81279 + Interview Questions in Mathematics QA in Mathematics Page 50 InterviewSolution

2451.	en pen5, 120Write the following polynomial in co2x33x4
Answer» Coefficient x^3=2x^2=1x=-3

Discussion

2452.	Simplify the following mathematicalexpressions.1. 18+9 - 15+ 32. 27 = 3 x 4 - 53. 90 + 2000 = 500 x 64. 35 + 12 - 4x8=25.25 x 5 - 12 + 18 + 96. 45678 - 3124 + 5670
Answer» 1. 152. 313. 114 4. 31 5. 1156. 48234 1=92=313=1144=315=616=

2452.

Simplify the following mathematicalexpressions.1. 18+9 - 15+ 32. 27 = 3 x 4 - 53. 90 + 2000 = 500 x 64. 35 + 12 - 4x8=25.25 x 5 - 12 + 18 + 96. 45678 - 3124 + 5670

Answer»

1. 152. 313. 114 4. 31 5. 1156. 48234

1=92=313=1144=315=616=

Discussion

2453.	Ifa person saves? 2 lakhs for twoyears in a fixed deposit account at 4 % interest rate per annum,what amount will he get on maturity if interest is compounded annually
Answer» let P = 200000, r = 4% = 4/100 = 0.04 and n = 2 years The amount A is given by A = p(1+r)ⁿ=> A = 200000(1+4/100)² = 200000(1+0.04)²=> A = 216320rs.

2453.

Ifa person saves? 2 lakhs for twoyears in a fixed deposit account at 4 % interest rate per annum,what amount will he get on maturity if interest is compounded annually

Answer»

let P = 200000, r = 4% = 4/100 = 0.04 and n = 2 years

The amount A is given by A = p(1+r)ⁿ=> A = 200000(1+4/100)² = 200000(1+0.04)²=> A = 216320rs.

Discussion

2454.	: 3TrFM 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39.
Answer» 3= 3,6,9,12,15,18,21,27,30,33,36,39,42,45,48,519 = 9,18,27,36,45,54,63,72,81,90

Discussion

2455.	If $ \frac{\cos 13^{\circ}+\sin 13^{\circ}}{\cos 13^{\circ}-\sin 13^{\circ}}=\tan A, $ then $ A= $
Answer» 2 3

2455.

If $ \frac{\cos 13^{\circ}+\sin 13^{\circ}}{\cos 13^{\circ}-\sin 13^{\circ}}=\tan A, $ then $ A= $

Answer»

2

3

Discussion

2456.	MATHS-Q.1 Simplify-1) (-9+7)X(-8)2) (-27)X(-16)+(-27)X153) (-2) X(-3)?X(-1)?4)-8-24+31-26-28+7+19-18-8+335) -26-20+33-(-33)+21+24-(-25)-26-14-34Ques 2 Simplify:-a) 8/25 X 55/22 % 21/11 X 4/9
Answer» 1)162)273)2884)-225)16question 233/35

Discussion

2457.	2.f Solve:1, 2118 1822 (e)15 155. 3(b) 413. Shubham painted 3 of the wall space in his room. His sister Madhavi hs12(0 88() 330) 3and painfedof the wall space. How much did they paint together?4. Fill in the missing fractions.
Answer» Bit g question has some misprintings.

Discussion

2458.	Prove that s ( \frac { \pi } { 4 } - \theta ) \operatorname { cos } ( \frac { \pi } { 4 } - \phi ) - \operatorname { sin } ( \frac { \pi } { 4 } - \theta ) \operatorname { sin } ( \frac { \pi } { 4 } - \phi ) = \operatorname { sin } ( \theta + \phi )
Answer»

Discussion

2459.	\operatorname { cos } ( \frac { \pi } { 4 } - A ) \operatorname { cos } ( \frac { \pi } { 4 } - B ) - \operatorname { sin } ( \frac { \pi } { 4 } - A ) \operatorname { sin } ( \frac { \pi } { 4 } - B ) = \operatorname { sin } ( A + B ) \quad [ N C )
Answer»

