SSLC Maths Question Bank & Solution All Chapters

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

SSLC

Problems (All Chapters) (English Medium) Prepared By Fassal Peringolam (Plus Two Maths & Science Teacher)

BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com http://brainsmoozhikkal.blogspot.com

Page

1

Mob: 9048332443

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Arithmetic Sequences Equations:  First term: f  Common difference: d  Nth term: 1 (a-common difference; a+b first term)  Common difference  Sum of sequence: 2

2

1

2 2 2 

etc. ;

Page

2



2

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Sample Questions PROBLEM

1.

For an arithmetic sequence , the 8th term is 35 and the 11th term is 47 a) Find 14th term b) Find 5th term c) To get the 17th term, how much is to be added to the 8th term?

Answer Common difference =d 8th term = 35 11th term = 47 8th term + 3d = 11th term 3d=11th term - 8th term = 47- 35= 12 a) 14th term = 11th term + 3d =47 + 12 = 59 b) 5th term = 8th term – 3d =35-12 = 23 c) 17th term = 8th term + 9d 17th term = 8th term + 3 x12 = 8th term + 36 PROBLEM

2.

For the arithmetic sequence 22, 26, 30... a) What is the common difference? b) Find the 7th term c) Will 50 be a term of this sequence? Why? d) Can the difference between any two terms of this sequence be 50? Justify your answer.

Page

3

Answer a) Common difference d = 26 – 22 = 4 b) 10 th term is 22 7 1 4 = 46

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions c) If any term of this sequence is divided by the common difference 4, the remainder is 2. 50 divided by 4 gives remainder 2. So, 50 is a term of this sequence d) The difference between any two terms of an arithmetic sequence will be the multiple of its common difference. Here, 50 is not a multiple of 4. So, for this sequence, 50 cannot be the difference of two terms. PROBLEM

3.

Consider the multiples of 7 in between 100 and 500. a) What are the first and last numbers? b) How many terms are ther in this sequence?

Answer To get the first term, first find the remainder when 100 is divided by 7, remainder is 2. Then subtract 2 from 100 and add 7 So, 1st term 100-2+7 105 To get the last term, find the remainder when 500 is divided by 7, remainder is 3. Subtract 3 from 500 So, last term 500-3 497 b) Common difference, 497 7 105 7 n (497-98)/7 57 So, number of term 57 First term of an arithmetic sequence is 6 and sum of first two terms is 10. (a) Find Common difference (b) Find the third term.

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

4.

4

PROBLEM

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer First term a 6 Sum of first two terms 10 a+a+d 10 d 10-12 -2 Third term a3 a+2d 2 PROBLEM

5.

Examine whether 685 is a term of the arithmetic sequence 7, 29, 51 …? Give reason?

Answer First term a = 7 Common difference d = 29 – 7 = 22 nth term = 685 a + (n-1)d = 685 7 + 22n - 22 = 685 n = 31.82 Thus 685 is not a term of the arithmetic sequence 7, 29, 51 … Because n is not a whole number PROBLEM

Answer First term, f =205 and Common difference d = 6 Let the nth term of the given AP be the first negative term. Then, xn < 0 f + (n-1) d < 0 205+ (n-1) x 6 < 0 205 + 6n - 6 < 0

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

5

Which term of the sequence, 205,199 … is the first negative term?

Page

6.

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 6n > 199 n > 33.17 i.e., n ≥ 34 Thus, the 34 th term of the given sequence is the first negative term. PROBLEM

7.

The 5th term of an arithmetic sequence is 34 and the 15th term is 9. (a) Find the common difference (b) Find the rth term of the arithmetic sequence.

Answer According to the given information, x5 = f + 4d = 34 and x15 = f + 14d = 9 Solving the two equations, we get f = 44 d = -2.5 Therefore, rth term = f + (r-1)d = 44 -(r-1)2.5 = 46.5 - 2.5 r PROBLEM

8.

Write the sequence got by adding 2 to the numbers got by multiplying 5 to the natural numbers starting from 1. (a) Check whether it is an Arithmetic sequence. (b) Will 100 be a term of this sequence?

CLASS X

Page

6

Answer Natural numbers starting from 1 are 1, 2, 3... Multiplying with 5; 5, 10, 15... Adding 2; 9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 7, 12, 17... (a) This is an Arithmetic sequence with common difference 5. (b) If any term of this sequence is divided by the common difference 5, the remainder is 2. 100 divided by 5 gives remainder 0, not 2. So, 100 is not a term of this sequence. PROBLEM

9.

a) Write the arithmetic sequence with 1st term 7 and common difference -2 b) Write the algebraic expression of this sequence. c) Will -30 be a term of this sequence? Justify your answer.

Answer a) 7, 5, 3... b) xn = f + (n-1) d = 7+ (n-1) × -2 = -2n +9 c) Take -30 as the nth term. If we get n as a natural number, it will be a term of the sequence. -2n + 9 = -30 n = -39 / -2 n = 19.5 Here, n is not a natural number. So, -30 is not a term of this sequence.

Page

10. For an arithmetic sequence , the 8th term is 67 and the 18th term is 147 a) Find the common difference b) Find the sum of 1st term and the 25th term c) Find 13th term d) Find the sum of the first 25 terms.

7

PROBLEM

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer a) Common difference,

18 8 (147-67)/(18-8) 80/10 8 b) 1st term + 25th term= 8th term + 18th term =67+147 = 214 c) 13th term = (12th term + 14th term)/2 = (1st term + 25th term)/2 = 214 / 2 = 107 d) Sum of the terms = Number of terms x Middle term Sum of first 25 terms = 25 x 13th term = 25 x 107 = 2675 PROBLEM

11. For what value of n, the nth terms of the arithmetic sequence 63, 65, 67… and 3, 10, 17… are equal?

CLASS X

Page

8

Answer The first AP is 63, 65, 67… Here, f=63, d=2 Therefore xn=63 + (n-1)2 The second AP is 3, 10, 17 … Here, f=3, d=7 Therefore, an= 3 + (n-1)7 Therfore, 63 + (n-1)2 = 3 + (n-1)7 60 = 5(n-1) 9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions n -1 = 12 n = 12+1 = 13 PROBLEM

12. Which term of the arithmetic sequence 5, 15, 25 ... will be 130 more than the 31st term? Answer Here, f = 5 and d = 10 x31= f + 30d = 5 + 30 x 10 = 305 Let the required term be the nth term. Then, xn = 130 + x31 f + (n-1)d = 130+305 5 + (n-1)10 = 435 (n-1)10 = 430 n-1 = 43 n = 44 Hence, the 44th term of the given AP is 130 more than its 31st term. PROBLEM

13. In an arithmetic sequence the first term is -4, the last term is 29 and the sum of all its term is 150. Find its common difference.

