## 11 Nov 2016

### IBPS RRB/Clerk 2016 - Practice Quantitative Aptitude Questions

IBPS RRB/Clerk 2016 - Practice Quantitative Aptitude Questions Set-36:
Dear Readers, Important Practice Aptitude Questions for IBPS Clerk and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.

Direction (Q.1-5): The following line graph shows the percentage increase in the population of two cities A and B over the period 2009 to 2013.

The table shows the population of these cities at the beginning of 2009.
 City Population (in lakh) A 12 B 10

1).What is the population of city A at the beginning of the year 2011?
a)    14.96 lakh
b)    15.18 lakh
c)    15.72 lakh
d)    16.04 lakh
e)    16.24 lakh

2).What is the difference between the population of city A and City B at the end of the year 2011?
a)    53400
b)    54600
c)    55300
d)    56100
e)    57500

3).What is the ratio of the population of City A to that of City B at the end of 2009?
a)    1 : 2
b)    6 : 5
c)    8 : 5
d)    11 : 10
e)    None of these

4).What will be the population of City B at the end of the year 2013?
a)    17.325 lakh
b)    18.464 lakh
c)    19.0575 lakh
d)    20.040 lakh
e)    None of these

5).The population of City B at the end of the year 2011 is approximately what per cent of the population of City A at the beginning of the year 2009?
a)    72.7%
b)    87.5%
c)    112.5%
d)    125%
e)    137.5%

Directions (Q.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 relation cannot be established between ‘x’ and  ‘y’.
6).I. x2 – x – 56 = 0
II.y=3√729
7).I.x2 + 7x + 10=0
II.y2 + 13y + 42 =0
8).I.14x2 - 55x + 50 = 0
II.2y2 – 13y + 20 = 0
9).I.7x + 8y = 5
II.6x-5y = 28
10).I.5x2 – 52x + 96 =0
II.5y2 + 3y-36 = 0

1).b)   2).d)  3).d)  4).c)  5).e)  6).c)  7).a)  8).d)  9).a)  10).b)

Solution:
1). Population of City A at the beginning of the year 2011
=12 × 110/100 × 115 / 100 =15.18 lakh

2). Population of City A at the end of 2011
=12 × 110/100 × 115/100 ×105/100 =15.939 lakh
Population of City B at the end of 2011
=10 × 120/100 × 125/100 × 105/100 = 16.5 lakh
Difference = 16.5 – 15.939 = 0.561 lakh = 56100

3). Ratio = [12 × (110/100)] /[10×1(20/100)] = (12× 11) / (10 × 12) = 11/10 = 11 : 10

4). Population of City B at the end of 2013
=10 × (120/100) × (125/100) × (110/100) × (105/100) × (110/100) =19.0575 lakh

5). Population of City B at the end of the year =10 × 1.20 × 1.25 ×1.1 = 16.5 lakh
Population of City A at the beginning of the year 2009 = 12 lakh
Reqd % = (16.5 /12) × 100 = 137.5 %

6). I . x2 – x – 56 = 0
Or,x2 + 7x – 8x-56 = 0
Or, x(x+7) -8(x+7) =0
Or,(x-8)(x+7)=0
Or, x= 8,-7
II. y=3√729
Or, y=9
Hence, x < y

7). I. x2 + 7x + 10 = 0
Or,x2 + 2x +5x + 10 = 0
Or, x(x+2)+ 5(x+2)=0
Or,(x+5)(x+2) = 0
or,x= -2,-5
II.y2 + 13y + 42 =0
Or,y2 + 6y+7y+42 = 0
Or, y(y+6)+7(y+6) = 0
Or,(y+6)(y+7)=0
Y = -6,-7
Hence, x > y

8). I. 14x2 – 55x+50 = 0
Or,14x2 -35x-20x+50 = 0
Or,7x(2x-5)-10(2x-5)=0
Or,(7x-10)(2x-5) = 0
X=10/7,5/2
II.2y2-13y + 20 = 0
Or,2y2-8y-5y+20 =0
Or,2y(y-4)-5(y-4) = 0
Or,(y-4)(2y-5) = 0
Or, y=4,5/2
Hence, x ≤ y

9). I. 7x + 8y = 5                            …..(i)
II.6x - 5y = 28                           …..(ii)
Now, eqn(i) × 5 + eqn (ii) × 8
35x + 40 y =25
48x – 40y = 224
Simplify above 2 eqns we get à83x = 249
X=3 and y=-2
Hence,x > y

10). I. 5x2-52x + 96 = 0
Or,5x2-40x-12x + 96 =0
Or,5x(x-8)-12(x-8)=0
Or,(x-8)(5x-12)=0
Or,x=8,12/5
II. 5y2 + 3y -36=0
Or,5y2 +15y -12y -36 =0
Or,5y(y + 3)-12(y+3) =0
Or,(5y-12)(y+3) = 0
Or,y=-3,12/5
Hence,x ≥ y