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Algebra Quantitative Aptitude Questions | Prasnya
Quantitative Aptitude > Algebra
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Difficulty:
All
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All
96Q
01
If a and b are integers of opposite signs such that
(
a
+
3
)
2
:
b
2
=
9
:
1
(a + 3)^2 : b^2 = 9 : 1
(
a
+
3
)
2
:
b
2
=
9
:
1
and
(
a
−
1
)
2
:
(
b
−
1
)
2
=
4
:
1
(a - 1)^2 : (b - 1)^2 = 4 : 1
(
a
−
1
)
2
:
(
b
−
1
)
2
=
4
:
1
, then the ratio a : b is
Single correct
HARD
3 marks
26 November 2017
02
Suppose,
log
3
x
=
log
12
y
=
a
\log_3 x = \log_{12} y = a
lo
g
3
x
=
lo
g
12
y
=
a
, where x, y are positive numbers. If G is the geometric mean of x and y, and
log
6
G
\log_6 G
lo
g
6
G
is equal to
Single correct
MEDIUM
3 marks
26 November 2017
03
If
x
+
1
=
x
2
x + 1 = x^2
x
+
1
=
x
2
and
x
>
0
x > 0
x
>
0
, then
2
x
4
2x^4
2
x
4
is
Single correct
HARD
3 marks
26 November 2017
04
The value of
log
0.008
5
+
log
3
81
−
7
\log_{0.008}\sqrt5 + \log_{\sqrt3}81 - 7
lo
g
0.008
5
+
lo
g
3
81
−
7
is equal to
Single correct
HARD
3 marks
26 November 2017
05
If
9
2
x
−
1
−
81
x
−
1
=
1944
9^{2x-1} - 81^{x-1} = 1944
9
2
x
−
1
−
8
1
x
−
1
=
1944
, then x is
Single correct
MEDIUM
3 marks
26 November 2017
06
The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is
Single correct
HARD
3 marks
26 November 2017
07
For how many integers n, will the inequality (n – 5) (n – 10) – 3(n – 2) ≤ 0 be satisfied?
Numerical
MEDIUM
3 marks
26 November 2017
08
If
f
1
(
x
)
=
x
2
+
11
x
+
n
f_1(x) = x^2 + 11x + n
f
1
(
x
)
=
x
2
+
11
x
+
n
and
f
2
(
x
)
=
x
f_2(x) = x
f
2
(
x
)
=
x
, then the largest positive integer n for which the equation
f
1
(
x
)
=
f
2
(
x
)
f_1(x) = f_2(x)
f
1
(
x
)
=
f
2
(
x
)
has two distinct real roots, is
Numerical
MEDIUM
3 marks
26 November 2017
09
If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of
(
a
−
b
)
2
+
(
a
−
c
)
2
+
(
a
−
d
)
2
(a - b)^2 + (a - c)^2 + (a - d)^2
(
a
−
b
)
2
+
(
a
−
c
)
2
+
(
a
−
d
)
2
is
Numerical
MEDIUM
3 marks
26 November 2017
10
If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Single correct
MEDIUM
3 marks
26 November 2017
11
If
f
(
x
)
=
5
x
+
2
3
x
−
5
f(x)=\dfrac{5x+2}{3x-5}
f
(
x
)
=
3
x
−
5
5
x
+
2
and
g
(
x
)
=
x
2
−
2
x
−
1
g(x)=x^2-2x-1
g
(
x
)
=
x
2
−
2
x
−
1
, then the value of
g
(
f
(
f
(
3
)
)
)
g(f(f(3)))
g
(
f
(
f
(
3
)))
is
Single correct
HARD
3 marks
26 November 2017
12
Let
a
1
,
a
2
,
…
…
,
a
3
n
a_1, a_2,……,a_{3n}
a
1
,
a
2
,
……
,
a
3
n
be an arithmetic progression with
a
1
=
3
a_1 = 3
a
1
=
3
and
a
2
=
7
a_2 = 7
a
2
=
7
. If
a
1
+
a
2
+
…
+
a
3
n
=
1830
a_1 + a_2 + …+a_{3n} = 1830
a
1
+
a
2
+
…
+
a
3
n
=
1830
, then what is the smallest positive integer m such that
m
(
a
1
+
a
2
+
…
+
a
n
)
>
1830
m(a_1 + a_2 + … + a_n) > 1830
m
(
a
1
+
a
2
+
…
+
a
n
)
>
1830
?
Numerical
HARD
3 marks
26 November 2017
13
Direction of the question: Solve the following question and mark the best possible option. If three sides of a rectangular park have a total length 400 ft, then the area of the park is maximum when the length (in ft) of its longer side is
Numerical
MEDIUM
3 marks
26 November 2017
14
Direction of the question: Solve the following question and mark the best possible option. If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is
Single correct
MEDIUM
3 marks
26 November 2017
15
Direction of the question: Solve the following question and mark the best possible option. If x is a real number such that
log
3
5
=
log
5
(
2
+
x
)
\log_3 5 = \log_5(2 + x)
lo
g
3
5
=
lo
g
5
(
2
+
x
)
, then which of the following is true?
Single correct
HARD
3 marks
26 November 2017
16
Direction of the question: Solve the following question and mark the best possible option. Let
f
(
x
)
=
x
2
f(x) = x^2
f
(
x
)
=
x
2
and
g
(
x
)
=
2
x
g(x) = 2^x
g
(
x
)
=
2
x
, for all real
x
x
x
. Then the value of
f
(
f
(
g
(
x
)
)
+
g
(
f
(
x
)
)
)
f(f(g(x)) + g(f(x)))
f
(
f
(
g
(
x
))
+
g
(
f
(
x
)))
at
x
=
1
x = 1
x
=
1
is
Single correct
MEDIUM
3 marks
26 November 2017
17
Direction of the question: Solve the following question and mark the best possible option. The minimum possible value of the sum of the squares of the roots of the equation
x
2
+
(
a
+
3
)
x
−
(
a
+
5
)
=
0
x^2 + (a + 3)x - (a + 5) = 0
x
2
+
(
a
+
3
)
x
−
(
a
+
5
)
=
0
is
Single correct
MEDIUM
3 marks
26 November 2017
18
Direction of the question: Solve the following question and mark the best possible option. If
9
x
−
1
2
−
2
2
x
−
2
=
4
x
−
3
2
x
−
3
9^{x - \frac{1}{2}} - 2^{2x-2} = 4^x - 3^{2x-3}
9
x
−
2
1
−
2
2
x
−
2
=
4
x
−
3
2
x
−
3
, then
x
x
x
is
Single correct
VERY_HARD
3 marks
26 November 2017
19
Direction of the question: Solve the following question and mark the best possible option. If
log
(
2
a
×
3
b
×
5
c
)
\log(2^a \times 3^b \times 5^c)
lo
g
(
2
a
×
3
b
×
5
c
)
is the arithmetic mean of
log
(
2
2
×
3
3
×
5
)
\log(2^2 \times 3^3 \times 5)
lo
g
(
2
2
×
3
3
×
5
)
,
log
(
2
6
×
3
×
5
7
)
\log(2^6 \times 3 \times 5^7)
lo
g
(
2
6
×
3
×
5
7
)
, and
log
(
2
×
3
2
×
5
4
)
\log(2 \times 3^2 \times 5^4)
lo
g
(
2
×
3
2
×
5
4
)
, then
a
a
a
equals
Numerical
MEDIUM
3 marks
26 November 2017
20
Direction of the question: Solve the following question and mark the best possible option. Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3. If the sum of the numbers in the new sequence is 450, then a5 is
Numerical
MEDIUM
3 marks
26 November 2017
21
Direction of the question: Solve the following question and mark the best possible option. If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(1) is
Numerical
MEDIUM
3 marks
26 November 2017
22
Direction of the question: Solve the following question and mark the best possible option. Let f(x) = 2x - 5 and g(x) = 7 - 2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if
Single correct
HARD
3 marks
26 November 2017
23
Direction of the question: Solve the following question and mark the best possible option. An infinite geometric progression
a
1
,
a
2
,
a
3
,
.
.
.
a_1, a_2, a_3,...
a
1
,
a
2
,
a
3
,
...
has the property that
a
n
=
3
(
a
n
+
1
+
a
n
+
2
+
.
.
.
)
a_n = 3(a_{n+1} + a_{n+2} + ...)
a
n
=
3
(
a
n
+
1
+
a
n
+
2
+
...
)
for every
n
≥
1
n \ge 1
n
≥
1
. If the sum
a
1
+
a
2
+
a
3
+
.
.
.
.
=
32
a_1 + a_2 + a_3 + .... = 32
a
1
+
a
2
+
a
3
+
....