9

Answer f = -4, xn = 29, Sn = 150 Given that the last term is 29. 29 -4 + (n - 1)d 33 (n - 1)d … (1) Also, given that the sum of all its term is 150. 150 2 4 1 n [-8 + 33]

[Using (1)]

CLASS X

Page

300

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 300 n (25) n 12 So, from (1), we get: d = 3 PROBLEM

14. Find the sum of first 24 terms of the list of numbers whose nth term is given by an= 3+2n. Answer xn= 3+2n Now, put n=1,2,3 x1= 3+2(1) = 5 x2= 3+2(2) = 7 x3= 3+2(3) = 9 Thus, the terms of the AP are 5,7,9 Here, f = 5 and d = 2 2 5 24 2 12[10+46] 12 56 672

1 2

PROBLEM

Page

Answer -5, -8, -11, … -230 forms an AP with f=-5, d=-8-(-5) =-3 Let -230 = xn = f+(n-1)d = -5+(n-1)(-3) -230 + 5 = (n-1)(-3) n -1 = 75 n = 76 2

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

10

15. Find the sum: -5+ (-8) + (-11) +..............+ (-230)

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions [(-5) + (-230)] 38 (-235) - 8930 PROBLEM

16. Find the sum of all two-digit odd positive numbers.

Answer Two digit odd positive numbers are 11, 13, 15, ...., 99. First term, f = 11 Common difference, d = 13 - 11 = 2 Last term, xn = l =99 Now, xn f + (n - 1)d 99 11 + (n - 1)2 99 11 + 2n - 2 2n 90 n 45 2 45 11 99 2 2475 Thus, the sum of all two-digit odd positive numbers is 2475.

2)

Find the 8th term of an arithmetic sequence whose 15th term is 47 and the common difference is 4. [19] Find the 31st term of an arithmetic sequence whose 11th term is 38 and the 16th term is 73. [178]

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

1)

11

PRACTICE EXERCISE

BIO-VISION VIDEO BLOG

SSLC Mathematics

4) 5)

6)

7)

8)

9)

10) 11)

12)

The 10th term of an arithmetic sequence is 52 and 16th term is 82. (a) Find the 32nd term (b) Obtain the general term. [x32 = 162, xn = 5n + 2] Which term of the arithmetic sequence 3, 8, 13, 18 ... is 248? [50] The 7th term of an arithmetic sequence is – 4 and its 13th term is –16. (a) What is common difference? (b) Find the arithmetic sequence. [-2; 8, 6, 4, ..... ] The 17th term of an arithmetic sequence exceeds its 10th term by 7. Find the common difference. [1] If the 10th term of an arithmetic sequence is 52 and 17th term is 20 more than the 13th term, find the arithmetic sequence. [7, 12, 17, 22.......] The 9th term of an arithmetic sequence is equal to 7 times the 2nd term and 12th term exceeds 5 times the 3rd term by 2. (a) Find the first term (b) Find the common difference. [a = 1, d = 6] Find the middle term of an arithmetic sequence with 17 terms whose 5th term is 23 and the common difference is –2. [15] Find 20th and 25th terms of an arithmetic sequence 2, 5,8,11… [59, 74] For what value of n the nth term of the arithmetic sequence 23, 25, 27, 29 ….and -17,-10,-3, 4 ….are equal? [9] The sixth term of an AP is -10 and tenth term is -26.Find its 15th term. [-46]

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

3)

12

Questions & Solutions

BIO-VISION VIDEO BLOG

SSLC Mathematics

14)

15) 16) 17)

18)

19)

20) 21)

22)

23)

The angles of a triangle are in arithmetic sequence. If the greatest angle equals the sum of the other two, find the angles. [30°, 60°, 90°] Three numbers are in arithmetic sequence. If the sum of these numbers is 27 and the product is 648, find the numbers. [6, 9, 12] Find sum of 1 + 3 + 5 + ....... to 50 terms. [2500] Find the sum of first 30 even natural numbers. [930] The 10th term of an arithmetic sequence is 29 and and sum of the first 20 terms is 610. Find the sum of the first 30 terms. [1365] How many terms of the arithmetic sequence 3, 5, 7, 9 … must be added to get the sum 120? [10] The 3rd term of arithmetic sequence is –40 and 13th term is 0. Find the common difference. [4] The sum of three numbers of an AP is 15. Find its first term. [5] The seats in a theatre are arranged in 20 rows. Each row contains 8seats more than the number of seats in the previous row. If the 1st row contains 70seats find the total number of seats in the [2920] theatre. In a flower decoration, flowers are arranged in 10 concentric circles, such that their numbers from an AP. There are 40 flowers in 3rd circle and 60 flowers in 5th circle. Find the total number of flowers. [650] If (x – 3), (x + 6), (2x – 3) are 1st three consecutive terms of an AP, find x. [18]

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

13)

13

Questions & Solutions

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Page

Points:  The ends of a diameter of a circle are joined to some point and we get a right angle at that point. At points within the circle we get an angle larger than a right angle. For points outside the circle, we get an angle smaller than aright angle.  Any diameter of the circle divides it into two equal arcs; and we get a pair of right angles by joining points on each to the ends of the diameter.  Two points on a circle divide it into a pair of arcs. The angle got by joining these two points to a point on one of these arcs is equal to half the central angle of the alternate arc.  The angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.  Every chord of a circle divides it into two pruts. Such a part is called a segment of a circle.  Angles in the same segment of a circle are equal.  A chord divides a circle into a pair of segments.  Angles in alternate segments are supplementary.  If all vertices of a quadrilateral are on a circle, then its opposite angles are supplementary.  Suppose a circle is drawn through three vertices of a quadrilateral. If the fourth vertex is outside this circle, then the sum of the angles at this vertex and the opposite vertex is less than 180°; if the fourth vertex is inside the circle, then this sum is greater than 180°.  A quadrilateral for which a circle can be draVvn through all the four vertices is called a cyclic quadrilateral.  All rectangles are cyclic quadrilaterals. Isosceles trapeziums are also cyclic quadrilaterals.  If the opposite angles of a quadrilateral are supplementary, then we can draw a circle through all four of its vertices.  The exterior angle at a vertex of a cyclic quadrilateral is equal to the interior opposite angle.

14

Circles

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Page

15

Questions & Solutions

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Sample Questions PROBLEM

Page

16

17. In the given figure, O is the centre of the circle. If OAC = 35o and OBC = 40o, find the value of x.

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer Join OP Since OA = OP ACO = ∠OAP = 35o Similarly, OB = OP and ∠OPB = ∠OBP = 40o APB = 35o + 40o = 75o ∠AOB 2x75 o 150 o PROBLEM

18. In figure, O is the centre of circle. If ∠PAO=35 and PCO=45,then Calculate ∠ APC and ∠AOC.

Answer Join OP. In ΔOQP, OQ = OP = r ∠OQP = ∠OPQ = 35o

17

In ΔOPR,

CLASS X

Page

OR = OP = r ∠ORP = ∠OPR = 45o ∠QPR = 35o + 45o = 80o 9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions ∠QOR = 2∠QPR = 2 QOR = 160o

80o

PROBLEM

19. In the figure find PQB, O is the centre.

Answer PQB = [180 - (2 x 42)]/2 = 48 PROBLEM

Page

18

20. In the figure PQ is the diameter of the circle,then (a) Find the value of R (b) Find the value of Q (c) If QR=6cm,then find PR

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer ∠R=90 ∠Q=90-30=60 The ratio of sides of triangle is 1:√3:2 Thus PR=6√3 PROBLEM

21. In the figure find APB and AQB where O is the centre of the circle and OAP = 32 and OBP = 47.

Answer Join OP. In ΔOAP, OA = OP = radius ∠OAP = ∠OPA = 32o In ΔOPR, OB = OP = radius

22. O is the centre of the circle as shown in the figure. Find ∠CBD

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

PROBLEM

19

∠OBP = ∠OPB = 47o ∠APB = 32o + 47o = 79o ∠AQB = 180 o -79 o =10 o

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Answer

20

Take a point E on the circle, join AE and CE. ∠AEC=100/2=50o ∠AEC + ∠ABC = 180o (Opposite Angles of a cyclic quadrilaterals) ∠ABC = 130o ABC + ∠CBD = 180o (linear pair) 130o + ∠CBD = 180o ∠ CBD = 50o PROBLEM

Page

23. Draw a squre of area 12cm 2

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer

PROBLEM

24. Using the following fiqure, (a) ∠B+∠E=......? (b) ∠ADC ∠E ......?

Answer (a) 180o (b) ∠EAD PRACTICE EXERCISE

[30]

25)

ABCD is a cyclic quadrilateral with DBC = 80, BAC = 40, find BCD.

