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Practice Aptitude Questions (Mixed Problems) With Solutions Set-51

Practice Aptitude Questions (Mixed Problems)
Practice Aptitude Questions (Mixed Problems) With Solutions Set-51:
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 x+ (1/x) =7, then the value of x5+ (1/x5) is
a)    16227
b)    17127
c)    15127
d)    16997
2). Find the smallest number among the following: ∛4, ∜5, 6√12, 12√25
a)    ∛4
b)    ∜5
c)    6√12
d)    12√25
3). The three vertices of ∆ABC are A (4, 5), B (6, 7) and C (-7, 8). The area of the triangle is:
a)    14 unit²
b)    16 unit²
c)    10 unit²
d)    12 unit²
4). Points (x, y), (4, 1) and (-13, 7) are lying on a straight line. Find the relation between x and y.
a)    6x+17y=41
b)    6x+17y=14
c)    6x+15y=14
d)    6x+9y=41
5). If the distances between the points (x, 0) and (-9, 0) be 15 units, then the value of x is
a)    6
b)    24
c)    -6
d)    -24
6). If 4x+y=12 and x=4 are equations of two lines, then the graph of the two equations meet at the point
a)    (4, 4)
b)    (4, 0)
c)    (4, -4)
d)    (-4, -4)
7). If the graph of the equation 3x+4y=12 forms a triangle with the coordinates axes, then the area of the triangle will be
a)    6 sq units
b)    10 sq units
c)    12 sq units
d)    3 sq units
8). Find the area of the greatest circle which can be inscribed in a rectangle with sides 36 cm and 14 cm.
a)    154cm²
b)    141cm²
c)    148cm²
d)    151cm²
9). There are two paths each 25m wide that intersect each other at perpendicular at the mid of a square ground having side of 180m. Find the area of the ground excluding the path.
a)    2552m²
b)    32400m²
c)    24025 m²
d)    8375m²
10). A spare with diameter 28 cm is moulded into a cone of 7 cm height. Find the radius of the cone.
a)    14√3 cm
b)    28√2 cm
c)    42√2 cm
d)    46√3 cm

Answers:
1). c) 2).d ) 3). a) 4).a ) 5).a ) 6). c) 7). a) 8).a ) 9). c) 10). b)

Solutions:

1). x+ (1/x) =7
On squaring both sides [x+ (1/x)]² =7²
or, x²+ (1/x²)=49-2=47
Again, cubing both sides [x+ (1/x)]3=73
or, x∴+ (1/x3) +3[x+ (1/x]=343
Or, x3+ (1/x3) +3×7=343
or, x3+ (1/x3) =343-21
or, x3+ (1/x3) =322
Now, [x²+ (1/x²)] [x3+ (1/x3)] =322×47
or, x5+ (1/x5) +[x+ (1/x)] =15134
Or, x5+ (1/x5) =15134-7
∴ x5+ (1/x5) =15127
Answer: c)

2). Given number are ∛4, ∜5, 6√12, 12√25
So, (4)1/3, (5)4/4, (12)1/6, (25)1/12
Taking LCM of 3, 4, 6, 12
(4)12/3, (5)12/4, (12)12/6, (25)12/12
or, (4)4, (5)3, (12)2,25
Hence, the smallest number is (25)1/12 or, 12√25
Answer: d)

3). Three vertices of the triangle are A(4, 5), B(6, 7), and C(-7, 8).
x1=4, y1=5, x2=6, y2=7, x3=-7, y3=8
Now, area of triangle=1/2 [x1(y2-y3) + x2(y3-y1) +x3(y1-y2)]
=1/2 [4(7-8) + 6(8-5) + (-7(5-7))]
= 1/2 [-4+6×3+ (-7) (-2)]
= 1/2 [-4+18+14] =1/2 [28] =14 unit²
Answer: a)

4). Points (x1, y1) = (x, y)
(x2, y2) = (4, 1)
(x3, y3) = (-13, 7)
For three points to be collinear (or on a straight line)
x1(y2-y3) +x2(y3-y1) + x3(y1-y2)=0
or, x(1-7) +4(7-y) +(-13) (y-1) =0
or, -6x+28-4y-13y+13=0
or, -6x-17y+41=0
or, 6x+17y=41
∴ Required relation is 6x+17y=41
Answer:a)

5).  (x1, y1) = (x,0) and (x2, y2) = (-9, 0)
D= √[(x1-x2)2+(y1-y2)2
or, 15=√[ (x+9)2+ (0-0)2]
or, 15=√(x+9)2
Or, x+9=15 or, x=6
Answer: a)

6). Given equation is
4x+y=12 …(ⅰ)
x=4 …(ⅱ)
On substituting x=4 in equation (ⅰ), we get 4×4+y=12 or, 16+y=12
∴ y=-4
Hence, meeting point= (x, y) = (4, -4)
Answer: c)

7). Given equation 3x+4y=12



Now, 3x+4y=12
or, (3x/12) + (4y/12) =1
or, (x/4)+(y/3) =1
In the given figure ∆OAB is formed by the axes and the line 3x+4y=12
Area of OAB=(1/2)×base×height
=(1/2)x4x3=6 sq units
Answer: a)

8). Radius of the circle inscribed in the
Rectangle = (breadth of rectangle/2) = (14cm/2) = 7cm
Area of the circle inscribed= πr²
=(22/7)×7×7cm²
=154cm²
Answer: a)

9).



Area of the ground=(180×180)m²=32400m²
Area of the path=25× { (180+180)-25}
=25(360-25)=25×335=8375m²
∴ Reqd area=area of the ground-area of the path = (32400-8375)m² =24025m²
Answer: c)

10). The sphere is moulded into a cone.
Volume of the sphere=4/3πr3
Volume of the cone=1/3πr²h
∴Diameter of the sphere=28cm
∴Radius=14cm
Now , the volume of the sphere=volume of the cone
or, (4/3)π×(14)3 =(1/3)πr²×h
or, (4/3)×(22/7)×(14×14×14)
=(1/3)×(22/7)×r²×7
or, r²=(4/3)× [ (2214×14×3×7×14) / (7×22×7) ]
or, r²=14×14×4×2
∴ r=28√2cm

                     
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