=
32
, then
a
5
a_5
a
5
is
Single correct
HARD
3 marks
26 November 2017
24
Direction of the question: Solve the following question and mark the best possible option. If a1 = 1/(2x5), a2 = 1/(5x8), a3 = 1/(8x11), ..., then a1 + a2 + a3 + ... + a100 is
Single correct
HARD
3 marks
26 November 2017
25
Direction of the question: Solve the following question and mark the best possible option. Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is
Single correct
MEDIUM
3 marks
25 November 2018
26
Direction of the question: Solve the following question and mark the best possible option. Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,... will be
Single correct
EASY
3 marks
25 November 2018
27
Direction of the question: Solve the following question and mark the best possible option. If x is a positive quantity such that
2
x
=
3
log
5
2
2^x = 3^{\log_5 2}
2
x
=
3
l
o
g
5
2
, then x is equal to
Single correct
MEDIUM
3 marks
25 November 2018
28
Direction of the question: Solve the following question and mark the best possible option. Given that
x
2018
y
2017
=
1
2
x^{2018}y^{2017}=\frac{1}{2}
x
2018
y
2017
=
2
1
and
x
2016
y
2019
=
8
x^{2016}y^{2019}=8
x
2016
y
2019
=
8
, the value of
x
2
+
y
3
x^2+y^3
x
2
+
y
3
is
Single correct
MEDIUM
3 marks
25 November 2018
29
Direction of the question: Solve the following question and key in your numerical answer. Let
f
(
x
)
=
min
{
2
x
2
,
52
−
5
x
}
f(x) = \min\{2x^2, 52-5x\}
f
(
x
)
=
min
{
2
x
2
,
52
−
5
x
}
, where x is any positive real number. Then the maximum possible value of f(x) is
Numerical
EASY
3 marks
25 November 2018
30
Direction of the question: Solve the following question and mark the best possible option. If
log
12
81
=
p
\log_{12} 81 = p
lo
g
12
81
=
p
, then
3
(
4
−
p
4
+
p
)
3\left(\dfrac{4-p}{4+p}\right)
3
(
4
+
p
4
−
p
)
is equal to
Single correct
HARD
3 marks
25 November 2018
31
Direction of the question: Solve the following question and key in your numerical answer. If
f
(
x
+
2
)
=
f
(
x
)
+
f
(
x
+
1
)
f(x+2) = f(x) + f(x+1)
f
(
x
+
2
)
=
f
(
x
)
+
f
(
x
+
1
)
for all positive integers x, and
f
(
11
)
=
91
f(11) = 91
f
(
11
)
=
91
,
f
(
15
)
=
617
f(15) = 617
f
(
15
)
=
617
, then
f
(
10
)
f(10)
f
(
10
)
equals
Numerical
MEDIUM
3 marks
25 November 2018
32
Direction of the question: Solve the following question and key in your numerical answer. While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is
Numerical
EASY
3 marks
25 November 2018
33
Direction of the question: Solve the following question and mark the best possible option. If
u
2
+
(
u
−
2
v
−
1
)
2
=
−
4
v
(
u
+
v
)
u^2+(u-2v-1)^2 = -4v(u+v)
u
2
+
(
u
−
2
v
−
1
)
2
=
−
4
v
(
u
+
v
)
, then what is the value of
u
+
3
v
u+3v
u
+
3
v
?
Single correct
MEDIUM
3 marks
25 November 2018
34
Direction of the question: Solve the following question and mark the best possible option. If
log
2
(
5
+
log
3
a
)
=
3
\log_2(5+\log_3 a) = 3
lo
g
2
(
5
+
lo
g
3
a
)
=
3
and
log
5
(
4
a
+
12
+
log
2
b
)
=
3
\log_5(4a+12+\log_2 b) = 3
lo
g
5
(
4
a
+
12
+
lo
g
2
b
)
=
3
, then
a
+
b
a+b
a
+
b
is equal to
Single correct
MEDIUM
3 marks
25 November 2018
35
Direction of the question: Solve the following question and mark the best possible option. How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?
Single correct
MEDIUM
3 marks
25 November 2018
36
Direction of the question: Solve the following question and mark the best possible option. If the sum of squares of two numbers is 97, then which one of the following cannot be their product?
Single correct
MEDIUM
3 marks
25 November 2018
37
Direction of the question: Solve the following question and key in your numerical answer. If a and b are integers such that
2
x
2
−
a
x
+
2
>
0
2x^2 - ax + 2 > 0
2
x
2
−
a
x
+
2
>
0
and
x
2
−
b
x
+
8
≥
0
x^2 - bx + 8 \geq 0
x
2
−
b
x
+
8
≥
0
for all real numbers x, then the largest possible value of
2
a
−
6
b
2a-6b
2
a
−
6
b
is
Numerical
MEDIUM
3 marks
25 November 2018
38
Direction of the question: Solve the following question and mark the best possible option. The value of the sum
7
×
11
+
11
×
15
+
15
×
19
+
.
.
.
+
95
×
99
7 \times 11 + 11 \times 15 + 15 \times 19 + ... + 95 \times 99
7
×
11
+
11
×
15
+
15
×
19
+
...
+
95
×
99
is
Single correct
HARD
3 marks
25 November 2018
39
Direction of the question: Solve the following question and mark the best possible option. The smallest integer n for which
4
n
>
17
19
4^n > 17^{19}
4
n
>
1
7
19
holds, is closest to
Single correct
MEDIUM
3 marks
25 November 2018
40
Direction of the question: Solve the following question and mark the best possible option. If
p
3
=
q
4
=
r
5
=
s
6
p^3 = q^4 = r^5 = s^6
p
3
=
q
4
=
r
5
=
s
6
, then the value of
log
s
(
p
q
r
)
\log_s(pqr)
lo
g
s
(
pq
r
)
is equal to
Single correct
MEDIUM
3 marks
25 November 2018
41
Direction of the question: Solve the following question and key in your numerical answer. Let
f
(
x
)
=
max
{
5
x
,
52
−
2
x
2
}
f(x)=\max\{5x, 52-2x^2\}
f
(
x
)
=
max
{
5
x
,
52
−
2
x
2
}
, where x is any positive real number. Then the minimum possible value of f(x) is
Numerical
MEDIUM
3 marks
25 November 2018
42
Direction of the question: Solve the following question and mark the best possible option.
1
log
2
100
−
1
log
4
100
+
1
log
5
100
−
1
log
10
100
+
1
log
20
100
−
1
log
25
100
+
1
log
50
100
=
?
\frac{1}{\log_2 100} - \frac{1}{\log_4 100} + \frac{1}{\log_5 100} - \frac{1}{\log_{10} 100} + \frac{1}{\log_{20} 100} - \frac{1}{\log_{25} 100} + \frac{1}{\log_{50} 100} = ?
l
o
g
2
100
1
−
l
o
g
4
100
1
+
l
o
g
5
100
1
−
l
o
g
10
100
1
+
l
o
g
20
100
1
−
l
o
g
25
100
1
+
l
o
g
50
100
1
=
?
Single correct
HARD
3 marks
25 November 2018
43
Direction of the question: Solve the following question and key in your numerical answer. Let
t
1
,
t
2
,
.
.
.
t_1, t_2, ...
t
1
,
t
2
,
...
be real numbers such that
t
1
+
t
2
+
.
.
.
+
t
n
=
2
n
2
+
9
n
+
13
t_1+t_2+...+t_n = 2n^2+9n+13
t
1
+
t
2
+
...
+
t
n
=
2
n
2
+
9
n
+
13
, for every positive integer n≥2. If
t
k
=
103
t_k=103
t
k
=
103
, then k equals
Numerical
MEDIUM
3 marks
25 November 2018
44
Direction of the question: Solve the following question and key in your numerical answer. The smallest integer n such that
n
3
−
11
n
2
+
32
n
−
28
>
0
n^3 - 11n^2 + 32n - 28 > 0
n
3
−
11
n
2
+
32
n
−
28
>
0
is
Numerical
MEDIUM
3 marks
25 November 2018
45
Direction of the question: Solve the following question and mark the best possible option. Let
a
1
,
a
2
,
.
.
.
,
a
52
a_1, a_2, ..., a_{52}
a
1
,
a
2
,
...
,
a
52
be positive integers such that
a
1
<
a
2
<
.
.
.
<
a
52
a_1<a_2<...<a_{52}
a
1
<
a
2
<
...
<
a
52
. Suppose, their arithmetic mean is one less than the arithmetic mean of
a
2
,
a
3
,
.
.
.
,
a
52
a_2, a_3, ..., a_{52}
a
2
,
a
3
,
...
,
a
52
. If
a
52
=
100
a_{52}=100
a
52
=
100
, then the largest possible value of
a
1
a_1
a
1
is
Single correct
HARD
3 marks
25 November 2018
46
Direction of the question: Solve the following question and key in your numerical answer. If N and x are positive integers such that
N
N
=
2
160
N^N = 2^{160}
N
N
=
2
160
and
N
2
+
2
N
N^2+2^N
N
2
+
2
N
is an integral multiple of
2
x
2^x
2
x
, then the largest possible x is
Numerical
HARD
3 marks
25 November 2018
47
If the equations
x
2
+
m
x
+
9
=
0
x^2 + mx + 9 = 0
x
2
+
m
x
+
9
=
0
,
x
2
+
n
x
+
17
=
0
x^2 + nx + 17 = 0
x
2
+
n
x
+
17
=
0
and
x
2
+
(
m
+
n
)
x
+
35
=
0
x^2 + (m + n) x + 35 = 0
x
2
+
(
m
+
n
)
x
+
35
=
0
have a common negative root, then the value of (2m + 3m) is
Numerical
VERY_HARD
3 marks
24 November 2024
48
If x is a positive real number such that
4
log
10
x
+
4
log
100
x
+
8
log
1000
x
=
13
4\log_{10} x + 4\log_{100} x + 8\log_{1000} x = 13
4
lo
g
10
x
+
4
lo
g
100
x
+
8
lo
g
1000
x
=
13
, then greatest integer not exceeding x, is
Numerical
HARD
3 marks
24 November 2024
49
Let x, y, and z be real numbers satisfying
4
(
x
2
+
y
2
+
z
2
)
=
a
4(x^2 + y^2 + z^2) = a
4
(
x
2
+
y
2
+
z
2
)
=
a
4
(
x
−
y
−
z
)
=
3
+
a
4(x - y - z) = 3 + a
4
(
x
−
y
−
z
)
=
3
+
a
Then a equals
Single correct
VERY_HARD
3 marks
24 November 2024
50
A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional 3 kg of these grains to the second customer. Finally, when the shop sells half of the remaining quantity and an additional 3 kg of these grains to the third customer, there are no grains left. The initial quantity, in kg, of grains is
Single correct
MEDIUM
3 marks
24 November 2024
51
If
(
a
+
b
n
)
(a + b\sqrt{n})
(
a
+
b
n
)
is the positive square root of
(
29
−
12
5
)
(29 - 12\sqrt{5})
(
29
−
12
5
)
, where a and b are integers, and n is a natural number, then the maximum possible value of (a + b + n) is
Single correct
VERY_HARD
3 marks
24 November 2024
52
The sum of all real values of k for which
(
1
8
)
k
×
(
1
32768
)
1
3
=
1
8
×
(
1
32768
)
1
k
\left(\frac{1}{8}\right)^{k} \times \left(\frac{1}{32768}\right)^{\frac{1}{3}} = \frac{1}{8} \times \left(\frac{1}{32768}\right)^{\frac{1}{k}}
(
8
1
)
k
×
(
32768
1
)
3
1
=
8
1
×
(
32768
1
)
k
1
, is
Single correct
HARD
3 marks
24 November 2024
53
Suppose
x
1
,
x
2
,
x
3
,
…
,
x
100
x_1, x_2, x_3, \ldots, x_{100}
x
1
,
x
2
,
x
3
,
…
,
x
100
are in arithmetic progression such that
x
5
=
−
4
x_5 = -4
x
5
=
−
4
and
2
x
6
+
2
x
9
=
x
11
+
x
13
2x_6 + 2x_9 = x_{11} + x_{13}
2
x
6
+
2
x
9
=
x
11
+
x
13
. Then,
x
100
x_{100}
x
100
equals.
