81279 + Interview Questions in Mathematics QA in Mathematics

1.	Rationalize the denominator 73-S
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2.	13. Rationalize the denominator of
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3.	20Rationalize the denominator of
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4.	s The tangent to the circle x2 + y? - 5 at the point (1,-2)also touches the circlea(B30()1.1x2 + y2 _ 8x + 6y + 20-oat
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5.	The coordinates of any point on the circle x2 y24are
Answer» A circle can be defined as the locus of all points that satisfy the equationsx = r cos(t) y = r sin(t)where x,y are the coordinates of any point on the circle, r is the radius of the circle and t is the parameter - the angle subtended by the point at the circle's center.Here r=2. So, x = 2 cos(t) y = 2 sin(t)

Discussion

6.	Up)yFind the centre of the circle x2 +y2 + 4x-8y=0.
Answer» Comparing with standard equation of circlehencex^2+y^2+2gx+2fy+c=0hence centre=(-g,-f)=(-2,4)

Discussion

7.	Q.6,Find the centre and radius of the circle x2 + y2-6x-8y-24 = 0
Answer» Comparing with standard equation hence,Standard equation is x^2+y^2+2gx+2fy+c=0hence radius is √g^2+f^2-c and centre is (-g,-f)hence g is -3f is -4c is -24hence radius is √(3)^2+(4)^2-(-24)=√9+16+24=√49=7Hence centre is (3,4)

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8.	tan θ1-cot θ, 1-tan θ[Hint: Write the expression in terms of sin 9 and cos e]cot θm)+-1 + sec θ cosec θ
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9.	\begin{array}{l}{\text { Compute the following: }} \\ {\text { (i) } \left[ \begin{array}{cc}{a} & {b} \\ {-b} & {a}\end{array}\right]+\left[ \begin{array}{ll}{a} & {b} \\ {b} & {a}\end{array}\right]}\end{array}
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10.	TRY THESEFind:\begin{array}{l}{\text { (i) } \frac{-3}{4} \times \frac{1}{7}} \\ {\text { (ii) } \frac{2}{3} \times \frac{-5}{9}}\end{array}
Answer» i) (-3)/4 × 1/7 = -3/28 ii) 2/3 × (-5)/9 = -10/27

10.

TRY THESEFind:\begin{array}{l}{\text { (i) } \frac{-3}{4} \times \frac{1}{7}} \\ {\text { (ii) } \frac{2}{3} \times \frac{-5}{9}}\end{array}

Answer»

i) (-3)/4 × 1/7 = -3/28

ii) 2/3 × (-5)/9 = -10/27

Discussion

11.	Show that 12" cannot end with the digit 0 or 5 for any natural number n.
Answer» If 12n ends with 0 then it must have 5 as a factor.But, 12n=(2×2×3)n which shows that only 2 and 3 are the prime factors of 12n.Also, we know from the fundamental theory of arithmetic that the prime factorization of each number is unique.So, 5 is not a factor of 12n.Hence, 12n can never end with the digit 0.

Discussion

12.	Show that 12" cannot end with the digit 0 or 5 for any natural number n.2
Answer» If 12n ends with 0 then it must have 5 as a factor. But, 12n=(2×2×3)n which shows that only 2 and 3 are the prime factors of 12n. Also, we know from the fundamental theory of arithmetic that the prime factorization of each number is unique. So, 5 is not a factor of 12n. Hence, 12n can never end with the digit 0. Hit like if you find it useful!

12.

Show that 12" cannot end with the digit 0 or 5 for any natural number n.2

Answer»

If 12n ends with 0 then it must have 5 as a factor.

But, 12n=(2×2×3)n which shows that only 2 and 3 are the prime factors of 12n.

Also, we know from the fundamental theory of arithmetic that the prime factorization of each number is unique.

So, 5 is not a factor of 12n.

Hence, 12n can never end with the digit 0.

