81279 + Interview Questions in Mathematics QA in Mathematics

1.	\left. \begin{array} { l } { 1.5 \times 10 ^ { 12 } \text { and } 3.7 \times 10 ^ { 11 } } \\ { 6 \times 10 ^ { 14 } \text { and } 3 \times 10 ^ { 24 } } \\ { 2 ^ { 3 } \times 10 \text { and } 2 ^ { 4 } \times 5 } \end{array} \right.
Answer»

Discussion

2.	. Two tangents PL and PM are drawn to a circlewith centre O from an external point P. Provethat ZLPM = 220LM.
Answer» thank you so much

Discussion

3.	1. From an external point P, tangents PA and PB are drawn to a circle with centre O. If <PAB = 50°, then findCAOB
Answer»

Discussion

4.	12. Two tangents PA and PB are drawn to a cirelewith centre O from an external point P. Prove that[2016LAPB2ZOAB
Answer» Let ∠APB= x° We know that the tangents to a circle from an external point are equal in length so PA= PB.PA =PB ∠PBA = ∠PAB[Angles opposite to the equal sides of a triangle are equal.] ∠APB+ ∠PBA +∠PAB= 180°[Sum of the angles of a triangle is 180°] x° + ∠PAB +∠PAB = 180° [∠PBA = ∠PAB] x° + 2∠PAB = 180° ∠PAB =½(180° - x°) ∠PAB =90° - x°/2∠OAB +∠PAB=90°∠OAB =90° - ∠PAB∠OAB =90° - (90° - x°/2)∠OAB =90° - 90° + x°/2∠OAB = x°/2∠OAB = ∠APB /2∠OAB = 1/2∠APB ∠APB = 2∠OAB

4.

12. Two tangents PA and PB are drawn to a cirelewith centre O from an external point P. Prove that[2016LAPB2ZOAB

Answer»

Let ∠APB= x°

We know that the tangents to a circle from an external point are equal in length so PA= PB.PA =PB

∠PBA = ∠PAB[Angles opposite to the equal sides of a triangle are equal.]

∠APB+ ∠PBA +∠PAB= 180°[Sum of the angles of a triangle is 180°]

x° + ∠PAB +∠PAB = 180°

[∠PBA = ∠PAB]

x° + 2∠PAB = 180°

∠PAB =½(180° - x°) ∠PAB =90° - x°/2∠OAB +∠PAB=90°∠OAB =90° - ∠PAB∠OAB =90° - (90° - x°/2)∠OAB =90° - 90° + x°/2∠OAB = x°/2∠OAB = ∠APB /2∠OAB = 1/2∠APB

∠APB = 2∠OAB

Discussion

5.	27. In the adjoiningfigure, from anexternal point Ptwo tangents PTand PS are drawn toa circle with centreO and radius r. IfOP 2r, then showthat LOTS = <OST= 30°CBSE 2016
Answer» AP is the tangent to the circleOA _\|_ AP (Radius is perpendicular to the tangent at the point of contact)∠OAP = 90 degreesIn angle OAP,Sin∠OPA= OA/OP = R/2R = 1/2∠OPA = 30In Angle ABP,AP = BP∠PAB =∠PBAso 60+∠PAB +∠PBA = 18060+2∠PAB = 180∠PAB = 180 - 60/2∠PAB = 60Butas∠OAP = OBP = 90OAP = OBPso,60 + x = 90x = 30therefore,∠OTS = OST= 30

Discussion

6.	EXAMPLE14 From an external point P, two tangents PA and PB are drawn to the circlewith centre O. Prove that OP is the perpendicular bisector of AB.
Answer» 1 2

6.

EXAMPLE14 From an external point P, two tangents PA and PB are drawn to the circlewith centre O. Prove that OP is the perpendicular bisector of AB.