Single correct
HARD
3 marks
24 November 2024
54
For any natural number n, let
a
n
a_n
a
n
be the largest integer not exceeding
n
\sqrt{n}
n
. Then the value of
a
1
+
a
2
+
…
+
a
50
a_1+a_2+\ldots+a_{50}
a
1
+
a
2
+
…
+
a
50
is
Numerical
HARD
3 marks
24 November 2024
55
If
(
x
+
6
2
)
1
/
2
−
(
x
−
6
2
)
1
/
2
=
2
2
\left(x+6\sqrt{2}\right)^{1/2} - \left(x-6\sqrt{2}\right)^{1/2} = 2\sqrt{2}
(
x
+
6
2
)
1/2
−
(
x
−
6
2
)
1/2
=
2
2
, then x equals
Numerical
HARD
3 marks
24 November 2024
56
A function
f
f
f
maps the set of natural numbers to whole numbers, such that
f
(
x
y
)
=
f
(
x
)
f
(
y
)
+
f
(
x
)
+
f
(
y
)
f(xy) = f(x)f(y) + f(x) + f(y)
f
(
x
y
)
=
f
(
x
)
f
(
y
)
+
f
(
x
)
+
f
(
y
)
for all
x
,
y
x, y
x
,
y
and
f
(
p
)
=
1
f(p) = 1
f
(
p
)
=
1
for every prime number
p
p
p
. Then, the value of
f
(
160000
)
f(160000)
f
(
160000
)
is
Single correct
HARD
3 marks
24 November 2024
57
The sum of the infinite series
1
5
(
1
5
−
1
7
)
+
(
1
5
)
2
[
(
1
5
)
2
−
(
1
7
)
2
]
+
(
1
5
)
3
[
(
1
5
)
3
−
(
1
7
)
3
]
+
⋯
\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\left(\dfrac{1}{5}\right)^2\left[\left(\dfrac{1}{5}\right)^2-\left(\dfrac{1}{7}\right)^2\right]+\left(\dfrac{1}{5}\right)^3\left[\left(\dfrac{1}{5}\right)^3-\left(\dfrac{1}{7}\right)^3\right]+\cdots
5
1
(
5
1
−
7
1
)
+
(
5
1
)
2
[
(
5
1
)
2
−
(
7
1
)
2
]
+
(
5
1
)
3
[
(
5
1
)
3
−
(
7
1
)
3
]
+
⋯
is equal to
Single correct
VERY_HARD
3 marks
24 November 2024
58
If
x
x
x
and
y
y
y
satisfy the equations
∣
x
∣
+
x
+
y
=
15
|x| + x + y = 15
∣
x
∣
+
x
+
y
=
15
and
x
+
∣
y
∣
−
y
=
20
x + |y| - y = 20
x
+
∣
y
∣
−
y
=
20
, then
(
x
−
y
)
(x-y)
(
x
−
y
)
equals
Single correct
HARD
3 marks
24 November 2024
59
The roots
α
,
β
\alpha, \beta
α
,
β
of the equation
3
x
2
+
λ
x
−
1
=
0
3x^2 + \lambda x - 1 = 0
3
x
2
+
λ
x
−
1
=
0
, satisfy
1
α
2
+
1
β
2
=
15
\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}=15
α
2
1
+
β
2
1
=
15
. The value of
(
α
3
+
β
3
)
2
(\alpha^3+\beta^3)^2
(
α
3
+
β
3
)
2
, is
Single correct
HARD
3 marks
24 November 2024
60
All the values of
x
x
x
satisfying the inequality
1
x
+
5
≤
1
2
x
−
3
\dfrac{1}{x+5} \le \dfrac{1}{2x-3}
x
+
5
1
≤
2
x
−
3
1
are
Single correct
HARD
3 marks
24 November 2024
61
If
a
,
b
a, b
a
,
b
and
c
c
c
are positive real numbers such that
a
>
10
≥
b
≥
c
a > 10 \ge b \ge c
a
>
10
≥
b
≥
c
and
log
8
(
a
+
b
)
log
2
c
+
log
27
(
a
−
b
)
log
3
c
=
2
3
\dfrac{\log_8(a+b)}{\log_2 c} + \dfrac{\log_{27}(a-b)}{\log_3 c} = \dfrac23
lo
g
2
c
lo
g
8
(
a
+
b
)
+
lo
g
3
c
lo
g
27
(
a
−
b
)
=
3
2
, then the greatest possible integer value of
a
a
a
is
Numerical
VERY_HARD
3 marks
24 November 2024
62
If
x
x
x
and
y
y
y
are real numbers such that
4
x
2
+
4
y
2
−
4
x
y
−
6
y
+
3
=
0
4x^2 + 4y^2 - 4xy - 6y + 3 = 0
4
x
2
+
4
y
2
−
4
x
y
−
6
y
+
3
=
0
, then the value of
(
4
x
+
5
y
)
(4x + 5y)
(
4
x
+
5
y
)
is
Numerical
HARD
3 marks
24 November 2024
63
If
3
a
=
4
3^a = 4
3
a
=
4
,
4
b
=
5
4^b = 5
4
b
=
5
,
5
c
=
6
5^c = 6
5
c
=
6
,
6
d
=
7
6^d = 7
6
d
=
7
,
7
e
=
8
7^e = 8
7
e
=
8
and
8
f
=
9
8^f = 9
8
f
=
9
, then the value of the product
a
b
c
d
e
f
abcdef
ab
c
d
e
f
is
Numerical
MEDIUM
3 marks
24 November 2024
64
The number of distinct integer solutions
(
x
,
y
)
(x, y)
(
x
,
y
)
of the equation
∣
x
+
y
∣
+
∣
x
−
y
∣
=
2
|x + y| + |x - y| = 2
∣
x
+
y
∣
+
∣
x
−
y
∣
=
2
, is
Numerical
MEDIUM
3 marks
24 November 2024
65
If
(
a
+
b
3
)
2
=
52
+
30
3
(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}
(
a
+
b
3
)
2
=
52
+
30
3
, where
a
a
a
and
b
b
b
are natural numbers, then
a
+
b
a + b
a
+
b
equals
Single correct
MEDIUM
3 marks
24 November 2024
66
The sum of all distinct real values of
x
x
x
that satisfy the equation
10
x
+
4
10
x
=
91
2
10^x + \dfrac{4}{10^x} = \dfrac{91}{2}
1
0
x
+
1
0
x
4
=
2
91
, is
Single correct
HARD
3 marks
24 November 2024
67
The number of distinct real values of
x
x
x
, satisfying the equation
max
{
x
,
2
}
−
min
{
x
,
2
}
=
∣
x
+
2
∣
−
∣
x
−
2
∣
\max\{x, 2\} - \min\{x, 2\} = |x + 2| - |x - 2|
max
{
x
,
2
}
−
min
{
x
,
2
}
=
∣
x
+
2∣
−
∣
x
−
2∣
, is
Numerical
HARD
3 marks
24 November 2024
68
For any non-zero real number
x
x
x
, let
f
(
x
)
+
2
f
(
1
x
)
=
3
x
f(x) + 2f\left(\dfrac{1}{x}\right) = 3x
f
(
x
)
+
2
f
(
x
1
)
=
3
x
. Then, the sum of all possible values of
x
x
x
for which
f
(
x
)
=
3
f(x) = 3
f
(
x
)
=
3
, is
Single correct
HARD
3 marks
24 November 2024
69
Consider the sequence
t
1
=
1
t_1 = 1
t
1
=
1
,
t
2
=
−
1
t_2 = -1
t
2
=
−
1
and
t
n
=
(
n
−
3
n
−
1
)
t
n
−
2
t_n = \left(\dfrac{n-3}{n-1}\right)t_{n-2}
t
n
=
(
n
−
1
n
−
3
)
t
n
−
2
for
n
≥
3
n \ge 3
n
≥
3
. Then, the value of the sum
1
t
2
+
1
t
4
+
1
t
6
+
…
+
1
t
2022
+
1
t
2024
\dfrac{1}{t_2} + \dfrac{1}{t_4} + \dfrac{1}{t_6} + \ldots + \dfrac{1}{t_{2022}} + \dfrac{1}{t_{2024}}
t
2
1
+
t
4
1
+
t
6
1
+
…
+
t
2022
1
+
t
2024
1
, is
Single correct
VERY_HARD
3 marks
24 November 2024
70
For some constant real numbers
p
p
p
,
k
k
k
and
a
a
a
consider the following system of linear equations in
x
x
x
and
y
y
y
:
p
x
−
4
y
=
2
px - 4y = 2
p
x
−
4
y
=
2
3
x
+
k
y
=
a
3x + ky = a
3
x
+
k
y
=
a
A necessary condition for the system to have no solution for
(
x
,
y
)
(x, y)
(
x
,
y
)
is
Single correct
HARD
3 marks
24 November 2024
71
The number of distinct integers
n
n
n
for which
log
1
/
4
(
n
2
−
7
n
+
11
)
>
0
\log_{1/4}(n^2 - 7n + 11) > 0
lo
g
1/4
(
n
2
−
7
n
+
11
)
>
0
, is
Single correct
MEDIUM
3 marks
30 November 2025
72
The number of distinct pairs of integers
(
x
,
y
)
(x, y)
(
x
,
y
)
satisfying the inequalities
x
>
y
≥
3
x > y \geq 3
x
>
y
≥
3
and
x
+
y
<
14
x + y < 14
x
+
y
<
14
is
Numerical
MEDIUM
3 marks
30 November 2025
73
Let
3
≤
x
≤
6
3 \leq x \leq 6
3
≤
x
≤
6
and
[
x
2
]
=
[
x
]
2
[x^2] = [x]^2
[
x
2
]
=
[
x
]
2
, where
[
x
]
[x]
[
x
]
is the greatest integer not exceeding
x
x
x
. If set
S
S
S
represents all feasible values of
x
x
x
, then a possible subset of
S
S
S
is
Single correct
HARD
3 marks
30 November 2025
74
The number of non-negative integer values of
k
k
k
for which the quadratic equation
x
2
−
5
x
+
k
=
0
x^2 - 5x + k = 0
x
2
−
5
x
+
k
=
0
has only integer roots, is
Numerical
EASY
3 marks
30 November 2025
75
Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is
Numerical
EASY
3 marks
30 November 2025
76
If
a
−
6
b
+
6
c
=
4
a - 6b + 6c = 4
a
−
6
b
+
6
c
=
4
and
6
a
+
3
b
−
3
c
=
50
6a + 3b - 3c = 50
6
a
+
3
b
−
3
c
=
50
, where a, b and c are real numbers, the value of
2
a
+
3
b
−
3
c
2a + 3b - 3c
2
a
+
3
b
−
3
c
is
Single correct
EASY
3 marks
30 November 2025
77
A value of c for which the minimum value of
f
(
x
)
=
x
2
−
4
c
x
+
8
c
f(x) = x^2 - 4cx + 8c
f
(
x
)
=
x
2
−
4
c
x
+
8
c
is greater than the maximum value of
g
(
x
)
=
−
x
2
+
3
c
x
−
2
c
g(x) = -x^2 + 3cx - 2c
g
(
x
)
=
−
x
2
+
3
c
x
−
2
c
, is
Single correct
HARD
3 marks
30 November 2025
78
In the set of consecutive odd numbers {1, 3, 5, ….., 57}, there is a number of
k
k
k
such that the sum of all the elements less than
k
k
k
is equal to the sum of all the elements greater than
k
k
k
. Then,
k
k
k
equals.