Hit like if you find it useful!

Discussion

13.	2. Show that 12" cannot end with the digit 0 or 5 for any natural number n.
Answer» If 12n ends with 0 then it must have 5 as a factor.But, 12n=(2×2×3)n which shows that only 2 and 3 are the prime factors of 12n.Also, we know from the fundamental theory of arithmetic that the prime factorization of each number is unique.So, 5 is not a factor of 12n.Hence, 12n can never end with the digit 0.

Discussion

14.	1. Rashmi is 14th from the right end ina row of 40 girls. What is her positionfrom the left end?
Answer» her position from left =40-14+1=26+1=27 thanks

14.

1. Rashmi is 14th from the right end ina row of 40 girls. What is her positionfrom the left end?

Answer»

her position from left =40-14+1=26+1=27

thanks

Discussion

15.	194tan θcot θ(iii) l-cot θ 1-tan θ1 + sec θ cosec θHint : Write the expression in terms of sin e and cos 81 + sec A sin2 A
Answer» Like my answer if you find it useful!

Discussion

16.	tan +cot θcot θ1-tan θHint . Write the expression .
Answer» GIVEN:- tanθ/(1 - cotθ) + cotθ/(1 - tanθ) => tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ) => tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1) => 1/(tanθ - 1) { tan²θ - 1/tanθ } => 1/(tanθ - 1) { (tan³θ - 1)/tanθ) [as, a³ - b³ = (a - b)(a² + b² + ab) => {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)} => tanθ + cotθ + 1 => sinθ/cosθ + cosθ/sinθ + 1 => (sin²θ + cos²θ)/sinθ . cosθ + 1 => 1/sinθ . cosθ + 1 => cosecθ . secθ + 1

16.

tan +cot θcot θ1-tan θHint . Write the expression .

Answer»

GIVEN:- tanθ/(1 - cotθ) + cotθ/(1 - tanθ)

=> tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ)

=> tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1)

=> 1/(tanθ - 1) { tan²θ - 1/tanθ }

=> 1/(tanθ - 1) { (tan³θ - 1)/tanθ)

[as, a³ - b³ = (a - b)(a² + b² + ab)

=> {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)}

=> tanθ + cotθ + 1

=> sinθ/cosθ + cosθ/sinθ + 1

=> (sin²θ + cos²θ)/sinθ . cosθ + 1

=> 1/sinθ . cosθ + 1

=> cosecθ . secθ + 1

Discussion

17.	tan θ , cot θ1-cot θ 1-tan θHint: Write the expression in terms of sin(iii)
Answer» Please hit the like button if this helped you

Discussion

18.	Q.6The expression, (1 + sec29) (1 +sec48) (1 + sec89) is equal totan 80(A) tanecot 8θ(B) cotetan 80cotecot 89tan θ
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19.	tan θcosec θ1-cote +1-tan. + sec θ[Hint: write the expression in terms of sin θ and cos θ]
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20.	tan θ1-cot θ 1-tan θ[Hint : Write the expression in terms of sin θ and cos θ]cot θ
Answer» tanθ/(1 - cotθ) + cotθ/(1 - tanθ) => tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ) => tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1) => 1/(tanθ - 1) { tan²θ - 1/tanθ } => 1/(tanθ - 1) { (tan³θ - 1)/tanθ) [as, a³ - b³ = (a - b)(a² + b² + ab) => {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)} => tanθ + cotθ + 1 => sinθ/cosθ + cosθ/sinθ + 1 => (sin²θ + cos²θ)/sinθ . cosθ + 1 => 1/sinθ . cosθ + 1 => cosecθ . secθ + 1

20.

tan θ1-cot θ 1-tan θ[Hint : Write the expression in terms of sin θ and cos θ]cot θ

Answer»

tanθ/(1 - cotθ) + cotθ/(1 - tanθ)

=> tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ)

=> tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1)