[60]

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

21

ABCD is a cyclic quadrilateral. If BCD = 100, ABD = 70, find ADB.

Page

24)

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 26)

In cyclic quadrilateral ABCD, ADBC. B = 70, find remaining angles. [110, 70, 110]

27) 28)

Draw a rectangle of sides 5cm and 3cm and draw a square of the same area. 2 chords AB and CD intersect at a point O. If AO = 3.5cm, CO = 5cm, DO = 7cm, find OB. [10cm]

29)

2 chords AB and CD intersect at O. If AO = 8cm, CO = 6cm, OD = 4cm, find OB. [3cm]

30)

31) 32)

Two circles intersect at two points A and B. AD and AC respectively are diameters to the two circles. Prove that B lies on the line segment DC. How do we draw a 22½ o angle EXERCISE In the picture below, O is centre of the circle. Find AOB

[1000]

Compute specified angle in each

Page

22

33)

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

What is the central angle of arc ABC?

35)

Find all angles of quadrilaterals.

Page

23

34)

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Second Degree Equations Points:  A quadratic equation in the variable x is of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0.  Different methods to findout the solutions of second degree equations  Completing the square method  Quadratic Formula (Shreedharacharya’s rule)method 4 √ 2 is the solutions of second degree equation ax2+bx+c=0,a≠0  b2‐4ac is the discriminant of ax2+bx+c=0  If b2‐4ac=0, then the equation has only one solution and the solution is –b/2a.  If b2‐4ac 0 (a positive number) then the equation has two different solutions Sample Questions

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

25. Find the roots of the second degree equation 2x2 - 7x +3 = 0 by the method of completing the square. Answer We have 7 3 0 2 7 3 0 2 2 7 3 2 4 2

24

PROBLEM

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 2 7 4 7 4 7 4 7 4 3

7 4 49 16 49 16 49 16 5 4 1 2

7 4 3 2 3 2 24 16

3 2

7 4

25 16

PROBLEM

26. Sum of the area of two squares is 500 m2. If the difference of their perimeters is 40 m, find the sides of the two squares. Answer Let the side of the squares be x and y meters. According to the condition, x2 + y2 = 500 ….(1)

CLASS X

Page

25

4x - 4y = 40 (x – y) = 10 y = x - 10 Substituting the value of y in (1), we get, x2 + (x - 10)2 = 500 2x2 - 20x - 400 = 0 x2 - 10x - 200 = 0 x = 20 or x = -10 As the side cannot be negative, x = 20 Hence, side of the first square, x = 20m 9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Side of the second square, y = (20 - 10) = 10 m PROBLEM

27. Three consecutive positive integers are taken such that the sum of the square of the first and the product of the other two is 232. Find the integers. Answer Let the three consecutive positive integers be x, x + 1, x + 2. x2 + (x + 1) (x + 2) = 232 x2 + (x2 + 3x + 2) = 232 2x2 + 3x - 230 = 0 4 √ 2 3 √1849 4 x = 10 or -11.5 But, x is a positive integer, so, x = 10. Thus, the numbers are 10, 11, and 12.

PROBLEM

28. The sum of the squares of two consecutive even numbers is 164. Find the numbers. Answer

CLASS X

26

2.

Page

Let the consecutive numbers be x, x x 2 2 164 x2 x2 x2 4x 4 164 2x2 4x ‐ 160 0 x2 2x ‐ 80 0 x 8 or x ‐10

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Neglecting the negative value, we get, x The numbers are 8 and 10.

8.

PROBLEM

29. 250 Rupees is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.

Answer Let the number of children be x. It is given that Rs 250 is divided amongst x children. So, money received by each child 250/x If there were 25 children more, then Money received by each child 250/ x 25 From the given information, 250 250 50 25 100 25 12500 0 = -125 or 100 Since, the number of children cannot be negative, so, x Hence, the number of children is 100.

100.

PROBLEM

30. By increasing the speed of a bus by 10 km/hr, it takes one and half hours less to cover a journey of 450 km. Find the original speed of the bus.

CLASS X

Page

27

Answer Let speed of the bus be x km/hr Time t = 450/x If speed is x + 10, then time T = 450/(x + 10) 9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions By question, 450/x - 450/(x + 10) =3/2 450 (x + 10) -450 x = (3/2) (x2 + 10x) 4500 2 = 3 x2 + 30x x2 + 10x - 3000 = 0 x = 50 or x = -60 x = 50 PROBLEM

31. A person has a rectangular garden whose area is 100 sq m. He fences three sides of the garden with 30 m barbed wire. On the fourth side, the wall of his of his house is constructed; find the dimensions of the garden. Answer Let the length and breadth of garden be x m and y m respectively. Area of the garden = 100 sq m xy = 100 m2 or y =100/x Suppose the person builds his house along the breadth of the garden. Then, we have: 2x + y = 30 2x + (100/x) =30 2x2 - 30x + 100 = 0 x = 10, x = 5 When x = 10 m, we have: y = 10 m When x = 5 m, we have: y = 20 m Thus, the dimensions of the garden are 10 m × 10 m or 5 m × 20 m.

Page

32. The hypotenuse of a right triangle is 20m. If the difference between the length of the other sides is 4m. Find the sides.

28

PROBLEM

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer x2 + y2 = 202 x2 + y2 = 400 Also x - y = 4 x=4+y (4 + y)2 + y2 = 400 2y2 + 8y – 384 = 0 y = 12 y = – 16 Sides are 12cm and 16cm

37)

38)

39)

40)

41) 42)

Perimeter of a rectangle is 40cm. If area is 96cm2, find the sides. [12, 8] If from a number, twice its reciprocal is subtracted we get 1. What is the number? [2 or -1] The sum of the squares of two consecutive odd positive integers is 290. Find them. [11, 13] Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164. [10, 6] The speed of a boat in still water is 8 km/hr. It can go 15 km upstream 22 km downstream in 5 hours. Find the speed of the stream. [3 km/hr] Find two consecutive numbers whose squares have the sum 85. [6, 7] Sum of 2 numbers is 12. If the sum of their squares is 90, find the numbers. [9, 3]

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

36)

29

PRACTICE EXERCISE

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 43)

44)

45)

46) 47) 48) 49) 50) 51) 52) 53)

Anu is 4 years older than Vinu. If 4 is added to the product of their ages, the result is 169. What are their ages? To the square of a natural number, four times the next natural number is added and the result is 36. What is number? The difference of two numbers is 6 and their product is 16. What are the numbers If from the squre of a number, six time the number is subtracted, we get 40. What is the number? How many terms of the arithmetic progression 3, 7, 11… must be added to get 253? If the product of a number with 6 more than the number is 160. What is the number? If the product of a number with 8 less than the number is 65, what is the number? How many terms of the arithmetic progression 4, 10, 16… starting from the first, are to be added to get 252? The width of a rectangle is 7metre more than its height and its area is 60 squre metres. Find the dimensions of rectangle.

