# FSO GM Mock Test

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# FSO GM Mock Test

### Paper - II

FSO recruitment process includes written exam (FSO GK, FSO GM), walking test and medical test. Practice and Mock tests will help the students in speed/time management by providing the candidates with an analysis of their performance in the practice and mock tests. Along with the speed and time management, the FSO GM Practice and FSO GM Mock Test will help the students to improve their accuracy in attempting questions correctly.FSO written exam is divided into two papers as Paper 1 and Paper 2.

Paper 2 – General Mathematics is an important section in the written exam of FSO recruitment. The FSO GM section includes 100 questions for 100 marks. The questions are primarily based on simple arithmetic, algebra, geometry, trigonometry, mensuration, and statistics. The sub topics included in the above mentioned topics are of the tenth grade level.

### Pattern

FSO General Mathematics section is as described below.

Exam Mode Paper Name No. of Questions Max Marks Duration
Objective Type Paper 2- General
Mathematics
100 100 90 minutes

### Syllabus

Detailed Syllabus for General Mathematics:

✦ Arithmetic
✦ Algebra
✦ Trigonometry
✦ Geometry
✦ Mensuration
✦ Statistics

### Questions

ARITHMETIC: Number System-Natural numbers, Integers, Rational and Real numbers, Fundamental operations, addition, subtraction, multiplication, division, Square roots, Decimal fractions. Unitary method-time and distance, time and work, percentages, applications to simple and compound interest, profit and loss, ratio and proportion, variation. Elementary Number Theory – Division algorithm. Prime and composite numbers. Tests of divisibility by 2, 3, 4, 5, 9 and 11. Multiples and factors. Factorization Theorem. H.C.F. and L.C.M. Euclidean algorithm. Logarithms to base 10, laws of logarithms, use of logarithmic tables.

Sample Questions:

1. If 2,4,6 are first three consecutive even numbers, then 12th even number is _________.

1. 22
2. 26
3. 24
4. 20

2. The sum of digits of a 2 digit odd integer is 9. If their difference is 3 them the odd number is _______.

1. 63
2. 54
3. 27
4. 72

Solution: The odd number is 63, 6 + 3 = 9(sum of digits) 6 – 3 = 3(different).

3. If N, N + 2 and N + 4 are prime number, then the number of possible solutions of N is ________.

1. 1
2. 2
3. 3
4. None

Solution: If X = 2, Y = 4 then Y-X is even. If x = 2, Y= 4 then Y X is even if X = 2, Y = 6 then  (X + Y)/X IS even. The correct option is 4.

4. a,b,c,d are first four odd prime numbers then which of the following is true?

1. a+b+c is not composite
2. b+c+d is prime
3. abc is even
4. a+b+c+d si not even compisite

Solution: a = 3, b = 5, c = 7, d = 11, b+ c + d = 5 + 7 + 11 += 23 is prime numbers.

5. If p = 5, and a = 2 then ap – a is divisible by _______.

1. p + 4
2. p + 2
3. p + 3
4. p

Solution: ap – a = 25 – 2 = 30 is divisible by 5.

6.Which of the following is divisible by 3?

1. 1536
2. 1535
3. 1534
4. 1532

7. Find the missing number in the given series, 2,12,36,80,150,?.

1. 194
2. 210
3. 252
4. 258

Solution:

8. Find the missing number in the given series, 625,5,125,25,25,?,5.

1. 5
2. 25
3. 125
4. 625

Solution: The given sequence is a combination of two series.

I. 625,125,25,5, and II. 5,25,?

The pattern in I is , while that in II is x 5.

So the missing term = 25 * 5 = 125.

9. 28x + 1426 = 3/4 of 2872, then the value of x is ______.

1. 576
2. 676
3. 1296
4. 1444

Solution: 28x + 1426 = 3 * 718 => 28x = 2154 – 1426

=> x = 26 => x = 262

=> (26)= 676

10. If 5x is a perfect cube, then the least number of ‘5’ s that must br contained in the prime factorization of x is _____.

