81279 + Interview Questions in Mathematics QA in Mathematics

1.	b^2 - a^2 %2B b^2 %2B a^2 %2B 2(ab)=0
Answer» a^2 + 2ab + b^2 - a^2 + b^2 = 0 2ab + 2b^2 = 0 2b(a + b) = 0 b = 0, - a

1.

b^2 - a^2 %2B b^2 %2B a^2 %2B 2(ab)=0

Answer»

a^2 + 2ab + b^2 - a^2 + b^2 = 0

2ab + 2b^2 = 0

2b(a + b) = 0

b = 0, - a

Discussion

2.	x^2 %2B x*(a/(a %2B b) %2B (a %2B b)/a) %2B 1=0
Answer» Let --> y = ( a / ( a + b )) ; --> x² + [ y + 1/y ]x + 1 = 0 => yx² + [ y² + 1 ] x + y = 0 => yx² + y²x + x + y = 0 => [ yx + 1 ][ x + y ] = 0 => x = [ - 1 / y ] \|\| x = - y => x = [ ( - a - b ) / a ], x = [ -a / ( a + b ) ]

2.

x^2 %2B x*(a/(a %2B b) %2B (a %2B b)/a) %2B 1=0

Answer»

Let --> y = ( a / ( a + b )) ;

--> x² + [ y + 1/y ]x + 1 = 0

=> yx² + [ y² + 1 ] x + y = 0

=> yx² + y²x + x + y = 0

=> [ yx + 1 ][ x + y ] = 0

=> x = [ - 1 / y ] || x = - y

=> x = [ ( - a - b ) / a ], x = [ -a / ( a + b ) ]

Discussion

3.	2 ( a x - b y ) %2B ( a %2B 4 b ) = 0,2 ( b x %2B a y ) %2B ( b - 4 a ) = 0
Answer» It's a system of 2 linear equations in variables x & y & constant terms a & b which need to be solved simultaneously. According to my knowledge, one can't solve simultaneous equations by cross multiplication. I solve it by equating coefficients. You can write these equations like :2ax - 2by = -a-4b ..... (1)2bx + 2ay = 4a-b .........(2)Now, multiplying (1) by a & (2) by b,2a^2x - 2aby = -a^2 - 4ab ......(3)2b^2x + 2aby = -b^2 + 4ab .....(4)Adding (3) & (4),2 (a^2+b^2) x = -(a^2+b^2).Obviously,x = -1/2. y= 2

3.

2 ( a x - b y ) %2B ( a %2B 4 b ) = 0,2 ( b x %2B a y ) %2B ( b - 4 a ) = 0

Answer»

It's a system of 2 linear equations in variables x & y & constant terms a & b which need to be solved simultaneously. According to my knowledge, one can't solve simultaneous equations by cross multiplication. I solve it by equating coefficients.

You can write these equations like :2ax - 2by = -a-4b ..... (1)2bx + 2ay = 4a-b .........(2)Now, multiplying (1) by a & (2) by b,2a^2x - 2aby = -a^2 - 4ab ......(3)2b^2x + 2aby = -b^2 + 4ab .....(4)Adding (3) & (4),2 (a^2+b^2) x = -(a^2+b^2).Obviously,x = -1/2.

y= 2

Discussion

4.	5111 5. + sin 3xe =lan 4x0055: + рео0532
Answer» LHS = (sin5x + sin3x)/(cos5x + cos3x) use the formula, sinC + sinD = 2sin(C + D)/2.cos(C-D)/2 cosC + cosD = 2cos(C + D)/2.cos(C - D)/2 = {2sin(5x+3x)/2.cos(5x-3x)/2}/{2cos(5x+3x)/2.cos(5x-3x)/2}= (2sin4x.cosx)/(2cos4x.cosx)= sin4x/cos4x = tan4x = RHS

4.

5111 5. + sin 3xe =lan 4x0055: + рео0532

Answer»

LHS = (sin5x + sin3x)/(cos5x + cos3x) use the formula, sinC + sinD = 2sin(C + D)/2.cos(C-D)/2 cosC + cosD = 2cos(C + D)/2.cos(C - D)/2

= {2sin(5x+3x)/2.cos(5x-3x)/2}/{2cos(5x+3x)/2.cos(5x-3x)/2}= (2sin4x.cosx)/(2cos4x.cosx)= sin4x/cos4x = tan4x = RHS

Discussion

5.	0055 3 + 0057ल्ण्म्[उ+ + 0८083 Cos‘(x——-’tj—ﬁz,%) T
Answer» Using this, cos²x = {1 + cos(2x)}/2cos²(x + π/3) = {1 + cos(2x + 2π/3)}/2 and cos²(x - π/3) = {1 + cos(2x - 2π/3)}/2 ii) Hence, left side of the given one is: = {1 + cos(2x)}/2 + {1 + cos(2x + 2π/3)}/2 + {1 + cos(2x - 2π/3)}/2 = (3/2) + (1/2)[cos(2x) + cos(2x + 2π/3) + cos(2x - 2π/3)] = (3/2) + (1/2)[cos(2x) + 2cos(2x)cos(2π/3)][Since cos(A+B) + cos(A-B) = 2cosAcosB] = (3/2) + (1/2)[cos(2x) - cos(2x)] [Since cos(2π/3) = -1/2] = 3/2 = Right side HENCE PROVED hit like if you find it useful

5.

0055 3 + 0057ल्ण्म्[उ+ + 0८083 Cos‘(x——-’tj—ﬁz,%) T

Answer»

Using this, cos²x = {1 + cos(2x)}/2cos²(x + π/3) = {1 + cos(2x + 2π/3)}/2 and cos²(x - π/3) = {1 + cos(2x - 2π/3)}/2

ii) Hence, left side of the given one is:

= {1 + cos(2x)}/2 + {1 + cos(2x + 2π/3)}/2 + {1 + cos(2x - 2π/3)}/2

= (3/2) + (1/2)[cos(2x) + cos(2x + 2π/3) + cos(2x - 2π/3)]

= (3/2) + (1/2)[cos(2x) + 2cos(2x)*cos(2π/3)][Since cos(A+B) + cos(A-B) = 2cosA*cosB]

= (3/2) + (1/2)[cos(2x) - cos(2x)] [Since cos(2π/3) = -1/2]

= 3/2 = Right side HENCE PROVED

hit like if you find it useful

Discussion

6.	3. On a graph paper, plot each of the following points:(vi) F (5, -3)
Answer»

Discussion

7.	curved surface area of a cone whose base radius is 5.6 cmand slant height is 9 cm.
Answer» curved surface area formula is πrl= 22×5.6×9/7= 158.4 cm^2 very nice answer

7.

curved surface area of a cone whose base radius is 5.6 cmand slant height is 9 cm.

Answer»

curved surface area formula is πrl= 22×5.6×9/7= 158.4 cm^2

very nice answer

Discussion

8.	Using a graphx.7 Copy and complete the table below for thi011141 1145146149 64 81ow for the equation y=ZAPlot the graph of y=vx,Use your graph to find the approximate values of the135 b Jas c es d 160e v2
Answer» option d is correct,please give positive rating ....... option c is correct answer because the question say that we can only count the value of y

8.

Using a graphx.7 Copy and complete the table below for thi011141 1145146149 64 81ow for the equation y=ZAPlot the graph of y=vx,Use your graph to find the approximate values of the135 b Jas c es d 160e v2

Answer»

option d is correct,please give positive rating .......

option c is correct answer because the question say that we can only count the value of y

Discussion

9.	:-MATHSASSIGNMElThe three vertices of AABC are AL, 4), B-2, 2) and C(3, 2). Plot these points on a graph pacalculate the area of NABC.DIWALIints on a graph paper and
Answer» Thanks it is very useful for me

Discussion

10.	10.A person is asked to choose a number n from -10 to 10. Find the probabilitythat the Inl <5.a. 9/20
Answer»

Discussion

11.	12. Choose the option which replaces the question mark:12(a) 8(c) 10(b) 9(d) 14
Answer» mechanism behind answer is to subtract the addition of lower 2 numbers from the addition of upper 3 numbersso in 1st figure addition of upper 3 numbers are 8+6+3=17 and addition of lower 2 numbers are =7+5=12so difference is =17-12=5same way in 2nd figure addition of upper 3 numbers are 2+4+9=15 and addition of lower 2 numbers are =3+6=9so difference is =15-9=6 same way in question figure addition of upper 3 numbers are 12+5+3=20 and addition of lower 2 numbers are =4+6=10so difference is =20-10=10 so answer is 10

11.

