IBPS Clerk Quantitative Aptitude Quiz: 19th November 2019

With the increasing level in exams, quantitative aptitude has become the ineluctable hitch. Generally, questions asked in this section are calculative and lengthy that consumes your time. This subject can do wonders if you always keep a check on your accuracy, speed and time. Accuracy is what matters the most.  Attempt this quantitative aptitude quiz and check your performance for the upcoming IBPS Clerk Prelims Study Plan 2019. Following is the quiz of 18th November, that inculcate the important topics of quantitative aptitude.

Directions (1-5): In the following questions, two quantities (I) and (II) are given. You have to solve and compare the numerical value of both the quantities and mark the appropriate option.

Q1. Volume and curved surface area of cylinder is 11550cm3 and 1320cm2 respectively.
Quantity I: Radius of cylinder.
Quantity II: Height of cylinder.
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≤ Quantity II
(d) Quantity I ≥ Quantity II
(e) Quantity I = Quantity II

Q5. Quantity I, (Difference of amount of discount allowed and profit earned): A seller marks his article 60% above cost price and he earned 22% profit on selling the article. Selling price of the article is Rs.1830.
Quantity II, (?): 16% of 2100= ?
(a) Quantity I > Quantity II
(b) Quantity I < Quantity II
(c) Quantity I ≤ Quantity II
(d) Quantity I ≥ Quantity II
(e) Quantity I = Quantity II

 

Directions (6-10) : In each of these questions, two equations (I) and (II) are given, You have to solve both the equations and give answer
(a) if x < y
(b) if x > y
(c) if x ≤ y
(d) if x ≥ y
(e) if x = y, or relationship between x and y can’t be established.

 

 

   

 

Directions (11-15): Find wrong number in given series.

Q11. 512, 517, 527, 542, 570, 607
(a) 517
(b) 527
(c) 542
(d) 607
(e) 570

Q12. 654, 660, 672, 690, 722, 764
(a) 690
(b) 654
(c) 660
(d) 722
(e) 764

Q13. 78, 90, 106, 128, 158, 198
(a) 78
(b) 90
(c) 106
(d) 158
(e) series is in correct order.

Q14. 12, 6, 4, 3, 2.4, 1.5
(a) 6
(b) 12
(c) 3
(d) 1.5
(e) 2.4

Q15. 12300, 11690, 11100, 10520, 9950, 9390
(a) 9390
(b) 12300
(c) 10520
(d) 11690
(e) 9950

 

                                               Solution:

S2. Ans.(d)
Sol. Quantity I:
x² – 15x + 56 = 0
x² – 8x – 7x + 56 = 0
x (x – 8) – 7 (x – 8) = 0
(x – 8) (x – 7) = 0
x = 8, 7

Quantity II:
y² – 12y + 35 = 0
y² – 7y – 5y + 35 = 0
y(y – 7 ) –5(y – 7) = 0
(y – 7) (y – 5) = 0
y = 5, 7
So, Quantity I ≥ Quantity II.

S3. Ans.(b)
Sol. Quantity I:
Let speed of boat in still water be ‘x km/hr’ and speed of stream be ‘y km/hr.’
Atq,
182/7 = x + y
x + y = 26 … (i)
And ((182 × 200/13 × 1/100))/7=x-y
x – y = 4 … (ii)
On solving (i) and (ii), we get:
x = 15 km/hr.
So, Quantity II > Quantity I.

S4. Ans.(b)
Sol. Quantity I:
44x² – 79x + 30 = 0
44x² – 55x – 24x + 30 = 0
11x (4x – 5) –6 (4x – 5) = 0
(4x – 5) (11x – 6) = 0
x = 5/4,6/11

Quantity II:
15y² – 59y + 56 = 0
15y² – 35y – 24y + 56 = 0
5y (3y – 7) –8 (3y – 7) = 0
(3y – 7) (5y – 8) = 0
y = 7/3 ,8/5
So, Quantity II > Quantity I.

S5. Ans.(b)
Sol. Quantity I:
Let cost price of the article be Rs. 100x.
So, marked price of the article = 100x × 160/100 = Rs. 160x
And selling price of the article = Rs. 1830
100x × 122/100 = 1830
x = 1830/122
x = 15
So, CP of article = 100x = Rs. 1500
And MP of article = 160x = Rs. 2400
Required difference = (2400 – 1830) – (1830 – 1500)
= 570 – 330 = Rs. 240

Quantity II:
16/100× 2100 = ?
?=336
So, Quantity II > Quantity I.

S8. Ans. (e)
Sol.
I. 6x^2-726=0
x^2-121=0
x=±11

II.y^2+y=56
y^2+y-56=0
y^2+8y-7y-56=0
(y-7)(y+8)=0
y=7,-8
So, no relation

S9. Ans. (b)
Sol.
I.2x+3y=7/12
II.3x+2y=2/3
On solving both equation
x=1/6 and y=1/12
So, x>y