1. 1
2. 2
3. 3
4. 5

Solution: Given 5x is a perfect cube, 25 = 5*5, 5x = 5*5*5 = 125 is perfect cube. 2 fives that must be contained in the prime factorization of x.

11. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

1. 4 days
2. 5 days
3. 6 days
4. 7 days

Solution:

Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.

Then, 6x + 8y = 1/10 and 26x + 48y = 1/2.
Solving these two equations, we get : x =1/100 and y = 1/200.
(15 men + 20 boy)’s 1 day’s work = 15/100 + 20/200 = 1/4.
=>15 men and 20 boys can do the work in 4 days.

12. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:

1. 50 km
2. 56 km
3. 70 km
4. 80 km

Solution:
Let the actual distance travelled be x km.
Then, x/10 = (x + 20)/14
14x = 10x + 200
=> 4x = 200
=> x = 50 km.

13. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?

1. Rs. 500
2. Rs. 1500
3. Rs. 2000
4. None of these

Solution:

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x – 3x = 1000
=> x = 1000.
B’s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

14. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

1. Rs. 650
2. Rs. 690
3. Rs. 698
4. Rs. 700

Solution:

S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
=> Principal = Rs. (815 – 117) = Rs. 698.

15. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?

1. 30%
2. 70%
3. 100%
4. 250%

Solution:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 – 125) = Rs. 295.
Required percentage = [(295/450)*100]% = [1475/21]%= 70% (approx)

16. Which of the following statements is not correct?

1. log10 10 = 1
2. log (2 + 3) = log (2 x 3)
3. log10 1 = 0
4. log (1 + 2 + 3) = log 1 + log 2 + log 3

Solution:

(a) Since loga a = 1, so log10 10 = 1.
(b) log (2 + 3) = log5 and log (2 x 3) = log 6 = log 2 + log 3
=> log (2 + 3) log (2 x 3)
(c) Since loga 1 = 0, so log10 1 = 0.
(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.
So, (b) is incorrect.

17. If log 27 = 1.431, then the value of log 9 is:

1. 0.934
2. 0.945
3. 0.954
4. 0.958

Solution:

log 27 = 1.431
log (33 ) = 1.431
3 log 3 = 1.431
log 3 = 0.477
=> log 9 = log(32 ) = 2 log 3 = (2 x 0.477) = 0.954.

18. A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

1. 3
2. 4
3. 5
4. 6

Solution:

C.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs. 6/5
For Rs. 6/5, toffees sold = 6.
5
For Re. 1, toffees sold = [6 x(5/6) = 5.

19. A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in:

1. 4 days
2. 6 days
3. 8 days
4. 12 days

Solution:

Suppose A, B and C take x, x/2, and x/3 days respectively to finish the work.
Then,[(1/x)+(2/x)+(3/x)] = 1/2
=> 6/c = 1/2
=> x = 1
So, B takes (12/2) = 6 days to finish the work.

20. Factors of the equation 4x2 + 20 x +25 is _______.

1. (2x +5)(2x +5)
2. (2x -5)(2x -5)
3. (3x +5)(5x +5)
4. (2x +3)(2x -5)

Solution:
a2 +2ab +b2 = ( a+b)2
(2x )2 + 2 (2x)(5) +(5)2
=(2x + 5)2
Factors are (2x +5)(2x +5)

ALGEBRA: Basic Operations, simple factors, Remainder Theorem, H.C.F., L.C.M. Theory of polynomials, solutions of quadratic equations, relation between its roots and coefficients (Only real roots to be considered). Simultaneous linear equations in two unknowns – analytical and Graphical solutions. Simultaneous linear in equations in two variables and their solutions. Practical problems leading to two simultaneous linear equations or in equations in two variables or quadratic equations in one variable and their solutions. Set language and set notation, Rational expressions and conditional identities, laws of indices.

Sample Questions:

1. If x=2, y=3 is a solutions of a pair of lines 2x-3y+a = 0 and 2x+3y-b+2 then_______.

1. a= 3b
2. a+3b = 0
3. 3a+b = 0
4. 3a = b

Solution:

put x=2, y=3 in 2x-3y+a = 0

2*2 – 3*3 + a = 0 => a= 5

put x=2, y=3 in 2x+3y-b+2

2 *2+3*3-b+2 = 0 => b= 15

3a = b.