Answer»

1

2

Discussion

7.	77. In Fig. PT and PS are tangents to a circle from a point P such that PT-5 cm andTPS60". Find the length of chord TS.
Answer» PS=PT. ( when two tangents met at a point externally then their lenghts are equal) therefore angle PTS= angle PST. (angle opposite to equal sides are equal) now in triangle PTS Angle PTS +Angle PST +Angle TPS=180 <PTS+<PTS+60=180 2<PTS=180-60 <PTS =60<PSt =60 therefore all angles are 60 degree it means it is an equilateral triangle and we know that sides of equilateral triangle are equal so the chord TS is of 5cm

7.

77. In Fig. PT and PS are tangents to a circle from a point P such that PT-5 cm andTPS60". Find the length of chord TS.

Answer»

PS=PT. ( when two tangents met at a point externally then their lenghts are equal)

therefore

angle PTS= angle PST. (angle opposite to equal sides are equal)

now in triangle PTS

Angle PTS +Angle PST +Angle TPS=180

<PTS+<PTS+60=180

2<PTS=180-60

<PTS =60<PSt =60

therefore all angles are 60 degree it means it is an equilateral triangle

and we know that sides of equilateral triangle are equal

so the chord TS is of 5cm

Discussion

8.	28. From an external point P, two tangents PA and PB are drawn to the circle with the centre O.Prove that OP is the perpendicular biscctor of AB.
Answer» Let OP intersect AB at C In ΔPAC and ΔPBC, we have PA = PB (Tangent from an external point are equal) ∠APC = ∠BPC (PA and PB are equally inclined to OP) and PC = PC ( Common) ∴ ΔPAC~= ΔPBC (by SAS congurency criteria) ⇒ AC = BC ......(1) and ∠ACP = ∠BCP ......(2) But ∠ACP + BCP = 180° .......(3) From (2) and (3) ∠ACP = ∠BCP = 90° Hence, from (1) and (2), we can conclude that OP is the perpendicular bisector of AB.

8.

28. From an external point P, two tangents PA and PB are drawn to the circle with the centre O.Prove that OP is the perpendicular biscctor of AB.

Answer»

Let OP intersect AB at C

In ΔPAC and ΔPBC, we have

PA = PB (Tangent from an external point are equal)

∠APC = ∠BPC (PA and PB are equally inclined to OP)

and PC = PC ( Common)

∴ ΔPAC~= ΔPBC (by SAS congurency criteria)

⇒ AC = BC ......(1)

and ∠ACP = ∠BCP ......(2)

But ∠ACP + BCP = 180° .......(3)

From (2) and (3)

∠ACP = ∠BCP = 90°

Hence, from (1) and (2), we can conclude that OP is the perpendicular bisector of AB.

Discussion

9.	हि हक न सर...; 2 haurn 354 rreved Yo ee a0\ \a\ A f‘\"\!
Answer» Time taken to cover 65000m = 60 minTime taken to cover 1m=65000÷60Time taken to cover 440m=60÷65000×440 =0.4 min approximately 1min=60 seconds 0.4 min = (0.4×60)seconds (i) =24 seconds ( Ans ) Distance traversed in 1 hour = 65 km (ii)Distance traversed in 15 hours = 65 × 15 = 975 km ( Ans ) Sorry it will be 60÷65000 instead of 65000÷60 it is wrrog oue

9.

हि हक न सर...; 2 haurn 354 rreved Yo ee a0\ \a\ A f‘\"\!

Answer»

Time taken to cover 65000m = 60 minTime taken to cover 1m=65000÷60Time taken to cover 440m=60÷65000×440 =0.4 min approximately 1min=60 seconds 0.4 min = (0.4×60)seconds (i) =24 seconds ( Ans )

Distance traversed in 1 hour = 65 km (ii)Distance traversed in 15 hours = 65 × 15 = 975 km ( Ans )

Sorry it will be 60÷65000 instead of 65000÷60

it is wrrog oue

Discussion

10.	3x 78 782 4 6 4
Answer» 3x(4)-(7)(-2)=(8)(4)-(7)(6)12x+14=32-42=-10so 12x=-10-14=-24so x=-2

Discussion

11.	Find the sum of:(a) 137 and 354(c) -312, 39 and 1924.(b)-52 and 52(d) 50, - 200 and 3005. Find the sum :
Answer» answer kis question ka chahiye