12. Choose the option which replaces the question mark:12(a) 8(c) 10(b) 9(d) 14

Answer»

mechanism behind answer is to subtract the addition of lower 2 numbers from the addition of upper 3 numbersso in 1st figure addition of upper 3 numbers are 8+6+3=17 and addition of lower 2 numbers are =7+5=12so difference is =17-12=5same way in 2nd figure addition of upper 3 numbers are 2+4+9=15 and addition of lower 2 numbers are =3+6=9so difference is =15-9=6

same way in question figure addition of upper 3 numbers are 12+5+3=20 and addition of lower 2 numbers are =4+6=10so difference is =20-10=10

so answer is 10

Discussion

12.	1. Ajoker's cap is in the form of right circular cone whose base radius is 7cm and heightis 24cm. Find the area of the sheet réquired to make 10 such caps.
Answer»

Discussion

13.	10..v..le.qurudullaAdverbs of time:boede,...nteleeAdverbs of place:B Choose the correct adverbs frompassage.
Answer» Adverbs of place includes (in, on, above,below,near,far).pls give me best

Discussion

14.	3.Find the curved surface area of a cone whose base radius is 5.6 cmand slant height is 9 cm.
Answer» curved surface area=πrl=22/7×5.6×9=158.4cm×cm 462 cm is right answer r=5.6cml=9cmCurved surface area=piy rl=22/7into5.6into9=22into0.8into9=158.4cmsquare CSA=22/7×r×l =22/7×5.6×9 = 22×0.8×9 =158.4 sq cm

14.

3.Find the curved surface area of a cone whose base radius is 5.6 cmand slant height is 9 cm.

Answer»

curved surface area=πrl=22/7×5.6×9=158.4cm×cm

462 cm is right answer

r=5.6cml=9cmCurved surface area=piy rl=22/7into5.6into9=22into0.8into9=158.4cmsquare

CSA=22/7×r×l =22/7×5.6×9 = 22×0.8×9 =158.4 sq cm

Discussion

15.	7.A Joker's cap is in the form of right circular cone, whose base radius is 7 cm andheight is 24 cm. Find the area of sheet required to make 10 such caps.
Answer» surface area of cone= πr^2h= 22/72449222473696cm^2for 10 sheets= 39690 cm^2

Discussion

16.	sinSxâ2sin3x +sin xŕĽŽ08 5. - (005 2=tan x
Answer» (sin5x-2sin3x+sinx)/(cos5x-cosx)=[{2sin(5x+x)/2cos(5x-x)/2}-2sin3x]/{2sin(5x+x)/2sin(x-5x)/2}=(2sin3xcos2x-2sin3x)/{2sin3xsin(-2x)}={2sin3x(cos2x-1)}/{-2sin3xsin2x}=-(cos2x-1)/sin2x=(1-cos2x)/sin2x=2sin²x/2sinxcosx=sinx/cosx=tanx (Proved)

Discussion

17.	. A joker's cap is in the form of rightcircular cone whose base radius is7 om and height is 24 cm. Find the areaof the sheet required to make 10 suchASJcaps
Answer»

Discussion

18.	Prove that :sin5x - 2sin3x +sinx0055: - 005 2=tanx
Answer» We use sin A + sin B = 2 sin (A+B)/2. cos (A-B)/2cos A - cos B = - 2 sin (A+B)/2 sin(A-B)/2 sin5x-2sin3x+sinx)/(cos5x-cosx) = [{2sin(5x+x)/2cos(5x-x)/2}-2sin3x]/{2sin(5x+x)/2sin(x-5x)/2} = (2sin3xcos2x-2sin3x)/{2sin3xsin(-2x)} = {2sin3x(cos2x-1)}/{-2sin3xsin2x} = -(cos2x-1)/sin2x = (1-cos2x)/sin2x = 2sin^2 x/2sinxcosx = sinx/cosx = tanx (Proved)

18.