2. The pair satisfying 2x+y=6 is ____.

1. (1,2)
2. (2,1)
3. (2,2)
4. (1,1)

Solution:

2(2)+(2)=4+2 = 6.

3. If -4 is a zero of the polynomial x2 – x (2 + 2k), then the value of K is_______.

1. 1
2. 2
3. 3
4. 4

Solution:

4. The number of zeroes of the polynomial (x2 – 2x) is ________.

1. 1
2. 2
3. 0
4. 3

5. If the graph of a polynomial does not intersect the x-axis then the number of zeroes of the polynomial is ______.

1. 0
2. 1
3. -1
4. 3

6.Which of the following is not a quadratic equation?

1. (x-2)2+1 = 2x-3
2. x(x+1)+8 =(x+2)(x-1)
3. x(2x+3) = x2+1
4. (x-2)2= x-4

Solution:

x(x+1)+8 = (x+2)(x-1)

x2+x+8=x-4

x+12=0

7. The roots of quadratic equation x-2x+1=0. are______.

1. 1,1
2. 1,-1
3. -1,-1
4. 2,2

Solution:

x-2x+1=0

(x-1)2=0 => x = 1,1

8.If p-1, p+3 and 3p-1 are three consecutive terms of an AP, then P is equal to ________.

1. -4
2. 4
3. -1
4. 1

Solution:

(p+3)-(p-1) = (3p-1) – (p+3)

4p = 29-4

=> p = 4

9. Common difference of the AP x+y, x-y, x-3y is ________.

1. -2y
2. 3y
3. -3y
4. 2y

10. For the value of k, k+2, 4k-6, 3k-2 are three consecutive terms of an AP?

1. -3
2. 3
3. -2
4. None

Solution:

(4k-6)-(k+2) = (3k-2)- (4k-6)

3k – 8 = -k + 4

=> k = 3

TRIGONOMETRY: Sine x, Cosine x, Tangent x when Oo = x = 90o values of sin x, cos x and tan x, for x= Oo , 30o , 45o , 60o and 90o.

• Simple trigonometric identities.
• Use of trigonometric tables.
• Simple cases of heights and distances.

Sample Questions:

1. The angle of the elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower is 30°. Find the height of the tower.

1. 10√3
2. 20√3
3. 30√3
4. 40√3

Solution:

In right triangle ABC
Tan 30° = AB/BC
1/√3 = AB/30
AB = 30/√3
AB = 10√3

2. A statue, 1.6m tall stands on the top of a pedestal. From a point the ground the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

1. 0.8(√3+1)
2. 0.8
3. √3+1
4. 8(√3+1)

Solution:

let the height of the pedestal be hm,
Then BC = hm
In right tingle ACP
Tan60° = AC/PC
√3= AB+BC / PC
In right tingle BCP
Tan45° = BC/PC
h = 0.8(√3+1)

3. If the height and lenght of the shadow of a man are the same, then the angle of elevation of the sum is _______.

1. 30°
2. 60°
3. 45°
4. 15°

Solution:

Tanθ = AB/BC = 5√3/5 = √3 = Tan60°

4. A ship is sighted at sea from the top oa light of 75m height. If the angle of depression is found to be 30° the distance of ship from the light house is_______.

1. 25√3
2. 75√3
3. 75/√2
4. 75√2

Solution:

Tan 30° = BC/AC
1/√3 = 75/AC
AC = 75√3m

5. A 1.8m tall girl stands at a distance of 4.6m from a lamp post and caste a shadow of 5.4m on the ground. Height the lamp post is ______.

1. 1.53m
2. 10/3m
3. 13.8m
4. 0.8m

Solution:

Tanθ = CP/PD
= 1.8/5.4 = 1/3
Tanθ = AB/PB
1/3 = AB/10
AB = 10/3m

6. The angle of the elevation of the top of a tower from a point P on the ground is α. After walking a distanced towards the foot of the tower angle of elevation is found to be β then ________.