=> 1/(tanθ - 1) { tan²θ - 1/tanθ }

=> 1/(tanθ - 1) { (tan³θ - 1)/tanθ)

[as, a³ - b³ = (a - b)(a² + b² + ab)

=> {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)}

=> tanθ + cotθ + 1

=> sinθ/cosθ + cosθ/sinθ + 1

=> (sin²θ + cos²θ)/sinθ . cosθ + 1

=> 1/sinθ . cosθ + 1

=> cosecθ . secθ + 1

Discussion

21.	If we subtract three from a number, it becomes a perfect square. The original number cannot end in which of the following?
Answer» Any (perfect square + 3) can not end with the number 0 0²+3= 31²+3= 42²+3= 73²+3= 124²+3= 175²+3= 286²+3= 397²+3=428²+3=679²+3=84 Every original number of this question will end with2, 3, 4, 7, 8, 9. I think 11is a perfect square i think 1 is the correct answer....like plz Yessss... I is correct answer 7

21.

If we subtract three from a number, it becomes a perfect square. The original number cannot end in which of the following?

Answer»

Any (perfect square + 3) can not end with the number 0

0²+3= 31²+3= 42²+3= 73²+3= 124²+3= 175²+3= 286²+3= 397²+3=428²+3=679²+3=84

Every original number of this question will end with2, 3, 4, 7, 8, 9.

I think 11is a perfect square

i think 1 is the correct answer....like plz

Yessss... I is correct answer

7

Discussion

22.	B5,8, 11,--14th teemstroom
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23.	Math: Section 1Question: 1 of 60C. 15/4
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24.	Question 1If the(1) A5(②) 5004(③)(4④) 606
Answer» 5 x⁴is dy/dx of x^5is the right answer

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25.	empi all questions from theQuestion 1(a) Rationalize the denominator:14
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26.	Question 1.Find all the angles of an equilateral triangle.
Answer» All the angles of an equilateral triangle are equal in measure. Let each angle be x degrees since the sum of all interior angles of a triangle is 180 degrees therefore, x+x+x =180=>3x =180=>x=60 degrees Hence, each angle will be 60 degrees.

26.

Question 1.Find all the angles of an equilateral triangle.

Answer»

All the angles of an equilateral triangle are equal in measure.

Let each angle be x degrees

since the sum of all interior angles of a triangle is 180 degrees

therefore, x+x+x =180=>3x =180=>x=60 degrees

Hence, each angle will be 60 degrees.

Discussion

27.	viangular field are 15 m. 16 m and 17 m. With the three comers of the Sedtals and a hoseare tied separately with ropes of enigth 7 m each sohe field. Find the area of the field which cannot be graned by three atinalsINCERT EXEMPLARP cot
Answer» Let ABC be the triangular field with sides Ac = 15 m, AB = 16 m and BC = 17 mAnd,Let the place where the cow, the buffalo and thehorse are tied, are three sectors i.e. sector ADE, sector BFG and sector CHIThe area of triangular field by Heron's formula =√s(s-a)(s-b)(s-c)s = (a+b+c)/2s = (15+16+17)/2s = 48/2s = 24 m√24(24-15)(24-16)(24-17)√24987√12096Area of triangular field = 109.98 sq mArea of the grazed part = Area of the triangular field = Area of the sector ADE + Area of sector BFG + Area of sector CHI= π7²∠A/360 + π7²∠B/360 +π7²∠C/360=π7²(∠ A + ∠ B + ∠ C)/360= 22/777*180/360= 154/2Area of the grazed part = 77 sq mNow, the area of the field which cannot be grazed by these animals= 109.98 - 77= 32.98 sq mAnswer.

