x , then the value of x is
(a) 3.5
(b) 4.5
(c) 6
(d) 7
Ans: (d)

Sol: 
(18)^3.5 ÷ (27)^3.5 × 6^3.5 = 2^x


⇒ (4)^3.5 = 2^x
⇒ (2)⁷ = 2^x
⇒ x =  7
Therefore, the value of x is 7.

Q5: If then find the value of n.
(a) 2
(b) 0
(c) 3
(d) 4
Ans: (b)

Sol: 
Given,

⇒ 3²ⁿ× 3⁵ × 3¹⁵= 3³ × 31 × 3¹⁶ 
⇒ 3²ⁿ⁺²⁰ = 3²⁰
⇒ 2n + 20 = 20 
⇒ 2n = 0 
⇒ n = 0

Q6: The value of   is;
(a) 3/7
(b) 7/3
(c) 1 (3/7)
(d) 2(2/7)
Ans: (a)

Sol: 

= 3/7

Q7: If , then find the value of
(a) abc
(b) 9abc
(c) 1/abc
(d) 1/9 ab
Ans: (a)

Sol:

We know, If p + q + r = 0 
Then, p³ + q³ + r³ = 3pqr 
Therefore

⇒ a + b + c = 3(abc) ^1/3
⇒ (a + b + c) ³ = 27abc 
⇒ (a + b + c/3) = abc

Q8: If 4^x = 5^y = 20^z, then z is equal to
(a) xy 
(b) (x + y)/xy
(c) 1/xy
(d) xy/((x + y)
Ans: (d)

Sol: 
Given, 4^x = 5^y = 20^z
Let 4^x = 5^y = 20^z = k
⇒ 4 = k^1/x, 5 = k^1/yand 20 = k^1/z
We know,
 20 = 4 × 5



⇒ z =  xy/x + y

Q9: If 2^x = 4^y = 8^z and then the value of z is
(a) 7/16
(b) 7/32
(c) 7/48
(d) 7/64
Ans: (c)

Sol: 
Given, 2^x = 4^y = 8^z
⇒ 2^x = (2²)^y = (2³)^z
⇒ 2^x = 2^2y = 2^3z
⇒ x = 2y = 3z .... (i)
Also,

⇒ 3/6z = 24/7 
⇒ 1/2z = 24/7 
⇒ z = 7/48

Q10: The simplified value of is
(a) a⁷ b⁷
(b) a⁵ b⁷
(c) a⁵ b⁵
(d) a⁷ b⁵
Ans: (b)

Sol:

Q11: What is the value of

(a) x^abc
(b) x ^{(a + b + c)}
(c) –1
(d) 1
Ans: (d)

Sol: 


x⁰ = 1 'or' cyclic order trick
Since, (b − c)[b + c − a] + (c − a)[c + a − b] + (a − b)[a + b − c] = 0

Q12: If is equal to
(a) y
(b) –1
(c) 1
(d) None of these
Ans: (c)

Sol: 

⇒ y⁰ = 1

Q13: Find the value of  log
(a) 0
(b) 1
(c) log pqr
(d) pqr
Ans: (a)

Sol:
log

= log (1) = 0 

Q14: If log_a b = 3 and log_b c = 2, then log_a c is 
 (a) 5
(b) 6
(c) 10 
(d) 4
Ans: (b)

Sol:
Given, log_a b = 3 and log_b c = 2
log_a b x  log_a c

⇒ log c/log a 
⇒ log_a c = log_a b × log_a c 
⇒ log_a c = 3 × 2 
⇒ log_a c = 6

Q15: The value of [log₁₀(5 log₁₀ 100)]² is
 (a) 1
(b) 2
(c) 10
(d) 25
Ans: (a)

Sol: 
[log₁₀(5 log₁₀ 100)]²
= [log₁₀(5 log₁₀ 10²)]²  
= [log₁₀(5 x 2 log₁₀ 10)]²
= [log₁₀(5 x 2(1))]²
= [log₁₀ (10)]²
= (1)² = 1

Q16: If log x = log 5 + 2 log 3 – (1/2) log 25, then the value of x is
(a) 8 
(b) 9
(c) 10
(d) None of these
Ans: (b)

Sol: 
Given, log x = log 5 + 2 log 3 – (1/2) log 25
⇒ log x = log 5 + log 3²- log(25)^1/2
⇒ log x = log 5 + log 9 - log 5 
⇒ log x = log 9 
⇒ x = 9

Q17: If 1 log_a find the value of a.
(a) 3
(b) 9
(c) 27
(d) 81
Ans: (c)

Sol: 
Given, log_a 

⇒ 1/2 log_a 3 = 1/6 
⇒ log_a 3 = 1/3
⇒ 3 = a^1/3
⇒ a = 3³ 
⇒ a = 27

Q18: Given that log₁₀ 2 = x and log₁₀ 3 = y, then the value of log₁₀ 60 is expressed as
 (a) x – y + 1
(b) x + y + 1
(c) x – y – 1
(d) None of these
Ans: (b)

Sol: 
Given, log₁₀ 2 = x and log₁₀ 3 = y
We know,
log₁₀ 60 = log₁₀ (2 × 3 × 10) 
= log₁₀ 2 + log₁₀ 3 + log₁₀ 10 
= x + y + 1

