IBPS Clerk/RRB Mains 2016 – Practice Quantitative Aptitude Questions (Data Interpretation& Quadratic Equation)

 IBPS Clerk 2016 - Practice Quantitative Aptitude Questions
IBPS Clerk/RRB Mains 2016 – Practice Quantitative Aptitude Questions (Data Interpretation& Quadratic Equation) Set-64:

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

 
Directions (Q. 1-5): Study the following information carefully and answer the questions given below:
Out of 800 students of a school 60.625% play cricket, 54.625% play football and 43.75% play hockey. 23.625% students play only cricket, 14% play only football and 16.875% students play only hockey.
Now answer the following questions based on this information.
 
1).How many students are there who play all three games?
a)   92
b)   108
c)   112
d)   124
e)   132
 
2).What is the percentage of students who play only football and hockey?
a)   7.2%
b)   8%
c)   8.5%
d)   9.6%
e)   12%
 
3).How many students are there who play cricket and hockey but don’t play football?
a)   39
b)   42
c)   48
d)   54
e)   57
 
4).What percentage of students play atmost one type of game?
a)   51.5%
b)   52.5%
c)   53.5%
d)   54.5%
e)   56.5%
 
5).How many students are there who play at least two games?
a)   352
b)   364
c)   372
d)   384
e)   256
 
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 can’t be established.
6). I. 7x2– 33x + 20 = 0
II.7y2 + 9y – 10 = 0
7). I.7x + 3y = 8
II.3x + 2y = 12
8). I.x2 = 289
II.y =3√2197
9). I.8x2 – 11x + 3 = 0
II.y2 – 8y + 7 = 0
10). I.x = √1089
II.y + y – 756 = 0
 
Solution:
Directions (Q. 1-5):

 

x + y + k = 485 -189 = 296 …(i)
x + z + k = 437 -112 = 325 …(ii)
y + z + k = 350 -135 = 215 …(iii)
x + y + z + k = 800 – (189 + 112 + 135) = 364     …(iv)
Solving all four eqns, we get
x = 149, y = 39, z = 68 and k = 108
 
1). 
Answer: b)
2).Reqd% = (68/ 800) x 100 = 8.5%
Answer: c)
3).39
Answer: a)
4).Reqd % = [(189+112+135) / 800] x 100 = 436 / 8 =54.5%
Answer: d)
5). Number of students playing at least two games = (149+39+68+108) = 364
Answer: b)
 
6). I.7x2 – 33x + 20 = 0
or, 7x2 – 28x – 5x + 20 = 0
or, 7x(x – 4) – 5(x —4)= 0
or (7x – 5) (x – 4) = 0
or, x = 4, (5/7)
II.7y2 + 9y -10 = 0
or, 7y2 + 14y – 5y – 10 = 0
or, 7y(y + 2) – 5(y + 2) = 0
 or, (7y – 5)(y + 2) = 0
 or, y = 5/7, -2
Hence, x ≥ y
Answer: d)
 
7).I. 7x + 3y = 8
II.3x + 2y = 12
Now, solving both the eqns, we get
eqn (i) x 2 – eqn (ii) x 3
(14x+ 6y=16) – (9x + 6y = 36)
 5x = -20
x = -4 and y = 12
 Hence, x<y
Answer: a)
 
8). I.x2 = 289
or x = ±17
II.y =3√2197
y = 13
Hence, we can’t find a relationship between x and y.
Answer: e)
 
9). 8x2 -11x + 3 = 0
or, 8x2 – 8x – 3x +3 = 0
or, 8x(x – 1) – 3(x – 1) = 0
or, (8x -3)(x – 1) = 0
or, x =1, 3/8
II.y2 – 8y + 7 = 0
or, y2 – y – 7y + 7 = 0
or, y(y – 1) – 7(y – 1) = 0
or,  (y – 1) (y – 7) = 0
or, y = 1, 7
Hence, x ≤ y
Answer: c)
 
10). I.x = √1089
or, x=33
II.y2 + y – 756 =0
or, y2 + 28y – 27y – 756 = 0
or, y(y +28) – 27(y + 28) = 0
or, (y – 27)(y + 28) = 0
or, y= 27, -28
Hence, x> y
Answer: b)

 

 

  
 

 

 

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