Exam Mode  Paper  No of Questions  Duration (min)  Max Marks 

OMR/Online  Mathematics Paper (Paper – II)(SSC/Higher School Standard) 
100 Questions  100 Minutes  100 Marks 
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. Factorisation Theorem. H.C.F. and L.C.M. Euclidean algorithm. Logarithms to base 10, laws of logarithms, use of logarithmic tables. 
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 â€“ 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. 
Trigonometry  Sine x, Cosine x, Tangent x when \(0^0\) = x = \(90^0\) values of sin x, cos x and tan x, for x= \(0^0\), \(30^0\), \(45^0\), \(60^0\) and \(90^0\). Simple trigonometric identities, use of trigonometric tables, simple cases of heights and distances. 
Geometry  Lines and angles, Plane and plane figures, Theorems on

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. 
Statistics  Collection and tabulation of statistical data, Graphical representation frequency polygons, histograms, bar charts, pie charts etc. Measures of central tendency. 
Number SystemNatural numbers, Integers, Rational and Real numbers, Fundamental operations, addition, subtraction, multiplication, division, Square roots, Decimal fractions. Unitary methodtime and distance, time and work, percentages, applications to simple and compound interest, profit and loss, ratio and proportion, variation. Elementary Number Theory â€“ 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.
1. 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:
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.
2. 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:
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.
3. 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?
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.
4. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
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.
5. 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?
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)
6. Which of the following statements is not correct?
Answer: 2
Solution:(a) Since log_{a}Â a = 1, so log_{10} 10 = 1.
(b) log (2 + 3) = log_{5}Â 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.
7.Â If log 27 = 1.431, then the value of log 9 is:
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.
8.Â A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?
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.
9.Â 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:
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.
10.Â Factors of the equation 4x^{2} + 20 x +25 is _______.
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)
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 â€“ 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.
1. If x=2, y=3 is a solutions of a pair of lines 2x3y+a = 0 and 2x+3yb+2 then_______.
Answer: 4
Solution: put x=2, y=3 inÂ 2x3y+a = 0
2*2 – 3*3 + a = 0 => a= 5
put x=2, y=3 inÂ 2x+3yb+2
2 *2+3*3b+2Â = 0 => b= 15
3a = b.
2. The pair satisfying 2x+y=6 is ____.
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_______.
Answer: 3
4. The number of zeroes of the polynomial (x^{2} – 2x) is ________.
Answer: 2
5. If the graph of a polynomial does not intersect the xaxis then the number of zeroes of the polynomial is ______.
Answer: 1
6.Which of the following is not a quadratic equation?
Answer: 2
Solution: x(x+1)+8 =Â (x+2)(x1)
x^{2}+x+8=x^{2Â }4
x+12=0
7. The roots of quadratic equationÂ x^{2Â }2x+1=0. are______.
Answer: 1
Solution: x^{2}2x+1=0
(x1)^{2}=0 => x = 1,1
8.If p1, p+3 and 3p1 are three consecutive terms of an AP, then P is equal to ________.
Answer: 2
Solution: (p+3)(p1) = (3p1) – (p+3)
4p = 294
=> p = 4
9. Common difference of the AP x+y, xy, x3y is ________.
Answer: 1
10. For the value of k, k+2, 4k6, 3k2 are three consecutive terms of an AP?
Answer: 2
Solution: (4k6)(k+2) = (3k2) (4k6)
3k – 8 = k + 4
=> k = 3
Sine x, Cosine x, Tangent x when O^{o}Â = x = 90^{o}Â values of sin x, cos x and tan x, for x= O^{o}Â , 30^{o}Â , 45^{o}Â , 60^{o}Â and 90^{o}.
1. The angle of the elevationÂ of theÂ top of a tower from a point on the ground, which is 30m away from the foot of the tower is 30Â°. Find the height of the tower.
Answer: 1
Solution:
Answer: 1
Solution:
let the height of the pedestal be hm,
Then BC = hm
In right tingle ACP
Tan60Â° = AC/PC
v3= AB+BC / PC
In right tingle BCP
Tan45Â° = BC/PC
h = 0.8(v3+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 _______.
Answer: 3
Solution:
Answer: 2
Solution:
Tan 30Â° = BC/AC
1/v3 = 75/AC
AC = 75âˆš3m
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 ______.
Answer: 2
Solution:
Tan? = CP/PD
= 1.8/5.4 = 1/3
Tan? = AB/PB
1/3 = AB/10
AB = 10/3m
6. The angle of the elevation of the top of a tower from a pointÂ P on the ground is a. After walking a distanced towards the foot of the tower angle of elevation is found to be ÃŸ then ________.
Answer: 1
Solution:
ÃŸ = a + PAQ
ÃŸ > a
a < ÃŸ
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 vab.
Answer: 1
8.Â A pole stands vertically inside a triangularÂ park ABC, If the angle of elevation of the top of the pole from each corner of the park is same the triangleÂ ABC the foot of the pole is at the __________.
Answer: 1
9.Â A lamp post 5v3 high carts a shadow 5m long on the ground the sun’s elevation at this point is __________.
Answer: 3
Solution:
Tan? = AB/BC = 5âˆš3/5 = âˆš3 = Tan60Â°
? = 60Â°
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 _________.
Answer: 3
Solution:
Tan? = AB/BC =15/15 = 1 =>Â Tan45Â°
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.