Page

30

54)

If the sum of the squares of 2 consecutive even natural numbers is 244. Find the numbers. [10, 12] Square of a number is 60 more than 7times the number. Find the number. [12 or -5] The sum of squares of 2 consecutive odd numbers is 74. Which is the smaller of the numbers? [5 or 7] EXERCISE

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Trigonometry Points:  If the angles of a triangle are 600,600,600; then its sides will be in the ratio 1:1:1  If the angles of a triangle are 450,450,900; then its sides will be in the ratio 1:1:√2  If the angles of a triangle are 300,600,900; then its sides will be in the ratio 1:√3:2  In triangle ABC Sin A=BC/AC Cos A= AB/AC Tan A=BC/AB  The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal when it is above the horizontal level  The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal when it is below the horizontal level Sample Questions PROBLEM

33. One angle of a right triangle is 300 and its hypotenuse is 4cm.What is its area? Answer

Page

31

Triangle side ratio is ratio 1:√3:2

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Altitude = hypotenuse × √3/2 = 4 × √3/2 = 3.46 cm PROBLEM

34. One angle of a triangle 600 and the length of its opposite side is 4cm.What is its circumradius. Answer

From figure, sin60 = 4/BD 0.8660 = BC/BD BD = 4.62 cm Therefore radius = 2.31 cm PROBLEM

35. Two sides of a triangle are 7 and 6 centimeters and the angle between them is 1200. Find length of third side?

Page

32

Answer

In triangle ADC

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions CD = AC sin60 = 6×0.8660 = 5.2 AD = AC cos60 = 6×0.5 = 3 BD = BA + AD = 3+7=10 In triangle BDC BC2 = BD2+CD2 = 102+5.22 = 100+27.04 = 127.04 BC = 11.3 Third side is 11.3cm PROBLEM

36. In the figure, ∠B = 90◦; also, AB = 10 cm and ∠C = 60◦ (a) What is the measure of ∠A (b) What are the lengths of AC and BC

Answer (a) (b)

∠A=300 sin 60 = AB/AC = 10/AC AC = 10/0.8660=11.55 cm tan 60 = AB/BC BC = 10/1.73 =5.78 cm

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

37. In the figure, ∠BAC = 90◦, AD=6cm, CD=9cm, ∠ACD = x (a) What is tan x? (b) How much is ∠BAD? (c) What is the length of BD?

33

PROBLEM

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Answer (a) tan x= AD/DC=6/9=2/3 (b) x (c) tan x = BD/AD=BD/6 2/3=BD/6 BD=(2/3)× 6=4 cm PROBLEM

38. In the figure, AQB is an arc of a circle centred at O. Also, ∠AOB = 120◦, ∠AOQ = 60◦, PQ = 3 cm What is the radius of the circle?

Answer cos 60 = OP/AO=(r-3)/r 0.5 r = r - 3 r - 0.5 r = 3 r = 3/0.5 = 6cm

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

39. The shadow of a tower standing on a level ground is found to be 45 m longer when the sun’s altitude is 30° than when it was 60°. Find the height of the tower.

34

PROBLEM

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Answer In Δ ABD, AB/BD = tan 30 h /(45+x)=1/√3 x = (√3h - 45) …..(1) In ΔABC, AB/BC = tan 60 h / x = √3 x = h/√3 …….(2) From equation (1) and (2), we get (√3h - 45) = h/√3 h = 38.97m PROBLEM

Page

35

40. The length of shadow of a tower is 24 m, when the sun is at an angle of elevation of 550. Find height of tower.

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer In the figure AB is tower. tan 50 = AB/BC AB = CB tan 55 AB = 24 × 1.4281 = 34.2744 Height of the tower = 34.3 m PROBLEM

41. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer

CLASS X

Page

36

In figure PR = 28.5 In ΔPAR, PR/AR = tan 30 28.5/AR = 0.5773 AR = 49.3634

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions In ΔPRB, PR/BR = tan 60 28.5/BR = √3 BR = 16.4545 ST = AR – BR = 49.3634 – 16. 3634 = 32.9089 m PROBLEM

42. The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building

CLASS X

Page

37

In ΔABC, AB/BC=tan 60° BC=50/√3 …..(1) In ΔDCB, DC/BC = tan 30 h /BC = 1/√3 h = BC/√3 h = 16.67m 9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

55)

PRACTICE EXERCISE In ΔABC, B = 90, C = 70, BC = 16cm, find AC. [Sin70 = 0.94; Cos 70 = 0.34; tan70 = 2.75] [47.05cm]

56)

One angle of a triangle is 110 and the side opposite to it is 4cm long. What is its circumradius? [2.13cm]

57)

The angle between the radius and slant height of a cone is 600. Find the radius of the cone, if its slant height is 14 cm [7cm]

58)

A man standing on the deck of a ship, which is 10 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. (a) Calculate the distance of the hill from the ship (b) Find the height of the hill. [10√3m, 40m]

59)

An observer in a lighthouse 100 m above the sea-level is watching the ship sailing towards the lighthouse. The angle of depression of the ship from the observer is 30°. How far is the ship from the lighthouse? [100√3m]

60)

A ladder is placed along a wall such that its upper end is touching the top of the wall. The foot of the ladder is 2m away from the wall and the ladder is making an angle of 60° with the level ground. Find the height of the wall. [3.46m]

The top of a tower is seen at an angle of elevation of 40 from a point 30m away from the base of the tower. What is the height of the tower? [Sin 40 = 0.64; Cos 40 = 0.77; tan 40 = 0.84]

Page

[25.20m]

38

61)

0

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 62)

The angle of elevation of a tower at a point is 45°. After going 40 m towards the foot of the tower, its angle of elevation becomes 60°. Find the height of the tower. [94.64m]

63)

Two men are on the opposite sides of a tower. They measure the angles of elevation of the top of the towers as 30° and 45°. If the height of the tower is 60 m, find the distance between them. [163.92 m]

64)

From the top of a building 60m high the angles of depression of the top and bottom are observed to be 30º and 60º.Find the height of the tower. [40m]

65)

The horizontal distance between 2 towers is 70m.The angle of depression of the top of first tower when seen from the top of second tower is 30º. If the height of the second tower is 120m, find the height of the first tower. [79.6m]

66)

An aero plane when 3000m high passes vertically above another aero plane at an instant when the angle of elevation of the two aero planes from the same point on the ground are 60 and 45 respectively. Find the vertical distance between the aero planes.

