SSC CHSL- Practice Aptitude Questions (Mixed Problems) With Solutions Set-47

SSC CHSL- Practice Aptitude Questions (Mixed Problems) With Solutions Set-47:
Dear Readers, Important Aptitude Questions for SSC CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1).A man standing on a railway platform noticed that a train took 18 seconds to cross him and it took 42 seconds to cross 176-meter-long station completely. The length of the train (in meters) was
a)    180
b)    121
c)    66
d)    132

2).If the average of 10 numbers is a, the average of 2 numbers is b and the average of the remaining 8 numbers is c, then
a)    3c-2a=b
b)    b-5a=4c
c)    4c-3a=b
d)    3b-2c=5a

3).Consider ∆ABD such that ∠ADB=30〫and C is a point on BD such that AB=Ac and CD=CA. Then the measure of ∠ABC is
a)    60〫
b)    75〫
c)    45〫
d)    30〫

4).∆ABC and ∆DEF are similar and their areas are 81 cm² and 100cm² respectively. If EF=20 cm then BC is equal to
a)    18 cm
b)    20 cm
c)    24 cm
d)    16 cm

5).A 12-cm-long perpendicular is drawn from the center of a circle to an 18-cm-long chord. Find the diameter of the circle.
a)    30 cm
b)    12 cm
c)    18 cm
d)    15 cm
6).PQRS is a cyclic quadrilateral and PQ is the diameter of the circle. If ∠QPR=55〫, then the value of ∠PRS is equal to
a)    55〫
b)    45〫
c)    145〫
d)    135〫

7).Find the co-ordinates of the point which divides internally the line segment joining the points (5, 8) and (3, 4) in the ratio of 8:9.
a)    [(69/17), (104/17)]
b)    [(104/17), (69/17)]
c)    [(47/17), (107/17)]
d)    [(107/17), (119/17)]

8).Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are (4, 4), (8, 8) and (4, 12).
a)    6 unit2
b)    5 unit2
c)    16 unit2
d)    4 unit2

9).Find the co-ordinates of the intersection of the line 6x+y+12=0 and 3x+y+6=0.
a)    (-2, -2)
b)    (-2, 1)
c)     (-2, 0)
d)    (4, 3)

10).Find the distance between the following given lines: 9x+12y+21=0 and 9x+12y+15=0.
a)    0.4 unit
b)    1 unit
c)    15 units
d)    4 units
Answers:
1). d) 2). b) 3).a ) 4). a) 5).a ) 6).c ) 7). a) 8). d) 9). c) 10). a)

Solutions:
1). Let the length of the train be xm.
The train takes 18 sec to cross the man.
∴ Speed of the train=Distance/Time= (x/18) m/sec     …..(ⅰ)
Now, length of the station=176m
Length of the train=xm
Time to cross the station=42 seconds
∴Speed of the train=Total distance/Time=(x+176)/42   …..(ⅱ)
By equation (ⅰ) and (ⅱ),
x/18= (x+176)/42
or, x/3= (x+176)/7
or, 7x=3x+528
or, 4x=528
∴ x=132 meters
Answer: d)

2). Average of 10 numbers is a.
Sum of 10 numbers=10a
Average of 2 numbers is b.
Sum of 2 numbers=2b
Average of 8 numbers is c.
Sum of 8 numbers=8c
Now, 2b+8c=10a
Or, b+4c=5a
Or, 4c-5a=b
∴ b-5a=4c
Answer: b)

3).a; Given AB=AC and CD=CA ∠ADB=30〫

In ∆ABD, ∠CAD=∠ADB (isosceles triangle)
In ∆ACD=30〫+30〫=60〫
∴ AB=AC
Hence, ∠ACD=∠ABC=60〫
Answer: a)

4). Given area of ∆ABC∼ Area of ∆DEF
(Area of ∆ABC)/ (Area of ∆DEF) = (AB/DE)² =   (BC/EF)²= (CA/FD)²
Given EF=20 cm
Now, 81/100= (BC/EF)²
or, 9/10=BC/EF   or, BC= (9*20)/10=18 cm
∴ BC=18 cm
Answer: a)

5).

Length of chord =18 cm
Length of perpendicular=12 cm
So in the right-angled triangle ∆OAC
(OA)² = (AC)²+ (OC)²
Or, r²= (12)²+ (9)²
or, r²=144+81
or, r²=225
∴ r=15 cm
∴ Diameter=15×2 =30 cm
Answer: a)
6). In the figure PQRS is a cyclic quadrilateral, where PQ is the diameter.

∠QPR=55〫And ∠PRQ=90〫(PQ is diameter)
∠PQR=90〫-55〫=35〫
Now, ∠RQP+∠RSP=180〫or, ∠PSR=180〫-∠PQR
or, ∠PSR=180〫-35〫
∴ ∠PSR=145〫
Answer: c)
7). Co-ordinate of points= (5, 8)
= (x1, y1) and (3, 4) = (x2, y2)
Ratio=8:9
m : n=8:9
Co-ordinates of dividing point(x, y)
= [((mx2+nx1)/m+ n), ((my2+ny1)/m+ n)]
= [((8×3+9×5)/8+9), ((8×4+9×8)/8+9)]
= [((24+45)/17), ((32+72)/17)]=[(69/17), (104/17)]
Answer: a)
8).

In the given figure P, Q, R are the mid-points of AB, BC and CA respectively.
Area of ∆ABC= ½[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]
= ½[4 (8-12) +8(12-4) +4(4-8)]
=1/2[4(-4) +8(8) +4(-4)]
=1/2[-16+64-16] =1/2[32] =16 unit2
There are 4 triangles of equal area-∆APQ= Area of ∆PBR =Area of ∆PQR
= Area  of  ∆QRC
∴ Area of ∆PQR=1/4 (Area of ∆ABC) = (1/4) × 16=4 unit²
Answer: d)

9). Equation of first line=6x+y+12=0   …….(ⅰ)
Equation of second line=3x+y+6=0 …….(ⅱ)
Subtracting equation (ⅱ) from equation (ⅰ),
We get 3x+6=0 or, 3x=-6
∴ x=-2
Putting x=-2 in equation (ⅰ), we get 6(-2)+ y+12=0
or,-12+y+12=0
y=0
∴ Intersection of lines=(x, y) = (-2, 0)
Answer: c)

10). Given lines are 9x+12y+21=0 And 9x+12y+15=0
Distance between the lines= (21-15)/√9²+12²
= (21-15)/√81+144
=6/√225
=6/15
=2/5 units=0.4 unit
Answer: a)

People Also Visited: