332354 + Interview Questions in Secondary School in Math

1.	If (3,2) is a solution of equation 3x-KY =5, then find the value of k
Answer» Hey Here is your ANSWER, Given that, (3,2) is a solution of the equation 3x -KY=5 so , x = 3 and y = 2 3x - ky = 5 3x3 - kx2 = 5 9-2k=5 -2k = 5-9 -2k = -4 k = -4/-2 k = 2 Hope it HELPS you!

Discussion

2.	Sudhir’s present age is 5 more than three times the age of Viru. Anil’s age is half the age of Sudhir. If the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6 , then find Viru’s age.
Answer» Let Viru 's PRESENT age = x yrs then Sudhir's present age = (5 + 3x) yrs Anil's present age = (5 + 3x)/2 yrs RATIO of the sum of Sudhir's and Viru's age to three times Anil's age is 5 : 6 e.g., (5 + 3x + x)/3(5 + 3x)/2 = 5/6 => 2(5 + 4x)/3(5 + 3x) = 5/6 => 12(5 + 4x) = 15(5 + 3x) => 4(5 + 4x) = 5(5 + 3x) => 20 + 16X = 25 + 15x => x = 5 $\textbf{hence, Viru's present age = 5 yrs}$

Discussion

3.	On selling an article for Rs 1020, a merchant loses 15%. For how much price should he sell the article to gain 12% on it ?
Answer» SELLING PRICE of the ARTICLE is 1344

Discussion

4.	A and B working together can mow field in 56 days and with the help of C they could have mowed it in 42 days how long would c take by himself
Answer» A and B take 56 DAYS ⇒ 1 DAY = 1/56 A and B + c take 42 ⇒ 1 day = 1/42 If c does it by himself: ⇒ 1 day = 1/42 - 1/56 = 1/168 Number of days NEEDED for c to do it by himself = 168 days Answer: 168 days

4.

A and B working together can mow field in 56 days and with the help of C they could have mowed it in 42 days how long would c take by himself

Answer»

A and B take 56 DAYS

⇒ 1 DAY = 1/56

A and B + c take 42

⇒ 1 day = 1/42

If c does it by himself:

⇒ 1 day = 1/42 - 1/56 = 1/168

Number of days NEEDED for c to do it by himself = 168 days

Answer: 168 days

Discussion

5.	Some tickets of ₹ 200 and some of ₹ 100, of a drama in a theatre were sold. The number of tickets of ₹ 200 sold was 20 more than the number of tickets of ₹ 100. The total amount received by the theatre by sale of tickets was ₹ 37000. Find the number of ₹ 100 tickets sold.
Answer» (X+20)200+x100=37000 (x+20)200+ x×100= 37000 200X+ 4000+ x ×100=37000 300x= 37000-4000 300x = 33000 x= 110 Number of tickets WHOSE cost is 100= 110 Number of tickets whose cost is 200= 120

Discussion

6.	In a two digit number, digit at the ten’s place is twice the digit at units’s place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number.
Answer» Let the tens digit be x and the UNITS digit be y. So the number=10x+y A.q. =>x=2y(equation 1) After interchanging the DIGITS we get 10y+x Also, =>(10x+y)+(10y+x)=66 =>10x+y+10y+x=66 =>11y+11x=66 =>11(y+x)=66 =>y+x=66/11 =>y+x=6 Frim equation 1 x=2y So, y+2y=6 =>3y=6 =>y=6/3 =>y=2 And x=2(2)=4 Hence the number is 42.

Discussion

7.	If 2 tables and 3 chairs cost rs. 3,500 and 3 tables and 2 chairs cost rs. 4,000, then how much does a table cost?
Answer» Take TABLES as X and chairs as y then acord2 to qs. 2x +3Y= 3500 3x +2y= 4000 by SOLVING these 2 equation you get the required ans

Discussion

8.	Mother is 25 year older than her son. Find son’s age if after 8 years ratio of son’s age to mother’s age will be 4/9.
Answer» Heya !! LET present age of son be X YEARS. Age of MOTHER = ( X + 25 ) years. After 8 years age of son= ( X + 8 ) years. After 8 years age of mother = ( X + 25 + 8 ) = ( X + 33 ) years. According to the question, X+ 8 / X + 33 = 4/9 9 ( X + 8 ) = 4 ( X + 33 ) 9X + 72 = 4X + 132 9X - 4X = 132 - 72 5X = 60 X = 60/5 X = 12 years. Present age of son = X = 12 Years. And, Present age of mother = X + 25 = 12 + 25 = 37 years.