Prove that :sin5x - 2sin3x +sinx0055: - 005 2=tanx

Answer»

We use sin A + sin B = 2 sin (A+B)/2. cos (A-B)/2cos A - cos B = - 2 sin (A+B)/2 sin(A-B)/2

sin5x-2sin3x+sinx)/(cos5x-cosx)

= [{2sin(5x+x)/2cos(5x-x)/2}-2sin3x]/{2sin(5x+x)/2sin(x-5x)/2}

= (2sin3xcos2x-2sin3x)/{2sin3xsin(-2x)}

= {2sin3x(cos2x-1)}/{-2sin3xsin2x}

= -(cos2x-1)/sin2x

= (1-cos2x)/sin2x

= 2sin^2 x/2sinxcosx

= sinx/cosx

= tanx (Proved)

Discussion

19.	Draw a number line and answer the following:(a) Which number will we reach if we move 4 numbers to the right of-2Which number will we reach if we move 5 numbers to the left of 1.(c) If we are at - 8 on the number line, in which direction should we move tareach - 13?d) If we are at- 6 on the number line, in which direction should we move treach - 1?
Answer»

Discussion

20.	s = o o il AN AR AE SR, N L Xglz 3), plu) अरे (g,=2)
Answer»

Discussion

21.	questions from the following151 in thedivision2 Solve any four sub-questions from the) Sohre : 2 - 8 = 12une cost * 700, how much will 12 cups cost?
Answer» 1) z-8=12 z=12+8 z=20 2) 20 cup costs 700 rs 1 cup cost 700/20 =35 rs.12 cup cost= 35 ×12= 420 rs. 20 cup cost rs 7001 cup cost=700/2012 cup cost 700/20*12=420 therefore 12 cup cost is rs 420

21.

questions from the following151 in thedivision2 Solve any four sub-questions from the) Sohre : 2 - 8 = 12une cost * 700, how much will 12 cups cost?

Answer»

1) z-8=12 z=12+8 z=20

2) 20 cup costs 700 rs 1 cup cost 700/20 =35 rs.12 cup cost= 35 ×12= 420 rs.

20 cup cost rs 7001 cup cost=700/2012 cup cost 700/20*12=420

therefore 12 cup cost is rs 420

Discussion

22.	t)A wire is bent in the form of a rectangle having length wlewire is bent in the form of a circle. It was found that the area of the circle is greaterthan that of the rectangle by 104.5 cm2. Find the length of the wire.ŃĐź".high r halls are red.
Answer» Step-by-step explanation: Define x: Let the breadth be x The length = 2x Find the area of the rectangle: Area = Length x Breadth Area = 2x² Find the perimeter of the rectangle: Perimeter = 2(Length + breadth) Perimeter = 2(2x + x) Perimeter = 6x Find the radius of the circle in term of x Circumference = 2πr 2πr = 6x r = 6x ÷2π r = 3x/π Find the area of the circle in term of x: Area = πr² Area = π(3x/π)² = 9x²/π Solve x: The area of the circle is greater than the rectangle by 104.5 cm² 9x²/π - 2x² = 104.5 x² (9/π - 2) = 104.5 19/22 x² = 104.5 x² = 104.5 ÷19/22 x² = 121 x = √121 x = 11 cm Find the length of the wire: Length = 6x = 6(11) = 66 cm Answer: The length is 66 cm

22.

t)A wire is bent in the form of a rectangle having length wlewire is bent in the form of a circle. It was found that the area of the circle is greaterthan that of the rectangle by 104.5 cm2. Find the length of the wire.ŃĐź".high r halls are red.

Answer»

Step-by-step explanation:

Define x:

Let the breadth be x

The length = 2x

Find the area of the rectangle:

Area = Length x Breadth

Area = 2x²

Find the perimeter of the rectangle:

Perimeter = 2(Length + breadth)

Perimeter = 2(2x + x)

Perimeter = 6x

Find the radius of the circle in term of x

Circumference = 2πr

2πr = 6x

r = 6x ÷2π

r = 3x/π

Find the area of the circle in term of x:

Area = πr²

Area = π(3x/π)² = 9x²/π

Solve x:

The area of the circle is greater than the rectangle by 104.5 cm²

9x²/π - 2x² = 104.5

x² (9/π - 2) = 104.5

19/22 x² = 104.5

x² = 104.5 ÷19/22

x² = 121

x = √121

x = 11 cm

Find the length of the wire:

Length = 6x = 6(11) = 66 cm

Answer: The length is 66 cm

Discussion

23.	3wich are as0.005 2 0.005
Answer» both are = eeee eee eeeee Both 0.005 and 0.005 are equal both are equal (0.005 and 0.005 both are equal stupid both are equal according to the problem. answer is: both are = ( equal)

23.