Discussion

12.	↑xample?Two tangents TP and TQ are drawnto a circle with centre O from an external point TProve that PTQ = 2 < OPQ.Solution : We are given a circle with centre O,an external point T and two tangents TP and TQto the circle, where P, Q are the points of contact(see Fig. 10.9). We need to prove that
Answer» another easy method for this question

Discussion

13.	the HCrO68, 2315, 36 andthe HCF o24 and 35414, 948 an
Answer» 1st

Discussion

14.	24. PQ is a chord of length 8 cm in a circle of radius 5 cm. Tangents at P and Q intersect atT. Find the length of PT.
Answer» Given radius, OP = OQ = 5 cmLength of chord, PQ = 4 cmOT ⊥ PQ,∴ PM = MQ =4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]In right ΔOPM, OP²= PM²+ OM²⇒ 52= 42+ OM²⇒ OM²= 25 – 16 = 9 Hence OM = 3cmIn right ΔPTM,PT²= TM²+ PM²→(1)∠OPT = 90º [Radius is perpendicular to tangent at point of contact] In right ΔOPT,OT2²= PT²+ OP²→(2) From equations (1) and (2), we get OT²= (TM²+ PM²) + OP2²⇒ (TM + OM)²= (TM²+ PM²) + OP²⇒ TM²+ OM²+ 2 × TM × OM = TM²+ PM²+ OP²⇒ OM²+ 2 × TM × OM = PM2+ OP²⇒ 32+ 2 × TM × 3 = 42+ 52⇒ 9 + 6TM = 16 + 25⇒ 6TM = 32⇒ TM =32/6 = 16/3 Equation (1) becomes, PT²= TM²+ PM² = (16/3)2+ 42 =(256/9) + 16 = (256 + 144)/9 = (400/9) = (20/3)2⇒PT = 20/3∴the length of tangent PT is(20/3) cm

14.

24. PQ is a chord of length 8 cm in a circle of radius 5 cm. Tangents at P and Q intersect atT. Find the length of PT.

Answer»

Given radius, OP = OQ = 5 cmLength of chord, PQ = 4 cmOT ⊥ PQ,∴ PM = MQ =4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]In right ΔOPM,

OP²= PM²+ OM²⇒ 52= 42+ OM²⇒ OM²= 25 – 16 = 9

Hence OM = 3cmIn right ΔPTM,PT²= TM²+ PM²→(1)∠OPT = 90º [Radius is perpendicular to tangent at point of contact]

In right ΔOPT,OT2²= PT²+ OP²→(2)

From equations (1) and (2), we get

OT²= (TM²+ PM²) + OP2²⇒ (TM + OM)²= (TM²+ PM²) + OP²⇒ TM²+ OM²+ 2 × TM × OM = TM²+ PM²+ OP²⇒ OM²+ 2 × TM × OM = PM2+ OP²⇒ 32+ 2 × TM × 3 = 42+ 52⇒ 9 + 6TM = 16 + 25⇒ 6TM = 32⇒ TM =32/6 = 16/3

Equation (1) becomes,

PT²= TM²+ PM² = (16/3)2+ 42 =(256/9) + 16 = (256 + 144)/9 = (400/9) = (20/3)2⇒PT = 20/3∴the length of tangent PT is(20/3) cm

Discussion

15.	Example 2: Two tangents TP and TQ are drawnto a circle with centre O from an external point TProve that < PTQ = 2 OPQ.Solution : We are given a circle with centre O,an external point T and two tangents TP and TQto the circle, where P, Q are the points of contact(see Fig. 10.9). We need to prove that
Answer»

Discussion

16.	Use the diagram shown tofind the m<B.Зroс 46°2x-1°в0000
Answer» 3x+2x-1+46=180; 5x=180-45=135; x= 135 /5=27 option C is correct answer

16.