Discussion

9.	A fruit vendor buys oranges at the rate of 5 for rs.40 and sells them at rs.9 per orangr. find his gain percent
Answer» He BUYS 5 oranges for₹40 . So cost of each is ₹8. SELLING PRICE of each is ₹9. he gets ₹1 PROFIT by buying and selling each . profit%=1/8×100=12.5% . hope this helps you.

Discussion

10.	X/2+y=4/5,x+y/2=7/10 Solve this by elimination method.
Answer» X/2+y=4/5 can be written as 2y+x=8/5-----(1) x+y/2=7/10 can be written as 2x+y=7/5 ----------(2) from (1) x=8/5-2y SUBSTITUTE in ---(2) we get 2(8/5-2y)+y=7/5 we get y=9/15 from this x =2/5

10.

X/2+y=4/5,x+y/2=7/10 Solve this by elimination method.

Answer»

X/2+y=4/5 can be written as 2y+x=8/5-----(1)

x+y/2=7/10 can be written as 2x+y=7/5 ----------(2)

from (1) x=8/5-2y SUBSTITUTE in ---(2) we get 2(8/5-2y)+y=7/5 we get y=9/15 from this x =2/5

Discussion

11.	(-5,3),Write the abscissae and ordinates of the given point.
Answer» Abscissa=-5 ordinate=3

Discussion

12.	Who is the govern of bihar
Answer» Hi dear Pawan.... ur ans is.................... The GOVERNOR of Bihar STATE is Satya PAL Malik since 30 September 2017. all the best. pls mark brainliest.

12.

Who is the govern of bihar

Answer»

Hi dear Pawan....

ur ans is....................

The GOVERNOR of Bihar STATE is Satya PAL Malik

since 30 September 2017.

all the best.

pls mark brainliest.

Discussion

13.	A coin is tossed 100 times and the following outcomes are recordedHead:45 times Tails:55 times from the experimenta) Compute the probability of each outcomes.b) Find the sum of probabilities of all outcomes.
Answer» A) PROBABILITY of heads=45/100 probability of tails=55/100 b) TOTAL probability of all OUTCOMES will be EQUAL to 1

Discussion

14.	Prove that secA(cosecA-SinA)+CosecA(SecA-CosA)=SecA cosecA
Answer» HENCE right HAND side is EQUAL to LEFT hand side

Discussion

15.	the population of town is 176400. If it increases at the rate of 5% per annum,what will be its population 2 years hence
Answer» Here is your ANSWER..

Discussion

16.	How much iron is needed to make a rod of length 90 cm and diameter 1.4 cm?
Answer» LENGTH of ROD , l = 90 cm diameter of rod , d = 1.4 cm so, radius of rod , r = 1.4/2 = 0.7 cm volume of rod = πr²l = 22/7 × 0.7 × 0.7 × 90 = 22 × 0.1 × 0.7 × 90 = 22 × 0.7 × 9 = 198 × 0.7 = 138.6 cm³ HENCE, 138.6 cm³ iron is needed to make a rod of length 90 cm and diameter 1.4 cm.

Discussion

17.	Prove that two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other trianglepls send the answer fast...........................I will mark it as brainliest
Answer» Given: Two triangles ABC and DEF such that B = E, C = F and BC = EF. To prove: ABC DEF Proof: Case I: If AB = DE then in ABC and DEF, AB = DE [by supposition] BC = EF [given] and B = E [given] Thus, ABC DEF [SAS criterion] Case II: If AB < DE TAKE a point G on ED such that EG = AB. Join GF. In ABC and GEF, we have AB = GE [by supposition] BC = EF [given] B = E [given] Thus, ABC GEF [SAS criterion] ACB = GFE [CORRESPONDING parts of congruent triangles are equal] But ACB = DFE [given] GFE =DFE, This is only POSSIBLE when FG coincides with FD or G coincides with D. AB must be equal to DE and hence, ABC DEF (by SAS) Case III: If AB > ED With a similar argument (as in case II), we may conclude that ABC DEF (by SAS) Thus, ABC DEF.

17.

Prove that two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other trianglepls send the answer fast...........................I will mark it as brainliest

Answer»

Given: Two triangles ABC and DEF such that B = E, C = F

and BC = EF. To prove: ABC DEF

Proof: Case I: If AB = DE then in

ABC and DEF, AB = DE [by supposition] BC = EF [given] and B = E [given]

Thus, ABC DEF [SAS criterion]

Case II: If AB < DE TAKE a point G on ED such that EG = AB.

Join GF. In ABC and GEF, we have AB = GE [by supposition] BC = EF [given] B = E [given] Thus, ABC GEF [SAS criterion]

ACB = GFE [CORRESPONDING parts of congruent triangles are equal]

But ACB = DFE [given] GFE =DFE,

This is only POSSIBLE when FG coincides with FD or G coincides with D.

AB must be equal to DE and hence, ABC DEF (by SAS) Case III: If AB > ED

With a similar argument (as in case II), we may conclude that ABC DEF (by SAS)

Thus, ABC DEF.

Discussion

18.	Write five different linear equations in two variables and find three solutions for each of them?
Answer» Hi, SOLUTION: A.T.Q. we ourselves take any 5 linear equation in two variables. $1) \: <klux>X</klux> + y = 5 \\ 2) \: x - y = 4 \\ 3) \: 2x + 3y = 10 \\ 4)3x + y = 9 \\ 5)2x + 5y = 1$ you can take any equation with any coefficient ,but remember don't put any coefficient zero,in that case equation will become linear equation in ONE variable. solution of these equations are ATTACHED . just put any value of y, for that value of x will be calculated. Hope this helps you.

Discussion

19.	Prove that root 2 + 3 by root 2 is an irrational number
Answer» If possible let root2 + 3/root2 be an rational number. Then root2 + 3/root2 = p/q where p and q are co primes. By TAKING LCM, (root2 × root2 + 3)/root2 = p/q (2+3)/root2 = p/q 5/root2 = p/q root2 = 5q/p .....(1) We know that root 2 is irrational but 5q/p is rational. But this it contradictory to the above equation (1). This CONTRADICTION arose because of our incorrect assumption of root2 + 3/root2 to be rational. This proves that root2 + 3/root2 is irrational. Hope it helps!

Discussion

20.	The sum of first 5 terms of an ap is 55 and sum of ist 10 terms of the ap is 235 , find the sum of ist 20 terms
Answer» The sum of first 5 terms of an ap is 55 first five terms of AP are a, (a+d), (a+2D), (a+3d), (a+4d) so, a + (a+d) + (a+2d) + (a+3d) + (a+4d) = 55 5A + 10D = 55 5(a + 2d) = 55 a + 2d = 55/5 a + 2d = 11................(1) where a = 11 - 2d sum of it's first 10 terms of AP is 235 first 10 terms of AP are a, (a+d), (a+2d), (a+3d), (a+4d), (a+5d), (a+6d), (a+7d), (a+8d), (a+9d) so, a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d)+(a+6d)+(a+7d)+(a+8d)+(a+9d) = 235 10a + 45d = 235 5(2a + 9d) = 235 2a + 9d = 235/5 2a + 9d = 47 2(11 - 2d) + 9d = 47 22 - 4d + 9d = 47 5d = 47 - 22 5d = 25 d = 25/5 d = 5 a = 11 - 2d a = 11 - 2(5) a = 11 - 10 a = 1 sum of first 20 terms, $S(n) = \frac{n}{2} (2a + (n - 1)d) \\ \\ where \: n = 20 \\a = 1 \\ d = 5 \\ S(20) = \frac{20}{2} (2 \times 1 + (20 - 1)5) \\ S(20) = 10(2 + 19 \times 5) \\ S(20) = 10(2 + 95) \\ S(20) = 10(97) \\ S(20) = 970$ #mark as BRAINLIEST

Discussion

21.	find the smallest number by which 17496 must be divided ,so that the quotient is a perfect cube.also find the cube root of the quotient .please give full solution
Answer» the smallest number by which 17496 must be divided ,so that the quotient is a perfect CUBE let smallest number be x and quotient number be y³ so, 17496/x which gives 0 as REMAINDER and y³ as quotient so, factors of 17496 = 2×2×2×3×3×3×3×3×3×3 no of 2's are 3 no of 3's are 7 so, factors of 17496 = 2³ × 3³ × 3³ × 3 = (2 × 3 × 3)³ × 3 so, smallest number is x = 3 y³= (2×3×3)³ y = 18 the perfect cube is y³ = 18³ cube root of 18³ = 18 #mark as brainliest