3wich are as0.005 2 0.005

Answer»

both are =

eeee

eee

eeeee

Both 0.005 and 0.005 are equal

both are equal (0.005 and 0.005

both are equal stupid

both are equal according to the problem.

answer is: both are = ( equal)

Discussion

24.	Q.(1)Find the roots of the following quadratic equations by factorization(x - 4)(x + 2) = 0(4,-2)(i)(2x + 3)(3x - 7) = 0(iii)9x2 - 3x - 2 = 0310x=3(v)413x? + 5x - 23 = 0(vi)S2x2 – 3x-202=0.(1.12)(1.73)(vii) a?x? - 3abx + 2b2 = 0(viii) x? - (12+1)x + V2 = 0(ix) x2 - (V3 + 1)x + V3 = 0(x) ax + (na? – 3b)x – 12ab = 0(xi) 2 - 2x + 6 = 04+3 3x-2(1232)
Answer» (x+4)(x-2)0either x+4=0 , x= -4or x-2=0 , x= 2roots are (4,2)2)(2x+3)(3x-7)0either 2x+3=02x= -3x= -3/2or 3x-7=03x= 7x= 3/7roots are ( -3/2,3/7)3)9x^2-3x-2=09x^2-6x+3x-2=03x(3x-2)+1(3x-2)=0(3x-2)(3x+1)=0either 3x-2=03x= 2x= 2/3or 3x+1=03x= -1x=. -1/3roots are ( 2/3, -1/3)4) 1)(x+4)(x-2); x=4; x=-2; 2)(2x+3)(3x-7); 2x=-3; x=-3/2; 3x=7; x=7/3 (-3/2, 7/3) 3)9x^2-3x-2=9x^2-6x+3x-2=0; 3x(3x-2)+1(3x-2); (3x+1)(3x-2); 3x=-1:;x-1 x=-1/3, 3x=2; x=2/3; 5)

24.

Q.(1)Find the roots of the following quadratic equations by factorization(x - 4)(x + 2) = 0(4,-2)(i)(2x + 3)(3x - 7) = 0(iii)9x2 - 3x - 2 = 0310x=3(v)413x? + 5x - 23 = 0(vi)S2x2 – 3x-202=0.(1.12)(1.73)(vii) a?x? - 3abx + 2b2 = 0(viii) x? - (12+1)x + V2 = 0(ix) x2 - (V3 + 1)x + V3 = 0(x) ax + (na? – 3b)x – 12ab = 0(xi) 2 - 2x + 6 = 04+3 3x-2(1232)

Answer»

(x+4)(x-2)0either x+4=0 , x= -4or x-2=0 , x= 2roots are (4,2)2)(2x+3)(3x-7)0either 2x+3=02x= -3x= -3/2or 3x-7=03x= 7x= 3/7roots are ( -3/2,3/7)3)9x^2-3x-2=09x^2-6x+3x-2=03x(3x-2)+1(3x-2)=0(3x-2)(3x+1)=0either 3x-2=03x= 2x= 2/3or 3x+1=03x= -1x=. -1/3roots are ( 2/3, -1/3)4)

1)(x+4)(x-2); x=4; x=-2; 2)(2x+3)(3x-7); 2x=-3; x=-3/2; 3x=7; x=7/3 (-3/2, 7/3) 3)9x^2-3x-2=9x^2-6x+3x-2=0; 3x(3x-2)+1(3x-2); (3x+1)(3x-2); 3x=-1:;x-1 x=-1/3, 3x=2; x=2/3; 5)

Discussion

25.	construct a quadrilateral ABCD in wichAem, ADDC 45.5 cm and B-120
Answer» Bhai phle 5.5ki line khich phir arc bna 60° phir angle leke 90 arc Lena phir 90 ke ark par line khic phir Tera 120° ka ho jayega question solved

Discussion

26.	\frac { ( 0 \cdot 137 + 0 \cdot 098 ) ^ { 2 } + ( 0 \cdot 137 - 0 \cdot 098 ) ^ { 2 } } { 0 \cdot 137 \times 0.137 + 0.098 \times 0.098 }
Answer» answer is not correct from the given options a. 0.2b. 2c. 4d. 0.4 b..2.07~2( round off)

26.

