## 22 Aug 2016

### IBPS PO/Upcoming Exams 2016-Aptitude Questions (Mixed)

IBPS PO/Upcoming Exams 2016-Aptitude Questions (Mixed Quiz) Set-21:
Dear Readers, Important Practice Aptitude Questions for IBPS PO and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.

Direction (Q. 1-5): Study the following table carefully and answer the given questions.
 Name Type of interest Principal Amount Year Rate % Anil Compound 12000 - - 6 Kamal Simple - 35360 4 - Sunil Compound 25000 - 3 5 Jalal Simple - - 5 - Bilal Compound - - 6 4

1). What is the amount of Sunil, if the interest is compounded yearly for 3 years?
a)   24385.625
b)   26900.615
c)   27500.565
d)   28490.625
e)   25148.169

2). At what Rate of Interest does the amount of Kamal become 5 times his principal?
a)   95%
b)   102%
c)   98%
d)   100%
e)   97%

3). What will be the amount of Bilal in two years when his principal is 30% more than Anil’s?
a)   18738.18
b)   16872.96
c)   19638.1
d)   19548.18
e)   19799.18

4). If the ratio of principal of Sunil to that of Jalal is 5 : 6 and the rate of interest of Jalal is 20% more than that of Sunil, then what is the interest of Jalal?
a)   9000
b)   8000
c)   9400
d)   7500
e)   8800

5). What is the principal of Kamal if the ratio of the rate of interest Anil to that of Kamal is 2 : 3?
a)   24500
b)   25500
c)   26000
d)   27500
e)   25600

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 relationship between x and y cannot be established.

6). I. 7x2 + 2x = 120
II. y2 + 11y + 30 = 0

7). I. 676x2 – 1 = 0
II. y = 1/3(13824)

8). I. x2 = 7x
II. (y + 7)2 = 0

9). I. 2x2 + 5x – 33 = 0
II. y2 – y – 6 = 0

10). I. 8x + 13y = 62
II. 13x – 17y + 128 = 0

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

Solution:
1). Amount of Sunil = P(1 + (r/100))t
= 25000 × (1+5/100)3
= 25000 ×(21/20) × (21/20) × (21/20)
= 28940.625

2). Let the principal be x
Then, SI = 5x – x = 4x
4x = x × r × 4/100
r = 100%

3). Principal of Bilal = 12000 × (130/100)= 15600
Amount of Bilal = 15600 × (1 + (4/100))
= 15600 × (26/25) × (26/25)= 16872.96

4). Principal of Jalal = 25000 × (6/5)= 30000
Rate of interest = 5 × (120/100)= 6%
Interest =(30000×6×5)/100 = 90000

5). Rate of interest of Kamal = 6 × (3/2)= 9%
Amount = 35360
Time = 4 years
Let the principal be x
Then, x + (x×9×4)/100 = x (1 + (36/100))= 35360
X × (136/100) = 35360
X = 26000
6). I. 7x2 – 28x + 30x – 120 = 0
or, 7x(x – 4) +30(x – 4) = 0
or, (x – 4) (7x + 30) = 0
x = 4, -30/7

II. y2 + 6y + 5y + 30 = 0
or, y(y + 6) + 5(y + 6) = 0
or, (y + 5) (y + 6) = 0
y = –5, –6
ie, x and y cannot be established.

7). I. 676x2 – 1 = 0
or, x2  = 1/676 = x = 1/26
II. y = 1/3(13824)= y = 1/24
ie, x < y

8). I. x2 = 7x
or, x2 – 7x = 0
or, x(x – 7) = 0
x = 0, 7

II. (y + 7)2 = 0
or, (y + 7) = 0
? y = –7
ie, x > y

9). I. 2x2 – 6x + 11x – 33 = 0
or, 2x(x – 3) +11(x – 3) = 0
or, (2x + 11) (x – 3) = 0
x = 3, -11/2

II. y2 – 3y + 2y – 6 = 0
or, y(y – 3) +2(y – 3) = 0
or, (y + 2) (y – 3) = 0
y = –2, 3
ie, x  y.