Discussion

2460.	\frac{\cos ^{2}\left(\frac{\pi}{4}-A\right)-\sin ^{2}\left(\frac{\pi}{4}-A\right)}{\cos ^{2}\left(\frac{\pi}{4}+A\right)+\sin ^{2}\left(\frac{\pi}{4}+A\right)}=
Answer» cos^2A-sin^2A=cos2Acos^2A+isn^2A=1hencehencecos(π/4-A+π/4-A)cos(π/2-2A)sin(2A)

Discussion

2461.	\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)
Answer»

Discussion

2462.	Ex7. Show that the roots of (x - a)(x - b) = ca are always(MAY-2009)real.
Answer» To solve this question, you require two key concepts: 1. For real roots, Discriminant >=O2. Sum of two square Real Numbers will always be a positive quantity. Because square makes everything positive and sum of two positives will always be >= 0.See picture below.

Discussion

2463.	\begin{array}{l}{\cos \left(\frac{\pi}{4}-x\right) \cdot \cos \left(\frac{\pi}{4}-y\right)-} \\ {\sin \left(\frac{\pi}{4}-x\right) \cdot \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)}\end{array}
Answer» LHS = cos(π/4 - x).cos(π/4-y)-sin(π/4 -x).sin(π/4-y) Let ( π/4 -x) = A (π/4 -y) = BThen, LHS = cosA.cosB -sinA.sinB But we Know, cos(A + B) = cosA.cosB-sinA.sinB use this, = cos(A + B) = cos{(π/4 -x)+(π/4 -y)}=cos(π/2 -(x +y)}We know, Cos(π/2 -∅) = sin∅ use this , = sin(x + y) = RHS

2463.

\begin{array}{l}{\cos \left(\frac{\pi}{4}-x\right) \cdot \cos \left(\frac{\pi}{4}-y\right)-} \\ {\sin \left(\frac{\pi}{4}-x\right) \cdot \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)}\end{array}

Answer»

LHS = cos(π/4 - x).cos(π/4-y)-sin(π/4 -x).sin(π/4-y)

Let ( π/4 -x) = A (π/4 -y) = BThen, LHS = cosA.cosB -sinA.sinB But we Know, cos(A + B) = cosA.cosB-sinA.sinB use this,

= cos(A + B) = cos{(π/4 -x)+(π/4 -y)}=cos(π/2 -(x +y)}We know, Cos(π/2 -∅) = sin∅ use this , = sin(x + y) = RHS

Discussion

2464.	( 0 , \frac { \pi } { 6 } ) \cup ( \frac { 5 \pi } { 6 } , 2 \pi ) ( b ) ( 0 , \frac { 5 \pi } { 6 } ) \cup ( \pi , 2 \pi ) \quad ( c ) ( 0 , \frac { \pi } { 6 } ) \cup ( \pi , 2 \pi )
Answer» Given the equation is. 2sin²∅-5sin∅+2 > 0=> 2sin²∅-4sin∅-sin∅+2 > 0=> 2sin∅(sin∅-2)-1(sin∅-2)>0=> (sin∅-2)(2sin∅-1) > 0=> sin∅ = [-1,1/2) but sin∅ has range from [-1,1] so, sin∅ belongs to. (0,π/6) U (5π/6,2π) nice solution

2464.

( 0 , \frac { \pi } { 6 } ) \cup ( \frac { 5 \pi } { 6 } , 2 \pi ) ( b ) ( 0 , \frac { 5 \pi } { 6 } ) \cup ( \pi , 2 \pi ) \quad ( c ) ( 0 , \frac { \pi } { 6 } ) \cup ( \pi , 2 \pi )

Answer»

Given the equation is.

