Practice Aptitude Questions (Mixed Problems) With Solutions Set-53:
Dear Readers, Important Aptitude Questions for SSC/FCI Exam was given here with solutions, candidates those who are preparing for those exams can use this material.
1).
If √19×√247=x√325, then the value of X is
a)
1
b)
3(4/5)
c)
√38
d)
√65
2).
153 typists typed 2295 pages in 1/12 hour. Find the number of pages typed per
minute by a typist
a)
2
b)
5
c)
3
d)
4
3).
The angles of a triangle are in the ratio of 3:5:7. Find the measurement of the
greatest angle.
a)
84°
b)
72°
c)
90°
d)
79°
4).
If x²-7x+1=0, then the value of x2-x+(1/x2)-(1/x) is
a)
35
b)
25
c)
40
d)
55
5).
If x=√13-2√3, then the value of [x-1/x]² is
a)
59
b)
36
c)
64
d)
48
6).
If (2x/3y)+ (3y/2x)-1=0, then the value of 8x3+27y3 is
a)
1
b)
0
c)
2
d)
1/2
7).
If x2-129x+1=0, then the value of 70x / (2x2+92x+2) is
a)
1/10
b)
1/15
c)
1/5
d)
1/18
8).
If
2x+3y+4z=0, then the value of [ (2x+3y)/z ] – [ (3y+4z)/x ] + [ (3x+4z)/y ] is
a)
1
b)
5
c)
-1
d)
-5
9).
If 25x²-100x-1=0, then the value of x²+ 1/625x² is
a)
17 (21/25)
b)
11 (13/25)
c)
15 (23/25)
d)
12 (23/25)
10).
If 2x=3, 3y=4 and 4z=2 then the value of xyz
is
a)
0
b)
1
c)
2
d)
3
Answers:
1).
b) 2).c ) 3).
a) 4). c) 5).
d) 6). b) 7).c
) 8). d) 9).c
) 10). b)
Solution:
1).
(√19×√247)=x√325 or, 19√13=x×5√13 or, x=(19/5)=3 (4/5)
Answer: b)
2).
1/12 hour=5 minutes.
Now, 153 typists can type 2295 pages in 5
minutes.
∴ In 1 minute 153 typists can type (2295/5)=459
pages
∴ In 1 minute, 1 typist can type (459/153)=3
pages
Answer: c)
3).
Suppose the angles are 3x, 5x and 7x.
Then, 3x+5x+7x=180 or, 15x=180°
∴x=12°
∴ Greatest angle=12×7=84°
Answer: a)
4). ∴ x²-7x+1=0 or, x-7+(1/x)=0 or, x+1/x=7
Now, x²-x+ 1/x²-1/x= [x²+1/x²] – [x+1/x]
=[x+1/x]² -2- [x+1/x]
=49-2-7=40
Answer: c)
5).
x=√13-2√3
or, 1/x= (1(1√13+2√3)) / [(√13-2√3) (√13+2√3)]
=(√13+2√3)/(13-12)=√13+2√3
Now, (x-1/x)²= (√13-2√3-√13-2√3)² = (-4√3)²
=48
Answer: d)
6).
(2x/3y)+(3y/2x)-1=0 or, 4x²+9y²-6xy=0
Now, 8x3+27y3= (2x)3
+ (3y)3
= (2x+3y) (4x2+9y2-6xy)
=0
∴ 8x3+27y3= (2x+3y) ×0=0
Answer: b)
7).
∴x²-129x+1=0 or, x+(1/x)=129
Now, 70x/ (2x²+92x+2)= (70x/x) / (2x²+92x+2/x)
= 70/ (2x+92+2/x)
=70/ 2(x+1/x) +92 = 70/ (2×129÷92) =(70/350)
=1/5
Answer: c)
8).
2x+3y=-4z
Again, 3y+4z=-2x
And 2x+4z=-3y
Now, (2x+3y/z) – (3y+4z/x) + (2x+4z/y)
=-4+2+ (-3) =-5
Answer: d)
9).
∵25x²-100x-1=0 or, x-4-(1/25x)=0 or, x-(1/25x)=4
Now, x²+(1/625)= [x-1/25x]² +2×x×1/25x
=16-(2/25) =(398/25) =15(23/25)
Answer: c)
10). ∵2x=3, 3y=4, 4z=2
Now, 2=4z = (3y)z =3yz = (2x)yz
=2xyz
∴xyz=1
Answer: b)
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