Dear Students,
Q1. A box contains 4 Green, ‘x’ red and 2 blue balls. Find the probability of selecting two balls such that color of both balls will be same if it is given that probability of selecting one red ball from the box is 1/3.
Q6. I. 15x² + 11x + 2 = 0
II. 24y² + 11y + 1 = 0
Q7. I. x² – 30x + 221 = 0
II. y² – 17y + 60 = 0
Q8. I. x² + 6x + 8 = 0
II. 8y² + 22y + 15 = 0
Q9. I. x² – 20x + 96 = 0
II. y² – 15y + 56 = 0
Q10. I. x² + 2x – 35 = 0
II. y² + 3y – 10 = 0
Directions (11-15): What comes at the place of question mark in the following number series?
Q11. 27, 1358, 1277, 1620, ? , 1622
Q12. 48, 72, 180, 810, ?, 69862.5
Q13. 8, 288, 512, 680, 792, ?
Quantitative Aptitude Quiz For SBI PO 2019
If talked about banking exams, Quant Section is considered to be one of the most difficult sections and so, gives heebie-jeebies to many. The questions asked in this section are calculative and very time-consuming and if you do not practice it well, it can make your blood run cold during the exam. Adda247 is here with practice questions on all the topics that are likely to be asked in the exam.
Q1. A box contains 4 Green, ‘x’ red and 2 blue balls. Find the probability of selecting two balls such that color of both balls will be same if it is given that probability of selecting one red ball from the box is 1/3.
1/3
2/3
4/9
13/18
5/18
Solution:
Q2. Ratio of A, B and C’s salary is 6 : 8 : 9 while ratio of A, B and C’s saving is 4 : 4 : 3. If A’s expenditure is 20% of his salary then find C’s expenditure is what percent of his salary? 60%
50%
40%
30%
20%
Solution:
Q3. Rahul invested money in scheme ‘A’ and scheme ‘B’ in the ratio 2 : 3. If scheme ‘A’ offer 10% p.a. at SI and scheme ‘B’ offer 10% at CI, then find the interest earned from Scheme ‘B’ is what percent more than interest earned from scheme ‘A’ after 2 years? 50%
52.5%
55%
57.5%
60%
Solution:
Q4. Sum of length of two train (A and B)is 540 m and ratio of speed of these two trains A and B is 1 : 2. If train A covers 90 m in 5 sec, then in what time they will cross each other when they travel in opposite direction. 11 seconds
8 seconds
12 seconds
10 seconds
15 seconds
Solution:
Q5. A boat takes 90 minutes less to travel 36 km downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 km/h then speed of the stream is: 4 km/h
3 km/h
2.5 km/h
2 km/h
3.5 km/h
Solution:
Directions (6-10): In each of these questions, two equations (I) and (II) are given. Solve both the equations and mark the correct option: Q6. I. 15x² + 11x + 2 = 0
II. 24y² + 11y + 1 = 0
if x>y
if x≥y
if x<y
if x≤y
if x = y or no relation can be established between x and y.
Solution:
I. 15x² + 5x + 6x + 2 = 0
5x (3x + 1) + 2 (3x + 1) = 0
(5x + 2) (3x + 1) = 0
x = -2/5, -1/3
II. 24y² + 8y + 3y + 1 = 0
8y (3y +1) + 1 (3y + 1) = 0
(8y + 1) (3y + 1) = 0
y = -1/3, -1/8
⇒ x ≤ y
5x (3x + 1) + 2 (3x + 1) = 0
(5x + 2) (3x + 1) = 0
x = -2/5, -1/3
II. 24y² + 8y + 3y + 1 = 0
8y (3y +1) + 1 (3y + 1) = 0
(8y + 1) (3y + 1) = 0
y = -1/3, -1/8
⇒ x ≤ y
Q7. I. x² – 30x + 221 = 0
II. y² – 17y + 60 = 0
if x>y
if x≥y
if x<y
if x≤y
if x = y or no relation can be established between x and y.