2sin²∅-5sin∅+2 > 0=> 2sin²∅-4sin∅-sin∅+2 > 0=> 2sin∅(sin∅-2)-1(sin∅-2)>0=> (sin∅-2)(2sin∅-1) > 0=> sin∅ = [-1,1/2)

but sin∅ has range from [-1,1]

so, sin∅ belongs to. (0,π/6) U (5π/6,2π)

nice solution

Discussion

2465.	\operatorname { cos } \frac { 5 \pi } { 3 } + \operatorname { cos } 4 \frac { 4 \pi } { 3 } + \operatorname { sin } \frac { 3 \pi } { 2 } \times \operatorname { sin } \frac { 5 \pi } { 8 } \times \operatorname { cos } 3 \pi
Answer» cos(5π/3) + cos(4π/3) + sin(3π/2)×sin(5π/8)×cos(3π/2) = 1/2 + (-1/2) + -1(sin5π/8)0= 1/2-1/2 +0= 0

2465.

\operatorname { cos } \frac { 5 \pi } { 3 } + \operatorname { cos } 4 \frac { 4 \pi } { 3 } + \operatorname { sin } \frac { 3 \pi } { 2 } \times \operatorname { sin } \frac { 5 \pi } { 8 } \times \operatorname { cos } 3 \pi

Answer»

cos(5π/3) + cos(4π/3) + sin(3π/2)×sin(5π/8)×cos(3π/2)

= 1/2 + (-1/2) + -1*(sin5π/8)*0= 1/2-1/2 +0= 0

Discussion

2466.	\frac { \pi } { 5 } \cdot \operatorname { sin } \frac { 2 \pi } { 5 } \operatorname { sin } \frac { 3 \pi } { 5 } \operatorname { sin } \frac { 4 \pi } { 5 } = \frac { 5 } { 16 }
Answer»

Discussion

2467.	\left. \begin{array} { l } { ( a ) x = - \frac { \pi } { 20 } + \frac { \pi } { 5 } n \quad ( b ) x = \frac { \pi } { 40 } + \frac { \pi } { 10 } n } \\ { ( c ) x = \frac { 3 \pi } { 20 } + \frac { \pi } { 5 } n } \end{array} \right.
Answer»

Discussion

2468.	Example-4. A boy observed the top of an electric pole at an angle of elevation of 60Â° when theobservation point is 8 meters away from the foot of the pole. Find the heightof the pole.triongle QAR
Answer»

Discussion

2469.	(a) A square can be thought of as a special rectangle.
Answer» Rectangle becomes a square if all the sides of the rectangle are equal.

Discussion

2470.	how a square can be thought of as a special rectangle
Answer»

Discussion

2471.	2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0
Answer» 2cos pi/13 cos 9pi/13+ cos 3pi/13 +cos 5pi/13 =cos 10 pi/13 +cos 8 pi/13 +cos 3pi/13 +cos 5pi/13 =cos 10 pi/13 +cos 3pi/13 +cos 8pi/13 +cos 5pi/13 =2 cos pi/2 .cos 7 pi/26 +2 cos pi/2 .cos 3 pi /26 =2 (0)cos 7 pi /26 + 2(0) cos 3pi/26 =0 =R.H.S.

2471.

2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0

Answer»

2cos pi/13 cos 9pi/13+ cos 3pi/13 +cos 5pi/13

=cos 10 pi/13 +cos 8 pi/13 +cos 3pi/13 +cos 5pi/13

=cos 10 pi/13 +cos 3pi/13 +cos 8pi/13 +cos 5pi/13

=2 cos pi/2 .cos 7 pi/26 +2 cos pi/2 .cos 3 pi /26

=2 (0)cos 7 pi /26 + 2(0) cos 3pi/26

=0 =R.H.S.

Discussion

2472.	\operatorname { tan } \frac { \pi } { 8 } \cdot \operatorname { tan } 3 \pi \cdot \operatorname { tan } 5 \pi \cdot \operatorname { tan } 7 \pi = 1
Answer» after putting these values it can be solved easily it will be 1

Discussion

2473.	(ii)A shop has 500 parts, out of which 5 are defective. What per cent are defective?
Answer» 5/500*1001 percentage is defective

Discussion

2474.	66 199
Answer» The number which comes immediately after a particular number is called its successor. The successor of a whole number is the number obtained by adding 1 to it.Clearly, the successor of 0 is 1; successor of 1 is 2; successor of 2 is 3 and so on.therefore successor of 100 199 are 101 200.