Single correct
EASY
3 marks
30 November 2025
79
For any natural number k, let
a
k
=
3
k
a_k = 3^k
a
k
=
3
k
. The smallest natural number
m
m
m
for which
{
(
a
1
)
1
×
(
a
2
)
2
×
…
.
×
(
a
20
)
20
}
<
{
a
21
×
a
22
×
…
.
.
×
a
(
20
+
m
)
}
\{(a_1)^1 \times (a_2)^2 \times ….\times (a_{20})^{20}\} < \{a_{21} \times a_{22} \times ….. \times a_{(20 + m)}\}
{(
a
1
)
1
×
(
a
2
)
2
×
…
.
×
(
a
20
)
20
}
<
{
a
21
×
a
22
×
…
..
×
a
(
20
+
m
)
}
, is
Single correct
HARD
3 marks
30 November 2025
80
The equations
3
x
2
−
5
x
+
p
=
0
3x^2-5x + p = 0
3
x
2
−
5
x
+
p
=
0
and
2
x
2
−
2
x
+
q
=
0
2x^2 - 2x + q = 0
2
x
2
−
2
x
+
q
=
0
have one common root. The sum of the other roots of these two equations is
Single correct
HARD
3 marks
30 November 2025
81
If
log
64
x
2
+
log
8
y
+
3
log
512
(
z
y
)
=
4
\log_{64} x^2 + \log_8 y + 3\log_{512}(z\sqrt y) = 4
lo
g
64
x
2
+
lo
g
8
y
+
3
lo
g
512
(
z
y
)
=
4
, where x, y and z are positive real numbers, then the minimum possible value of (x + y + x) is
Single correct
HARD
3 marks
30 November 2025
82
The set of all real values of x for which
(
x
2
−
∣
x
+
9
∣
+
x
)
>
0
(x^2 - |x + 9| + x) > 0
(
x
2
−
∣
x
+
9∣
+
x
)
>
0
, is
Single correct
HARD
3 marks
30 November 2025
83
Let
a
n
a_n
a
n
be the nth term of a decreasing infinite geometric progression. If
a
1
+
a
2
+
a
3
=
52
a_1+a_2+a_3=52
a
1
+
a
2
+
a
3
=
52
and
a
1
a
2
+
a
2
a
3
+
a
3
a
1
=
624
a_1a_2+a_2a_3+a_3a_1=624
a
1
a
2
+
a
2
a
3
+
a
3
a
1
=
624
, then the sum of this geometric progression is
Single correct
HARD
3 marks
30 November 2025
84
Let
f
(
x
)
=
x
2
x
−
1
f(x)=\dfrac{x}{2x-1}
f
(
x
)
=
2
x
−
1
x
and
g
(
x
)
=
x
x
−
1
g(x)=\dfrac{x}{x-1}
g
(
x
)
=
x
−
1
x
. Then, the domain of the function
h
(
x
)
=
f
(
g
(
x
)
)
+
g
(
f
(
x
)
)
h(x) = f(g(x)) + g(f(x))
h
(
x
)
=
f
(
g
(
x
))
+
g
(
f
(
x
))
is all real numbers except
Single correct
HARD
3 marks
30 November 2025
85
If a, b, c and d are integers such that their sum is 46, then the minimum possible value of
(
a
−
b
)
2
+
(
a
−
c
)
2
+
(
a
−
d
)
2
(a - b)^2 + (a- c)^2 + (a - d)^2
(
a
−
b
)
2
+
(
a
−
c
)
2
+
(
a
−
d
)
2
is
Numerical
MEDIUM
3 marks
30 November 2025
86
If
9
x
2
+
2
x
−
3
−
4
⋅
3
x
2
+
2
x
−
2
+
27
=
0
9^{x^2+2x-3} - 4\cdot3^{x^2+2x-2} + 27 = 0
9
x
2
+
2
x
−
3
−
4
⋅
3
x
2
+
2
x
−
2
+
27
=
0
, then the product of all possible values of x is
Single correct
HARD
3 marks
30 November 2025
87
If m and n are integers such that
(
m
+
2
n
)
(
2
m
+
n
)
=
27
(m + 2n)(2m + n) = 27
(
m
+
2
n
)
(
2
m
+
n
)
=
27
, then the maximum possible value of
2
m
−
3
n
2m - 3n
2
m
−
3
n
is
Numerical
HARD
3 marks
30 November 2025
88
Direction of the question: If
x
2
+
1
x
2
=
25
x^2 + \dfrac{1}{x^2} = 25
x
2
+
x
2
1
=
25
and
x
>
0
x > 0
x
>
0
, then the value of
x
7
+
1
x
7
x^7 + \dfrac{1}{x^7}
x
7
+
x
7
1
is
Single correct
HARD
3 marks
30 November 2025
89
Direction of the question: For real values of
x
x
x
, the range of the function
f
(
x
)
=
2
x
−
3
2
x
2
+
4
x
−
6
f(x) = \dfrac{2x-3}{2x^2+4x-6}
f
(
x
)
=
2
x
2
+
4
x
−
6
2
x
−
3
is
Single correct
HARD
3 marks
30 November 2025
90
Direction of the question: For a 4-digit number (greater than 1000), sum of the digits in the thousands, hundreds, and tens places is 15. Sum of the digits in the hundreds, tens, and units places is 16. Also, the digit in the tens place is 6 more than the digit in the units place. The difference between the largest and smallest possible value of the number is
Single correct
MEDIUM
3 marks
30 November 2025
91
Direction of the question: If
f
(
x
)
=
(
x
2
+
3
x
)
(
x
2
+
3
x
+
2
)
f(x)=(x^2+3x)(x^2+3x+2)
f
(
x
)
=
(
x
2
+
3
x
)
(
x
2
+
3
x
+
2
)
, then the sum of all real roots of the equation
f
(
x
)
+
1
=
9701
\sqrt{f(x)+1}=9701
f
(
x
)
+
1
=
9701
, is
Single correct
MEDIUM
3 marks
30 November 2025
92
Direction of the question: In an arithmetic progression, if the sum of fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is
Numerical
MEDIUM
3 marks
30 November 2025
93
Direction of the question: Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose value lies between 150 and 500. If p is not less than 0.3q and not more than 0.7q, then the sum of the maximum and minimum possible values of p is
Numerical
HARD
3 marks
30 November 2025
94
Direction of the question: If
12
12
x
×
4
24
x
+
12
×
5
2
y
=
84
z
×
20
12
x
×
243
3
x
−
6
12^{12x} \times 4^{24x+12} \times 5^{2y} = 84^z \times 20^{12x} \times 243^{3x-6}
1
2
12
x
×
4
24
x
+
12
×
5
2
y
=
8
4
z
×
2
0
12
x
×
24
3
3
x
−
6
, where x, y and z are natural numbers, then x + y + z equals
Numerical
VERY_HARD
3 marks
30 November 2025
95
Direction of the question: In a school with 1500 students, each student chooses any one of the streams out of science, arts, and commerce, by paying a fee of Rs 1100, Rs 1000, and Rs 800, respectively. The total fee paid by all the students is Rs 15,50,000. If the number of science students is not more than the number of arts students, then the maximum possible number of science students in the school is
Numerical
HARD
3 marks
30 November 2025
96
Direction of the question: The sum of all possible real values of x for which
log
x
−
3
(
x
2
−
9
)
=
log
x
−
3
(
x
+
1
)
+
2
\log_{x-3}(x^2-9) = \log_{x-3}(x+1) + 2
lo
g
x
−
3
(
x
2
−
9
)
=
lo
g
x
−
3
(
x
+
1
)
+
2
, is
Single correct
VERY_HARD
3 marks
30 November 2025
Showing 96 questions.