Use the diagram shown tofind the m<B.Зroс 46°2x-1°в0000

Answer»

3x+2x-1+46=180; 5x=180-45=135; x= 135 /5=27

option C is correct answer

Discussion

17.	Use the diagram shown to find x.3xooooNN
Answer» 3x = 12; x=12/3=4

Discussion

18.	9+6-78+61X35
Answer» 9+6-78+61×35 Using BODMAS, =9+6-78+2135=2072.

Discussion

19.	:3c. 78-2d.
Answer» the answer is of 78 divided by 2 is 39

Discussion

20.	Use the diagram shown to find x.4x-22x+6
Answer» 4x-2+2x+6=6x+4; 6x=4; x=4/6=1/3 when two angle are equal so opposite sides will equal4x-2= 2x+64x-2x= 6+22x= 8x= 8/2=4 X=4 is correct answer 4x-2=2x+6; 4x-2x=6+2; 2x=8; x=8/2=4

20.

Use the diagram shown to find x.4x-22x+6

Answer»

4x-2+2x+6=6x+4; 6x=4; x=4/6=1/3

when two angle are equal so opposite sides will equal4x-2= 2x+64x-2x= 6+22x= 8x= 8/2=4

X=4 is correct answer

4x-2=2x+6; 4x-2x=6+2; 2x=8; x=8/2=4

Discussion

21.	78 is a multiple of 6.
Answer» 78 is a multiple of 613 , answer YES 78 is a multiple of 6 78/613is correct answeryes 78 is multiple of 6 yes 78 is a multiple of 6 Yes it is the right answer Yes it is right answer. 613 is the correct answer of the given question yes . . 78/613is correct answeryes 78 is a multiple of 6

21.

78 is a multiple of 6.

Answer»

78 is a multiple of 6*13 , answer

YES 78 is a multiple of 6

78/613is correct answeryes 78 is multiple of 6

yes 78 is a multiple of 6

Yes it is the right answer

Yes it is right answer.

6*13 is the correct answer of the given question

yes . .

78/613is correct answeryes 78 is a multiple of 6

Discussion

22.	3)16 1 (4) 13रद यदि न८+ कर है, तो ८+८ का मान है(UPTET June 2e !51Py T LRl \e (2) T 3 o (4)न हैं
Answer»

Discussion

23.	3x 78 72 4 6 4
Answer» (3x)(4)-(7)(-2)=(8)(4)-(6)(7)12x+14=32-42=-1012x=-10-14=-24so x=-2

Discussion

24.	he base area of the cone is 616 sq.cits height is 48 cm. Find its total suJune-20.ndce area?
Answer»

Discussion

25.	TTT. Read the number and write the number name.a) 84,290: Eighly forb) 29,354c) 60,032:
Answer» eighty-four thousand two hundred neighty eighty four thousand two hundred ninetytwenty nine thousand three hundred fifty foursixty thousand thirty two eighty four thousand two hundred and ninety onlytwenty nine thousand three hundred and fifty four only sixty thousand and thirty two

25.

TTT. Read the number and write the number name.a) 84,290: Eighly forb) 29,354c) 60,032:

Answer»

eighty-four thousand two hundred neighty

eighty four thousand two hundred ninetytwenty nine thousand three hundred fifty foursixty thousand thirty two

eighty four thousand two hundred and ninety onlytwenty nine thousand three hundred and fifty four only sixty thousand and thirty two

Discussion

26.	211.If a, b, c are in А.Р., then (a-c)-(a) 1(c) 3- ac(b) 2(d) 4
Answer» As A, B and C are in AP,b-a=c-b(a-c)^2 = (2b-2c)^2= 4(b^-2bc+c^2)b^2-ac= b^2 –c(2b-c)= b^2- 2bc +c^2(a-c)^2 / (b^2 –ac) = 4

Discussion

27.	Write down the decimal expansions of those rational numbersiwhich have terminating decimal expansions.
Answer»

Discussion

28.	(a) If n(4) 3, n(B)-5 and Ac B. Then n(An B)
Answer» Here n(A∩B)=n(A) =3 as A€B