Discussion

28.	A steamer goes downstream and covers the distance between two parts in 4 hours whilecovers the same distance upstream in 5 hours. If the speed of the steamer is 2km/hr. Find ihspeed of the steamer in still water.s.the2.n0 mm
Answer» correct

Discussion

29.	A steamer goes downstream and covers the distance between two ports in 4 hours, while it covers thesame distance upstream in 5 hours. If the speed of the stream is 2 km/h, find the speed of the steamer in Cstill water
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30.	12. A steamer goes downstream from one point to another in 9 hours. It covers the same distanceupstream in 10 hours. If the speed of the stream be 1 km/hr, find the speed of the steamer in stillwater and the distance between the ports,
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Discussion

31.	It m times the mh term of an A.P. is equal to n times its nth term, then show that the(m+nih term of the A.P. is zero.
Answer» Let the first term of AP = acommon difference = dWe have to show that (m+n)th term is zero ora + (m+n-1)d = 0 mth term= a + (m-1)dnth term = a + (n-1) d Given thatm{a +(m-1)d} = n{a + (n -1)d}⇒am +m²d -md = an + n²d - nd⇒am - an +m²d - n²d -md + nd = 0⇒a(m-n) + (m²-n²)d-(m-n)d = 0⇒a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0⇒ a(m-n) + {(m-n)(m+n) -(m-n)} d = 0⇒a(m-n) +(m-n)(m+n -1) d = 0⇒ (m-n){a+ (m+n-1)d} = 0⇒a + (m+n -1)d = 0/(m-n)⇒a + (m+n -1)d = 0

31.

It m times the mh term of an A.P. is equal to n times its nth term, then show that the(m+nih term of the A.P. is zero.

Answer»

Let the first term of AP = acommon difference = dWe have to show that (m+n)th term is zero ora + (m+n-1)d = 0

mth term= a + (m-1)dnth term = a + (n-1) d

Given thatm{a +(m-1)d} = n{a + (n -1)d}⇒am +m²d -md = an + n²d - nd⇒am - an +m²d - n²d -md + nd = 0⇒a(m-n) + (m²-n²)d-(m-n)d = 0⇒a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0⇒ a(m-n) + {(m-n)(m+n) -(m-n)} d = 0⇒a(m-n) +(m-n)(m+n -1) d = 0⇒ (m-n){a+ (m+n-1)d} = 0⇒a + (m+n -1)d = 0/(m-n)⇒a + (m+n -1)d = 0

Discussion

32.	long and 64-m-broad lawn has two roads at right angles, one 2 m wide, rumningo its leneth, and the other 2.5 m wide, running parallel to its breadth. Find the115-
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Discussion

33.	f min and n tennissum of n terms of an A.P. is 3n? +5n and its mh term is 164, find the valuelcof m.ng ceauences an A p
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34.	the new rectangle.A steamer travels 90 km downstream in the same time as it takes to travelupstream. If the speed of the stream is 5 km/h, find the speed of the steamer60 kstillwaters the distance between two ports in 5 ho
Answer» Let x be the speed of the steamer in still water Speed of the stream = 5km/h (Given) Downstream: Speed = (x + 5) Time = Distance ÷speed Time = 90 /( x + 5) Upstream: Speed = ( x - 5) Time = Distance ÷Speed Time = 60/( x - 5) Since the time taken for upstream and downstream is the same: 90 /( x + 5) = 60/( x - 5) 90 (x - 5) = 60 (x + 5) 90x - 450 = 60x + 300 30x = 750 x = 25 km/h Answer: The speed of the steamer in still water is 25km/h

34.

the new rectangle.A steamer travels 90 km downstream in the same time as it takes to travelupstream. If the speed of the stream is 5 km/h, find the speed of the steamer60 kstillwaters the distance between two ports in 5 ho

Answer»

Let x be the speed of the steamer in still water

Speed of the stream = 5km/h (Given)

Downstream:

Speed = (x + 5)

Time = Distance ÷speed

Time = 90 /( x + 5)

Upstream:

Speed = ( x - 5)

Time = Distance ÷Speed

Time = 60/( x - 5)