Discussion

22.	What is the probability that a randomly thrown dart that hits the square board in shaded region.(Take ???? = 22/7 and express in percentage)
Answer» Hi , It is given that , i ) radius of the circle ( r ) = 2 cm Area of the circle ( A1 ) = πr² A1 = 22/7 × 2² A1 = 88/7 square cm ----( 1 ) ii ) Side of the square( a ) = 2 × radius a = 2 × 2 = 4 cm Area of the Square ( A2 ) = a² A2 = 4² A2 = 16 square cm ---( 2 ) III ) Area of the SHADED region ( A ) = A2 - A1 A = 16 - 88/7 A = ( 112 - 88 )/7 A = 24/7 square cm ---( 3 ) Probability of hitting the shaded region = ( favourable area )/total area P( E ) = ( 3 )/( 2 ) = ( 22/7 )/( 16 ) = 24/( 7 × 16 ) = 3/14 As Percentage = 3/14 × 100% = 300%/14 = 21.428 % I hope this helps you. : )

Discussion

23.	Find the area of the sheet required to make a cylindrical container which is open at one side and whose diameter is 28 cm and height is 20 cm. Find the approximate area of the sheet required to make a lid of height 2 cm for this container.
Answer» I ) DIMENSIONS of the CYLINDRICAL CONTAINER: diameter ( d ) = 28 cm radius ( r ) = d/2 = 28/2 = 14 cm Height ( H ) = 20 cm Area of the sheet required to make the container ( A ) = base area + L.S.A = πr² + 2πrH = πr( r + 2H ) = ( 22/7 ) × 14 [ 14 + 2 × 20 ] = 22 × 2 × 54 = 2376 square centimetres ii ) Dimensions of the lid : radius( r ) = 14 cm height ( h ) = 2 cm Area of the sheet required to make the lid = πr( r + 2h ) = ( 22/7 ) × 14 [ 14 + 2 × 2 ] = 22 × 2 × 18 = 792 cm² I hope this helps you. : )

Discussion

24.	Big essay about come out of the well to ocean
Answer» Come out of the well to the ocean It is absolutely true we live in the narrow well of circumstances created by our false ego. Living in the comfort zone of this narrow well appears to be safe and convenient. However, if you really intend and aspire to accomplish something worthwhile in LIFE or develop a skill upto a LEVEL of expertise, you will have to come out of this well. As soon as you come out of it, you will realize there is an ocean of opportunities waiting for you out of it. You will realize there is a series of steps LEADING to infinite higher levels. For example if you are an excellent player in your colony admired for your playing skills, you will realize that there are far better players like you at the city, state, zonal, national, and international levels. In the limited colony you were in a well; outside you are in the ocean. Similarly, in all spheres there is no end to betterment and advancement. Everyday people who are daring to come out of their narrow well of existence and exploring new horizons, are finding something new. All great men achieved greatness by coming out of the well of their comfort zone. In the ocean of life they found innumerable opportunities that they made best use of. Lincoln, Einstein, Gandhi, Louis Pasteur, Helen Keller, Charlie Chaplin, and all the legends in various fields, they achieved unparalleled success, name and fame only by coming out of the well of their narrow world. If Columbus had not DARED to come out of the security and safety of Spain, would he ever have found America? If Louis Pasteur had not dared to come out his narrow FEAR of well, would he have ever found cure to rabies? So, if you too intend to be great, come out of your narrow world and live in the ocean.

24.

Big essay about come out of the well to ocean

Answer»

Come out of the well to the ocean

It is absolutely true we live in the narrow well of circumstances created by our false ego. Living in the comfort zone of this narrow well appears to be safe and convenient. However, if you really intend and aspire to accomplish something worthwhile in LIFE or develop a skill upto a LEVEL of expertise, you will have to come out of this well. As soon as you come out of it, you will realize there is an ocean of opportunities waiting for you out of it.

You will realize there is a series of steps LEADING to infinite higher levels. For example if you are an excellent player in your colony admired for your playing skills, you will realize that there are far better players like you at the city, state, zonal, national, and international levels. In the limited colony you were in a well; outside you are in the ocean. Similarly, in all spheres there is no end to betterment and advancement. Everyday people who are daring to come out of their narrow well of existence and exploring new horizons, are finding something new.

All great men achieved greatness by coming out of the well of their comfort zone. In the ocean of life they found innumerable opportunities that they made best use of. Lincoln, Einstein, Gandhi, Louis Pasteur, Helen Keller, Charlie Chaplin, and all the legends in various fields, they achieved unparalleled success, name and fame only by coming out of the well of their narrow world. If Columbus had not DARED to come out of the security and safety of Spain, would he ever have found America? If Louis Pasteur had not dared to come out his narrow FEAR of well, would he have ever found cure to rabies?