Q19: Find the value of log(x⁶) if log x + 2 log (x² ) + 3 log (x³) = 14
(a) 3 
(b) 4
(c) 5
(d) 6
Ans: (d)

Sol: 
Given, log x + 2 log (x² ) + 3log (x³) = 14 
⇒ log x + 2 × 2 log (x) + 3 × 3 log (x) = 14 
⇒ log x + 4 log x + 9 log x = 14 
⇒ 14 log x = 14 
⇒ log x = 1 
Therefore, log (x⁶) = 6 log x = 6(1) = 6

Q20: If log₁₀ x = m + n – 1 and log₁₀ y = m – n, then the value of ₁₀  (100x/y²) expressed in terms of m and n is
 (a) 1 – m + 3n
(b) m – 1 + 3n
(c) m + 3n + 1
(d) m 2 – n 2
Ans: (a)

Sol:
Given, log₁₀ x = m + n – 1 and log₁₀ y = m – n 
Now, the given expression can be simplified as;
log₁₀ (100x /y²) = log₁₀(100) + log₁₀(x) - log₁₀(y²)
⇒ log₁₀(100x/y²) = log₁₀(10²) + log₁₀(x) - log₁₀(y²)
⇒ log₁₀(100x/y²) = 2log₁₀(10) + log₁₀(x) - 2log₁₀(y)
⇒ log₁₀(100x/y²) = 2(1) + (m + n - 1) - 2(m - n)
⇒ log₁₀(100x/y²) = 2 + m + n - 1 - 2m + 2n
⇒ log₁₀(100x /y²) = 1 - m + 3n

Q21:
(a) 1
(b) 2
(c) 3
(d) None of these
Ans: (b)

Sol:
​

= log_xyz (xy) + log_xyz yz + log_xyz (xz) 
= log_xyz (xy × yz + xz) 
= log_xyz (xyz²) 
= 2log_xyz (xyz) 
= 2(1) = 2

Q22: log4 (x² + x) – log₄ (x + 1) = 2. Find x.
(a) 16
(b) 0
(c) –1
(d) None of thes
Ans: (a)

Sol: 
log4 (x² + x) – log₄ (x + 1) = 2
⇒ log₄ (x² + x)/(x + 1) = 2 
⇒ log₄ x(x + 1)/x + 1 = 2 
⇒ log₄ x = 2 
⇒ x = 4²
⇒ x = 16

Q23: log₂ log₂ log₄ 256 + 2log _√2 2 log 2 is equal to
 (a) 2
(b) 3
(c) 5
(d) 7
Ans: (c)

Sol: 
log₂ log₂ log₄ 256 + 2log _√2 2 
= log₂ log₂ log₄ 4⁴+ 2 log log ₂2
= log₂ log₂ (4 log4 ₄) +log₂ 2
= log₂ log₂ (2² ) + 4(1) 
= log₂ 2 + 4 = 1 + 4 = 5

Q24: The value of log₅ + log₅+ ....... + log_{5
(a)2 
(b) 3
(c) 5 
(d) Cannot be determined
Ans: (b)}

Sol:
_{log5 + log5+ ....... + log5}

= log₅ (6/5)  + log₅  (7/6)+ ....... + log₅ (625/624)

= log₅ (625/5)
= log₅ (125) = log₅ (5³) = 3

Q25: If log (log a + log b), then the value of a/b + b/a will be
(a) 12
(b) 14
(c) 16
(d) 8
Ans: (b)

Sol: 
Given, log (log a + log b)
⇒log a + b/4 = 1/2 log (ab)
⇒log a + b/4 = log (ab)^1/2
⇒ a + b/4 = log (ab)^1/2
⇒ (a + b/4) ²= ab 
⇒ (a² + b² + 2ab) /16 = ab 
⇒ a² + b² + 2ab = 16ab 
⇒ a² + b² = 14ab
 ⇒ (a² + b²)/ab = 14
 ⇒ a/b + b/a = 14

Q26: If log_a (ab) = x, then log_b(ab) is 
(a) 1/x 
(b) x/1 + x
(c) x/x −1
(d) None of these
Ans: (c)

Sol: 
Given, log_a (ab) = x


logb/ loga=  x - 1 ......(i) 
Therefore,
log_b (ab) = log ab/ log b
⇒ log_b (ab) = (log a + log b)/log b 
⇒ log_b (ab) = log a/log b + 1 

Q27: If x = log₂₄ 12, y = log₃₆ 24, z = log₄₈ 36, then xyz + 1 = ?
 (a) 2xy
(b) 2xz
(c) 2yz
 (d) 2
Ans: (c)

Sol: 
Given, x = log₂₄ 12, y = log₃₆ 24, z = log₄₈ 36
Thus, xyz + 1

= log₄₈ 12 + log₄₈ 48 
= log₄₈ (12 × 48)
= log₄₈ (576) 
= log₄₈ (24)² 
= 2 log₄₈ (24)
For option (c): 2yz 
= 2(log₃₆ 24)(log₄₈ 36)


= 2 log₄₈(24) 
Therefore, xyz + 1 = 2yz