Quantitative Aptitude Quiz for IBPS, SBI and other bank exams
Directions (1 – 5): In the following question, two
equations are given. You have to solve these equations and find out the values
of x and y and determine the relation between them.
1. I. 17x2 + 48x = 9; II. 13y2 –
32y –21 = 0
A) x < y
B) x > y
C) x ≥ y
D) x ≤ y
E) x = y or the relationship cannot
be established
2. I. 10a2 – 21a – 10 = 0; II. 40b2
– 19b – 14 = 0
A) a < b
B) a > b
C) a = b OR the relationship cannot
be determined
D) a ≥ b
E) a ≤ b
3. I. 17x2 – 153x + 238 = 0; II. 2y2
+ 3y – 14 = 0
A) x < y
B) x > y
C) x = y OR the relationship cannot
be determined
D) x ≥ y
E) x ≤ y
4. I. 13x2 – 156x – 364 = 0; II. y2
– 6y – 16 = 0
A) x < y
B) x > y
C) x = y OR the relationship cannot
be determined
D) x ≥ y
E) x ≤ y
5. I. √(899x) + √1297 = 0; II. (255)1/4
y + (215)1/3 = 0
A) x > y
B) x ≥ y
C) x < y
D) x ≤ y
E) x = y Or the relationship cannot
be established.
6. Quantity A: A is 75% of B then what % of A is B?
Quantity B: Isaac scores 73 marks in Mathematics, 98 marks in
Physics and 75 in Chemistry. If the maximum marks in Mathematics, Physics and
Chemistry are 75,100,90 respectively. What will be the % of total marks
obtained with respect to the sum of maximum marks of Mathematics and Chemistry?
A) Quantity A > Quantity B
B) Quantity A < Quantity B
C) Quantity A ≥ Quantity B
D) Quantity A ≤ Quantity B
E) Quantity A = Quantity B
7. Quantity A: Walking at 5/4 of his usual pace, a man reaches his office 25
minutes earlier. Find his usual time.
Quantity B: Two horses are running in an opposite direction on a circular track of
9858 m. Speed of horses are 32 km/h and 42 km/h respectively. When will they
meet for the first time?
A) Quantity A > Quantity B
B) Quantity A < Quantity B
C) Quantity A ≥ Quantity B
D) Quantity A ≤ Quantity B
E) Quantity A = Quantity B
8. Quantity A: The mean of 45 observations was 40. It was found later that
an observation 36 was wrongly taken as 63. The corrected new mean is
Quantity B: The average age of a husband and his wife was 26 years at the time of
marriage. After five years they have a one-year child. The twice of average age
of the family now is
A) Quantity A > Quantity B
B) Quantity A < Quantity B
C) Quantity A ≥ Quantity B
D) Quantity A ≤ Quantity B
E) Quantity A = Quantity B
9. Based on the given information, determine the
relation between the two quantities
Quantity A: In a certain A.P., 5 times the 5th term is equal to 8 times the 8th term,
then the 13th term is:
Quantity B: That value of n, for which (an+1 + bn+1)/(an
+ bn) is the arithmetic mean of a and b?
A) Quantity A > Quantity B
B) Quantity A < Quantity B
C) Quantity A ≥ Quantity B
D) Quantity A ≤ Quantity B
E) Quantity A = Quantity B
10. Based on the given information, determine the
relation between the two quantities
A function is defined as f(x) = x2 –
3x. Then
Quantity A: The value of f(– 1).
Quantity B: The value of f(4).
A) Quantity A > Quantity B
B) Quantity A < Quantity B
C) Quantity A ≥ Quantity B
D) Quantity A ≤ Quantity B
E) Quantity A = Quantity B
Solutions:
1. E) I. 17x2 + 48x = 9
⇒ 17x2 + 48x – 9 = 0
⇒ 17x2 + 51x – 3x – 9 = 0
⇒ 17x(x + 3) – 3(X + 3) = 0
⇒ (x + 3)(17x – 3) = 0
Then, x = - 3 or x = +
3/17
II. 13y2 –
32y –21 = 0
⇒ 13y2 – 39y + 7y – 21 = 0
⇒ 13y( y – 3) + 7(y – 3) = 0
⇒ (y – 3)(13y + 7) =
0
Then, y = + 3 or y = -
7/13
So, when x = - 3, x <
y for y = + 3 and x < y for y = - 7/13
And when x = + 3/17, x
< y for y = + 3 and x > y for y = - 7/13
∴ So, we can observe that no clear relationship cannot be
determined between x and y.