2474.

66 199

Answer»

The number which comes immediately after a particular number is called its successor.

The successor of a whole number is the number obtained by adding 1 to it.Clearly, the successor of 0 is 1; successor of 1 is 2; successor of 2 is 3 and so on.therefore successor of 100 199 are 101 200.

Discussion

2475.	3057 \text { and } 199
Answer» Incomplete Question.please submit complete question.

Discussion

2476.	Due to heavy floods in a state, thousandfor 1500expenditure equally. The lower part of each tent is cylindrical of base ras were rendered homeless. 50 schoolse oered to the state government to provide place and the canvastents to be fixed by the government and decided to share the wholeupper part of same base radius but ofper sq.m, find2.8 m and height 3.5 m, with conicalthe amount shared by each school to set up the tents. What value is generatedby the above problem ? (Use Ď=-)height 2.1 m. Ifthe cunvas used to made the tents costs ? 120 per sq m, find
Answer» C.S.A of cylinder =2πrh =2×22/7×2.8×3.5 =61.6m^2 Slant height of cone = √2.8^2+2.1^2 = 3.5m C.S.A of cone = πrl =22/7×2.8×3.5 =30.8m^2 S.A of a tent = 92.4m^2 Cost of one tent = 92.4×120 =11088 Cost of 1500 tents = 1500×11088 = 16632000 Cost shared by each school = 16632000/50 = ₹332640

2476.

Due to heavy floods in a state, thousandfor 1500expenditure equally. The lower part of each tent is cylindrical of base ras were rendered homeless. 50 schoolse oered to the state government to provide place and the canvastents to be fixed by the government and decided to share the wholeupper part of same base radius but ofper sq.m, find2.8 m and height 3.5 m, with conicalthe amount shared by each school to set up the tents. What value is generatedby the above problem ? (Use Ď=-)height 2.1 m. Ifthe cunvas used to made the tents costs ? 120 per sq m, find

Answer»

C.S.A of cylinder =2πrh

=2×22/7×2.8×3.5

=61.6m^2

Slant height of cone = √2.8^2+2.1^2

= 3.5m

C.S.A of cone = πrl

=22/7×2.8×3.5

=30.8m^2

S.A of a tent = 92.4m^2

Cost of one tent = 92.4×120

=11088

Cost of 1500 tents = 1500×11088 = 16632000

Cost shared by each school = 16632000/50 = ₹332640

Discussion

2477.	5. The birth rate per thousand in five countries over a period oftimeisshownbelowChinaIndia GermanyUKSwedenCountryBirth rate per thousand4235142821Represent the above data by a bar graph.
Answer»

Discussion

2478.	The following data on the number of girls (to the nearest ten) per thousand boys indifferent sections of Indian society is given below.SectionScheduled Caste (SC)Scheduled Tribe (ST)Non SC/STBackward districtsNon-backward districtsRuralUrbanNumber of girls per thousand boys940970920950920930
Answer» thanks

Discussion

2479.	14. Estimate the sale amount of a Mall which has 73 shops and the minimum sale isaround 81 thousand per shop. .
Answer» 1 shop=8100073 shop=8100073=rs 5913000 Given:Sales for 1 shop=Rs 81,000No of shops =73Thus total sales =73×81,000=Rs 5913000 1 shop = 81000 rupees73 shops = 73810000 =5913000

2479.

14. Estimate the sale amount of a Mall which has 73 shops and the minimum sale isaround 81 thousand per shop. .

Answer»

1 shop=8100073 shop=81000*73=rs 5913000

Given:Sales for 1 shop=Rs 81,000No of shops =73Thus total sales =73×81,000=Rs 5913000

1 shop = 81000 rupees73 shops = 73*810000 =5913000

Discussion

2480.	A lane 180 m long and 5 m wide is to be pavedwith bricks, of length 20 cm and breadth 15cm. Find the cost of the bricks that are required,at the rate of 750 per thousand..
Answer»

Discussion

2481.	16. A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm.Find the cost of bricks that are required, at the rate of Rs 750 per thousand.
Answer» We indicate with L length of the lane, with l breadth of the lane. we calculate the area of a lane:A₁=L×l=180×5=900 m² we indicate with b length of the brick, with h breadthof the brick. we calculate the areaof a brick:A₂=b×h=20×15=300 cm² we convert the cm² in m² (1m²=10000 cm²)300 cm²=0,03 m²thenA₂=0,03 m² we calculate the required number of bricks:n=A₁/A₂=900/0,03=30000 brickswe calculate the total cost of bricks:tot=(750×30000):1000=22500 ruppes

2481.