Discussion

29.	Write down the decimal expansions of those rational numbersin Question 1 above which have terminating decimalexpansions.(vi)
Answer»

Discussion

30.	LCM of 100 and 200 is
Answer» 100=2×2×5×5200=2×2×2×5×5LCM=2×2×2×5×5answer the LCM of 100 and 200 is 200 LCM of 100 and 200=200 Lcm of 100 and 200 is right answer hai Lcm of 100 and 200 is right answer hhhhh 200 is the lcm of 100 and 200 LCM OF 100 , 200 =200 LCM of 100and 200 is correct answer The LCM of 100,200 is 200 Lcm(100, 200)=200 answer 100=2255 200=22255LCM=22255is the best correct answer 100=2×2×5×5200=2×2×2×5×5 LCM stands for lowest common method So,when 100,200 LCM is 200 .because 200 divided by 100 and 200 its answer is 200 . you are happy 100=2255200=22255LCM-22255=200 100=2×2×5×5200=2×2×2×5×5LCM=2×2×2×5×5 =200 100=2×2×5×5200=2×2×2×5×5 100=2×2×5×5200=2×2×2×5×5lcm=2×2×2×5×5 100=2×2×5×5200=2×2×2×5×5LCM=2×2×2×5×5answer me lcm of 100and 200 is 100 💯=2×2×5×5 200=2×2×2×5×5 LCM= 2×2×2×5×5 lcm= 200 10 and 20 respectively 100=2×2×5×5200=2×2×2×5×5LCM=2×2×5×5is right correct answer LCM= 2×2×2×5×5 is the correct answer

30.

LCM of 100 and 200 is

Answer»

100=2×2×5×5200=2×2×2×5×5LCM=2×2×2×5×5answer

the LCM of 100 and 200 is 200

LCM of 100 and 200=200

Lcm of 100 and 200 is right answer hai

Lcm of 100 and 200 is right answer hhhhh

200 is the lcm of 100 and 200

LCM OF 100 , 200 =200

LCM of 100and 200 is correct answer

The LCM of 100,200 is 200

Lcm(100, 200)=200 answer

100=2*2*5*5

200=2*2*2*5*5LCM=2*2*2*5*5is the best correct answer

100=2×2×5×5200=2×2×2×5×5

LCM stands for lowest common method So,when 100,200 LCM is 200 .because 200 divided by 100 and 200

its answer is 200 . you are happy

100=2*2*5*5200=2*2*2*5*5LCM-2*2*2*5*5=200

100=2×2×5×5200=2×2×2×5×5LCM=2×2×2×5×5 =200

100=2×2×5×5200=2×2×2×5×5

100=2×2×5×5200=2×2×2×5×5lcm=2×2×2×5×5

100=2×2×5×5200=2×2×2×5×5LCM=2×2×2×5×5answer me

lcm of 100and 200 is 100

💯=2×2×5×5 200=2×2×2×5×5 LCM= 2×2×2×5×5 lcm= 200

10 and 20 respectively

100=2×2×5×5200=2×2×2×5×5LCM=2×2×5×5is right correct answer

LCM= 2×2×2×5×5 is the correct answer

Discussion

31.	'Ac. B, then show that AUB ," B (use Veno diagram
Answer»

Discussion

32.	presentations (expansions). Can you guess whariree numbers whose decimal expansions are non-terminating non-recuring
Answer» Here are the three numbers whose decimal expansion are non terminating and non recurring:- 1) √2= 1.4142......... 2) √3= 1.73205...... 3) √5 = 2.236...... these values are non terminating and non recurring. thus, we say them as irrational numbers.Like if you find it useful

32.

presentations (expansions). Can you guess whariree numbers whose decimal expansions are non-terminating non-recuring

Answer»

Here are the three numbers whose decimal expansion are non terminating and non recurring:-

1) √2= 1.4142.........

2) √3= 1.73205......