Since the time taken for upstream and downstream is the same:

90 /( x + 5) = 60/( x - 5)

90 (x - 5) = 60 (x + 5)

90x - 450 = 60x + 300

30x = 750

x = 25 km/h

Answer: The speed of the steamer in still water is 25km/h

Discussion

35.	ing equaType IV. Problems based on speed8. A steamer covers the distance between two parts down the stream in 2 hours. It covers the same distance up theIf the speed of the stream be 3 km/hour. then find the speed of the steamer in still water.
Answer» Let speed of the steamer in still water=x=xkm/hr Speed downstream=(x+3)=(x+4)km/hrDistance between the ports=2(x+3)=2(x+3)km ---(1) Speed upstream=(x−3)=(x−4)km/hrDistance between the ports=3(x−3)=3(x−3)km ---(2) From (1) and (2)2(x+3)=3(x−3)2x+6=3x−9x=15

35.

ing equaType IV. Problems based on speed8. A steamer covers the distance between two parts down the stream in 2 hours. It covers the same distance up theIf the speed of the stream be 3 km/hour. then find the speed of the steamer in still water.

Answer»

Let speed of the steamer in still water=x=xkm/hr

Speed downstream=(x+3)=(x+4)km/hrDistance between the ports=2(x+3)=2(x+3)km ---(1)

Speed upstream=(x−3)=(x−4)km/hrDistance between the ports=3(x−3)=3(x−3)km ---(2)

From (1) and (2)2(x+3)=3(x−3)2x+6=3x−9x=15

Discussion

36.	Sire + cos8 =VBो सिह कीजिए किtanot Cet 871
Answer» sinA+cosA=√3squaring both sidessin^2A+cos^2A+2sinAcosA=31+2sinAcosA=32sinAcosA=3-1=2sinAcosA=2/2=1 tanA+cotAsinA/cosA + cosA/sinA=sin^2A+cos^2A/sinAcosA=1/1=1 1 is the right answer

36.

Sire + cos8 =VBो सिह कीजिए किtanot Cet 871

Answer»

sinA+cosA=√3squaring both sidessin^2A+cos^2A+2sinAcosA=31+2sinAcosA=32sinAcosA=3-1=2sinAcosA=2/2=1

tanA+cotAsinA/cosA + cosA/sinA=sin^2A+cos^2A/sinAcosA=1/1=1

1 is the right answer

Discussion

37.	hind the AP which has its sed teemus 16 and 7th Herm us 12 more thanits 5th derm.
Answer» Given:a3 = 16a7 = a5 + 12 ............ (1) a7 = a5 + d + d = a5 + 2d............(2) From Equation (1) & (2) a5 + 12 = a5 + 2d 2d = 12 d = 6 From Given, a3 = 16 a3 = a + 2d = 16 a + ( 2× 6 ) = 16 [ We know that d = 6 ] a + 12 = 16 a = 4 a1=4 a2=4+6=10a3=10+6=16.... So,AP is 4, 10, 16, 22, 28....... a3 = 16=> a + 2d = 16 . . . . . (1)a7 = a5 + 12=> a + 6d = a + 4d + 12=> 2d = 12=> d = 6Substitute in (1)a + 2d = 16=> a = 16 - 2(6) = 16 - 12 = 4Thus, required A. P is 4, 10, 16, . . . . .

37.

hind the AP which has its sed teemus 16 and 7th Herm us 12 more thanits 5th derm.

Answer»

Given:a3 = 16a7 = a5 + 12 ............ (1)

a7 = a5 + d + d = a5 + 2d............(2)

From Equation (1) & (2)

a5 + 12 = a5 + 2d

2d = 12

d = 6

From Given,

a3 = 16

a3 = a + 2d = 16

a + ( 2× 6 ) = 16 [ We know that d = 6 ]

a + 12 = 16

a = 4

a1=4 a2=4+6=10a3=10+6=16....