Quantitative Aptitude > Algebra
All PYQs
Topic-Wise PYQs
Start Weekly Test
Type:
All
Difficulty:
All
Year:
All
96Q
01
If a and b are integers of opposite signs such that
(
a
+
3
)
2
:
b
2
=
9
:
1
(a + 3)^2 : b^2 = 9 : 1
(
a
+
3
)
2
:
b
2
=
9
:
1
and
(
a
−
1
)
2
:
(
b
−
1
)
2
=
4
:
1
(a - 1)^2 : (b - 1)^2 = 4 : 1
(
a
−
1
)
2
:
(
b
−
1
)
2
=
4
:
1
, then the ratio a : b is
Single correct
HARD
3 marks
26 November 2017
02
Suppose,
log
3
x
=
log
12
y
=
a
\log_3 x = \log_{12} y = a
lo
g
3
x
=
lo
g
12
y
=
a
, where x, y are positive numbers. If G is the geometric mean of x and y, and
log
6
G
\log_6 G
lo
g
6
G
is equal to
Single correct
MEDIUM
3 marks
26 November 2017
03
If
x
+
1
=
x
2
x + 1 = x^2
x
+
1
=
x
2
and
x
>
0
x > 0
x
>
0
, then
2
x
4
2x^4
2
x
4
is
Single correct
HARD
3 marks
26 November 2017
04
The value of
log
0.008
5
+
log
3
81
−
7
\log_{0.008}\sqrt5 + \log_{\sqrt3}81 - 7
lo
g
0.008
5
+
lo
g
3
81
−
7
is equal to
Single correct
HARD
3 marks
26 November 2017
05
If
9
2
x
−
1
−
81
x
−
1
=
1944
9^{2x-1} - 81^{x-1} = 1944
9
2
x
−
1
−
8
1
x
−
1
=
1944
, then x is
Single correct
MEDIUM
3 marks
26 November 2017
06
The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is
Single correct
HARD
3 marks
26 November 2017
07
For how many integers n, will the inequality (n – 5) (n – 10) – 3(n – 2) ≤ 0 be satisfied?
Numerical
MEDIUM
3 marks
26 November 2017
08
If
f
1
(
x
)
=
x
2
+
11
x
+
n
f_1(x) = x^2 + 11x + n
f
1
(
x
)
=
x
2
+
11
x
+
n
and
f
2
(
x
)
=
x
f_2(x) = x
f
2
(
x
)
=
x
, then the largest positive integer n for which the equation
f
1
(
x
)
=
f
2
(
x
)
f_1(x) = f_2(x)
f
1
(
x
)
=
f
2
(
x
)
has two distinct real roots, is
Numerical
MEDIUM
3 marks
26 November 2017
09
If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of
(
a
−
b
)
2
+
(
a
−
c
)
2
+
(
a
−
d
)
2
(a - b)^2 + (a - c)^2 + (a - d)^2
(
a
−
b
)
2
+
(
a
−
c
)
2
+
(
a
−
d
)
2
is
Numerical
MEDIUM
3 marks
26 November 2017
10
If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Single correct
MEDIUM
3 marks
26 November 2017
11
If
f
(
x
)
=
5
x
+
2
3
x
−
5
f(x)=\dfrac{5x+2}{3x-5}
f
(
x
)
=
3
x
−
5
5
x
+
2
and
g
(
x
)
=
x
2
−
2
x
−
1
g(x)=x^2-2x-1
g
(
x
)
=
x
2
−
2
x
−
1
, then the value of
g
(
f
(
f
(
3
)
)
)
g(f(f(3)))
g
(
f
(
f
(
3
)))
is
Single correct
HARD
3 marks
26 November 2017
12
Let
a
1
,
a
2
,
…
…
,
a
3
n
a_1, a_2,……,a_{3n}
a
1
,
a
2
,
……
,
a
3
n
be an arithmetic progression with
a
1
=
3
a_1 = 3
a
1
=
3
and
a
2
=
7
a_2 = 7
a
2
=
7
. If
a
1
+
a
2
+
…
+
a
3
n
=
1830
a_1 + a_2 + …+a_{3n} = 1830
a
1
+
a
2
+
…
+
a
3
n
=
1830
, then what is the smallest positive integer m such that
m
(
a
1
+
a
2
+
…
+
a
n
)
>
1830
m(a_1 + a_2 + … + a_n) > 1830
m
(
a
1
+
a
2
+
…
+
a
n
)
>
1830
?
Numerical
HARD
3 marks
26 November 2017
13
Direction of the question: Solve the following question and mark the best possible option. If three sides of a rectangular park have a total length 400 ft, then the area of the park is maximum when the length (in ft) of its longer side is
Numerical
MEDIUM
3 marks
26 November 2017
14
Direction of the question: Solve the following question and mark the best possible option. If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is
Single correct
MEDIUM
3 marks
26 November 2017
15
Direction of the question: Solve the following question and mark the best possible option. If x is a real number such that
log
3
5
=
log
5
(
2
+
x
)
\log_3 5 = \log_5(2 + x)
lo
g
3
5
=
lo
g
5
(
2
+
x
)
, then which of the following is true?
Single correct
HARD
3 marks
26 November 2017
16
Direction of the question: Solve the following question and mark the best possible option. Let
f
(
x
)
=
x
2
f(x) = x^2
f
(
x
)
=
x
2
and
g
(
x
)
=
2
x
g(x) = 2^x
g
(
x
)
=
2
x
, for all real
x
x
x
. Then the value of
f
(
f
(
g
(
x
)
)
+
g
(
f
(
x
)
)
)
f(f(g(x)) + g(f(x)))
f
(
f
(
g
(
x
))
+
g
(
f
(
x
)))
at
x
=
1
x = 1
x
=
1
is
Single correct
MEDIUM
3 marks
26 November 2017
17
Direction of the question: Solve the following question and mark the best possible option. The minimum possible value of the sum of the squares of the roots of the equation
x
2
+
(
a
+
3
)
x
−
(
a
+
5
)
=
0
x^2 + (a + 3)x - (a + 5) = 0
x
2
+
(
a
+
3
)
x
−
(
a
+
5
)
=
0
is
Single correct
MEDIUM
3 marks
26 November 2017
18
Direction of the question: Solve the following question and mark the best possible option. If
9
x
−
1
2
−
2
2
x
−
2
=
4
x
−
3
2
x
−
3
9^{x - \frac{1}{2}} - 2^{2x-2} = 4^x - 3^{2x-3}
9
x
−
2
1
−
2
2
x
−
2
=
4
x
−
3
2
x
−
3
, then
x
x
x
is
Single correct
VERY_HARD
3 marks
26 November 2017
19
Direction of the question: Solve the following question and mark the best possible option. If
log
(
2
a
×
3
b
×
5
c
)
\log(2^a \times 3^b \times 5^c)
lo
g
(
2
a
×
3
b
×
5
c
)
is the arithmetic mean of
log
(
2
2
×
3
3
×
5
)
\log(2^2 \times 3^3 \times 5)
lo
g
(
2
2
×
3
3
×
5
)
,
log
(
2
6
×
3
×
5
7
)
\log(2^6 \times 3 \times 5^7)
lo
g
(
2
6
×
3
×
5
7
)
, and
log
(
2
×
3
2
×
5
4
)
\log(2 \times 3^2 \times 5^4)
lo
g
(
2
×
3
2
×
5
4
)
, then
a
a
a
equals
Numerical
MEDIUM
3 marks
26 November 2017
20
Direction of the question: Solve the following question and mark the best possible option. Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3. If the sum of the numbers in the new sequence is 450, then a5 is
Numerical
MEDIUM
3 marks
26 November 2017
21
Direction of the question: Solve the following question and mark the best possible option. If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(1) is
Numerical
MEDIUM
3 marks
26 November 2017
22
Direction of the question: Solve the following question and mark the best possible option. Let f(x) = 2x - 5 and g(x) = 7 - 2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if
Single correct
HARD
3 marks
26 November 2017
23
Direction of the question: Solve the following question and mark the best possible option. An infinite geometric progression
a
1
,
a
2
,
a
3
,
.
.
.
a_1, a_2, a_3,...
a
1
,
a
2
,
a
3
,
...
has the property that
a
n
=
3
(
a
n
+
1
+
a
n
+
2
+
.
.
.
)
a_n = 3(a_{n+1} + a_{n+2} + ...)
a
n
=
3
(
a
n
+
1
+
a
n
+
2
+
...
)
for every
n
≥
1
n \ge 1
n
≥
1
. If the sum
a
1
+
a
2
+
a
3
+
.
.
.
.
=
32
a_1 + a_2 + a_3 + .... = 32
a
1
+
a
2
+
a
3
+
....