68)

69) 70)

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

67)

EXERCISE A long pole leans against a short wall, making a 400 angle with the ground. The foot of the pole is 2metres away from the bottom of the wall. What is the height of the wall? A man 1.7metres tall standing 10 metres away from a tree sees the top of the tree at an angle of elevation 500. What is the height of the tree? When the sun is at an angle of elevation 480, the shadow of a tree is 18metres long. What is the height of the tree? The difference in the lengths of the shadow of a tower when the sun is at angles of elevations 300 and 600 is 45 metres. Compute the height of the tower.

39

[1268m]

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Solids Points: Square Pyramid:  Lateral surface area = 4×½(Base edge × Slant height) L.S.A 2  Total Surface Area = Base Area × Lateral surface area 2 T.S.A  Relations connecting base edge a, lateral edge e, slant height l, height h and base diagonal d: 1 4 1 4 1 4  Volume = Base area × height  Volume Cones  The radius of the sector becomes the slant height of the cone; the arc length of the sector becomes the base circumference of the cone.  Suppose that a cone of base radius r and slant height l, radius of the sector l and the central angle x ,then

40 Page

 L.S.A of Cone  T.S.A of Cone   Volume

360

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Spheres  T.S.A 4  Volume = Hemispheres  L.S.A 2  T.S.A 3  Volume =

Sample Questions PROBLEM

43. A toy in the shape of a square pyramid has base edge 16cm and slant height 10cm. (a) Find lateral surface area. (b) Find Total surface area. (c) Calculate its volume. Answer L.S.A = 2 =2×16×10=320 cm2 T.S.A= 2 = 16×16 + 320=576 cm2 =100 – 64 6 Volume =

= 256 × 6=1536 cm3

PROBLEM

44.

41

Height of a cone is 40cm. Slant height is 41cm. (a) Find diameter of its base. (b) Find Volume

CLASS X

Page

41 40 =81 9 ∴ Diameter = 18cm 9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Volume 9 40 = 3394.3 cm3 Answer PROBLEM

45. Diameter of a football is 30cm. (a) What is the least area of leather required to make 50 such footballs? (b) Also find volume of air inside 50 such footballs Answer Diameter of a football is 30cm. Surface area of a football 4 4 15 900 The least area of leather required to make 50 footballs 50 900 45000 Volume of air inside 50 such footballs = 50

50

= 225000

15

cm3

PROBLEM

46. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19cm and diameter of the cylinder is 7cm. (a) Find volume. (b) Total Surface Area of solid. Answer 2

12

641.67 cm3 Total Surface Area of solid 2 2

3.5 4 3.5 12

42

3.5

4

CLASS X

3.5 9048332443

BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

Volume of solid

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions = 418 cm2 PROBLEM

47. A circus tent is made of canvas and is in the form of a right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126m and 5m respectively. The total height of the tent is 21m. Find the total cost of the tent if the canvas used costs Rs.12 per sq.m Answer Diameter and height of the cylindrical part of the tent are 126m and 5m. Total height of the tent is 21m. 16 63 256 3969 = 4225 65m The total surface area 2 2 63 5 63 65 4725 The total cost of the tent 4725 12 56700 178200 PROBLEM

48. A cylindrical jar of radius 6cm contains oil. Iron spheres each of radius 1.5cm are immersed in the oil. How many spheres are necessary to raise the oil by 2cm? Answer Volume of cylinder with height 2cm = n

6

2

4 3

Volume of iron spheres

1.5

CLASS X

Page

43

72 4.5 72 16 4.5 Number of spheres are 16. 9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions PROBLEM

49. A Gulab Jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found is 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Answer Volume of one gulab jamun = volume of cylindrical part + 2 × (volume of hemispherical part) 2 2 3

[1:√3]

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

71)

PRACTICE EXERCISE A solid is hemispherical at the bottom and conical above. If the curved surface area of the two parts are equal, then from the ratio of the radius and height of the conical part.

44

1.4 2.2 1.4 Volume of such 45 gulab jamun 7.97 Volume of Syrup 30% Volume of such 45 gulab jamun 338 cm3

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 72)

A toy is in form of a cone mounted on a hemi-sphere of radius 3.5 cm. The total height of the toy is 15.5 cm. (a) Find slant height. (b) Find its total surface area. [12.5 cm, 214.5 cm2]

73)

A tent is of the shape of a right circular cylinder upto a height of 3 metres and conical above it. The total height of the tent is 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per square metre, if the radius of the base is 14 metres. [2068]

74)

A solid sphere of radius 3cm is melted and then cast into small spherical balls each of diameter 0.6cm.Find the number of small balls thus obtained. [1000]

75)

A cylinder of radius 12 cm contains water to a depth of 20cm, a spherical iron ball is dropped into the cylinder and thus the level of water is raised to 6.75cm.Find the radius of the ball. [9cm]

76)

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. [572 cm2]

77)

A glass cylinder with diameter 20 cm has water to the height of 9 cm. A metal cube of 8 cm edge is immersed in it completely. Find the height by which the water will rise in the cylinder. The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder. [8/3 cm]

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

78)

45

[1.6 cm]

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 79)

If the number of square centimetres on the surface of a sphere is equal to the number of cubic centimetres in its volume, what is the diameter of the sphere? [12 cm]

80)

The diameter of a metallic solid sphere is 12 cm. It is melted and drawn into a wire having diameter of the cross-section 0.2 cm. Find the length of the wire. [1188 cm2]

81)

A semi-circular thin sheet of metal of diameter 28 cm, is bent to make an open conical cap. Find the capacity of the cap.

83) 84)

85)

86) 87) 88) 89) 90)

An ice-cream cone has a hemispherical top. If the height of the cone is 9 cm and base radius is 2.5 cm, find the volume of ice cream [91.7 cm3] cone. EXERCISE The volume of a square pyramid is 720 cubic centimetres and its base edge is12 centimetres. Find its height. Two square pyramids are of equal volume and the base edge of the first is double that of the second. What fraction of the height of the second is the height of the first? Two square pyramids are of equal volume and the height of the first is double that of the second. What fraction of the base edge of the second is the base edge of the first? Compute the curved surface area of a cone of base radius 12 cm and slant height 25cm. What is the surface area of a cone with the diameter of the base 30cm and height 40cm? If the volumes of two spheres are 27 cubic centimetres and 64 cubic centimetres, what is the ratio of their radii? The ratio of the surface areas of two spheres is 3: 5. What is the ratio of their volumes? Compute the surface area of the largest sphere that can be cut out of a cube of side 15 centimetres.

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

82)

46

[622.38 cm3]

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

The base edges of two square pyramids are in the ratio 1: 2 and their heights are in the ratio 1: 3. The volume of the first pyramid is 180 cubic centimetres. Compute the volume of the second. 92) A cylindrical vessel of base radius 10 centimetres and height 85 centimetres is completely filled with water. Spheres of radius 10 centimetres, as many as can be completely immersed in water, are put into the vessel. Find the volume of water remaining in the vessel. 93) A cylindrical rod of length 4 centimetres and diameter 4 centimetres is melted and recast into spheres of radius 2 centimetres. How many such spheres can be made? 94) A metal sphere of diameter 24 centimetres is melted and recast into cones of base radius and height 6 centimetres. How many such cones are made? 95) A metal sphere of diameter 24 centimetres is melted and recast into cones of base radius and height 6 centimetres. How many such cones are made? 96) The cost of painting a hemispherical paper weight was 80 rupees. What will be the cost to paint a hemisphere of triple the radius at the same rate? 97) If the surface area of a solid hemisphere is 432π square centimetres, what is its radius? 98) If the inner radius of a hemispherical bowl is 60 centimetres, how many litres of water can it contain? 99) A petrol tank is in the shape of a cylinder with hemispheres of the same radius attached to both ends. If the total length of the tank is 6 metres and the radius is 1 metre, what is the capacity of the tank in litres? 100) A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them of the same radius of 1.5 metre. The total length of the rocket is 7 metres and the height of the cone is 2 metres. Compute the volume of the rocket.

