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.

Exam Mode | Paper Name | No. of Questions | Max Marks | Duration |
---|---|---|---|---|

Objective Type | Paper 2- General Mathematics |
100 | 100 | 90 minutes |

✦ Arithmetic

✦ Algebra

✦ Trigonometry

✦ Geometry

✦ Mensuration

✦ Statistics

**For FBO GM Practice: CLICK HERE**

**For FBO GM Mock Test: CLICK HERE**

**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 _________.**

- 22
- 26
- 24
- 20

**Correct Answer: 3**

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

- 63
- 54
- 27
- 72

**Correct Answer: 1**

**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
- 2
- 3
- None

**Correct Answer: 1**

**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?**

- a+b+c is not composite
- b+c+d is prime
- abc is even
- a+b+c+d si not even compisite

**Correct Answer: 2**

**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 a ^{p} – a is divisible by _______.**

- p + 4
- p + 2
- p + 3
- p

**Correct Answer: 4**

**Solution:** a^{p} – a = 2^{5} – 2 = 30 is divisible by 5.

**6.Which of the following is divisible by 3?**

- 1536
- 1535
- 1534
- 1532

**Correct Answer: 1**

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

- 194
- 210
- 252
- 258

**Correct Answer: 3**

**Solution:**

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

- 5
- 25
- 125
- 625

**Correct Answer: 3**

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

- 576
- 676
- 1296
- 1444

**Correct Answer: 2**

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

=> √*x = 26 => x = 26 ^{2}*

=> (26)^{2 }= 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
- 2
- 3
- 5

**Correct Answer: 2**

**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:**

- 4 days
- 5 days
- 6 days
- 7 days

**Correct Answer: 1**

**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:**

- 50 km
- 56 km
- 70 km
- 80 km

**Correct Answer: 1**

**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?**

- Rs. 500
- Rs. 1500
- Rs. 2000
- None of these

**Correct Answer: 3**

**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:**

- Rs. 650
- Rs. 690
- Rs. 698
- Rs. 700

**Correct Answer: 3**

**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?**

- 30%
- 70%
- 100%
- 250%

**Correct Answer: 2**

**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?**

- log
_{10}10 = 1 - log (2 + 3) = log (2 x 3)
- log
_{10}1 = 0 - log (1 + 2 + 3) = log 1 + log 2 + log 3

**Correct Answer: 2**

**Solution:**

(a) Since log_{a} a = 1, so log_{10} 10 = 1.

(b) log (2 + 3) = log_{5} 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:**

- 0.934
- 0.945
- 0.954
- 0.958

**Correct Answer: 3**

**Solution:**

log 27 = 1.431

log (3^{3} ) = 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%?**

- 3
- 4
- 5
- 6

**Correct Answer: 3**

**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:**

- 4 days
- 6 days
- 8 days
- 12 days

**Correct Answer: 2**

**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 4x ^{2} + 20 x +25 is _______.**

- (2x +5)(2x +5)
- (2x -5)(2x -5)
- (3x +5)(5x +5)
- (2x +3)(2x -5)

**Correct Answer: 1**

**Solution:**

a^{2} +2ab +b^{2} = ( 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_______.**

- a= 3b
- a+3b = 0
- 3a+b = 0
- 3a = b

**Correct Answer: 4**

**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 ____.**

- (1,2)
- (2,1)
- (2,2)
- (1,1)

**Correct Answer: 3**

**Solution:**

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

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

- 1
- 2
- 3
- 4

**Correct Answer: 3**

**Solution:**

**4. The number of zeroes of the polynomial (x ^{2} – 2x) is ________.**

- 1
- 2
- 0
- 3

**Correct Answer: 2**

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

- 0
- 1
- -1
- 3

**Correct Answer: 1**

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

- (x-2)
^{2}+1 = 2x-3 - x(x+1)+8 =(x+2)(x-1)
- x(2x+3) = x
^{2}+1 - (x-2)
^{2}= x^{3 }-4