=
32
, then
a
5
a_5
a
5
is
Single correct
HARD
3 marks
26 November 2017
24
Direction of the question: Solve the following question and mark the best possible option. If a1 = 1/(2x5), a2 = 1/(5x8), a3 = 1/(8x11), ..., then a1 + a2 + a3 + ... + a100 is
Single correct
HARD
3 marks
26 November 2017
25
Direction of the question: Solve the following question and mark the best possible option. Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is
Single correct
MEDIUM
3 marks
25 November 2018
26
Direction of the question: Solve the following question and mark the best possible option. Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,... will be
Single correct
EASY
3 marks
25 November 2018
27
Direction of the question: Solve the following question and mark the best possible option. If x is a positive quantity such that
2
x
=
3
log
5
2
2^x = 3^{\log_5 2}
2
x
=
3
l
o
g
5
2
, then x is equal to
Single correct
MEDIUM
3 marks
25 November 2018
28
Direction of the question: Solve the following question and mark the best possible option. Given that
x
2018
y
2017
=
1
2
x^{2018}y^{2017}=\frac{1}{2}
x
2018
y
2017
=
2
1
and
x
2016
y
2019
=
8
x^{2016}y^{2019}=8
x
2016
y
2019
=
8
, the value of
x
2
+
y
3
x^2+y^3
x
2
+
y
3
is
Single correct
MEDIUM
3 marks
25 November 2018
29
Direction of the question: Solve the following question and key in your numerical answer. Let
f
(
x
)
=
min
{
2
x
2
,
52
−
5
x
}
f(x) = \min\{2x^2, 52-5x\}
f
(
x
)
=
min
{
2
x
2
,
52
−
5
x
}
, where x is any positive real number. Then the maximum possible value of f(x) is
Numerical
EASY
3 marks
25 November 2018
30
Direction of the question: Solve the following question and mark the best possible option. If
log
12
81
=
p
\log_{12} 81 = p
lo
g
12
81
=
p
, then
3
(
4
−
p
4
+
p
)
3\left(\dfrac{4-p}{4+p}\right)
3
(
4
+
p
4
−
p
)
is equal to
Single correct
HARD
3 marks
25 November 2018
31
Direction of the question: Solve the following question and key in your numerical answer. If
f
(
x
+
2
)
=
f
(
x
)
+
f
(
x
+
1
)
f(x+2) = f(x) + f(x+1)
f
(
x
+
2
)
=
f
(
x
)
+
f
(
x
+
1
)
for all positive integers x, and
f
(
11
)
=
91
f(11) = 91
f
(
11
)
=
91
,
f
(
15
)
=
617
f(15) = 617
f
(
15
)
=
617
, then
f
(
10
)
f(10)
f
(
10
)
equals
Numerical
MEDIUM
3 marks
25 November 2018
32
Direction of the question: Solve the following question and key in your numerical answer. While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is
Numerical
EASY
3 marks
25 November 2018
33
Direction of the question: Solve the following question and mark the best possible option. If
u
2
+
(
u
−
2
v
−
1
)
2
=
−
4
v
(
u
+
v
)
u^2+(u-2v-1)^2 = -4v(u+v)
u
2
+
(
u
−
2
v
−
1
)
2
=
−
4
v
(
u
+
v
)
, then what is the value of
u
+
3
v
u+3v
u
+
3
v
?
Single correct
MEDIUM
3 marks
25 November 2018
34
Direction of the question: Solve the following question and mark the best possible option. If
log
2
(
5
+
log
3
a
)
=
3
\log_2(5+\log_3 a) = 3
lo
g
2
(
5
+
lo
g
3
a
)
=
3
and
log
5
(
4
a
+
12
+
log
2
b
)
=
3
\log_5(4a+12+\log_2 b) = 3
lo
g
5
(
4
a
+
12
+
lo
g
2
b
)
=
3
, then
a
+
b
a+b
a
+
b
is equal to
Single correct
MEDIUM
3 marks
25 November 2018
35
Direction of the question: Solve the following question and mark the best possible option. How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?
Single correct
MEDIUM
3 marks
25 November 2018
36
Direction of the question: Solve the following question and mark the best possible option. If the sum of squares of two numbers is 97, then which one of the following cannot be their product?
Single correct
MEDIUM
3 marks
25 November 2018
37
Direction of the question: Solve the following question and key in your numerical answer. If a and b are integers such that
2
x
2
−
a
x
+
2
>
0
2x^2 - ax + 2 > 0
2
x
2
−
a
x
+
2
>
0
and
x
2
−
b
x
+
8
≥
0
x^2 - bx + 8 \geq 0
x
2
−
b
x
+
8
≥
0
for all real numbers x, then the largest possible value of
2
a
−
6
b
2a-6b
2
a
−
6
b
is
Numerical
MEDIUM
3 marks
25 November 2018
38
Direction of the question: Solve the following question and mark the best possible option. The value of the sum
7
×
11
+
11
×
15
+
15
×
19
+
.
.
.
+
95
×
99
7 \times 11 + 11 \times 15 + 15 \times 19 + ... + 95 \times 99
7
×
11
+
11
×
15
+
15
×
19
+
...
+
95
×
99
is
Single correct
HARD
3 marks
25 November 2018
39
Direction of the question: Solve the following question and mark the best possible option. The smallest integer n for which
4
n
>
17
19
4^n > 17^{19}
4
n
>
1
7
19
holds, is closest to
Single correct
MEDIUM
3 marks
25 November 2018
40
Direction of the question: Solve the following question and mark the best possible option. If
p
3
=
q
4
=
r
5
=
s
6
p^3 = q^4 = r^5 = s^6
p
3
=
q
4
=
r
5
=
s
6
, then the value of
log
s
(
p
q
r
)
\log_s(pqr)
lo
g
s
(
pq
r
)
is equal to
Single correct
MEDIUM
3 marks
25 November 2018
41
Direction of the question: Solve the following question and key in your numerical answer. Let
f
(
x
)
=
max
{
5
x
,
52
−
2
x
2
}
f(x)=\max\{5x, 52-2x^2\}
f
(
x
)
=
max
{
5
x
,
52
−
2
x
2
}
, where x is any positive real number. Then the minimum possible value of f(x) is
Numerical
MEDIUM
3 marks
25 November 2018
42
Direction of the question: Solve the following question and mark the best possible option.
1
log
2
100
−
1
log
4
100
+
1
log
5
100
−
1
log
10
100
+
1
log
20
100
−
1
log
25
100
+
1
log
50
100
=
?
\frac{1}{\log_2 100} - \frac{1}{\log_4 100} + \frac{1}{\log_5 100} - \frac{1}{\log_{10} 100} + \frac{1}{\log_{20} 100} - \frac{1}{\log_{25} 100} + \frac{1}{\log_{50} 100} = ?
l
o
g
2
100
1
−
l
o
g
4
100
1
+
l
o
g
5
100
1
−
l
o
g
10
100
1
+
l
o
g
20
100
1
−
l
o
g
25
100
1
+
l
o
g
50
100
1
=
?
Single correct
HARD
3 marks
25 November 2018
43
Direction of the question: Solve the following question and key in your numerical answer. Let
t
1
,
t
2
,
.
.
.
t_1, t_2, ...
t
1
,
t
2
,
...
be real numbers such that
t
1
+
t
2
+
.
.
.
+
t
n
=
2
n
2
+
9
n
+
13
t_1+t_2+...+t_n = 2n^2+9n+13
t
1
+
t
2
+
...
+
t
n
=
2
n
2
+
9
n
+
13
, for every positive integer n≥2. If
t
k
=
103
t_k=103
t
k
=
103
, then k equals
Numerical
MEDIUM
3 marks
25 November 2018
44
Direction of the question: Solve the following question and key in your numerical answer. The smallest integer n such that
n
3
−
11
n
2
+
32
n
−
28
>
0
n^3 - 11n^2 + 32n - 28 > 0
n
3
−
11
n
2
+
32
n
−
28
>
0
is
Numerical
MEDIUM
3 marks
25 November 2018
45
Direction of the question: Solve the following question and mark the best possible option. Let
a
1
,
a
2
,
.
.
.
,
a
52
a_1, a_2, ..., a_{52}
a
1
,
a
2
,
...
,
a
52
be positive integers such that
a
1
<
a
2
<
.
.
.
<
a
52
a_1<a_2<...<a_{52}
a
1
<
a
2
<
...
<
a
52
. Suppose, their arithmetic mean is one less than the arithmetic mean of
a
2
,
a
3
,
.
.
.
,
a
52
a_2, a_3, ..., a_{52}
a
2
,
a
3
,
...
,
a
52
. If
a
52
=
100
a_{52}=100
a
52
=
100
, then the largest possible value of
a
1
a_1
a
1
is
Single correct
HARD
3 marks
25 November 2018
46
Direction of the question: Solve the following question and key in your numerical answer. If N and x are positive integers such that
N
N
=
2
160
N^N = 2^{160}
N
N
=
2
160
and
N
2
+
2
N
N^2+2^N
N
2
+
2
N
is an integral multiple of
2
x
2^x
2
x
, then the largest possible x is
Numerical
HARD
3 marks
25 November 2018
47
If the equations
x
2
+
m
x
+
9
=
0
x^2 + mx + 9 = 0
x
2
+
m
x
+
9
=
0
,
x
2
+
n
x
+
17
=
0
x^2 + nx + 17 = 0
x
2
+
n
x
+
17
=
0
and
x
2
+
(
m
+
n
)
x
+
35
=
0
x^2 + (m + n) x + 35 = 0
x
2
+
(
m
+
n
)
x
+
35
=
0
have a common negative root, then the value of (2m + 3m) is
Numerical
VERY_HARD
3 marks
24 November 2024
48
If x is a positive real number such that
4
log
10
x
+
4
log
100
x
+
8
log
1000
x
=
13
4\log_{10} x + 4\log_{100} x + 8\log_{1000} x = 13
4
lo
g
10
x
+
4
lo
g
100
x
+
8
lo
g
1000
x
=
13
, then greatest integer not exceeding x, is
Numerical
HARD
3 marks
24 November 2024
49
Let x, y, and z be real numbers satisfying
4
(
x
2
+
y
2
+
z
2
)
=
a
4(x^2 + y^2 + z^2) = a
4
(
x
2
+
y
2
+
z
2
)
=
a
4
(
x
−
y
−
z
)
=
3
+
a
4(x - y - z) = 3 + a
4
(
x
−
y
−
z
)
=
3
+
a
Then a equals
Single correct
VERY_HARD
3 marks
24 November 2024
50
A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional 3 kg of these grains to the second customer. Finally, when the shop sells half of the remaining quantity and an additional 3 kg of these grains to the third customer, there are no grains left. The initial quantity, in kg, of grains is
Single correct
MEDIUM
3 marks
24 November 2024
51
If
(
a
+
b
n
)
(a + b\sqrt{n})
(
a
+
b
n
)
is the positive square root of
(
29
−
12
5
)
(29 - 12\sqrt{5})
(
29
−
12
5
)
, where a and b are integers, and n is a natural number, then the maximum possible value of (a + b + n) is
Single correct
VERY_HARD
3 marks
24 November 2024
52
The sum of all real values of k for which
(
1
8
)
k
×
(
1
32768
)
1
3
=
1
8
×
(
1
32768
)
1
k
\left(\frac{1}{8}\right)^{k} \times \left(\frac{1}{32768}\right)^{\frac{1}{3}} = \frac{1}{8} \times \left(\frac{1}{32768}\right)^{\frac{1}{k}}
(
8
1
)
k
×
(
32768
1
)
3
1
=
8
1
×
(
32768
1
)
k
1
, is
Single correct
HARD
3 marks
24 November 2024
53
Suppose
x
1
,
x
2
,
x
3
,
…
,
x
100
x_1, x_2, x_3, \ldots, x_{100}
x
1
,
x
2
,
x
3
,
…
,
x
100
are in arithmetic progression such that
x
5
=
−
4
x_5 = -4
x
5
=
−
4
and
2
x
6
+
2
x
9
=
x
11
+
x
13
2x_6 + 2x_9 = x_{11} + x_{13}
2
x
6
+
2
x
9
=
x
11
+
x
13
. Then,
x
100
x_{100}
x
100
equals.