Solution:
I. x² – 13x – 17x + 221 = 0
x (x – 13) – 17 (x – 13) = 0
(x – 17) (x – 13) = 0
x = 13, 17
II. y² – 12y – 5y + 60 = 0
y (y – 12) – 5 (y – 12) = 0
(y – 5) (y – 12) = 0
y = 5, 12
⇒ x > y
x (x – 13) – 17 (x – 13) = 0
(x – 17) (x – 13) = 0
x = 13, 17
II. y² – 12y – 5y + 60 = 0
y (y – 12) – 5 (y – 12) = 0
(y – 5) (y – 12) = 0
y = 5, 12
⇒ x > y
Q8. I. x² + 6x + 8 = 0
II. 8y² + 22y + 15 = 0
if x>y
if x≥y
if x<y
if x ≤y
if x = y or no relation can be established between x and y.
Solution:
I. x² + 6x + 8 = 0
x² + 2x + 4x + 8 = 0
x (x + 2) + 4 (x + 2) = 0
(x + 4) (x + 2) = 0
x = –2, –4
II. 8y² + 22y + 15 = 0
8y² + 10y + 12y + 15 = 0
2y (4y + 5) +3(4y + 5) = 0
(2y + 3) (4y + 5) = 0
y=(-3)/2,-5/4
⇒ x < y
x² + 2x + 4x + 8 = 0
x (x + 2) + 4 (x + 2) = 0
(x + 4) (x + 2) = 0
x = –2, –4
II. 8y² + 22y + 15 = 0
8y² + 10y + 12y + 15 = 0
2y (4y + 5) +3(4y + 5) = 0
(2y + 3) (4y + 5) = 0
y=(-3)/2,-5/4
⇒ x < y
Q9. I. x² – 20x + 96 = 0
II. y² – 15y + 56 = 0
if x>y
if x≥y
if x<y
if x ≤y
if x = y or no relation can be established between x and y.
Solution:
I. x² – 20x + 96 = 0
x² – 8x – 12x + 96 = 0
x (x – 8) – 12 (x – 8) = 0
(x – 12) (x – 8) = 0
x = 12, 8
II. y² – 15y + 56 = 0
y² – 7y – 8y + 56 = 0
(y – 7) (y – 8) = 0
y = 7, 8
⇒ x ≥ y
x² – 8x – 12x + 96 = 0
x (x – 8) – 12 (x – 8) = 0
(x – 12) (x – 8) = 0
x = 12, 8
II. y² – 15y + 56 = 0
y² – 7y – 8y + 56 = 0
(y – 7) (y – 8) = 0
y = 7, 8
⇒ x ≥ y
Q10. I. x² + 2x – 35 = 0
II. y² + 3y – 10 = 0
if x>y
if x≥y
if x<y
if x ≤y
if x = y or no relation can be established between x and y.
Solution:
I. x² + 2x – 35 = 0
x² + 7x – 5x – 35 = 0
x (x + 7) – 5 (x + 7) = 0
(x – 5) (x + 7) = 0
x = 5, –7
II. y² + 3y – 10 = 0
y² + 5y – 2y– 10 = 0
(y + 5) (y – 2) = 0
y= –5, 2
⇒ no relation can be established between x and y
x² + 7x – 5x – 35 = 0
x (x + 7) – 5 (x + 7) = 0
(x – 5) (x + 7) = 0
x = 5, –7
II. y² + 3y – 10 = 0
y² + 5y – 2y– 10 = 0
(y + 5) (y – 2) = 0
y= –5, 2
⇒ no relation can be established between x and y
Directions (11-15): What comes at the place of question mark in the following number series?
Q11. 27, 1358, 1277, 1620, ? , 1622
1546
1536
1596
1598
1595
Solution:
pattern is
27+11³=1358
1358-9²=1277
1277+7³=1620
1620-5²=1595
1595+3³=1622
So, ?=1620 – 5² = 1595
27+11³=1358
1358-9²=1277
1277+7³=1620
1620-5²=1595
1595+3³=1622
So, ?=1620 – 5² = 1595
Q12. 48, 72, 180, 810, ?, 69862.5
6078
6075
6077
6080
6085
Solution:
Pattern is –
48×1.5=72
72×2.5=180
180×4.5=810
810×7.5=6075
So,?= 810 × 7.5 = 6075
48×1.5=72
72×2.5=180
180×4.5=810
810×7.5=6075
So,?= 810 × 7.5 = 6075
Q13. 8, 288, 512, 680, 792, ?
848
840
876
890
896
Solution:
Q14. 57, 65, 74, 138, ?, 379 169
164
156
163
166
Solution:
Q15. 16 ?, 32, 128, 64, 256 56
64
68
72
78
Solution:
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