2. C) I.
10a2 – 21a – 10 = 0
⇒ 10a2 – 25a + 4a – 10 = 0
⇒ 5a(2a – 5) + 2(2a – 5) = 0
⇒ (2a – 5)(5a + 2) =
0
Then, a = + 5/2 or a = -
2/5
II. 40b2 –
19b – 14 = 0
⇒ 40b2 – 35b + 16b – 2 = 0
⇒ 5b(8b – 7) + 2(8b – 7) = 0
⇒ (8b – 7)(5b + 2) =
0
Then, b = + 7/8 or b
= - 2/5
So, when a = + 5/2, a
> b for b = + 7/8 and a > b for b = - 2/5
And when a = - 2/5 , a
< b for b = + 7/8 and a = b for b = - 2/5
∴ So, we can observe that no clear relationship cannot be
determined between a and b.
3. D) I.
17x2 – 153x + 238 = 0
⇒ x2 – 9x + 14 = 0
[Dividing both sides by 17 ]
⇒ x2 – 7x – 2x + 14 = 0
⇒ x(x – 7) – 2(x – 7) = 0
⇒ (x – 7)(x – 2) = 0
Then, x = + 7 or x = + 2
II. 2y2 +
3y – 14 = 0
⇒ 2y2 – 4y + 7y – 14 = 0
⇒ 2y(y – 2) + 7(y – 2) = 0
⇒ (y – 2)(2y + 7) = 0
Then, y = + 2 or y = -
7/2
So, when x = + 7, x >
y for y = + 2 and x > y for y = - 7/2
And when x = + 2, x = y
for y = + 2 and x > y for y = - 7/2
∴ So, we can clearly observe that x ≥ y.
4. C) I.
13x2 – 156x – 364 = 0
⇒ x2 – 12x – 28 = 0 [Dividing both
sides by 13 ]
⇒ x2 – 14x + 2x – 28 = 0
⇒ x(x – 14) + 2(x – 14) = 0
⇒ (x – 14)(x + 2) =
0
Then, x = + 14 or x = - 2
II. y2 –
6y – 16 = 0
⇒ y2 – 8y + 2y – 16 = 0
⇒ y(y – 8) + 2(y – 8) = 0
⇒ (y – 8)(y + 2) = 0
Then, y = + 8 or y = - 2
So, when x = + 14, x >
y for y = + 8 and x > y for y = - 2
And when x = - 2, x <
y for y = + 8 and x = y for y = - 2
∴ So, we can observe that no clear relationship cannot be
determined between x and y.
5. A) According
to the given equations,
√(899x) + √1297 = 0
899 is not a perfect
square, but 900 is a perfect square of 30. Similarily, 1297 is not a perfect
square but 1296 is a perfect square of 36. So, we are going to consider √899 as
30 and √1297 as 36, because it is not going to affect the question.
30x + 36 = 0
30x = -36
x = -6/5
II. (255)1/4 y
+ (215)1/3 = 0
255 does not have a
perfect fourth root, but 256 is a perfect fourth root of 4. Similarily, 215
does not have a perfect cube root but 216 is a cube root of 6. So, we are going
to consider (255)1/4 as 4 and (215)1/3 as 6,
because it is not going to affect the question.
4y + 6 = 0
y = -6/4
y = -3/2
It is clearly seen that y
˂ x.
Hence, (A) is correct.