Single correct
HARD
3 marks
24 November 2024
54
For any natural number n, let
a
n
a_n
a
n
be the largest integer not exceeding
n
\sqrt{n}
n
. Then the value of
a
1
+
a
2
+
…
+
a
50
a_1+a_2+\ldots+a_{50}
a
1
+
a
2
+
…
+
a
50
is
Numerical
HARD
3 marks
24 November 2024
55
If
(
x
+
6
2
)
1
/
2
−
(
x
−
6
2
)
1
/
2
=
2
2
\left(x+6\sqrt{2}\right)^{1/2} - \left(x-6\sqrt{2}\right)^{1/2} = 2\sqrt{2}
(
x
+
6
2
)
1/2
−
(
x
−
6
2
)
1/2
=
2
2
, then x equals
Numerical
HARD
3 marks
24 November 2024
56
A function
f
f
f
maps the set of natural numbers to whole numbers, such that
f
(
x
y
)
=
f
(
x
)
f
(
y
)
+
f
(
x
)
+
f
(
y
)
f(xy) = f(x)f(y) + f(x) + f(y)
f
(
x
y
)
=
f
(
x
)
f
(
y
)
+
f
(
x
)
+
f
(
y
)
for all
x
,
y
x, y
x
,
y
and
f
(
p
)
=
1
f(p) = 1
f
(
p
)
=
1
for every prime number
p
p
p
. Then, the value of
f
(
160000
)
f(160000)
f
(
160000
)
is
Single correct
HARD
3 marks
24 November 2024
57
The sum of the infinite series
1
5
(
1
5
−
1
7
)
+
(
1
5
)
2
[
(
1
5
)
2
−
(
1
7
)
2
]
+
(
1
5
)
3
[
(
1
5
)
3
−
(
1
7
)
3
]
+
⋯
\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\left(\dfrac{1}{5}\right)^2\left[\left(\dfrac{1}{5}\right)^2-\left(\dfrac{1}{7}\right)^2\right]+\left(\dfrac{1}{5}\right)^3\left[\left(\dfrac{1}{5}\right)^3-\left(\dfrac{1}{7}\right)^3\right]+\cdots
5
1
(
5
1
−
7
1
)
+
(
5
1
)
2
[
(
5
1
)
2
−
(
7
1
)
2
]
+
(
5
1
)
3
[
(
5
1
)
3
−
(
7
1
)
3
]
+
⋯
is equal to
Single correct
VERY_HARD
3 marks
24 November 2024
58
If
x
x
x
and
y
y
y
satisfy the equations
∣
x
∣
+
x
+
y
=
15
|x| + x + y = 15
∣
x
∣
+
x
+
y
=
15
and
x
+
∣
y
∣
−
y
=
20
x + |y| - y = 20
x
+
∣
y
∣
−
y
=
20
, then
(
x
−
y
)
(x-y)
(
x
−
y
)
equals
Single correct
HARD
3 marks
24 November 2024
59
The roots
α
,
β
\alpha, \beta
α
,
β
of the equation
3
x
2
+
λ
x
−
1
=
0
3x^2 + \lambda x - 1 = 0
3
x
2
+
λ
x
−
1
=
0
, satisfy
1
α
2
+
1
β
2
=
15
\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}=15
α
2
1
+
β
2
1
=
15
. The value of
(
α
3
+
β
3
)
2
(\alpha^3+\beta^3)^2
(
α
3
+
β
3
)
2
, is
Single correct
HARD
3 marks
24 November 2024
60
All the values of
x
x
x
satisfying the inequality
1
x
+
5
≤
1
2
x
−
3
\dfrac{1}{x+5} \le \dfrac{1}{2x-3}
x
+
5
1
≤
2
x
−
3
1
are
Single correct
HARD
3 marks
24 November 2024
61
If
a
,
b
a, b
a
,
b
and
c
c
c
are positive real numbers such that
a
>
10
≥
b
≥
c
a > 10 \ge b \ge c
a
>
10
≥
b
≥
c
and
log
8
(
a
+
b
)
log
2
c
+
log
27
(
a
−
b
)
log
3
c
=
2
3
\dfrac{\log_8(a+b)}{\log_2 c} + \dfrac{\log_{27}(a-b)}{\log_3 c} = \dfrac23
lo
g
2
c
lo
g
8
(
a
+
b
)
+
lo
g
3
c
lo
g
27
(
a
−
b
)
=
3
2
, then the greatest possible integer value of
a
a
a
is
Numerical
VERY_HARD
3 marks
24 November 2024
62
If
x
x
x
and
y
y
y
are real numbers such that
4
x
2
+
4
y
2
−
4
x
y
−
6
y
+
3
=
0
4x^2 + 4y^2 - 4xy - 6y + 3 = 0
4
x
2
+
4
y
2
−
4
x
y
−
6
y
+
3
=
0
, then the value of
(
4
x
+
5
y
)
(4x + 5y)
(
4
x
+
5
y
)
is
Numerical
HARD
3 marks
24 November 2024
63
If
3
a
=
4
3^a = 4
3
a
=
4
,
4
b
=
5
4^b = 5
4
b
=
5
,
5
c
=
6
5^c = 6
5
c
=
6
,
6
d
=
7
6^d = 7
6
d
=
7
,
7
e
=
8
7^e = 8
7
e
=
8
and
8
f
=
9
8^f = 9
8
f
=
9
, then the value of the product
a
b
c
d
e
f
abcdef
ab
c
d
e
f
is
Numerical
MEDIUM
3 marks
24 November 2024
64
The number of distinct integer solutions
(
x
,
y
)
(x, y)
(
x
,
y
)
of the equation
∣
x
+
y
∣
+
∣
x
−
y
∣
=
2
|x + y| + |x - y| = 2
∣
x
+
y
∣
+
∣
x
−
y
∣
=
2
, is
Numerical
MEDIUM
3 marks
24 November 2024
65
If
(
a
+
b
3
)
2
=
52
+
30
3
(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}
(
a
+
b
3
)
2
=
52
+
30
3
, where
a
a
a
and
b
b
b
are natural numbers, then
a
+
b
a + b
a
+
b
equals
Single correct
MEDIUM
3 marks
24 November 2024
66
The sum of all distinct real values of
x
x
x
that satisfy the equation
10
x
+
4
10
x
=
91
2
10^x + \dfrac{4}{10^x} = \dfrac{91}{2}
1
0
x
+
1
0
x
4
=
2
91
, is
Single correct
HARD
3 marks
24 November 2024
67
The number of distinct real values of
x
x
x
, satisfying the equation
max
{
x
,
2
}
−
min
{
x
,
2
}
=
∣
x
+
2
∣
−
∣
x
−
2
∣
\max\{x, 2\} - \min\{x, 2\} = |x + 2| - |x - 2|
max
{
x
,
2
}
−
min
{
x
,
2
}
=
∣
x
+
2∣
−
∣
x
−
2∣
, is
Numerical
HARD
3 marks
24 November 2024
68
For any non-zero real number
x
x
x
, let
f
(
x
)
+
2
f
(
1
x
)
=
3
x
f(x) + 2f\left(\dfrac{1}{x}\right) = 3x
f
(
x
)
+
2
f
(
x
1
)
=
3
x
. Then, the sum of all possible values of
x
x
x
for which
f
(
x
)
=
3
f(x) = 3
f
(
x
)
=
3
, is
Single correct
HARD
3 marks
24 November 2024
69
Consider the sequence
t
1
=
1
t_1 = 1
t
1
=
1
,
t
2
=
−
1
t_2 = -1
t
2
=
−
1
and
t
n
=
(
n
−
3
n
−
1
)
t
n
−
2
t_n = \left(\dfrac{n-3}{n-1}\right)t_{n-2}
t
n
=
(
n
−
1
n
−
3
)
t
n
−
2
for
n
≥
3
n \ge 3
n
≥
3
. Then, the value of the sum
1
t
2
+
1
t
4
+
1
t
6
+
…
+
1
t
2022
+
1
t
2024
\dfrac{1}{t_2} + \dfrac{1}{t_4} + \dfrac{1}{t_6} + \ldots + \dfrac{1}{t_{2022}} + \dfrac{1}{t_{2024}}
t
2
1
+
t
4
1
+
t
6
1
+
…
+
t
2022
1
+
t
2024
1
, is
Single correct
VERY_HARD
3 marks
24 November 2024
70
For some constant real numbers
p
p
p
,
k
k
k
and
a
a
a
consider the following system of linear equations in
x
x
x
and
y
y
y
:
p
x
−
4
y
=
2
px - 4y = 2
p
x
−
4
y
=
2
3
x
+
k
y
=
a
3x + ky = a
3
x
+
k
y
=
a
A necessary condition for the system to have no solution for
(
x
,
y
)
(x, y)
(
x
,
y
)
is
Single correct
HARD
3 marks
24 November 2024
71
The number of distinct integers
n
n
n
for which
log
1
/
4
(
n
2
−
7
n
+
11
)
>
0
\log_{1/4}(n^2 - 7n + 11) > 0
lo
g
1/4
(
n
2
−
7
n
+
11
)
>
0
, is
Single correct
MEDIUM
3 marks
30 November 2025
72
The number of distinct pairs of integers
(
x
,
y
)
(x, y)
(
x
,
y
)
satisfying the inequalities
x
>
y
≥
3
x > y \geq 3
x
>
y
≥
3
and
x
+
y
<
14
x + y < 14
x
+
y
<
14
is
Numerical
MEDIUM
3 marks
30 November 2025
73
Let
3
≤
x
≤
6
3 \leq x \leq 6
3
≤
x
≤
6
and
[
x
2
]
=
[
x
]
2
[x^2] = [x]^2
[
x
2
]
=
[
x
]
2
, where
[
x
]
[x]
[
x
]
is the greatest integer not exceeding
x
x
x
. If set
S
S
S
represents all feasible values of
x
x
x
, then a possible subset of
S
S
S
is
Single correct
HARD
3 marks
30 November 2025
74
The number of non-negative integer values of
k
k
k
for which the quadratic equation
x
2
−
5
x
+
k
=
0
x^2 - 5x + k = 0
x
2
−
5
x
+
k
=
0
has only integer roots, is
Numerical
EASY
3 marks
30 November 2025
75
Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is
Numerical
EASY
3 marks
30 November 2025
76
If
a
−
6
b
+
6
c
=
4
a - 6b + 6c = 4
a
−
6
b
+
6
c
=
4
and
6
a
+
3
b
−
3
c
=
50
6a + 3b - 3c = 50
6
a
+
3
b
−
3
c
=
50
, where a, b and c are real numbers, the value of
2
a
+
3
b
−
3
c
2a + 3b - 3c
2
a
+
3
b
−
3
c
is
Single correct
EASY
3 marks
30 November 2025
77
A value of c for which the minimum value of
f
(
x
)
=
x
2
−
4
c
x
+
8
c
f(x) = x^2 - 4cx + 8c
f
(
x
)
=
x
2
−
4
c
x
+
8
c
is greater than the maximum value of
g
(
x
)
=
−
x
2
+
3
c
x
−
2
c
g(x) = -x^2 + 3cx - 2c
g
(
x
)
=
−
x
2
+
3
c
x
−
2
c
, is
Single correct
HARD
3 marks
30 November 2025
78
In the set of consecutive odd numbers {1, 3, 5, ….., 57}, there is a number of
k
k
k
such that the sum of all the elements less than
k
k
k
is equal to the sum of all the elements greater than
k
k
k
. Then,
k
k
k
equals.