47

91)

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Coordinates Points  The coordinates of a point are the distances of the point from x-axis and y-axis.  The coordinates of a point on the x-axis are of the form (x, 0) and of a point on the y-axis are of the form (0, y). Sample Questions PROBLEM

50. Find the coordinates of the other three vertices of the rectangle in the figure below.

Answer

T

The coordinates of rectangle are (0,0),(5,0),(5,4) and (0,4)

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

51. A circle is drawn with centre at (0, 0) and radius 6 units in a coordinate system. (a) What are the coordinates of the points at which it cuts the x-axis? (b) And the points where it cuts the y-axis?

48

PROBLEM

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer (a) (b) Answer

(6,0) and (-6,0) (0,6) and (0,-6)

PROBLEM

52. From the points given below, find the pair which are on a line parallel to the x-axis and the pair which are on a line parallel to the y-axis A(4, 3), B(3, 5), C(−6, 3), D(3,−2), E(5, 4) Answer Parallel to the x-axis A (4, 3) and C (−6, 3) Parallel to the y-axis B (3, 5) and D (3,−2), PROBLEM

53.

What is the distance between the points (−3, 2) and (4, 2)?

Answer Distance =|-3-4|=7 PROBLEM

54. The coordinates of a point on a line parallel to the y-axis are (5, 2). (a) What is the distance between this line and the y-axis? (b) Find the coordinates of the point where this line meets the x-axis. (c) What is the distance between these two points? Answer (a) 5 (b) (5,0) (c) Distance = |2-0|=2

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

55. One end of a line of length 10 units is at the point (-3, 2). If the y coordinate of the other end is 10, then find the x coordinate of the other end.

49

PROBLEM

BIO-VISION VIDEO BLOG

SSLC Mathematics

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

Answer Let the x coordinate of other end be x. Given, distance of A(-3,2) from B(x,10) is 10 units, i.e., AB =10. AB2 = (-3-x)2 + (2-10)2 102 =9 + x2 +6x + 64 or 100 = x2 + 6x + 73 x2 + 6x - 27 = 0 x = -9 or 3 Thus, the x coordinate of the other end is -9 or 3. PRACTICE EXERCISE 101) Find a relation between x and y such that the point (x, y) is equidistant from the points A (7,1) and B(3,5). [x - y = 2] 102) Find the point on y – axis which is equidistant from (-5, -2) and (3, 2). [-2, 0] 103) Find the point on y axis which is equidistant from (-5,2) and (3,2) [(0,-2)] 104) Draw x and y axis then mark the following points. (4,3),(-4,7),(-4,-6),(5,9),(6,-4) 105) Draw x and y axis and mark the points P (-1, 6) and Q (6, 6) then join PQ. Test whether the following points are on the line PQ (3, 4), (-6, 6), (4, 6), (5, 6) [(-6, 6), (4, 6), (5, 6)] 106) Find the distance from x axis (4, 4), (4, 3), (5, 7),(4,-3) [4, 4, 5, 4] 107) Draw a circle passing through the points (0,-3),(0,-2),(2,3),(0,1) 108) Mark the following points without drawing axis [(5, 7),(3,4)]

50

Questions & Solutions

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions EXERCISE 109) Draw a triangle joining the points with coordinates (0,0),(4,0) and (2,5) 110) Draw a quadrilaterals joining the points with coordinates (0,0),(3,2),(4,6),(1,4) 111) Draw a quadrilaterals joining the points with coordinates (0,0),(1,4),(5,4) and (6,0)

Mathematics of Chance Points  An outcome of a random experiment is called an elementary event.  The probability of an event = Number of outcomes/ Number of all possible outcomes  The probability of an impossible event is 0, and that of sure Sample Questions PROBLEM

56. Two coins are tossed together. Find the probability of getting at least one tail. Answer Total possibilities = {HH, HT, TH, TT} Number of possible outcomes = 4 The probability of getting at least one tail =3/4 PROBLEM

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

A bag contains 6 red balls, 8 green balls and 8 white balls. One ball is drawn at random from the bag, find the probability of getting (i) A white or green ball (ii) Neither green ball nor a red ball

51

57.

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer Red balls=6 Green balls=8 White balls=8 Total number of balls= 6 + 8 + 8 = 22 The probability of getting a white or green ball =16/20 The probability of getting neither green ball nor a red ball =8/20 PROBLEM

58. A bag contains 4 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag. Answer Number of red balls = 4 Number of blue balls = x Total Number of balls = 4 + x Probability of drawing a red ball from the bag = 5/ (4 + x) Probability of drawing a blue ball from the bag = x/ (4 + x) By condition 4 3 4 4 8 48 0 4 12 12 Hence the number of blue balls in the bag = 12

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

59. 20 cards numbered 1, 2, 3, 4… 19, 20 are put in a box. One boy draws a card from the box. Find the probability that the number on the card is: (i) Prime (ii) Divisible by 3 (iii) Divisible by 5

52

PROBLEM

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer Total number of outcomes = 20 (i) Prime numbers from 1 to 17 are 2,3,5,7,11,13,17,19. Number of outcomes = 8 The probability that the the card drawn is prime number = 8/20 (ii) Numbers are divisible by 3 are 3,6,9,12,15,18. Number of outcomes = 6 The probability that the card drawn is divisible by 3 =6/20 (iii) Numbers are divisible by 5 are 5,10,15,20. Number of outcomes = 4 The probability that the card drawn is divisible by 3 = 4/20 PROBLEM

60. Three coins are tossed simultaneously. Find the probability of getting (a) Three heads (b) Three heads (c) Exactly 2 heads (d) At least 2 heads. Answer Possible outcomes = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Number of total outcomes = 8 (a) 1/8 (b) 1/8 (c) 3/8 (d) 4/8 = 1/2 PROBLEM

Answer Total probable pairs=6 6=36 (1,4),(2,3),(3,2) and (4,1) are the four pairs whose sum is 5

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

53

61. Two dice rolled simultaneously. Find the probability of getting sum 5.

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions The probability of getting sum 5=4/36=1/9. PROBLEM

62. Find the probability that a leap year selected at random will contain 53 sundays. Answer In a leap year, there are 366 days. We have, 366 days = 52 weeks + 2 days. Thus, a leap year has always 52 sundays. probability = 2/7 PROBLEM

63. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number of the card is (i) an even number (ii) a number less than 16 (iii) a number which is a perfect square (iv) a prime number less than 25. Answer

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

1/2 14/100=7/50 The perfect squares are 4, 9, 16.....100 Probability=9/100 (iv) The prime numbers less than 25 are 2, 3, 5, 7, 11, 13, 17,19,23 Probability=9/100 PRACTICE EXERCISE 112) A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card is neither a red card nor a queen. [6/19] 113) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from a bag. What is the probability that the ball drawn is (i) red (ii) not red