1. The sum of interior angles of a triangle is Â ________.
Answer: 1
2. In the fig, AOB is a line OC?AB, OP bisects ?BOCÂ and OQ bisectsÂ ?AOC thenÂ ?POQ is =______.
Answer: 3
Solution:Â 2x + 210Â° = 260 => 2x = 50Â° => x = 25Â°
Complement of 25Â° is 90Â° – 25Â° = 65Â°
3. What value of x would make AOB is a line ifÂ ?AOC = 4x,Â ?BOC = 6x+30Â°.
Answer: 4
4. In the figureÂ ?XYZ=64Â°, If YQ bisects ?ZYP then the measure of reflex ?QYP is _______.
Answer: 4
5. If AB/CD and P is any point as show in the figure, thenÂ ?ABP +Â ?BOD +Â ?CDP is ________.
Answer: 3
Solution:Â Draw PX // AB // CD
?ABP +Â ?BPX = 180Â° …………….(1)
?CDP +Â ?DPX = 180Â° …………….(2)
Now adding (1) and (2)
?ABP +Â ?BPX +Â ?DPX +Â ?CDP = 360Â°
=>Â ?ABP +Â ?BOD +Â ?CDP = 360Â°
6. If two hands of a clock shows the time of 7am and, then the angle between them is _______.
Answer: 2
7. If ?AOB = 100Â°, where ?BOC:?COD = 3:5 then ?AOD = ______.
Answer: 3
8. If an angle XÂ° is equal to its supplement then the angle is ______.
Answer: 2
9. One angle forming a linear pair is twice the order. The larger angle is ________.
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 _______.
Answer: 4
11.Â ?ABC is right angled at A. If AB = 24mm and AC =mm then BC Â = ?
Answer: 1
Solution:
By pythagoras theorem, we have BC^{2} = AB^{2} + AC^{2}
24^{2}Â + 7^{2} = 576 + 49 = 625
v625 = 25mm
12. In the parallelogram PQRS Â OS = 5cm and PR is 7cm more than QS find QP.
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
Answer: 3
Solution:
vOAP = 90Â°
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?
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_______.
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.
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.
1. What is the are of an equilateral triangle of side 16 cm?
Answer: 4
Solution:
Area of an equilateral triangle = v3/4 S^{2}
If S = 16, Area of triangle = v3/4 * 16 * 16 = 64v3 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.
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.Â Find the area of a parallelogram with base 24 cm and height 16 cm.
Answer: 2
Solution:
Area of a parallelogram = base * height = 24 * 16 = 384 cm^{2}
4.Â 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?
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.Â 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?
Answer: 4
Solution:
Let the sides of the rectangle be l and b respectively.
From the given data,
(vl^{2} + b^{2}) = (1 + 108 1/3 %)lb
=> l^{2Â }+ 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.Â 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?
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.Â The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles.
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.Â 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?
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.Â 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?
Answer: 1
Solution:
Given that the area of the square = 1225 sq.units
=> Side of square = v1225 = 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.Â The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface areas?
Answer: 1
Solution:
Ratio of the sides = Â³v729 : Â³v1331 = 9 : 11
Ratio of surface areas = 9^{2}Â : 11^{2}Â = 81 : 121
Collection and tabulation of statistical data, Graphical representation frequency polygons, histograms, bar charts, pie charts etc. Measures of central tendency.
1. Find the class marks of classes 1025 and 3555.
Answer: 4
Solution:
class marks = (10+25)/2 , (35+55)/2 =>17.5,45
2. The mean of first natural numbers is ______.
Answer: 1
3.The means of the frequency distribution is ______.
x  1  2  3  4  5  6 
f  45  25  19  8  2  1 
Answer: 2
4. The modal class of data given below is 1015 then ________.
x  05  510  1015  1520  2025 
Frequency  7  6  f  4  3 
Answer: 2
5. The number of family members of 30 families of a village is according to the following table. Find the mode.
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.
Answer: 3
7. The cumulative frequency of the class 5558 is how much greater than the frequency of the class 5861 in the following distribution.
Height  5255  5558  5861  6164 
No of students  10  20  25  10 
Answer: 4
Solution:
CumulativeÂ frequencyÂ of the class 5558 = 10+20 = 30
Frequency of the class 5861 = 25
8. The mean and median of a data are respectively 20 and 22. The value of mode is ________.
Answer: 2
Solution:
3 median =Â mode + 2 mean
3 * 22 = Mode + 2*20
Mode = 26
9. If mean = (3medain – mode)/k, then K = ______.
Answer: 3
10. Find the mode of the following distribution.
Size of the shows  4.5  5.0  5.5  6.0  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 
Answer: 3
Tip 1: Learn the material covering all the topics mentioned in the syllabus. Learn the concepts first and then practice relevant sample questions which will ensure that the concepts are never forgotten and can help to solve any type of questions.
Tip 2: Work on Time management, Speed and Accuracy.
Tip 3: Prepare for a difficult level exam to avoid any last minute surprises and to avoid getting anxious during the exam.
Tip 4: Work on Mock Tests to have a better understanding of the exact exam pattern, online exam, and expected level of questions. Online fulllength mock tests will help to overcome exam paranoia and the candidates will be able to perform more efficiently. Mock tests will also help in enhancing the 3 parameters: Time management, Speed, and Accuracy.
Tip 5: Identify and analyze the weaknesses/ weakest topics in a timely fashion.
FRO Mathematics Practice: Click Here
FRO Mathematics Mock Test : Click Here