1. α < β
2. α > β
3. α = β
4. None of these

Solution:

β = α + ∠PAQ
β > α
α < β

7. Statement 1: The angles of elevation of a tower from two points which are at distances 9m and 64mts from the foot of the tower on the opposite sides are complementary. The height of the tower is 24mts.
Statement 2: The angles of the elevation of a tower from the points which are at distance a and b from the foot of the tower are complementary then the height of the tower is √ab.

1. Both statements 1 and 2 are true.
2. Both statements 1 and 2 are false.
3. Statement 1 is true but statement 2 is false.
4. Statement 1 is false but statement 2 is true.

8. A pole stands vertically inside a triangular park ABC, If the angle of elevation of the top of the pole from each corner of the park is same the triangle ABC the foot of the pole is at the __________.

1. Circumcenter
2. incenter
3. orthocenter
4. centroid

9. A lamp post 5√3 high carts a shadow 5m long on the ground the sun’s elevation at this point is __________.

1. 30°
2. 45°
3. 60°
4. 90°

Solution:

Tanθ = AB/BC = 5√3/5 = √3 = Tan60°
θ = 60°

10. The angle of elevation of the top of a 15m high tower at a point 15m away from the base of the tower is _________.

1. 30°
2. 60°
3. 45°
4. 75°

Solution:

Tanθ = AB/BC =15/15 = 1 => Tan45°

Geometry: Lines and angles, Plane and plane figures, Theorems on (i) Properties of angles at a point, (ii) Parallel lines, (iii) Sides and angles of a triangle, (iv) Congruency of triangles, (v) Similar triangles, (vi) Concurrence of medians and altitudes, (vii) Properties of angles, sides and diagonals of a parallelogram, rectangle and square, (viii) Circles and its properties including tangents and normal, (ix) Loci.

Sample Questions:

1. The sum of interior angles of a triangle is  ________.

1. 180°
2. 360°
3. 270°
4. 300°

2. In the fig, AOB is a line OC⊥AB, OP bisects ∠BOC  and OQ bisects  ∠AOC then  ∠POQ is =______.

1. 60°
2. 75°
3. 65°
4. 90°

Solution:  2x + 210° = 260 => 2x = 50° => x = 25°

Complement of 25° is 90° – 25° = 65°

3. What value of x would make AOB is a line if ∠AOC = 4x, ∠BOC = 6x+30°.

1. 36°
2. 30°
3. 20°
4. 15°

4. In the figure ∠XYZ=64°, If YQ bisects ∠ZYP then the measure of reflex ∠QYP is _______.

1. 142°
2. 98°
3. 244°
4. 302°

5. If AB/CD and P is any point as show in the figure, then ∠ABP + ∠BOD + ∠CDP is ________.

1. 72°
2. 180°
3. 360°
4. 54°

Solution: Draw PX // AB // CD

∠ABP + ∠BPX = 180° …………….(1)

∠CDP + ∠DPX = 180° …………….(2)

∠ABP + ∠BPX + ∠DPX + ∠CDP = 360°

=> ∠ABP + ∠BOD + ∠CDP = 360°

6. If two hands of a clock shows the time of 7am and, then the angle between them is _______.

1. 180°
2. 210°
3. 270°
4. 240°

7. If ∠AOB = 100°, where ∠BOC:∠COD = 3:5 then ∠AOD = ______.

1. 140°
2. 150°
3. 130°
4. 125°

8. If an angle X° is equal to its supplement then the angle is ______.

1. 1/3 right angle
2. 1/4 complete angle
3. 1/4 fourth angle
4. 2/5 right angle

9. One angle forming a linear pair is twice the order. The larger angle is ________.

1. 120°
2. 60°
3. 160°
4. 100°

>10. From the given diagram of parallel lines EF, GH, IJ are intercepted by transversals l and m. Where EG = 2cm, GI = 4cm, FH = 3cm, then HJ is _______.

1. 8cm
2. 9cm
3. 12cm
4. 6cm

11. ΔABC is right angled at A. If AB = 24mm and AC =mm then BC  = ?