16. A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm.Find the cost of bricks that are required, at the rate of Rs 750 per thousand.

Answer»

We indicate with L length of the lane, with l breadth of the lane.

we calculate the area of a lane:A₁=L×l=180×5=900 m²

we indicate with b length of the brick, with h breadthof the brick.

we calculate the areaof a brick:A₂=b×h=20×15=300 cm²

we convert the cm² in m² (1m²=10000 cm²)300 cm²=0,03 m²thenA₂=0,03 m²

we calculate the required number of bricks:n=A₁/A₂=900/0,03=30000 brickswe calculate the total cost of bricks:tot=(750×30000):1000=22500 ruppes

Discussion

2482.	S. A lane 180 m long and 5 m wide is to be pavedwith bricks, of length 20 cm and breadth 15 cmm.Find the cost of the bricks that are required, atthe rate of R 750 per thousand.
Answer»

Discussion

2483.	Q. 5. Solve the given equation and verify youranswer:1.5x +0.3 33x
Answer» ( 1.5x + 0.3 )/3x = 3/10 1.5x + 0.3 = 3x × 0.3 1.5x + 0.3 = 0.9x 1.5x - 0.9x = - 0.3 0.6x = -0.3 x = -0.3/0.6 x = -1/2 = -0.5

2483.

Q. 5. Solve the given equation and verify youranswer:1.5x +0.3 33x

Answer»

( 1.5x + 0.3 )/3x = 3/10

1.5x + 0.3 = 3x × 0.3

1.5x + 0.3 = 0.9x

1.5x - 0.9x = - 0.3

0.6x = -0.3

x = -0.3/0.6

x = -1/2 = -0.5

Discussion

2484.	9.A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm.Find the cost of bricks that are required, at the rate of Rs 750 per thousand. bAick
Answer» 40 40

2484.

9.A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm.Find the cost of bricks that are required, at the rate of Rs 750 per thousand. bAick

Answer»

40

Discussion

2485.	Internal fertilisation occurs
Answer» This occurs in most mammals, some cartilaginous fish, and a few reptiles, making these animals viviparous. Internal fertilization has the advantage of protecting the fertilized egg from dehydration on land. The embryo is isolated within the female, which limits predation on the young.

Discussion

2486.	Write the coefficients of x in each of the following:1 2+*+xm2-8erimg+x(iv) Sex-1
Answer» 1. 12. -13. π/24. 0

Discussion

2487.	9. Form a quadratic equation whose roots are 2 and 3.20. Form a quadratic equation whose roots are -3 and 4
Answer» 19. Equation:(x-2)(x-3)=0 => x²-5x+6=0

2487.

9. Form a quadratic equation whose roots are 2 and 3.20. Form a quadratic equation whose roots are -3 and 4

Answer»

19. Equation:(x-2)(x-3)=0

=> x²-5x+6=0

Discussion

2488.	Solve the quadratic equation 2x 3x+I0using the formulSelve the quadratic equation 2 3x+0formul
Answer» a= 2 b= -3 c= 1Discriminant = b^2-4ac9-8= 1now roots will be

Discussion

2489.	) The roots of the quadratic equation are (1+2) and (1 - 2). Form the quadratic equation.3) Draw a hist
Answer» Sum of roots is 1+√2+1-√2 = 2Product of roots = (1-√2)(1+√2) = 1 - 2 = -1x2 - (sum of roots)x + product of roots = 0 x2 - 2x - 1 = 0

Discussion

2490.	8. Find the roots of the quadratic equation 22 0.he roots of the quadratic equation v2 x4 7x 5V20
Answer»