So, if you too intend to be great, come out of your narrow world and live in the ocean.

Discussion

25.	For rain water harvesting a tank of length 10 m, breadth 6 m and depth 3m is built. What is the capacity of the tank ? How many litre of water can it hold?
Answer» Length of tank , l = 10m BREADTH of tank , b = 6M depth of tank , h = 3M $\textbf{capacity of tank means,}\\\textbf{ <klux>VOLUME</klux> of tank or the amount}\\\textbf{ of space occupied by tank}$ so, volume of tank = length × breadth × depth = 10m × 6m × 3m = 180 m³ we know, 1 L = 1000cm³ = 10^-3 m³ or , 1 m³ = 1000 L so, 180m³ = 180, 000 L = 1,80, 000 L hence, water can hold 1,80,000 L

Discussion

26.	Find dy/dx...................
Answer» If HELPFUL MARK as BRAINLEIST

Discussion

27.	find the area of a Trapezium whose parallel sides are 11m and 25 m long and the non parallel sides are 15 M and 13m long
Answer» this is the EXPLANATION for the QUESTION

Discussion

28.	In a 3-digit number units digit is one more than the hundreds digit and tens digit is one less than the hubdreds digit if the sum of the original 3-digit number and numbers obtained by changing the orders of digits cyclically is 2664 find the number
Answer» this is the RIGHT ANS.

Discussion

29.	Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer» TOTAL no of bags = 11 no of bags CONTAINS more than 5kg of WHEAT flour = 7 that are 5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07 probability of these bags chosen at RANDOM contains more than 5kg of flour = 7/11

Discussion

30.	A farmer has 14 cows 21 buffalos and 35 goats. He gave maximum number of equal animals of same type to his servants then no animal left with the farmer. Find the number of servants farmer has. Also write values.
Answer» ANSWER: Check the above 3 photos and you'll get your answer. SWIPE on the previous photo TOWARDS LEFT to go to next photo.

Discussion

31.	Quadrilateral HELP he= 6 cm el= 4.5 cm angle h= 60 degree angle e 105 and angle p = 120 degree
Answer» The ROUGH DIAGRAM is above-mentioned

Discussion

32.	Find the hcf of 847 and 2160 by euclid's division algorithm
Answer» HEYA! Here is yr answer..... HCF of 847 & 2160 by EUCLID'S DIVISION ALGORITHM.... According to Euclid's postulate... a = bq + r a = divident - 2160 b = divisor - 847 q = quotient - ? r = remainder - ? 2160 = 847 × 2 + 466 847 = 466 × 1 + 381 466 = 381 × 1 + 85 381 = 85 × 4 + 41 85 = 41 × 2 + 3 41 = 3 × 13 + 2 3 = 2 × 1 + 1 2 = 1 × 2 + 0 here, r = 0 ....so HCF = 1 Therefore, 847 and 2160 are co-primes... Hope it hlpz..

Discussion

33.	A racing track is a circular ring with inner diameter 140 m and track 7 m wide how much further does motorist on the outside rim travel when he goes round the circuit once than another who goes round the circuit on the side rim
Answer» Given that the path is 7 M, the diameter of the OUTSIDE rim is : 140 + (7 × 2) = 154M The circumference of the OUTER rim is : 22/7 × 154 = 484meters The circumference of the INSIDE rim: 22/7 × 140 = 440 meters 484 - 440 = 44 meters

Discussion

34.	For any positive integer n,prove that n3 -n is divisible by 6
Answer» HEY!!☺☺ Refer to the ATTACHMENT HOPE it will HELPS you ☺☺

Discussion

35.	A letter is chosen from English alphabet. Find the probability of the letters being with A vowel
Answer» Solution : Total possible outcomes = { A,B,C,...,Z } Number of total possible outcomes = 26 ----( 1 ) Favourable outcomes = { A,E,I,O,U } Number of favourable outcomes = 5 --( 2 ) Probability of the letters being with a VOWEL = ( 2 )/( 1 ) = 5/26 ••••

Discussion

36.	A man is facing south. He turns 135° anti-clockwise and then 180° clockwise. Which direction he is facing now ?(a) North-East (b) West(c) South-East (d) South-West
Answer» ANSWER: South - West Step-by-step explanation: It is suggested to refer the attached image. Firstly the PERSON is facing the south DIRECTION. From there when it turns 135 degree anticlockwise then the direction is NORTH East. Now again it is said that the person turns 180 degree clockwise so the direction will be southwest.