3) √5 = 2.236......

these values are non terminating and non recurring.

thus, we say them as irrational numbers.Like if you find it useful

Discussion

33.	Find the decimal expansions ofand3
Answer»

Discussion

34.	The base and the height of a triangle arein the ratio 4:5. If the area of the triangleis 40 m2; find its base and height5.
Answer» Let base of triangle=4xLet height of triangle=5x Area of triangle=(1/2)baseheight=(1/2)4x5x 10x²=40x²=4x=2 Therefore base=4x=42=8 cmheight=5x=52=10 cm

34.

The base and the height of a triangle arein the ratio 4:5. If the area of the triangleis 40 m2; find its base and height5.

Answer»

Let base of triangle=4xLet height of triangle=5x

Area of triangle=(1/2)*base*height=(1/2)*4x*5x

10x²=40x²=4x=2

Therefore base=4x=4*2=8 cmheight=5x=5*2=10 cm

Discussion

35.	ChallengeA single discount equivalent to three successive discounts of 20%, 25% and 10% is(a) 55%(b) 50%(c) 48%(d) 46%
Answer» (100-x)/100 = (100-20)/100 * (100-25)/100 * (100-10)/100=> 100-x = (4/5)(3/4)90=> 100-x = 54=> x = 46% (100-x)/100 = (100-20)/100 * (100-25)/100 * (100-10)/100=> 100-x = (4/5)(3/4)90=> 100-x = 54=> x = 46% thank you thank you

35.

ChallengeA single discount equivalent to three successive discounts of 20%, 25% and 10% is(a) 55%(b) 50%(c) 48%(d) 46%

Answer»

(100-x)/100 = (100-20)/100 * (100-25)/100 * (100-10)/100=> 100-x = (4/5)*(3/4)*90=> 100-x = 54=> x = 46%

thank you

Discussion

36.	factorize 6xsquar -13x-5
Answer» 6xx-13x-5 = 6xx-15x+2x-5 = 3x(2x-5)+1(2x-5) = (3x+1)(2x-5)

Discussion

37.	1+1.0 -0.33In a test a student got 30% marks and failedby 25 marks. In the same test anotherstudent got 40% marks and secured 25%marks more than the essential minimumpass marks. The maximum pass marks forthe test were: ।किसी टेस्ट में एक विद्यार्थी को 30% अंक प्राप्त हुए औरवह 25 अंक से अनुत्तीर्ण हो गया. उसी टेस्ट में दुसरे विद्यार्थीको 40% अंक प्राप्त हुए और आवश्यक न्यूनतम पास अंकोंसे 25% अधिक अंक प्राप्त हुए. टेस्ट के लिए अधिकतमपास अंक कितने थे?A) 400B480c500D) 580
Answer» B) 480 is the correct answer B is the correct answer 480 is the righy answer B) 480 is the correct answer B) 480 is the right answer

37.

1+1.0 -0.33In a test a student got 30% marks and failedby 25 marks. In the same test anotherstudent got 40% marks and secured 25%marks more than the essential minimumpass marks. The maximum pass marks forthe test were: ।किसी टेस्ट में एक विद्यार्थी को 30% अंक प्राप्त हुए औरवह 25 अंक से अनुत्तीर्ण हो गया. उसी टेस्ट में दुसरे विद्यार्थीको 40% अंक प्राप्त हुए और आवश्यक न्यूनतम पास अंकोंसे 25% अधिक अंक प्राप्त हुए. टेस्ट के लिए अधिकतमपास अंक कितने थे?A) 400B480c500D) 580

Answer»

B) 480 is the correct answer

B is the correct answer

480 is the righy answer

B) 480 is the correct answer

B) 480 is the right answer

Discussion

38.	13. In an election, there were only two candidates for the post of president. The winning candidate got 53% ofthe total votes. His opponent got 31,000 votes which represent 31% of the total votes. Find (a) the numberof votes and (b) the winning margin.
Answer» 31x/100 = 31000 total number of votes x = 100000 votes winning margin = 100000-31000 = 69000 votes

38.

13. In an election, there were only two candidates for the post of president. The winning candidate got 53% ofthe total votes. His opponent got 31,000 votes which represent 31% of the total votes. Find (a) the numberof votes and (b) the winning margin.

