Daily Practice Questions Questions for CAT with Answers PDF

The given equations in the simplest form can be written as:

From equation (ii),

⇒ 0.6x = 3
⇒ x = 5
And 0.4y = 4/5
⇒ y = 2
So, the sum of the values of x and y is given as 7.

We know that

Put x = log_e(z)

Initially at 10 a.m., i.e. at t = 0, population of the bacteria was 20.
At 12 p.m., t = 2 and at 3 p.m., t = 5
Now, P(0) = ae^b(0) = ae⁰ = a = 20 (given)
Now, at t = 2, P(t) = 40
⇒ 40 = 20e^b(2)
⇒ e^2b = 2
Taking log of both sides, we get
2b = log_e 2 (log e = 1)
2b = 0.693
b = = 0.346
∴ P(t) = 20e^0.346t
∴ P(5) = 20e^{0.346 x 5}
= 20 x 5.64
= 112.8
Hence, population of bacteria at 3 p.m. = 113

10^x = x50
Take log of both L.H.S. and R.H.S.
L.H.S. = log (10^x) = x (log 10) = x (because log 10 = 1)
R.H.S = log (x⁵⁰) = 50 log x
L.H.S. = R.H.S.
x = 50 log x

Case I:
If x = 1, then
1 = 50 (log 1) (because log 1 = 0)
1 = 0 (Not possible)

Case II:
If x = 10, then
10 = 50 log (10)
10 = 50 (Not possible) (because log 10 = 1)

Case III:
If x = √10, then
√10= 50 log (√10)
√10 = 25 (Not possible) (because log √10 = 1/2)

Case IV:
If x = √10, then
√10 = 50 log (100) = 50 log [(10)²]
= 50 × 2 log (10) (because log a = x log a)
= 50 × 2 × 1 (because log 10 = 1)
100 = 100
L.H.S. = R.H.S.
Hence, x = 100

log2² < log2³ < log2⁴
i.e. 1 < P < 2
Similarly, log6⁶ < log6⁹ < log6³⁶
1 < Q < 2
Both P and Q belong to the interval (1, 2).
We shall now compare both P and Q with the average of 1 and 2, i.e. 3/2.
Let us assume that log2 3 > 3/2
Then 3 > 2^3/2 ⇒ 3² > 2³
i.e. 9 > 8. This is true
P > 3/2 is correct.
We shall also suppose that log6 9 > 3/2
i.e. 9 > 6^3/2
9² > 6³
i.e. 81 > 216 is not true
Hence Q > 3/2 is not correct.
Q < 3/2
Therefore P > Q.

f(x) + f(y) = 
f(x) + f(y) = log
f(x) + f(y) = log 
f(x) + f(y) = f

Thus, x = 7, -8
Since x cannot be negative value, x = 7 is the answer.

The radioactive decay is given as per the exponential function A(t) = A0 × e^at.
The half-life of plutonium-80 is 20 years.
After 20 years, the amount of plutonium-80 left will be half.
∴ 0.5 A₀ = A₀ × e^{a × 20}
0.5 = e^{a × 20}
On taking log of both sides, the equation becomes:
log_e (0.5) = a × 20 (log e = 1)
a = 1/20 (loge 0.5)
= - = -0.0346
As it is given, after 20 years, radioactivity is reduced to one-twentieth of the initial activity.Hence, A(t) = 1/20 A0
Substituting this in the activity equation, we get
A₀ = A₀ e^-0.346t
0.05 = e^-0.346t
On taking log of both the sides, the equation becomes:
loge (0.05) = -0.0346t
-2.9957 = -0.346 t
t = = 8.66 years

Now, log_a31 = 1
a = 31
So, (a + 1) = 32
(a + 1)^(1/5) = 32^(1/5) = 2

|log10 e + loge 10|

If = x, then the given expression can be written as:

Hence, the value of the given expression is always greater than 2.