1. 25 mm
2. 35 mm
3. 45 mm
4. 10 mm

Solution:

By pythagoras theorem, we have BC2 = AB2 + AC2

242 + 72 = 576 + 49 = 625

625 = 25mm

12. In the parallelogram PQRS  OS = 5cm and PR is 7cm more than QS find QP.

1. 8.5 cm
2. 8 cm
3. 10 cm
4. 9 cm

13. PA is a triangle to a circle from a point O with centre O. Find the radius OA if PA = 4cm and OP = 5cm

1. 1 cm
2. 2 cm
3. 3 cm
4. 4 cm

Solution:

√OAP = 90°

OA2 + AP2 = OP

OA2 + 42 = 5

OA = 3cm

14. Assertion (A): A(0,2), B(0,-2) and PA + PB = 6. The locus of P is an ellipse.
Reason: Locus of a point, the sum of whose distance from two feild points A and B is always constant and greater then AB is an ellipse.
Which of the following is correct?

1. A is true, R is true and R is correct explanation of A
2. A is true, R is also true but R is not correct explanation of A
3. A true, R false
4. A false, R true

Solution:
PA +PB = 6> AB(=4)
∴ Because of R, locus of P is an ellipse with A,B as its foci.

15. If a point P moves such that its distance from the point (1,1) and the line x + y + 2 = 0 are equal, then locus of P is_______.

1. a parabola
2. pair of straight lines
3. straight line
4. an ellipse

Solution:
P is the moving point and A(1,1) is a fixes point PM is the perpendicular drawn from P to x+y+2 = 0. PA = PM and so locus of P is a parabola.

Mensuration: Areas of squares, rectangles, parallelograms, triangle and circle. Areas of figures which can be split up into these figures (Field Book), Surface area and volume of cuboids, lateral surface and volume of right circular cones and cylinders, surface area and volume of spheres.

Sample Questions:

1. What is the are of an equilateral triangle of side 16 cm?

1. 48√3 cm2
2. 128√3 cm2
3. 9.6√3 cm2
4. 64√3 cm2

Solution:
Area of an equilateral triangle = √3/4 S2
If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm2

2. Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.

1. 225 cm2
2. 275 cm2
3. 285 cm2
4. 315 cm2

Solution:
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm2

3. Find the area of a parallelogram with base 24 cm and height 16 cm.

1. 262 cm2
2. 384 cm2
3. 192 cm2
4. 131 cm2

Solution:
Area of a parallelogram = base * height = 24 * 16 = 384 cm2

4. The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle?

1. 1 : 96
2. 1 : 48
3. 1 : 84
4. 1 : 68

Solution:

Let the length and the breadth of the rectangle be 4x cm and 3x respectively.
(4x)(3x) = 6912
12x2 = 6912
x2 = 576 = 4 * 144 = 22 * 122 (x > 0)
=> x = 2 * 12 = 24
Ratio of the breadth and the areas = 3x : 12x2 = 1 : 4x = 1: 96.

5. The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?

1. 8
2. 12
3. 6
4. 2

Solution:

Let the sides of the rectangle be l and b respectively.
From the given data,
(√l2 + b2) = (1 + 108 1/3 %)lb
=> l+ b2 = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l2 + b2)/lb = 25/12
12(l2 + b2) = 25lb
12l2 + 12b2 + 24lb = 49lb
12(l2 + b2 + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)2 = 49lb
=> 12(14)2 = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.

6. An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m?

1. Rs. 3642.40
2. Rs. 3868.80
3. Rs. 4216.20
4. Rs. 4082.40

Solution:

Length of the first carpet = (1.44)(6) = 8.64 cm
Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)
= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m
Cost of the second carpet = (45)(12.96 * 7) = 315 (13 – 0.04) = 4095 – 12.6 = Rs. 4082.40

7. The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles.

1. 4192 sq m
2. 4304 sq m
3. 4312 sq m
4. 4360 sq m

Solution:

Let the radii of the smaller and the larger circles be s m and l m respectively.
2∏s = 264 and 2∏l = 352
s = 264/2∏ and l = 352/2∏
Difference between the areas = ∏l2 – ∏s2
= ∏{1762/∏2 – 1322/∏2}
= 1762/∏ – 1322/∏
= (176 – 132)(176 + 132)/∏
= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m

8. A cube of side one meter length is cut into small cubes of side 10 cm each. How many such small cubes can be obtained?