Discussion

2491.	ue of the discriminant for the quadratic equation 2
Answer» x² + 10x - 7 = 0 D = b² - 4ac (10)² - 4 × (-7) 100 + 28 128

2491.

ue of the discriminant for the quadratic equation 2

Answer»

x² + 10x - 7 = 0

D = b² - 4ac

(10)² - 4 × (-7)

100 + 28

128

Discussion

2492.	1) Find the quadratic equation whose sum &s product of the rootsFind the quadratic equation whose sum & product of the roots4 respectively.are 3 and
Answer» Tqsm

Discussion

2493.	Find the value of K for the Quadratic equation 2 x^{2}+k x+3=0, so that it has real and equal roots.
Answer»

Discussion

2494.	II, vut it is not a quadratic equation.(11) dove, the giIn (iv) above, the given equation appeTegree 3) and not a quadratic equation. Butou can see, often we need to simplify thequadratic or not.EXERC1. Check whether the following are quadra(i) (x+1)2 = 2(x-3)(1) (x-2)(x+1)=(x-1(x+3)
Answer»

Discussion

2495.	8. A coin is tossed 200 times. The head appears 79 times. Find the probability of a tail.
Answer»

Discussion

2496.	25. A coin is tossed 100 times and head appears 52 times. Now if wetoss a coin at random then find the probability of getting-a. A headb. A tail
Answer» 1) Probability of getting a head is 52/100 (Given) 2) Probability of getting a tail is 1 - 52/100 = 48/100 Explanation: A probability of one means that the event is certain. If you toss a coin, it will come up a head or a tail. So there is a probability of one that either of these will happen. A probability of zero means that an event is impossible.

2496.

25. A coin is tossed 100 times and head appears 52 times. Now if wetoss a coin at random then find the probability of getting-a. A headb. A tail

Answer»

1) Probability of getting a head is 52/100 (Given)

2) Probability of getting a tail is 1 - 52/100 = 48/100

Explanation:

A probability of one means that the event is certain. If you toss a coin, it will come up a head or a tail.

So there is a probability of one that either of these will happen.

A probability of zero means that an event is impossible.

Discussion

2497.	2. If roots of the quadratic equation x2 + 2px + mn=0 are real and equal, show that the roots of thequadratic equation x2 - 2m+n)x+(m? + n2 + 2p) = 0 are also equal.
Answer» x² + 2px + mn = 0Roots are real and equal => D = 0 => (2p)² - 4 (1) (mn) = 0 => 4p² - 4mn = 0 => p² - mn = 0 => p² = mn Substituting p² = mn in the second equation, x² - 2(m + n)x + (m² + n² + 2p²) = 0 => x² - 2(m + n)x + (m² + n² + 2mn) = 0=> x² - 2(m + n)x + (m + n)² = 0 D = [-2(m + n)]² - 4(1)(m + n)² = 4(m + n)² - 4(m + n)² = 0 Since D = 0, the roots are real and equal.

2497.

2. If roots of the quadratic equation x2 + 2px + mn=0 are real and equal, show that the roots of thequadratic equation x2 - 2m+n)x+(m? + n2 + 2p) = 0 are also equal.

Answer»

x² + 2px + mn = 0Roots are real and equal => D = 0 => (2p)² - 4 (1) (mn) = 0 => 4p² - 4mn = 0 => p² - mn = 0 => p² = mn

Substituting p² = mn in the second equation, x² - 2(m + n)x + (m² + n² + 2p²) = 0

=> x² - 2(m + n)x + (m² + n² + 2mn) = 0=> x² - 2(m + n)x + (m + n)² = 0

D = [-2(m + n)]² - 4(1)(m + n)² = 4(m + n)² - 4(m + n)² = 0

Since D = 0, the roots are real and equal.