36.

A man is facing south. He turns 135° anti-clockwise and then 180° clockwise. Which direction he is facing now ?(a) North-East (b) West(c) South-East (d) South-West

Answer»

ANSWER:

South - West

Step-by-step explanation:

It is suggested to refer the attached image. Firstly the PERSON is facing the south DIRECTION. From there when it turns 135 degree anticlockwise then the direction is NORTH East.

Now again it is said that the person turns 180 degree clockwise so the direction will be southwest.

Discussion

37.	The average age of 50 pupils in a class is 10.5. If the class teacher is included, the average increases by 0.5. Then the age of the class teacher is
Answer» Here is your ANSWER....

Discussion

38.	101*99, Evaluate the given product without actual multiplication.
Answer» ANSWER: The product is $101\times 99=9999$ Step-by-step EXPLANATION: Given : Expression $101\times 99$ To FIND : Evaluate the given product without actual MULTIPLICATION ? Solution : We can write product as, $101\times 99=(100+1)\times (100-1)$ We know, $(a+b)(a-b)=a^2-b^2$ $101\times 99=(100+1)(100-1)$ here a=100 , b=1 $101\times 99=(100)^2-(1)^2$ $101\times 99=10000-1$ $101\times 99=9999$ Therefore, The product is $101\times 99=9999$

38.

101*99, Evaluate the given product without actual multiplication.

Answer»

ANSWER:

The product is $101\times 99=9999$

Step-by-step EXPLANATION:

Given : Expression $101\times 99$

To FIND : Evaluate the given product without actual MULTIPLICATION ?

Solution :

We can write product as,

$101\times 99=(100+1)\times (100-1)$

We know, $(a+b)(a-b)=a^2-b^2$

$101\times 99=(100+1)(100-1)$ here a=100 , b=1

$101\times 99=(100)^2-(1)^2$

$101\times 99=10000-1$

$101\times 99=9999$

Therefore, The product is $101\times 99=9999$

Discussion

39.	For an AP if a25-a20=45,then d is equal to
Answer» ANSWER: The value of d is 9. Step-by-step EXPLANATION: Given : For an A.P if $a_{25}-a_{20}=45$ To find : The value of d ? SOLUTION : The nth term of AP is $a_n=a+(n-1)d$ $a_{25}=a+(25-1)d=a+24d$ $a_{20}=a+(20-1)d=a+19d$ Substitute in the given expression, $a+24d-(a+19d)=45$ $a+24d-a-19d=45$ $5d=45$ $d=\frac{45}{5}$ $d=9$ Therefore, the value of d is 9.

39.

For an AP if a25-a20=45,then d is equal to

Answer»

ANSWER:

The value of d is 9.

Step-by-step EXPLANATION:

Given : For an A.P if $a_{25}-a_{20}=45$

To find : The value of d ?

SOLUTION :

The nth term of AP is $a_n=a+(n-1)d$

$a_{25}=a+(25-1)d=a+24d$

$a_{20}=a+(20-1)d=a+19d$

Substitute in the given expression,

$a+24d-(a+19d)=45$

$a+24d-a-19d=45$

$5d=45$

$d=\frac{45}{5}$

$d=9$

Therefore, the value of d is 9.

Discussion

40.	If the selling price of flower vase is 4/5 of its cost price find the loss percent
Answer» Selling PRICE is 4/5 of COST price let CP = x SP = 4/5 of x = 4x/5 Loss= CP - SP = x - 4x/5 = x/5 Loss% = loss/CP * 100 = (x/5) / x * 100 = 100/5 = 20%

40.

If the selling price of flower vase is 4/5 of its cost price find the loss percent

Answer»

Selling PRICE is 4/5 of COST price

let CP = x

SP = 4/5 of x = 4x/5

Loss= CP - SP = x - 4x/5 = x/5

Loss% = loss/CP * 100

= (x/5) / x * 100

= 100/5

= 20%

Discussion

41.	A villager Ramayya has a plot of land in the shape of a quadrilateral. The grampanchayat of the village decided to take over some portion of his plot from one of the corners to construct a school. Ramayya agrees to the above proposal with the condition that he should be given equal amount of land in exchange of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Answer» Two Triangles on the same BASE and between the same parallels are equal in area Let ABCD be the plot of the land of the shape of a quadrilateral. Construction, Diagonal BD is joined. AE is drawn parallel BD. BE is joined which INTERSECTED AD at O. △BCE is the shape of the original field and △AOB is the area for constructing health CENTRE. Also, △DEO land joined to the plot. To prove: ar(△DEO) = ar(△AOB) Proof: △DEB and △DAB lie on the same base BD and between the same parallel lines BD and AE. ar(△DEB) = ar(△DAB) ar(△DEB) – ar△DOB) = ar(△DAB) – ar(△DOB) [ On subtracting ar(△DOB) from both sides] ar(△DEO) = ar(△AOB) Hope this will HELP you...