**Correct Answer: 2**

**Solution:**

x(x+1)+8 = (x+2)(x-1)

x^{2}+x+8=x^{2 }-4

x+12=0

**7. The roots of quadratic equation x ^{2 }-2x+1=0. are______.**

- 1,1
- 1,-1
- -1,-1
- 2,2

**Correct Answer: 1**

**Solution:**

x^{2 }-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 ________.**

- -4
- 4
- -1
- 1

**Correct Answer: 2**

**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 ________.**

- -2y
- 3y
- -3y
- 2y

**Correct Answer: 1**

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

- -3
- 3
- -2
- None

**Correct Answer: 2**

**Solution:**

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

3k – 8 = -k + 4

=> k = 3

**TRIGONOMETRY:** Sine x, Cosine x, Tangent x when O^{o} = x = 90^{o} values of sin x, cos x and tan x, for x= O^{o} , 30^{o} , 45^{o} , 60^{o} and 90^{o}.

- 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.**

- 10√3
- 20√3
- 30√3
- 40√3

**Correct Answer: 1**

- 0.8(√3+1)
- 0.8
- √3+1
- 8(√3+1)

**Correct Answer: 1**

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)

- 30°
- 60°
- 45°
- 15°

**Correct Answer: 3**

- 25√3
- 75√3
- 75/√2
- 75√2

**Correct Answer: 2**

- 1.53m
- 10/3m
- 13.8m
- 0.8m

**Correct Answer: 2**

**Solution:**

Tanθ = CP/PD

= 1.8/5.4 = 1/3

Tanθ = AB/PB

1/3 = AB/10

AB = 10/3m

- α < β
- α > β
- α = β
- None of these

**Correct Answer: 1**

**Solution:**

β = α + ∠PAQ

β > α

α < β

7.

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

**Correct Answer: 1**

**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 __________.**

- Circumcenter
- incenter
- orthocenter
- centroid

**Correct Answer: 1**

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

- 30°
- 45°
- 60°
- 90°

**Correct Answer: 3**

**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 _________.**

- 30°
- 60°
- 45°
- 75°

**Correct Answer: 3**

**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 ________.**

- 180°
- 360°
- 270°
- 300°

**Correct Answer: 1**

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

- 60°
- 75°
- 65°
- 90°

**Correct Answer: 3**

**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°.**

- 36°
- 30°
- 20°
- 15°

**Correct Answer: 4 **

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

- 142°
- 98°
- 244°
- 302°

**Correct Answer: 4 **

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

- 72°
- 180°
- 360°
- 54°

**Correct Answer: 3**

**Solution:** Draw PX // AB // CD

∠ABP + ∠BPX = 180° …………….(1)

∠CDP + ∠DPX = 180° …………….(2)

Now adding (1) and (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 _______.**

- 180°
- 210°
- 270°
- 240°

**Correct Answer: 2**

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

- 140°
- 150°
- 130°
- 125°

**Correct Answer: 3**

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

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

**Correct Answer: 2**

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

- 120°
- 60°
- 160°
- 100°

**Correct Answer: 1**

>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 _______.

- 8cm
- 9cm
- 12cm
- 6cm

**Correct Answer: 4**

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

- 25 mm
- 35 mm
- 45 mm
- 10 mm

**Correct Answer: 1**

**Solution:**

By pythagoras theorem, we have BC^{2} = AB^{2} + AC^{2}

24^{2} + 7^{2} = 576 + 49 = 625

√625 = 25mm

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

- 8.5 cm
- 8 cm
- 10 cm
- 9 cm

**Correct Answer: 1**

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 cm
- 2 cm
- 3 cm
- 4 cm

**Correct Answer: 3**

**Solution:**

√OAP = 90°

OA^{2} + AP^{2} = OP

OA^{2} + 4^{2} = 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?**

- A is true, R is true and R is correct explanation of A
- A is true, R is also true but R is not correct explanation of A
- A true, R false
- A false, R true