So,AP is 4, 10, 16, 22, 28.......

a3 = 16=> a + 2d = 16 . . . . . (1)a7 = a5 + 12=> a + 6d = a + 4d + 12=> 2d = 12=> d = 6Substitute in (1)a + 2d = 16=> a = 16 - 2(6) = 16 - 12 = 4Thus, required A. P is 4, 10, 16, . . . . .

Discussion

38.	\|; ¢ 5 e\| किए e guos of e felechy_pigomtal - 5 [Fx +505 बेदीo e
Answer» 5√5x²+30x+8√5=5√5x²+20x+10x+8√5=5x(√5x+4)+2√5(√5x+4)=(5x+2√5)(√5x+4)x=-2√5/5, -4/√5

Discussion

39.	tano -1- CatotCet 0- tano
Answer» option Kya hai is question ka o

39.

tano -1- CatotCet 0- tano

Answer»

option Kya hai is question ka

o

Discussion

40.	A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to thenumber, the digits are reversed. Find the number[CBSE 2001C)
Answer»

Discussion

41.	12. A 115-m-long and 64-m-broad lawn has two roads at right angles, one 2 m wide, running parallel to itslength, and the other 2.5 m wide, running parallel to its breadth. Find the cost of gravelling the roads at ks4.60 per m
Answer»

Discussion

42.	1x—3 y-5 77 il vl :v () —= 2___- =5 aR T4 e 1 AT27 WS spEen Y F कह । 8R g e e L5 s e e L SR
Answer»

Discussion

43.	what is MH?
Answer» Malignant hyperthermia (MH) is a disease that causes a fast rise in body temperature and severe muscle contractions when someone withMHgets general anesthesia.MHis passed down through families. Hyperthermia means high body temperature. Malignant hyperthermia (MH) is a disease that causes a fast rise in body temperature and severe muscle contractions when someone with MH gets general anesthesia .MH is passed down through families. Hyperthermia means high body temperature. Maharashtra State of India Description Maharashtra, a state spanning west-central India, is best known for its fast-paced capital, Mumbai (formerly Bombay). This sprawling metropolis is the seat of the Bollywood film industry. It also has sites like the British Raj-era Gateway of India monument and cave temples at Elephanta Island Malignant hyperthermia, in medicine. Megahenry (MH), an SI unit of inductance. Millihenry (mH), an SI unit of inductance. MH, a symbol for a silt of high plasticity in the Unified Soil Classification System. Metal-halide lamp, a type of electrical gas-discharge lamp. Malignant hyperthermia (MH) is a disease that causes a fast rise in body temperature and severe muscle contractions when someone withMHgets general anesthesia.MHis passed down through families. Hyperthermia means high body temperature.

43.

what is MH?

Answer»

Malignant hyperthermia (MH) is a disease that causes a fast rise in body temperature and severe muscle contractions when someone withMHgets general anesthesia.MHis passed down through families. Hyperthermia means high body temperature.

Malignant hyperthermia (MH) is a disease that causes a fast rise in body temperature and severe muscle contractions when someone with MH gets general anesthesia .MH is passed down through families. Hyperthermia means high body temperature.

Maharashtra

State of India

Description

Maharashtra, a state spanning west-central India, is best known for its fast-paced capital, Mumbai (formerly Bombay). This sprawling metropolis is the seat of the Bollywood film industry. It also has sites like the British Raj-era Gateway of India monument and cave temples at Elephanta Island

Malignant hyperthermia, in medicine. Megahenry (MH), an SI unit of inductance. Millihenry (mH), an SI unit of inductance.

MH, a symbol for a silt of high plasticity in the Unified Soil Classification System. Metal-halide lamp, a type of electrical gas-discharge lamp.

Malignant hyperthermia (MH) is a disease that causes a fast rise in body temperature and severe muscle contractions when someone withMHgets general anesthesia.MHis passed down through families. Hyperthermia means high body temperature.

