## Quantitative Aptitude Quiz for IBPS, SBI and other bank exams

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. 13y– 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. 13y– 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.