6. B) First
we will find Quantity A,
Quantity A:
A = 75% of B
⇒ A = 0.75B
⇒ B = A/0.75 = 1.33A = 133.33% of A
∴ B = 133.33% of A
Now,
Quantity B:
Sum of maximum marks of
Mathematics and Chemistry = 75 + 90 = 165
Total marks obtained by
Isaac = 73 + 98 + 75 = 246
Percentage of total marks
obtained with respect to the sum of maximum marks of Mathematics and Chemistry
= (246/165) × 100 = 149.09%
∴ Percentage of total marks obtained with respect to the sum
of maximum marks of Mathematics and Chemistry = 149.09%
∴ Quantity A < Quantity B
7. A) First
we will find Quantity A,
Quantity A:
Let the original speed
and time be s km/hr and t hrs respectively.
∵ Distance = Speed × Time
⇒ The distance travelled by the man = s × t = st km
At 5/4 of his usual
speed, the time taken by the man = t – (25/60) = [t – (5/12)]
hrs
(∵ 1 hour = 60 min)
⇒ The distance traveled = (5s/4) × [t – (5/12)] = 1.25st – (25s/48)
But the distance traveled
in both the cases will be the same.
⇒ st = 1.25st – (25s/48)
⇒ 0.25st = 25s/48
⇒ t = 100/48 = 2.08 hrs
∴ His usual time = 2.08 hrs
Now,
Quantity B:
Let both horses meet
after T minutes.
32000 m is covered by
first horse in 60
min. (∵ 1 km = 1000 m and 1 hr = 60 min)
Distance covered by the
first horse in T min = (32000 × T)/60
Likewise, Distance
covered by the second horse in T min = (42000 × T)/60
Given that the two horses
are running in an opposite direction on a circular track of 9858 m.
⇒32000×T/60+42000×T/60=9858
⇒ T ≈ 7.993 minutes
∴ The two horses will meet for the first time in 7.993
minutes.
∴ Quantity A > Quantity B
8. B) First
we will find Quantity A,
Quantity A:
Sum of 45 observations =
40 × 45 = 1800
Correct sum of the 45
observations = 1800 – 63 + 36 = 1773
Corrected mean = 1773/45
= 39.4
∴ Corrected mean = 39.4
Now,
Quantity B:
Sum of the ages of husband
and wife at time of marriage = 26 × 2 = 52
Sum of the ages of the
family after 5 years when they have a one-year old child = 52 + 5 + 5 + 1 = 63
Average age of the family
after 5 years when they have a one-year old child = 63/3 = 21
∴ Twice of average age of the family now = 21 × 2 = 42
∴ Quantity A < Quantity B
9. E) First
we will find Quantity A,
Quantity A:
We know that for an
A.P.(Arithmetic progression),
The nth term,
Tn = a + (n – 1)d
Where,
a = First term
d = Common difference
According to the
question,
5T5 = 8T8
⇒ 5[a + (5 – 1)d] = 8[a +
(8 – 1)d]
⇒ 5a + 20d = 8a + 56d
⇒ 3a + 36d = 0
⇒ 3(a + 12d) = 0
⇒ a + (13 – 1)d = 0
⇒ T13 = 0
∴ T13 = 0
Now,
Quantity B:
∵ Arithmetic mean of a and b is (a + b)/2.
(an+1 + bn+1)/(an +
bn) = (a + b)/2
⇒ 2(an+1 + bn+1) = (an +
bn)(a + b)
⇒ 2an+1 + 2bn+1 = an+1 +
anb + bna + bn+1
⇒ an+1 + bn+1 = anb
+ bna
⇒ an(a – b) = bn(a – b)
⇒ an = bn
⇒ (a/b)n = 1
⇒ (a/b)n = (a/b)0
(∵ m0 = 1)
∵ Bases are equal, equate the powers.
⇒ n = 0
∴ Quantity A = Quantity B.
10. E) Given,
f(x) = x2 –
3x
First we will find
Quantity A,
Quantity A:
f(– 1) = (– 1)2 –
3(–1) = 1 + 3 = 4
Quantity B:
f(4) = 42 –
3(4) = 16 – 12 = 4
∴ f(– 1) = 4 = f(4)
∴ Quantity A = Quantity B.