Discussion

44.	TRY Iside included between the angles M and N of AMNPh ADEEAMNP using the ASA coneruenl. What is theincluded
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Discussion

45.	sasteamer goes downstream from one port to another in 9 hours. It covers thedistance upstream in 10 hours. If the speed of the stream be 1 km/h. find the speed of esteamer in still water and the distance between the ports少A
Answer»

Discussion

46.	A steamer goes downstream from one port to another in 9 hours. It covers the samedistance upstream in 10 hours. If the speed of the stream be 1 km/h, find the speed of thesteamer in still water and the distance between the ports.
Answer»

Discussion

47.	1s A steamer goes downstream from one port to another in 9 hours. It covers thedistance upstream tn 10 hours. II the speed of the stream be 1 kun/h. nnd the speed of thsteamer tn still water and the distance between the ports
Answer»

Discussion

48.	-a2-2ab+ b2 = (a-b)2= (2x-5)2= (2x-5) (2x-5)爾-a2-b? = (α + b) (a-b)= (X + 3) (x-3)X2-9 = 0(x + 3) (x-3)= 0
Answer» x^2-3^2= (x+3)(x-3)अतः शून्यक होंगेx= 3 and -3

Discussion

49.	८9 e ST (05 e Ov © €5 e LE L e 66013 Do 1 पड. सन 1% ऐ7 रह. (मान्य हुनर शक (दर 06 ०14० 14 \|
Answer» D....x=25 or y=65.....

Discussion

50.	3. A park 20 m long and 15 m broad has 2 m wide roads20 mcrossing cach other as shown. The sides of the roadswhich have been covered with stones are shown by dottedlines. Find the total length to be paved by stones.[HOTS]E-TenWhat will be the length of each side if the
Answer» here is a similar question- Area of the rectangle =Length×Breadth=65×50=3250m2Length×Breadth=65×50=3250m2 Given Width of the Path = 2m Length of the Path is Parellel to the side , Hence Length = 65 m Area of the Path 1 =Length×Width=65×2=130m2Length×Width=65×2=130m2 Given Breadth of the Path = 2m Length of the Path2 is Parellel to the other side , Hence Length = 50 m Area of the Path2 =Length×Breadth=50×2=100m2Length×Breadth=50×2=100m2 Add Both the Area of the Path =130+100=230m2130+100=230m2 Area of the Common Field ABCD =2×2=4m22×2=4m2 As both path are overlapping each other, so to calculate the area of shaded portion area of one square should be deducted from the area of the total path. ∴∴Total Area for the Path Construction =230−4=226m2230−4=226m2 ∴∴Cost of Constructing the Path =230−4=226×69=15594

50.

3. A park 20 m long and 15 m broad has 2 m wide roads20 mcrossing cach other as shown. The sides of the roadswhich have been covered with stones are shown by dottedlines. Find the total length to be paved by stones.[HOTS]E-TenWhat will be the length of each side if the

Answer»

here is a similar question-

Area of the rectangle =Length×Breadth=65×50=3250m2Length×Breadth=65×50=3250m2

Given Width of the Path = 2m

Length of the Path is Parellel to the side , Hence Length = 65 m

Area of the Path 1 =Length×Width=65×2=130m2Length×Width=65×2=130m2

Given Breadth of the Path = 2m

Length of the Path2 is Parellel to the other side , Hence Length = 50 m

Area of the Path2 =Length×Breadth=50×2=100m2Length×Breadth=50×2=100m2

Add Both the Area of the Path =130+100=230m2130+100=230m2

Area of the Common Field ABCD =2×2=4m22×2=4m2

As both path are overlapping each other, so to calculate the area of shaded portion area of one square should be deducted from the area of the total path.

∴∴Total Area for the Path Construction =230−4=226m2230−4=226m2

∴∴Cost of Constructing the Path =230−4=226×69=15594

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