Single correct
EASY
3 marks
30 November 2025
79
For any natural number k, let
a
k
=
3
k
a_k = 3^k
a
k
=
3
k
. The smallest natural number
m
m
m
for which
{
(
a
1
)
1
×
(
a
2
)
2
×
…
.
×
(
a
20
)
20
}
<
{
a
21
×
a
22
×
…
.
.
×
a
(
20
+
m
)
}
\{(a_1)^1 \times (a_2)^2 \times ….\times (a_{20})^{20}\} < \{a_{21} \times a_{22} \times ….. \times a_{(20 + m)}\}
{(
a
1
)
1
×
(
a
2
)
2
×
…
.
×
(
a
20
)
20
}
<
{
a
21
×
a
22
×
…
..
×
a
(
20
+
m
)
}
, is
Single correct
HARD
3 marks
30 November 2025
80
The equations
3
x
2
−
5
x
+
p
=
0
3x^2-5x + p = 0
3
x
2
−
5
x
+
p
=
0
and
2
x
2
−
2
x
+
q
=
0
2x^2 - 2x + q = 0
2
x
2
−
2
x
+
q
=
0
have one common root. The sum of the other roots of these two equations is
Single correct
HARD
3 marks
30 November 2025
81
If
log
64
x
2
+
log
8
y
+
3
log
512
(
z
y
)
=
4
\log_{64} x^2 + \log_8 y + 3\log_{512}(z\sqrt y) = 4
lo
g
64
x
2
+
lo
g
8
y
+
3
lo
g
512
(
z
y
)
=
4
, where x, y and z are positive real numbers, then the minimum possible value of (x + y + x) is
Single correct
HARD
3 marks
30 November 2025
82
The set of all real values of x for which
(
x
2
−
∣
x
+
9
∣
+
x
)
>
0
(x^2 - |x + 9| + x) > 0
(
x
2
−
∣
x
+
9∣
+
x
)
>
0
, is
Single correct
HARD
3 marks
30 November 2025
83
Let
a
n
a_n
a
n
be the nth term of a decreasing infinite geometric progression. If
a
1
+
a
2
+
a
3
=
52
a_1+a_2+a_3=52
a
1
+
a
2
+
a
3
=
52
and
a
1
a
2
+
a
2
a
3
+
a
3
a
1
=
624
a_1a_2+a_2a_3+a_3a_1=624
a
1
a
2
+
a
2
a
3
+
a
3
a
1
=
624
, then the sum of this geometric progression is
Single correct
HARD
3 marks
30 November 2025
84
Let
f
(
x
)
=
x
2
x
−
1
f(x)=\dfrac{x}{2x-1}
f
(
x
)
=
2
x
−
1
x
and
g
(
x
)
=
x
x
−
1
g(x)=\dfrac{x}{x-1}
g
(
x
)
=
x
−
1
x
. Then, the domain of the function
h
(
x
)
=
f
(
g
(
x
)
)
+
g
(
f
(
x
)
)
h(x) = f(g(x)) + g(f(x))
h
(
x
)
=
f
(
g
(
x
))
+
g
(
f
(
x
))
is all real numbers except
Single correct
HARD
3 marks
30 November 2025
85
If a, b, c and d are integers such that their sum is 46, then the minimum possible value of
(
a
−
b
)
2
+
(
a
−
c
)
2
+
(
a
−
d
)
2
(a - b)^2 + (a- c)^2 + (a - d)^2
(
a
−
b
)
2
+
(
a
−
c
)
2
+
(
a
−
d
)
2
is
Numerical
MEDIUM
3 marks
30 November 2025
86
If
9
x
2
+
2
x
−
3
−
4
⋅
3
x
2
+
2
x
−
2
+
27
=
0
9^{x^2+2x-3} - 4\cdot3^{x^2+2x-2} + 27 = 0
9
x
2
+
2
x
−
3
−
4
⋅
3
x
2
+
2
x
−
2
+
27
=
0
, then the product of all possible values of x is
Single correct
HARD
3 marks
30 November 2025
87
If m and n are integers such that
(
m
+
2
n
)
(
2
m
+
n
)
=
27
(m + 2n)(2m + n) = 27
(
m
+
2
n
)
(
2
m
+
n
)
=
27
, then the maximum possible value of
2
m
−
3
n
2m - 3n
2
m
−
3
n
is
Numerical
HARD
3 marks
30 November 2025
88
Direction of the question: If
x
2
+
1
x
2
=
25
x^2 + \dfrac{1}{x^2} = 25
x
2
+
x
2
1
=
25
and
x
>
0
x > 0
x
>
0
, then the value of
x
7
+
1
x
7
x^7 + \dfrac{1}{x^7}
x
7
+
x
7
1
is
Single correct
HARD
3 marks
30 November 2025
89
Direction of the question: For real values of
x
x
x
, the range of the function
f
(
x
)
=
2
x
−
3
2
x
2
+
4
x
−
6
f(x) = \dfrac{2x-3}{2x^2+4x-6}
f
(
x
)
=
2
x
2
+
4
x
−
6
2
x
−
3
is
Single correct
HARD
3 marks
30 November 2025
90
Direction of the question: For a 4-digit number (greater than 1000), sum of the digits in the thousands, hundreds, and tens places is 15. Sum of the digits in the hundreds, tens, and units places is 16. Also, the digit in the tens place is 6 more than the digit in the units place. The difference between the largest and smallest possible value of the number is
Single correct
MEDIUM
3 marks
30 November 2025
91
Direction of the question: If
f
(
x
)
=
(
x
2
+
3
x
)
(
x
2
+
3
x
+
2
)
f(x)=(x^2+3x)(x^2+3x+2)
f
(
x
)
=
(
x
2
+
3
x
)
(
x
2
+
3
x
+
2
)
, then the sum of all real roots of the equation
f
(
x
)
+
1
=
9701
\sqrt{f(x)+1}=9701
f
(
x
)
+
1
=
9701
, is
Single correct
MEDIUM
3 marks
30 November 2025
92
Direction of the question: In an arithmetic progression, if the sum of fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is
Numerical
MEDIUM
3 marks
30 November 2025
93
Direction of the question: Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose value lies between 150 and 500. If p is not less than 0.3q and not more than 0.7q, then the sum of the maximum and minimum possible values of p is
Numerical
HARD
3 marks
30 November 2025
94
Direction of the question: If
12
12
x
×
4
24
x
+
12
×
5
2
y
=
84
z
×
20
12
x
×
243
3
x
−
6
12^{12x} \times 4^{24x+12} \times 5^{2y} = 84^z \times 20^{12x} \times 243^{3x-6}
1
2
12
x
×
4
24
x
+
12
×
5
2
y
=
8
4
z
×
2
0
12
x
×
24
3
3
x
−
6
, where x, y and z are natural numbers, then x + y + z equals
Numerical
VERY_HARD
3 marks
30 November 2025
95
Direction of the question: In a school with 1500 students, each student chooses any one of the streams out of science, arts, and commerce, by paying a fee of Rs 1100, Rs 1000, and Rs 800, respectively. The total fee paid by all the students is Rs 15,50,000. If the number of science students is not more than the number of arts students, then the maximum possible number of science students in the school is
Numerical
HARD
3 marks
30 November 2025
96
Direction of the question: The sum of all possible real values of x for which
log
x
−
3
(
x
2
−
9
)
=
log
x
−
3
(
x
+
1
)
+
2
\log_{x-3}(x^2-9) = \log_{x-3}(x+1) + 2
lo
g
x
−
3
(
x
2
−
9
)
=
lo
g
x
−
3
(
x
+
1
)
+
2
, is
Single correct
VERY_HARD
3 marks
30 November 2025
Showing 96 questions.