1. 10
2. 100
3. 1000
4. 10000

Solution:

Along one edge, the number of small cubes that can be cut
= 100/10 = 10
Along each edge 10 cubes can be cut. (Along length, breadth and height). Total number of small cubes that can be cut = 10 * 10 * 10 = 1000

9. The length of a rectangle is two – fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 1225 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units?

1. 140
2. 156
3. 175
4. 214

Solution:

Given that the area of the square = 1225 sq.units

=> Side of square = √1225 = 35 units
The radius of the circle = side of the square = 35 units Length of the rectangle = 2/5 * 35 = 14 units
Given that breadth = 10 units
Area of the rectangle = lb = 14 * 10 = 140 sq.units
10. The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface areas?
1. 81 : 121
2. 9 : 11
3. 729 : 1331
4. 27 : 121

Solution:

Ratio of the sides = ³√729 : ³√1331 = 9 : 11

Ratio of surface areas = 92 : 112 = 81 : 121

Statistics: Collection and tabulation of statistical data, Graphical representation frequency polygons, histograms, bar charts, pie charts etc. Measures of central tendency.

Sample Questions:

1. Find the class marks of classes 10-25 and 35-55.

1. 10,35
2. 25,55
3. 15,20
4. 17.5,45

Solution:

class marks = (10+25)/2 , (35+55)/2 =>17.5,45

2. The mean of first natural numbers is ______.

1. (n+1)/2
2. n(n+1)/2
3. (n-1)/2

3.The means of the frequency distribution is ______.

 x 1 2 3 4 5 6 f 45 25 19 8 2 1
1. 1
2. 2
3. 3
4. 4

4. The modal class of data given below is 10-15 then ________.

 x 0-5 5-10 10-15 15-20 20-25 Frequency 7 6 f 4 3
1. f < 8
2.  ≥ 8
3. f > 8 only
4. f < 7

5. The number of family members of 30 families of a village is according to the following table. Find the mode.

1. 2
2. 4
3. 3
4. 6

6. The wickets taken over by a bowler in 10 cricket matches are as follows 2 6 4 5 0 1 2 1 3 2 3 . Find the mode of this data.

1. 0
2. 1
3. 2
4. 3

7. The cumulative frequency of the class 55-58 is how much greater than the frequency of the class 58-61 in the following distribution.

 Height 52-55 55-58 58-61 61-64 No of students 10 20 25 10
1. 2
2. 3
3. 4
4. 5

Solution:

Cumulative frequency of the class 55-58 = 10+20 = 30

Frequency of the class 58-61 = 25

8. The mean and median of a data are respectively 20 and 22. The value of mode is ________.

1. 20
2. 26
3. 22
4. 21

Solution:

3 median =  mode + 2 mean

3 * 22 = Mode + 2*20

Mode = 26

9. If mean = (3medain – mode)/k, then K = ______.

1. 1
2. 1/2
3. 2
4. None of these

10. Find the mode of the following distribution.

 Size of the shows 4.5 5 5.5 <6.0/td> 6.5 7 7.5 8 8.5 9 No of shoes 1 2 3 4 5 15 30 60 95 82 75
1. 6
2. 7
3. 8
4. 9

### Tips

Candidates can crack the exam only with right approach and correct strategy, following are few strategies to crack the FSO exam:

Tip 1: Read News papers daily for improving general awareness.

Tip 2: Work on your Time management, Speed and Accuracy.

Tip 3: Prepare yourself for a high level of the exam to avoid any last minute surprises. Prepare with a mindset of a difficult level of exam, and attempting questions in the exam will be comparatively easy for you.

Tip 4: Attempt as many mock tests as you can. Online full-length tests will help you overcome your exam paranoia and you’ll be able to prepare more efficiently. Mock tests will also help you work on the 3 parameters i.e. Time management, Speed and Accuracy.

Tip 5: Focus more on Telangana related issues for GK section.