54

(i) (ii) (iii)

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

116) A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is ; (i) red or white (ii) not black (iii) neither white nor black [13/20, 13/20,1/4] 117) Find the probability that a number selected from the number 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected. [16/25] 118) Find the probability that a number selected at a random from the numbers 1, 2, 3, ...., 35 is a (i) prime number (ii) multiple of 7 (iii) a multiple of 3 or 5 [11/35,1/7,16/35] 119) A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is : (i) white or blue (ii) red or black (iii) not white

55

[3/8; 5/8] 114) A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is four times that of a red ball, find the number of blue balls in the bag. [20] 115) A bag contains 12 balls out of which x are white. (i) If one ball is drawn at random, what is the probability that it will be a white ball? (ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be double than that in (i). Find x. [x/12; x=3]

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions (iv) neither white nor black [1/2,13/20,2/5,9/10] 120) A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is : (i) a card of spade or an ace (ii) a red king (iii) neither a king nor a queen (iv) either a king or a queen [4/13,1/26,11/13,2/13] 121) Cards marked with numbers 3, 4, 5, ...., 50 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the card drawn is : (i) divisible by 7 (ii) a number which is a perfect square [7/48,1/8]

Page

122) A box contains 100 bulbs out of which 10 are defective. What is the probability that if a bulb is drawn, it is (i) defective (ii) nondefective? 123) An integer is chosen from 1 to 15. Find the probability that the integer chosen is divisible by 4. 124) A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is: (i) red or white (ii) neither white nor black. 125) A bag contains 5 red balls and some black balls. If the probability of drawing a black ball is double that of a red ball, find the number of black balls in a bag. 126) A pair of dice is thrown once. Find the probability of getting the sum of numbers on two dice as 11.

56

EXERCISE

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Tangents Points:  Tangent to a circle at a point is perpendicular to the radius through the point of contact.  From a point, lying outside a circle, two and only two tangents can be drawn to it.  The tangent at a point on a circle is perpendicular to the radius through the point.  Each of the two angles made by a tangent to a circle and chord through the point of contact is equal to an angle in the segment on the other side of the chord.  The lengths of the tangents from a point outside a circle are equal.  The bisectors of the three angles of a triangle intersect at a point.

Sample Questions PROBLEM

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

Answer Let PQ be a tangent to the circle from point P and OQ be the radius at the point of contact. ∠OQP = 90° OP2 = OQ2 + PQ2 OQ2 = OP2 – PQ2 = 262 – 102 = (26 + 10) (26 – 10) = 36 × 16 OQ = 6 × 4 = 24

57

64. A point P is 26 cm away from the centre O of a circle and the length PQ of the tangent segment drawn from P to the circle is 10 cm. Find the radius of the circle.

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Radius of the circle = 24 cm. PROBLEM

65. A circle is touching the side BC of ΔABC at P and touching AB and AC produced at Q and R respectively. Prove that AQ Δ1/2 (Perimeter of ΔABC) Answer Since tangents from an exterior point to a circle are equal in length. BP = BQ ... (1) CP = CR ... (2) and, AQ = AR ... (3) from eqn. (3), we have AQ = AR AB + BQ = AC + CR AB + BP = AC + CP ... (4) Now, Perimeter of ΔABC = AB + BC + AC = AB + (BP + PC) + AC = (AB + BP) + (AC + PC) Using (4) = 2 (AB + BP) Perimeter of ΔABC = 2 AQ AQ =1/2 (Perimeter of ΔABC). PROBLEM

Page

58

66. C In the figure below,the tangents at A and B intersect at P.Prove that PA=PB

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Answer Join the points A, B and P to the centre O of the circle.

Look at the triangles ΔOAPand ΔOBP. Since the tangents PA and PB are perpendicular to the radii OA and OB these triangles are right angled. Both share the hypotenuse OP. Also OA = OB being radii of the circle. 2 So by Pythagoras Theorem PA2=OP2-OA2=OP2-OA2=PB2 From which we have PA = PB. PROBLEM

67. Find all the angles of the triangle in the figure below

Page

59

Answer From figure, ∠C=∠PAB=600 ∠B=∠QAC=400

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions ∠BAC+180-(600+400)=800 PROBLEM

68. AB is a diameter of a circle. BC is the tangent at B. Show that ∠PBC = ∠BAP. Answer ∠ABC=90o Since AB being diameter is perpendicular to tangent BC at the point of contact. (1) So ∠ABP +∠PBC =90o Also∠APB =90o (angle in the semi-circle) So ∠BAP+∠ABP = 90o (2) From (1) and (2), ∠PBC = ∠BAP

PRACTICE EXERCISE 127) A point P is 13 cm from the centre of the circle. The length of tangent drawn from P to the circle is 12 cm. Find the radius of the circle. [5cm] 128) The tangents at the points A and B on a circle centred at a point O meet at P. Prove that the line OP bisects ∠APB 129) The tangents at the points A and B on a circle centred at the point O meet at P . Prove that the line OP bisects the line AB. 130) In the figure below the tangent to the circle at A and the

Page

60

chord BC extended meet at P . Prove that PB × PC = PA2.

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Hint:

∠PAB=∠PCA ∠PBA=∠CAB+∠PCA ∠PBA=∠CAB+∠PAB=∠PAC ΔPAB and ΔPAC are similar PB/PA=PA/PC PB × PC = PA2 131) In the figure AB is a diameter of the circle and PQ is the tangent at A. Find the angles of ΔABC

Page

133) In the figure PQ is the tangent to the circle at A. Find the angles of the quadrilateral ABCD

61

132) In the figure AB is a diameter of the circle and PQ is the tangent at B. Prove that AP × AR = AQ × AB

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

135) Compute the length of the tangent from a point 29 cm away from the centre of a circle of radius 20 cm. [21cm] 136) Draw a circle of radius 3 centimetres and mark a point P which is 7 centimetres away from its centre. Draw the tangents from P to the circle. 137) The tangents from a point P to a circle centred at 0 with radius r touch the circle at A and B. The line cuts 0P at Q. Prove that OP × OQ = r2 138) The length of the tangent from a point to a circle of radius 12 centimetres is 16 centimetres. How far away is this point from the centre of the circle? [20cm]

62

134) Drawing the figures below according to the specifications.

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 139) The length of the tangent from a point 61 centimetres away from the centre of a circle is 60 centimetres. What is the radius of the circle [11cm] 140) Draw triangle ABC with the specifications given below and draw the incircle of each (a) AB = 5 cm, BC = 6 cm, CA = 7 cm (b) AB = 7 cm ∠A = 70 ∠B = 50 (c) AB = BC = 6 cm ∠B = 40 141) Draw an equilateral triangle with each side 6 centimetre and draw its circumcircle and incircle. 142) Draw the pictures given below according to the specifications.

Polynomials Points:  The reminder on dividing the polynomial p(x) by the polynomial x-a is p(a)  If p(a)=0,then x-a is a factor of polynomial p(x).  If p(a)≠0,then x-a is not a factor of polynomial p(x) Sample Questions 69.