41.

A villager Ramayya has a plot of land in the shape of a quadrilateral. The grampanchayat of the village decided to take over some portion of his plot from one of the corners to construct a school. Ramayya agrees to the above proposal with the condition that he should be given equal amount of land in exchange of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Answer»

Two Triangles on the same BASE and between the same parallels are equal in area

Let ABCD be the plot of the land of the shape of a quadrilateral.

Construction,

Diagonal BD is joined. AE is drawn parallel BD. BE is joined which INTERSECTED AD at O. △BCE is the shape of the original field and △AOB is the area for constructing health CENTRE. Also, △DEO land joined to the plot.

To prove:

ar(△DEO) = ar(△AOB)

Proof:

△DEB and △DAB lie on the same base BD and between the same parallel lines BD and AE.

ar(△DEB) = ar(△DAB)

ar(△DEB) – ar△DOB) = ar(△DAB) – ar(△DOB)

[ On subtracting ar(△DOB) from both sides]

ar(△DEO) = ar(△AOB)

Hope this will HELP you...

Discussion

42.	A company has 2819 employees who work for 9 hours everyday. Assuming that employees take a brake of approximate 1 hour everyday (from 9 hours) then find the approximate number of hours that employees devote for work everyday
Answer» 9-1=8; ANSWER is 8 HOURS

Discussion

43.	If the cost price of 4 toffees be equal to the selling price of 3 toffees then the gain %is
Answer» COST price of 4 toffee = SP of 3 toffee LET CP of 1 toffee = Re 1 Then CP of 4 toffee = RS 4 = SP of 3 toffee SP of 3 toffee = Rs 4 CP of 3 toffee = Rs 3 Profit = SP - CP = 4-3 = 1 Profit % = 1/3 * 100 = 33.33 %

43.

If the cost price of 4 toffees be equal to the selling price of 3 toffees then the gain %is

Answer»

COST price of 4 toffee = SP of 3 toffee

LET CP of 1 toffee = Re 1

Then CP of 4 toffee = RS 4 = SP of 3 toffee

SP of 3 toffee = Rs 4

CP of 3 toffee = Rs 3

Profit = SP - CP = 4-3 = 1

Profit % = 1/3 * 100 = 33.33 %

Discussion

44.	Which is the smallest positive number that must be added to 2,010 in order to obtain a perfect square?
Answer» Here is the ANSWER...

Discussion

45.	solve using substitution method : 0.2x+0.3y=1.3 and 0.4x+ 0.5y=2.3. pair of linear equation. ch3 class 10 ncert
Answer» ANSWER: X = 2 and y = 3 Step-by-step explanation: Given : $0.2x+0.3y=1.3$ $0.4x+ 0.5y=2.3$ To Find : solve USING substitution method SOLUTION : $0.2x+0.3y=1.3$ -----1 $0.4x+ 0.5y=2.3$ ------2 Substitute the value of x from 1 in 2 $0.4(\frac{1.3 - 0.3 y}{0.2})+ 0.5y=2.3$ $y=3$ $0.2x+0.3(3)=1.3$ $0.2x=1.3-0.9$ $x=\frac{1.3-0.9}{0.2}$ $x=2$ So, x = 2 and y = 3

45.

solve using substitution method : 0.2x+0.3y=1.3 and 0.4x+ 0.5y=2.3. pair of linear equation. ch3 class 10 ncert

Answer»

ANSWER:

X = 2 and y = 3

Step-by-step explanation:

Given : $0.2x+0.3y=1.3$

$0.4x+ 0.5y=2.3$

To Find : solve USING substitution method

SOLUTION :