**Correct Answer: 1**

**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_______.**

- a parabola
- pair of straight lines
- straight line
- an ellipse

**Correct Answer: 1**

**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?**

- 48√3 cm
^{2} - 128√3 cm
^{2} - 9.6√3 cm
^{2} - 64√3 cm
^{2}

**Correct Answer: 4**

**Solution:**

Area of an equilateral triangle = √3/4 S^{2}

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.**

- 225 cm
^{2} - 275 cm
^{2} - 285 cm
^{2} - 315 cm
^{2}

**Correct Answer: 3**

**Solution:**

Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm^{2}

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

- 262 cm
^{2} - 384 cm
^{2} - 192 cm
^{2} - 131 cm
^{2}

**Correct Answer: 2**

**Solution:**

Area of a parallelogram = base * height = 24 * 16 = 384 cm^{2}

**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 : 96
- 1 : 48
- 1 : 84
- 1 : 68

**Correct Answer: 1**

**Solution:**

Let the length and the breadth of the rectangle be 4x cm and 3x respectively.

(4x)(3x) = 6912

12x^{2} = 6912

x^{2} = 576 = 4 * 144 = 2^{2} * 12^{2} (x > 0)

=> x = 2 * 12 = 24

Ratio of the breadth and the areas = 3x : 12x^{2} = 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?**

- 8
- 12
- 6
- 2

**Correct Answer: 4**

**Solution:**

Let the sides of the rectangle be l and b respectively.

From the given data,

(√l^{2} + b^{2}) = (1 + 108 1/3 %)lb

=> l^{2 }+ b^{2} = (1 + 325/3 * 1/100)lb

= (1 + 13/12)lb

= 25/12 lb

=> (l2 + b2)/lb = 25/12

12(l2 + b2) = 25lb

Adding 24lb on both sides

12l^{2} + 12b^{2} + 24lb = 49lb

12(l^{2} + b^{2} + 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?**

- Rs. 3642.40
- Rs. 3868.80
- Rs. 4216.20
- Rs. 4082.40

**Correct Answer: 4**

**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.**

- 4192 sq m
- 4304 sq m
- 4312 sq m
- 4360 sq m

**Correct Answer: 3**

**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 = ∏l^{2} – ∏s^{2}

= ∏{176^{2}/∏^{2} – 1322/∏^{2}}

= 176^{2}/∏ – 132^{2}/∏

= (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?**

- 10
- 100
- 1000
- 10000

**Correct Answer: 3**

**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?**

- 140
- 156
- 175
- 214

**Correct Answer: 1**

**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

- 81 : 121
- 9 : 11
- 729 : 1331
- 27 : 121

**Correct Answer: 1**

**Solution:**

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

Ratio of surface areas = 9^{2} : 11^{2} = 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.**

- 10,35
- 25,55
- 15,20
- 17.5,45

**Correct Answer: 4**

**Solution:**

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

**2. The mean of first natural numbers is ______.**

- (n+1)/2
- n(n+1)/2
- (n-1)/2
- n²

**Correct Answer: 1**

**3.The means of the frequency distribution is ______.**

x |
1 | 2 | 3 | 4 | 5 | 6 |

f |
45 | 25 | 19 | 8 | 2 | 1 |

- 1
- 2
- 3
- 4

**Correct Answer: 2**

**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 |

- f < 8
- f ≥ 8
- f > 8 only
- f < 7

**Correct Answer: 2**

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

- 2
- 4
- 3
- 6

**Correct Answer: 4**

**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.**

- 0
- 1
- 2
- 3

**Correct Answer: 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 |

- 2
- 3
- 4
- 5

**Correct Answer: 4**

**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 ________.**

- 20
- 26
- 22
- 21

**Correct Answer: 2**

**Solution:**

3 median = mode + 2 mean

3 * 22 = Mode + 2*20

Mode = 26

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

- 1
- 1/2
- 2
- None of these

**Correct Answer: 3**

**10. Find the mode of the following distribution.**

Size of the shows |
4.5 | 5.0 | 5.5 | <6.0/td> | 6.5 | 7.0 | 7.5 | 8.0 | 8.5 | 9.0 | ||

No of shoes |
1 | 2 | 3 | 4 | 5 | 15 | 30 | 60 | 95 | 82 | 75 |

- 6
- 7
- 8
- 9

**Correct Answer: 3**

**✦ 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.