\frac { ( 0 \cdot 137 + 0 \cdot 098 ) ^ { 2 } + ( 0 \cdot 137 - 0 \cdot 098 ) ^ { 2 } } { 0 \cdot 137 \times 0.137 + 0.098 \times 0.098 }

Answer»

answer is not correct from the given options a. 0.2b. 2c. 4d. 0.4

b..2.07~2( round off)

Discussion

27.	In the following, upper triangular matrix is2100(A) o 2 030 35 4 2(B) 0 0 3[O23](D) 0 30
Answer» (B) An upper triangular matrix is one which has non zero elements above the diagonal.

Discussion

28.	9 x ^ 2 - 9 ( a %2B b ) x %2B ( 2 a ^ 2 %2B 5 a b %2B 2 b ^ 2 ) = 0
Answer» Consider, 9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0 9x^2 - 9(a + b)x + (2a^2 + 4ab + ab + 2b^2) = 0 9x^2 - 9(a + b)x + [2a(a + 2b) + b(a + 2b)] = 0 9x^2 - 9(a + b)x + [(a + 2b)(2a + b)] = 0 9x^2 – 3[(a + 2b) + (2a + b)]x + [(a + 2b)(2a + b)] = 0 9x^2 – 3(a + 2b)x - 3(2a + b)x + [(a + 2b)(2a + b)] = 0 3x[3x–(a + 2b)] - (2a + b)[3x + (a – 2b)] = 0 [3x – (a + 2b)][3x - (2a + b)] = 0 [3x – (a + 2b)] = 0 or [3x - (2a + b)] = 0 3x = (a + 2b) or 3x = (2a + b) x = (a + 2b)/3 or x = (2a + b)/3

28.

9 x ^ 2 - 9 ( a %2B b ) x %2B ( 2 a ^ 2 %2B 5 a b %2B 2 b ^ 2 ) = 0

Answer»

Consider, 9x^2-9(a+b)x+(2a^2+5ab+2b^2)=0

9x^2 - 9(a + b)x + (2a^2 + 4ab + ab + 2b^2) = 0

9x^2 - 9(a + b)x + [2a(a + 2b) + b(a + 2b)] = 0

9x^2 - 9(a + b)x + [(a + 2b)(2a + b)] = 0

9x^2 – 3[(a + 2b) + (2a + b)]x + [(a + 2b)(2a + b)] = 0

9x^2 – 3(a + 2b)x - 3(2a + b)x + [(a + 2b)(2a + b)] = 0

3x[3x–(a + 2b)] - (2a + b)[3x + (a – 2b)] = 0

[3x – (a + 2b)][3x - (2a + b)] = 0

[3x – (a + 2b)] = 0 or [3x - (2a + b)] = 0

3x = (a + 2b) or 3x = (2a + b)

x = (a + 2b)/3 or x = (2a + b)/3

Discussion

29.	2b^2 %2B 2a^2 %2B 5(ab) - 9x(a %2B b) %2B x/((9*x^2))=0
Answer» 9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0 9x^2 - 9(a + b)x + (2a^2 + 4ab + ab + 2b^2) = 0 9x^2 - 9(a + b)x + [2a(a + 2b)+ b(a + 2b)] = 0 9x^2 - 9(a + b)x + (2a + b)(a + 2b) = 0 9x^2 - 3(2a + b)x - 3(a + 2b)x + (2a + b)(a + 2b) = 0 3x(3x - 2a - b) - (a + 2b)(3x - 2a - b) = 0 (3x - a - 2b)(3x - 2a - b) = 0 x = - (a + 2b)/3, - (2a + b)/3

29.

2b^2 %2B 2a^2 %2B 5(ab) - 9x(a %2B b) %2B x/((9*x^2))=0

Answer»

9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0

9x^2 - 9(a + b)x + (2a^2 + 4ab + ab + 2b^2) = 0

9x^2 - 9(a + b)x + [2a(a + 2b)+ b(a + 2b)] = 0

9x^2 - 9(a + b)x + (2a + b)(a + 2b) = 0

9x^2 - 3(2a + b)x - 3(a + 2b)x + (2a + b)(a + 2b) = 0

3x(3x - 2a - b) - (a + 2b)(3x - 2a - b) = 0

(3x - a - 2b)(3x - 2a - b) = 0

x = - (a + 2b)/3, - (2a + b)/3

Discussion

30.	-2/3 - 3/2 %2B 2/2 %2B 2/9
Answer» = 5/2 - 7/2 - 7/3 + 19/9= (45-63-42+38)/18= (83-105)/18= -22/18 = -11/9 ans. 5/2-7/2-7/3+19/9 =45-63-42+38/18 =-22/18 =-11/9 5/2-7/2-7/2+19/9=(45-63-42+38)/18=(-18-4)/18=-22/18=-11/9

30.

