### Practice Aptitude Questions (Mixed Problems) With Solutions Set-53

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

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)

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

3). Suppose the angles are 3x, 5x and 7x.
Then, 3x+5x+7x=180 or, 15x=180°
∴x=12°
∴ Greatest angle=12×7=84°

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

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

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

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

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

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)