Discussion

2498.	Solve the quadratic equation (x2-502-7(x2-50 + 6-0
Answer» please like my answer if you find it useful

Discussion

2499.	79 \times 192
Answer» by basic multiplication79×192=15168

Discussion

2500.	If roots of the quadratic equation x2 +2px +mn =0 are real andequal, show that the roots of the quadratic equationx2 -2(m+n)x +(m2+n2 +2p2) =0 are also equal=5cm
Answer»

Discussion

Explore topic-wise InterviewSolutions in .

en pen5, 120Write the following polynomial in co2x33x4

Simplify the following mathematicalexpressions.1. 18+9 - 15+ 32. 27 = 3 x 4 - 53. 90 + 2000 = 500 x 64. 35 + 12 - 4x8=25.25 x 5 - 12 + 18 + 96. 45678 - 3124 + 5670

Ifa person saves? 2 lakhs for twoyears in a fixed deposit account at 4 % interest rate per annum,what amount will he get on maturity if interest is compounded annually

: 3TrFM 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39.

If $ \frac{\cos 13^{\circ}+\sin 13^{\circ}}{\cos 13^{\circ}-\sin 13^{\circ}}=\tan A, $ then $ A= $

MATHS-Q.1 Simplify-1) (-9+7)X(-8)2) (-27)X(-16)+(-27)X153) (-2) X(-3)?X(-1)?4)-8-24+31-26-28+7+19-18-8+335) -26-20+33-(-33)+21+24-(-25)-26-14-34Ques 2 Simplify:-a) 8/25 X 55/22 % 21/11 X 4/9

2.f Solve:1, 2118 1822 (e)15 155. 3(b) 413. Shubham painted 3 of the wall space in his room. His sister Madhavi hs12(0 88() 330) 3and painfedof the wall space. How much did they paint together?4. Fill in the missing fractions.

Prove that s ( \frac { \pi } { 4 } - \theta ) \operatorname { cos } ( \frac { \pi } { 4 } - \phi ) - \operatorname { sin } ( \frac { \pi } { 4 } - \theta ) \operatorname { sin } ( \frac { \pi } { 4 } - \phi ) = \operatorname { sin } ( \theta + \phi )

\operatorname { cos } ( \frac { \pi } { 4 } - A ) \operatorname { cos } ( \frac { \pi } { 4 } - B ) - \operatorname { sin } ( \frac { \pi } { 4 } - A ) \operatorname { sin } ( \frac { \pi } { 4 } - B ) = \operatorname { sin } ( A + B ) \quad [ N C )

\frac{\cos ^{2}\left(\frac{\pi}{4}-A\right)-\sin ^{2}\left(\frac{\pi}{4}-A\right)}{\cos ^{2}\left(\frac{\pi}{4}+A\right)+\sin ^{2}\left(\frac{\pi}{4}+A\right)}=

\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)

Ex7. Show that the roots of (x - a)(x - b) = ca are always(MAY-2009)real.

\begin{array}{l}{\cos \left(\frac{\pi}{4}-x\right) \cdot \cos \left(\frac{\pi}{4}-y\right)-} \\ {\sin \left(\frac{\pi}{4}-x\right) \cdot \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)}\end{array}

( 0 , \frac { \pi } { 6 } ) \cup ( \frac { 5 \pi } { 6 } , 2 \pi ) ( b ) ( 0 , \frac { 5 \pi } { 6 } ) \cup ( \pi , 2 \pi ) \quad ( c ) ( 0 , \frac { \pi } { 6 } ) \cup ( \pi , 2 \pi )

\operatorname { cos } \frac { 5 \pi } { 3 } + \operatorname { cos } 4 \frac { 4 \pi } { 3 } + \operatorname { sin } \frac { 3 \pi } { 2 } \times \operatorname { sin } \frac { 5 \pi } { 8 } \times \operatorname { cos } 3 \pi

\frac { \pi } { 5 } \cdot \operatorname { sin } \frac { 2 \pi } { 5 } \operatorname { sin } \frac { 3 \pi } { 5 } \operatorname { sin } \frac { 4 \pi } { 5 } = \frac { 5 } { 16 }

\left. \begin{array} { l } { ( a ) x = - \frac { \pi } { 20 } + \frac { \pi } { 5 } n \quad ( b ) x = \frac { \pi } { 40 } + \frac { \pi } { 10 } n } \\ { ( c ) x = \frac { 3 \pi } { 20 } + \frac { \pi } { 5 } n } \end{array} \right.