Answer»

31x/100 = 31000

total number of votes x = 100000 votes

winning margin = 100000-31000

= 69000 votes

Discussion

39.	The points A(x, y), B( x2, y2) and C(x3, y3)are the vertices of a triangle ABC. What arethe coordinates of the centroid of thetriangle ABC?
Answer»

Discussion

40.	imA student got 177 marksalindample 7 : 40% of the maximum marks are required to pass an exam. A student got 14by 23 marks. Find the maximum marks.
Answer» 500 500

40.

imA student got 177 marksalindample 7 : 40% of the maximum marks are required to pass an exam. A student got 14by 23 marks. Find the maximum marks.

Answer»

500

Discussion

41.	a student got 325 marks and failed by 35 marks.ifv40% is the minimum percentage of passing marks ,find the maximum marks
Answer»

Discussion

42.	EXAMPLES BEFORE EXERhe slope of the tangent to the curve y = x3-3 at x
Answer» slope = y' = 3x²-1 and slope at x = 2 is y'(2) = 3(2)²-1 = 12-1 = 11

42.

EXAMPLES BEFORE EXERhe slope of the tangent to the curve y = x3-3 at x

Answer»

slope = y' = 3x²-1 and slope at x = 2 is

y'(2) = 3(2)²-1 = 12-1 = 11

Discussion

43.	Find the point on the curve y = x3-11x + 5 at which the tangent is y = x-11..
Answer»

Discussion

44.	qual, ten(A) b = ac() b = ac(B)b= 5 (a+c)D) none of these22. 1-0 = ?22. -**(A)..?cB(D)4ab23. If x=-a+bX +2a X+2b+X-Pay -2a1s
Answer» Option c is correct.

Discussion

45.	l t0 a-3Q.2. If the matrix A2 0 -1 is skew symmetric, find the values of 'a' and 'b'
Answer»

Discussion

46.	uctions.Find the.IHCEx.LCM forthe lumbers 100-an90.0Find the numher of o
Answer» From which book have you taken this question? Please tell us so that we can provide you faster answer. Hcflcm=ab (a &b are two given no.)so hcf * lcm = 100*190 = 19000

46.

uctions.Find the.IHCEx.LCM forthe lumbers 100-an90.0Find the numher of o

Answer»

From which book have you taken this question? Please tell us so that we can provide you faster answer.

Hcf*lcm=a*b (a &b are two given no.)so hcf * lcm = 100*190 = 19000

Discussion

47.	adea picture ofan aeroplane with coloured paper as shown in Fig 12.15, Finthe total area of the paper used.5cmctn6cmiV1.5cm6.5cmIcm2cm
Answer»

Discussion

48.	The base and the height of a triangle arein the ratio 5:3. If the area of the triangle6.is 67.5 m2 find its base and height.1S
Answer» Let the base of triangle b = 5xLet height of triangle h = 3xGiven, Area of triangle = 67.5 m^2 Area of triangle = (1/2)bh Then,(1/2)5x3x = 67.515x^2 = 135x^2 = 135/15 = 9x = 3 Base of triangle = 5x = 53 = 15 mHeight of triangle = 3x = 33 = 9 m

48.

The base and the height of a triangle arein the ratio 5:3. If the area of the triangle6.is 67.5 m2 find its base and height.1S

Answer»

Let the base of triangle b = 5xLet height of triangle h = 3xGiven, Area of triangle = 67.5 m^2

Area of triangle = (1/2)*b*h

Then,(1/2)*5x*3x = 67.515x^2 = 135x^2 = 135/15 = 9x = 3

Base of triangle = 5x = 5*3 = 15 mHeight of triangle = 3x = 3*3 = 9 m

Discussion

49.	26. If the sum of a certain number of terms of the AP25, 22, 1S27. Find the sum of odd integers from 1 to 2001
Answer»

Discussion

50.	22. Find three numbers in G.P. whose sum is 52 and the sum of whose products in1Spairs is 624.hors in GPare subtracted and the
Answer»

Discussion

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