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

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

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

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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.
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1).How many students are there who play all three games?
a)ย ย ย 92
b)ย ย ย 108
c)ย ย ย 112
d)ย ย ย 124
e)ย ย ย 132
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2).What is the percentage of students who play only football and hockey?
a)ย ย ย 7.2%
b)ย ย ย 8%
c)ย ย ย 8.5%
d)ย ย ย 9.6%
e)ย ย ย 12%
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3).How many students are there who play cricket and hockey but don’t play football?
a)ย ย ย 39
b)ย ย ย 42
c)ย ย ย 48
d)ย ย ย 54
e)ย ย ย 57
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4).What percentage of students play atmost one type of game?
a)ย ย ย 51.5%
b)ย ย ย 52.5%
c)ย ย ย 53.5%
d)ย ย ย 54.5%
e)ย ย ย 56.5%
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5).How many students are there who play at least two games?
a)ย ย ย 352
b)ย ย ย 364
c)ย ย ย 372
d)ย ย ย 384
e)ย ย ย 256
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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)ย ย ย if x< y
b)ย ย ย if x> y
c)ย ย ย if x โค y
d)ย ย ย if x โฅ y
e)ย ย ย if x = y, or relationship between x and y can’t be established.
6).ย I. 7x2– 33x + 20 = 0
II.7y2 + 9y โ 10 = 0
7). I.7x + 3y = 8
II.3x + 2y = 12
8). I.x2 = 289
II.y =3โ2197
9). I.8x2 – 11x + 3 = 0
II.y2 – 8y + 7 = 0
10). I.x = โ1089
II.y + y โ 756 = 0
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Solution:
Directions (Q. 1-5):

x + y + k = 485 -189 = 296ย โฆ(i)
x + z + k = 437 -112 = 325ย โฆ(ii)
y + z + k = 350 -135 = 215ย โฆ(iii)
x + y + z + k = 800 – (189 + 112 + 135) = 364ย ย ย ย  …(iv)
Solving all four eqns, we get
x = 149, y = 39, z = 68 and k = 108
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1).ย
2).Reqd% = (68/ 800) x 100 = 8.5%
3).39
4).Reqd % = [(189+112+135) / 800] x 100 = 436 / 8 =54.5%
5).ย Number of students playing at least two games = (149+39+68+108) = 364
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6).ย I.7x2 – 33x + 20 = 0
or, 7x2 – 28x – 5x + 20 = 0
or, 7x(x – 4) – 5(x โ4)= 0
or (7x – 5) (x – 4) = 0
or, x = 4, (5/7)
II.7y2 + 9y -10 = 0
or, 7y2 + 14y – 5y โ 10 = 0
or, 7y(y + 2) – 5(y + 2) = 0
ย or, (7y – 5)(y + 2) = 0
ย or, y = 5/7, -2
Hence, x โฅ y
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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) โ (9x + 6y = 36)
ย 5x = -20
x = -4 and y = 12
ย Hence, x<y
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8). I.x2 = 289
or x = ยฑ17
II.y =3โ2197
y = 13
Hence, we canโt find a relationship between x and y.
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9).ย 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,ย  (y – 1) (y – 7) = 0
or, y = 1, 7
Hence, x โค y
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10). I.x = โ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

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