Example-4. A boy observed the top of an electric pole at an angle of elevation of 60Â° when theobservation point is 8 meters away from the foot of the pole. Find the heightof the pole.triongle QAR

(a) A square can be thought of as a special rectangle.

how a square can be thought of as a special rectangle

2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0

\operatorname { tan } \frac { \pi } { 8 } \cdot \operatorname { tan } 3 \pi \cdot \operatorname { tan } 5 \pi \cdot \operatorname { tan } 7 \pi = 1

(ii)A shop has 500 parts, out of which 5 are defective. What per cent are defective?

66 199

3057 \text { and } 199

5. The birth rate per thousand in five countries over a period oftimeisshownbelowChinaIndia GermanyUKSwedenCountryBirth rate per thousand4235142821Represent the above data by a bar graph.

The following data on the number of girls (to the nearest ten) per thousand boys indifferent sections of Indian society is given below.SectionScheduled Caste (SC)Scheduled Tribe (ST)Non SC/STBackward districtsNon-backward districtsRuralUrbanNumber of girls per thousand boys940970920950920930

14. Estimate the sale amount of a Mall which has 73 shops and the minimum sale isaround 81 thousand per shop. .

A lane 180 m long and 5 m wide is to be pavedwith bricks, of length 20 cm and breadth 15cm. Find the cost of the bricks that are required,at the rate of 750 per thousand..

16. A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm.Find the cost of bricks that are required, at the rate of Rs 750 per thousand.

S. A lane 180 m long and 5 m wide is to be pavedwith bricks, of length 20 cm and breadth 15 cmm.Find the cost of the bricks that are required, atthe rate of R 750 per thousand.

Q. 5. Solve the given equation and verify youranswer:1.5x +0.3 33x

9.A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm.Find the cost of bricks that are required, at the rate of Rs 750 per thousand. bAick

Internal fertilisation occurs

Write the coefficients of x in each of the following:1 2+*+xm2-8erimg+x(iv) Sex-1

9. Form a quadratic equation whose roots are 2 and 3.20. Form a quadratic equation whose roots are -3 and 4

Solve the quadratic equation 2x 3x+I0using the formulSelve the quadratic equation 2 3x+0formul

) The roots of the quadratic equation are (1+2) and (1 - 2). Form the quadratic equation.3) Draw a hist

8. Find the roots of the quadratic equation 22 0.he roots of the quadratic equation v2 x4 7x 5V20

ue of the discriminant for the quadratic equation 2

1) Find the quadratic equation whose sum &amp;s product of the rootsFind the quadratic equation whose sum &amp; product of the roots4 respectively.are 3 and

Find the value of K for the Quadratic equation 2 x^{2}+k x+3=0, so that it has real and equal roots.

II, vut it is not a quadratic equation.(11) dove, the giIn (iv) above, the given equation appeTegree 3) and not a quadratic equation. Butou can see, often we need to simplify thequadratic or not.EXERC1. Check whether the following are quadra(i) (x+1)2 = 2(x-3)(1) (x-2)(x+1)=(x-1(x+3)

8. A coin is tossed 200 times. The head appears 79 times. Find the probability of a tail.

25. A coin is tossed 100 times and head appears 52 times. Now if wetoss a coin at random then find the probability of getting-a. A headb. A tail

2. If roots of the quadratic equation x2 + 2px + mn=0 are real and equal, show that the roots of thequadratic equation x2 - 2m+n)x+(m? + n2 + 2p) = 0 are also equal.

Solve the quadratic equation (x2-502-7(x2-50 + 6-0

79 \times 192

If roots of the quadratic equation x2 +2px +mn =0 are real andequal, show that the roots of the quadratic equationx2 -2(m+n)x +(m2+n2 +2p2) =0 are also equal=5cm

1) Find the quadratic equation whose sum &s product of the rootsFind the quadratic equation whose sum & product of the roots4 respectively.are 3 and