-2/3 - 3/2 %2B 2/2 %2B 2/9

Answer»

= 5/2 - 7/2 - 7/3 + 19/9= (45-63-42+38)/18= (83-105)/18= -22/18 = -11/9 ans.

5/2-7/2-7/3+19/9 =45-63-42+38/18 =-22/18 =-11/9

5/2-7/2-7/2+19/9=(45-63-42+38)/18=(-18-4)/18=-22/18=-11/9

Discussion

31.	IS CG0S PAR
Answer»

Discussion

32.	3. एक घड़ी को आइने के सामने रखने पर लगता है कि 7 बजकर। 12 मिनट हुआ है। घड़ी में सही समय क्या हो रहा होगा?[RRB-2005)(a) 4 बजकर 10 मिनट (b) 4 बजकर 45 मिनट\| (c) 4 बजकर 48 मिनट (d) 4 बजकर 50 मिनट
Answer» c). 4 bajkar 48 min. C option is the right ans

32.

3. एक घड़ी को आइने के सामने रखने पर लगता है कि 7 बजकर। 12 मिनट हुआ है। घड़ी में सही समय क्या हो रहा होगा?[RRB-2005)(a) 4 बजकर 10 मिनट (b) 4 बजकर 45 मिनट| (c) 4 बजकर 48 मिनट (d) 4 बजकर 50 मिनट

Answer»

c). 4 bajkar 48 min.

C option is the right ans

Discussion

33.	Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Answer»

Discussion

34.	3)Mr. Yadav bought kg of sweets. After his children had eaten some, there wasKg left. How much they had eaten.
Answer» Let kg of sweets eaten by children=XX+1/10=1/2X=1/2-1/10X=10-2/20hence, X =2/5 let kg of sweets eaten by children =x X+1/10=1/2 X=1/2-1/10 X=10-2/20 hence, X =2/5 2/5 is the best answer

34.

3)Mr. Yadav bought kg of sweets. After his children had eaten some, there wasKg left. How much they had eaten.

Answer»

Let kg of sweets eaten by children=XX+1/10=1/2X=1/2-1/10X=10-2/20hence, X =2/5

let kg of sweets eaten by children =x X+1/10=1/2 X=1/2-1/10 X=10-2/20 hence, X =2/5

2/5 is the best answer

Discussion

35.	A table has two drawers. First drawerhas 3 goid coins & second has 4 silvercoins. Open a drawer randomly & takea coin. What is the probability that itis a gold coin?
Answer»

Discussion

36.	i)Name the group of islands lying in the Arabian sea.
Answer» LakshadweepIslands lie in the Arabian Sea

Discussion

37.	\begin{array} { l } { \text { The length, breadth and height of cuboid are } } \\ { \text { in the ratio } 6 : 5 : 3 \text { . If its total surface area is } } \\ { 504 \mathrm { cm } ^ { 2 } \text { , find its volume. } } \end{array}
Answer»

Discussion

38.	find the following product (2x-1)(x^2+x+1)
Answer»

Discussion

39.	The letters of the word"FLOWER" are taken4 at a time and arranged in all possible ways.The number of arrangements which begin with'F' and end with 'R' is1) 20 2) 18 3) 14 4) 12
Answer»

Discussion

40.	A cubical block of side 7 cm is surmounted by a hemisphere. What is the greater diameter the hemisphere can have? Find the surface area of the solid?
Answer» The hemisphere is mounted on the cubical blockThe diameter of the hemisphere = the side of the cube= 7cm TSA of the solid = TSA of cube - Area of base of hemisphere + CSA of hemisphere ⇒ TSA = 6×7² -π×(3.5)² + 2×π×(3.5)² ⇒ TSA = 6×7² + π×(3.5)² ⇒ TSA = 332. 465 cm² Thank you

40.

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greater diameter the hemisphere can have? Find the surface area of the solid?