Check whether x -1 is a factor of 3x2-2x2-3x+2

CLASS X

Page

Answer P (1) =3×13-2×12-3×1+2 = 0 9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

63

PROBLEM

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Therefore x -1 is a factor PROBLEM

70. Write 2x2+5x+3 as a product of two first degree polynomials. Answer P(x) =2x2+5x+3 2x2+5x+3=0 x =( -5 ± √1)/2= -1 or -3/2 Then the polynomial x + 2 , x + (3/2) are factors of 2x2+5x+3 (x + 2) (x + 3/2) = x2 + (5/2)x + 3 = ½(2x2+5x+3) From this 2x2+5x+3 = 2(x + 2) (x + 3/2)=(2x+3)(x+1) PROBLEM

71. Find the value of k if remainder when 5x3 + 4x2 – 11x + k is divided by (x – 1) is 0. Answer P(1) = 5x3+4x2-11x +k = 0 5×13+4×12-11×1 +k = 0 k=2

Page

72. When x3 – 2x2 + kx + 7 is divided by (x – 4) remainder is 11. Find k. Answer P(x) = x3-2x2+kx +7 = 11 P(4) = 43-2×42+k×4 +7 = 11 4k = -64+32-7+11=13-8=-28 k = -7

64

PROBLEM

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions PRACTICE EXERCISE 143) When x3 – 3x2 – 5x + n is divided by (x + 3) remainder is 5. Find n. [-10] 3 2 144) Find the value of k if (x + 4) is a factor of 2x + 11x + kx – 36. [3] 3 2 145) If (x + 1) and (x – 1) are factors of x + 2x + px + q, find p and q. [-1, -2] 146) When 3x3 – 7x2 + px + q is divided by (x – 2) and (x – 3), the remainders are 3 and 17 respectively. Find p and q. [-2, 5] 147) When P(x) = 3x3 + kx2 – 12x + 8 is divided by (x – 1), the remainder is 35. Find k and test whether (3x – 2) is a factor of P(x)? [36, no]

Page

148) If (x + 2) and (x – 2) are factors of 2x3 – 3x2 – 8x + 12, find 3rd factor. 149) If (3x + 1) and (x + 3) are 2 factors of 3x3 + 13x2 + 13x +3, find 3rd factor. 150) Find the remainder when 2x3 – 11x2 + 3x + 5 is divided by (x – 3) 151) 3x3 – 2x2 – 7x + 5 is divided by (x – 4) 152) 5x3 + 7x2 + x – 1 is divided by (x + 1) 153) 2x3 + 7x2 – 11x – 8 is divided by (x – 2) 3 2 154) 4x + 5x + 2x – 10 is divided by (x + 2) 155) Is (x + 2) a factor of 3x3 – x2 – 20 x – 12? 156) Is (x – 2) a factor of 2x3 – 11x 2 + 17x -6? 157) Is (x + 1) a factor of x3 – 5x 2 + 2x + 3? 158) Prove that (x + 2) is a factor of x3 + 3x2 – 4x – 12.

65

EXERCISE

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions

Geometry and Algebra  Distance  Slope of a line  Equation of a line

Sample Questions PROBLEM

73. The centre of a circle is (3, 3) and it passes through the point (2, 6). What is its radius? Answer Radius r = Distance between (3, 3) and (2, 6) r2 = (3-2)2 + (3-6)2 r = √10 PROBLEM

74. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle. Answer AB2= (6 -5)2 + (4 - 2)2 = 12 + 62 = 1+36 =37 BC2 = (7 + 6)2 + (-2 - 4)2 = 12 + (-6)2 = 1+ 36 = 37 AC2 = (7 + 5)2 + (-2 +2)2 = 22 + 02 = 4 Since, AB = BC = √37 We observe that A, B, C are the vertices of an isosceles triangle.

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

75. Prove the points A(2,3),B(7,5),C(9,8) and D(4,6) are the vertices of parallelogram. Answer Slope AB = 2/5

66

PROBLEM

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Slope BC = 3/2 Slope CD = 2/5 Slope AD = 3/2 AB = √29; BC = √29; CD = √13; AD = √13 Since the length of opposite sides and their slopes are equal the points are the vertices of parallelogram. PROBLEM

76. What is the equation of the line joining the points (1,3) and (2,7)

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

= 2/1=2 If (x,y) be any point on this line, then (y - 4)/(x - 1) = 2 y – 4 = 2 (x - 1) y – 4 = 2x – 2 2x + y +2 =0 PRACTICE EXERCISE 159) Find the value of y if the distance between the points (2, –3) and (10, y) be 10 units. [y = 3 or –9] 160) Find the point on x-axis which is equidistant from the points (–4, 6) and (5, 9). [3, 0] 161) Find the area of rhombus ABCD, if A(2, 0), B(5, -5), C(8, 0), D(5, 5) [30] 162) Show that A (-5, 5), B (7, 10), C (10, 6), D (-2, 1) are vertices of a parallelogram. 163) Show that the points (2,-2) ,(14,10) (11,13)and (-1,1) are vertices of a rectangle.

67

Answer Slope of the the line joining the points (1, 4) and (2, 6)

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions Statistics Points:  Mean :  Median : is median,

/2

Sample Questions PROBLEM

77.

Find mean Rainfall(mm) 54 56 58 55 50 47 44 41

Number of Days 3 5 6 3 2 4 5 2

Number of Days 3 5 6 3 2 4 5 2 30

Total Rainfall(mm) 162 280 348 165 100 188 220 82 1545

Page

Rainfall(mm) 54 56 58 55 50 47 44 41 Total Mean = 1545/30=51.5mm

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

68

Answer

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions PROBLEM

78.

Find median Age Below 30 Below 35 Below 40 Below 45 Below 50 Below 55

Number of teachers 6 8 12 20 16 6

Answer Age Below 30 Below 35 Below 40 Below 45 Below 50 Below 55

c.f 6 14 26 46 62 68

Median =? y = 68/2=34 Using Proportionality assumption, (x – 40)/(45-40)=(34-26)/(46-26) x = 42 Median = 42 PRACTICE EXERCISE

[6]

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

Page

69

164) If the mean of the following data is 5, find p. x 2 3 5 p 9 f 9 4 6 3 8

BIO-VISION VIDEO BLOG

SSLC Mathematics

Questions & Solutions 165) Find the mean of the following distribution x 1 3 5 7 8 f 7 8 1 1 1 [55]

166) Find the median of the following frequency distribution x 0-10 10-20 20-30 30-40 40-50 50-60 f 5 3 10 6 4 2 [27]

167) Calculate the median income : Income No of employees

600700 40

700800 68

800900 86

9001000 120

10001100 90

11001200 40

1200 26

[Rs. 934.17]

168) Calculate the median for the following data : x 1 2 3 4 5 6 f 6 1 2 4 6 7 [35]

Fasaludeen.P.K (Fassal Peringolam) Parayarukandiyil House Peringolam http://sciencetablet.blogspot.com http://brainsmoozhikkal.blogspot.com Mob: 9048332443

IX, X CBSE Tuition; +1, +2 Tuition + Entrance; +2 SAY Batch

Page

(Mathematics, Physics and Chemistry) Chapter wise Detailed Class Notes Editing & Design Fassal Peringolam

CLASS X

9048332443 BRAINS MOOZHIKKAL, KOZHIKKODE http://sciencetablet.blogspot.com

70

Special Tuition Batch (Tutor Fassal Peringolam)