$0.2x+0.3y=1.3$ -----1

$0.4x+ 0.5y=2.3$ ------2

Substitute the value of x from 1 in 2

$0.4(\frac{1.3 - 0.3 y}{0.2})+ 0.5y=2.3$

$y=3$

$0.2x+0.3(3)=1.3$

$0.2x=1.3-0.9$

$x=\frac{1.3-0.9}{0.2}$

$x=2$

So, x = 2 and y = 3

Discussion

46.	3a+5b /3a-5b =5 than a:b
Answer» $\frac{3a + 5b}{3a - 5b} = 5$ By COMPONENDO and DIVIDENDO, we GET $\frac{3a + 5b + 3a - 5b}{3a + 5b - 3a + 5b} = \frac{5 + 1}{5 - 1} \\ \\ \\ => \frac{6a}{10b} = \frac{6}{4} \\ \\ => \frac{a}{b} = \frac{6 \times 10}{4 \times 6} \\ \\ = > \frac{a}{b} = \frac{10}{4} \\ \\ => \frac{a}{b} = \frac{5}{2}$

Discussion

47.	The volume of a cube is 46656 cubic metres what is its length
Answer» HEY MATE HERE'S YOUR ANSWER volume of a cube = LENGTH×breadth× height since, the Length, breadth and height of a cube are EQUAL, therefore let each of the Length, breadth and height be x metres. CLEARLY, $46656 \: \: cubic \: metres = x {}^{3}$ => $46656 = x {}^{3}$ => x = 36 HENCE length of the cube is 36 m. HAVE A GREAT DAY

Discussion

48.	A reduction of 21% in the price of an item enables a person to buy 3 kg more for Rs 100.The reduced price of Item per kg is?
Answer» Let's take the price of 1kg of ITEM be x. then kgs of item bought for ₹100 is 100/x. when reduced by21% , price BECOMES 0.79x. kgs of item bought for ₹100 is 10000/79x. The difference in kgs of item bought is 3 KG. so 10000/79x-100/x=2100/79x =3. x=700/79=8.86 RUPPEES per kg that is the ANSWER .

Discussion

49.	From a sum one third is invested at 3% one sixth at 6% and the rest at 8%. if the simple interest for 2 years from all these investments is rs 600. find the sum
Answer» SUM = Rs 5000 <P> Step-by-step explanation: Let say Sum = 6P Rs one third is invested at 3% for 2 Years => 6P/3 = 2P Interest = 2P * 3 * 2 /100 = 12P/100 Rs one Sixth is invested at 6% for 2 Years => 6P/6 = P Interest = P * 6 * 2 /100 = 12P/100 Rs REST at 8 % for 2 Years Rest = 6P - 2P - P = 3P Rs Interest = 3P * 8 * 2 /100 = 48P/100 Rs 12P/100 + 12P/100 + 48P/100 = 600 => 72P = 60000 Dividing by 12 on both sides => 6P = 5000 Sum = Rs 5000 Learn more: A person invested part of Tk 45000 at 4% and the rest at 6%. if his ... brainly.in/question/13288931 Divide 35400 into two parts such that if one part is invested in 9 ... brainly.in/question/8958305 hey @amitnrw PLEASE help me - Brainly.in brainly.in/question/7645846

49.

From a sum one third is invested at 3% one sixth at 6% and the rest at 8%. if the simple interest for 2 years from all these investments is rs 600. find the sum

Answer»

SUM = Rs 5000

<P>

Step-by-step explanation:

Let say Sum = 6P Rs

one third is invested at 3% for 2 Years

=> 6P/3 = 2P

Interest = 2P * 3 * 2 /100 = 12P/100 Rs

one Sixth is invested at 6% for 2 Years

=> 6P/6 = P

Interest = P * 6 * 2 /100 = 12P/100 Rs

REST at 8 % for 2 Years

Rest = 6P - 2P - P = 3P Rs

Interest = 3P * 8 * 2 /100 = 48P/100 Rs

12P/100 + 12P/100 + 48P/100 = 600

=> 72P = 60000

Dividing by 12 on both sides

=> 6P = 5000

Sum = Rs 5000

Learn more:

A person invested part of Tk 45000 at 4% and the rest at 6%. if his ...

brainly.in/question/13288931

Divide 35400 into two parts such that if one part is invested in 9 ...

brainly.in/question/8958305

hey @amitnrw PLEASE help me - Brainly.in

brainly.in/question/7645846

Discussion

50.	Rina lost her weight in the ratio 5 :3 .Her original weight was 80kg. what is her new weight?
Answer» LET the NEW WEIGHT be X $\frac{5}{3} = \frac{80}{x} \\ solve 5x = 240 \\ x = \frac{240}{5} \\ x = 48$

Discussion

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