Answer»

The hemisphere is mounted on the cubical blockThe diameter of the hemisphere = the side of the cube= 7cm

TSA of the solid = TSA of cube - Area of base of hemisphere + CSA of hemisphere

⇒ TSA = 6×7² -π×(3.5)² + 2×π×(3.5)²

⇒ TSA = 6×7² + π×(3.5)²

⇒ TSA = 332. 465 cm²

Thank you

Discussion

41.	The non- parallel sides of a trapezium are of length 6 cm each, and the parallel sides are of length 14 cmand 20 cm. Find the height of the trapezium.
Answer» thank u

Discussion

42.	ana uCHALLENGING QUESTIONdes are of length 14 cmThe non-parallel sides of a trapezium are of length 6 cm each, and the parallel sides are oand 20 cm. Find the height of the trapezium.
Answer»

Discussion

43.	A triangle has sides of length 6 cm, 7.5 cm and 4.5 cm. It is a right-angled triangle.Why?
Answer» to prove this check whether the square of longest side is equal to sum of square of other two sideshence 7.5^2=56.25and 6^2+4.5^=56.25hence they are equal so it is a right angled triangle by Pythagoras theorem Thanks

43.

A triangle has sides of length 6 cm, 7.5 cm and 4.5 cm. It is a right-angled triangle.Why?

Answer»

to prove this check whether the square of longest side is equal to sum of square of other two sideshence 7.5^2=56.25and 6^2+4.5^=56.25hence they are equal so it is a right angled triangle by Pythagoras theorem

Thanks

Discussion

44.	Word Problemsn aeroplane, there are 120 pasengers. If 20 are children, find hownany percent are children?
Answer» among 120, 20 are children so percent of children are 20/120 * 100 = 16.67%

Discussion

45.	6. मैरी ने एक मोबाइल फोन र 1950 में बेचाऔर 25% नुकसान उठाया । वह 30% लाभपाने के लिए किस विक्रयमूल्य पर बेचे ?assa
Answer» 1950 में मेरी को मोबाइल बेचने पर 25% का नुकसान हुआ यानी 1950 रूपए उस मोबाइल के क्रय मूल्य का 75% है इसका 100% निकालने के लिए हम 1950 में 75 का भाग देने पर 1% बराबर ₹28 आएंगे इसी प्रकार इसी प्रकार अगर उसे 30% प्रतिशत लाभ कमाना है तो 28 का गुना 130 से करने पर उसका मूल्य प्राप्त होगा उसका मूल्य 3640 रुपए प्राप्त होंगे 3380sygxjvddjjfucffscjhdfg

45.

6. मैरी ने एक मोबाइल फोन र 1950 में बेचाऔर 25% नुकसान उठाया । वह 30% लाभपाने के लिए किस विक्रयमूल्य पर बेचे ?assa

Answer»

1950 में मेरी को मोबाइल बेचने पर 25% का नुकसान हुआ यानी 1950 रूपए उस मोबाइल के क्रय मूल्य का 75% है इसका 100% निकालने के लिए हम 1950 में 75 का भाग देने पर 1% बराबर ₹28 आएंगे इसी प्रकार इसी प्रकार अगर उसे 30% प्रतिशत लाभ कमाना है तो 28 का गुना 130 से करने पर उसका मूल्य प्राप्त होगा उसका मूल्य 3640 रुपए प्राप्त होंगे

3380sygxjvddjjfucffscjhdfg

Discussion

46.	Example 5 : The height of a cone is 16 cm and its base radius is 12 cm. Find tcurved surface area and the total surface area of the cone (Use π = 3.14).
Answer» PLEASE LIKE IT, IF YOU FIND THE ANSWER HELPFUL.

Discussion

47.	(v) 6+3) 02-3)2. Evaluate the following product
Answer» identity 3 will be applied here which is:a2 -b2 = ( a+b ) ( a-b ) ( y^2+3/2)( y^2-3/2)= ( y^4-3/2y^2+3/2y^2+9/4)= (y^4+9/4)

47.

(v) 6+3) 02-3)2. Evaluate the following product

Answer»

identity 3 will be applied here which is:a2 -b2 = ( a+b ) ( a-b )

( y^2+3/2)( y^2-3/2)= ( y^4-3/2y^2+3/2y^2+9/4)= (y^4+9/4)

Discussion

48.	points (4. 8, 1o in which the line joining theplane.Find the ratio in1) and (6, 10, -8) is divided by YZ
Answer»

Discussion

49.	3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Fin(i) radius of the base and (ii) total surface area of the cone.
Answer»

Discussion

50.	The surface area of a sphere of radius 5 cm is five times the curved surface area of a cone of radius 4cm. Find the height and volume of the cone.
Answer»

Discussion

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