The distance (in km) of 40 engineers -Maths 9th

The distance (in km) of 40 engineers -Maths 9th
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Answer :

Frequency distribution of above data in tabular form is given as : It is observed that 36 engineers out of 40 live at a distance not more than 20 km from their residence.
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Boston society of Civil Engineer's formula Q = 0.0056 × (D/t) in cumecs/square km is based upon (A) Rainfall and drainage area(B) Total run off and drainage area (C) Drainage area and its shape (D) Drainage area

Description : Boston society of Civil Engineer's formula Q = 0.0056 × (D/t) in cumecs/square km is based upon (A) Rainfall and drainage area (B) Total run off and drainage area (C) Drainage area and its shape (D) Drainage area

Last Answer : Answer: Option B

Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Description : Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Last Answer : Convert all to metres: 5 km = 5000 m 14 cm = 0.14 m 7 cm = 0.07 m Find the radius: Radius = Diameter 2 Radius = 0.14 2 = 0.07 m Find the amount of water that flowed out in an hour: Volume ... hours needed: Number of hours = 154 77 = 2 hours It takes 2 hours to fill up the tank to rise by 7 cm

Water flows in a tank 150 m × 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm, at a speed of 15 km per hour. -Maths 9th

Description : Water flows in a tank 150 m × 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm, at a speed of 15 km per hour. -Maths 9th

Last Answer : Volume of water discharged through the pipe = Volume increase of the tank First, consider the pipe. Volume discharge through pipe = length breadth speed time Let the time taken to fill the tank to 3 m depth be t. ... the tank V=150 100 3 cu. m ⇒ V=450 100 Therefore, 450t=450 100 ⇒ t=100 hours

A train covers the first half of the distance between two stations with a speed of 40 km/h and the other half with 60 km/h. Then its average speed is

Description : A train covers the first half of the distance between two stations with a speed of 40 km/h and the other half with 60 km/h. Then its average speed is

Last Answer : A train covers the first half of the distance between two stations with a speed of 40 km/h and the other half with ... 48 km/h C. 52 km/h D. 100 km/h

How much distance was traveled by first Train of India? A. 30 km B. 34 km C. 40 km D. 44 km

Description : How much distance was traveled by first Train of India? A. 30 km B. 34 km C. 40 km D. 44 km

Last Answer : B [34 km] Explanation: India's first train ran from Mumbai to Thane in 1853. The first train on the Indian sub-continent ran from Bori Bunder in Mumbai to Thane at 3.30 pm on April 16, 1853.

For a space wave transmission, the radio horizon distance of a transmitting antenna with a height of 100 meters is approximately:A. 10 km B. 40 km C. 100 km D. 400 km

Description : For a space wave transmission, the radio horizon distance of a transmitting antenna with a height of 100 meters is approximately: A. 10 km B. 40 km C. 100 km D. 400 km

Last Answer : B. 40 km

A person covered a certain distance by bus at the rate of 40 kmph and walked back to the initial point at the rate of 6 kmph. The whole journey took 13 hours and 48 minutes. What distance did he walk? a) 60km b) 64 km c) 70km d) 72km e) 80 km

Description : A person covered a certain distance by bus at the rate of 40 kmph and walked back to the initial point at the rate of 6 kmph. The whole journey took 13 hours and 48 minutes. What distance did he walk? a) 60km b) 64 km c) 70km d) 72km e) 80 km

Last Answer : Let the distance be x km. then (x/40) + (x/6) = 13 + (48/60) = 69/5 Answer is: d)

A train starts from Delhi and reaches Lucknow in 24.5 hours. If it travels the first half of journey at 30 kmph and second half at 40 kmph then what is the total distance it travelled? 1) 750 km 2) 800 km 3) 840 km 4) 900 km 5) 920 km

Description : A train starts from Delhi and reaches Lucknow in 24.5 hours. If it travels the first half of journey at 30 kmph and second half at 40 kmph then what is the total distance it travelled? 1) 750 km 2) 800 km 3) 840 km 4) 900 km 5) 920 km

Last Answer : 3) 840 km

A bus starts at bus stop and reaches a destination in 4 hours. If it travels first and second half at the speed of 40 km/hr and 50 km/hr respectively then the distance between bus stop and destination is a) 177.7 km b) 154.2 km c) 163.5 km d) 147.3 km

Description : A bus starts at bus stop and reaches a destination in 4 hours. If it travels first and second half at the speed of 40 km/hr and 50 km/hr respectively then the distance between bus stop and destination is a) 177.7 km b) 154.2 km c) 163.5 km d) 147.3 km

Last Answer : A Let the distance between bus stop and destination be X. The total time taken by the bus to cover X = 4 hours Since X/2 by 40km/hr and remaining X/2 by 50km/hr Then by Time = distance/speed, we have ... = 1600 9X = 1600 X = 1600/9 = 177.7 Hence 177.7 km is the required answer.

Raghu starts a bike at 40 km/hr and he increases his speed in every hour by 4 km/hr. Then the maximum distance covered by him in 12 hours is: a) 744 km b) 658 km c) 436 km d) 512 km

Description : Raghu starts a bike at 40 km/hr and he increases his speed in every hour by 4 km/hr. Then the maximum distance covered by him in 12 hours is: a) 744 km b) 658 km c) 436 km d) 512 km

Last Answer : A Speed of the rider = 40km/hr. Distance covered in 1st hour = 40 km. He increased his speed in every 1 hour by 4 km/hr. Distance covered in every 1 hours will be, 44, 48, ,... upto 12 terms.(for 12 hours). ... (2(40)+(11)4) = 6(80 + 44) = 6(124) = 744. Hence, he covers 744 km in 12 hours.

Anitha travels 400 km by bus at 70 km/hr, 600 km by train at 80km/hr, 250 km by bike at 60 km/hr and 50 kn by car at 40 km/hr. what is the average speed for the total distance? a) 68.45 b) 69.45 c) 67.45 d) 69.77

Description : Anitha travels 400 km by bus at 70 km/hr, 600 km by train at 80km/hr, 250 km by bike at 60 km/hr and 50 kn by car at 40 km/hr. what is the average speed for the total distance? a) 68.45 b) 69.45 c) 67.45 d) 69.77

Last Answer : D Total distance = (400+250+50+600)km =1300 km Total time taken = (400/70 + 600/80+250/60+50/40) hrs =(40/7 +60/8+25/6+5/4) =(24*40)+(21*60)+(28*25)+(42*5) / 168 =960+1260+700+210 / 168 =(3130 / 168) Average speed =(1300*168 / 3130) km/hr =218400/3130 =69.77 km/hr

A girl covered a definite distance at same speed. Had she moved 6 km/hr faster, she would have taken 30 mins less. If she had moved 4 km/hr slower, she would have taken 30 mins more. The distance (in km) is. a) 50 b) 40 c) 70 d) 60

Description : A girl covered a definite distance at same speed. Had she moved 6 km/hr faster, she would have taken 30 mins less. If she had moved 4 km/hr slower, she would have taken 30 mins more. The distance (in km) is. a) 50 b) 40 c) 70 d) 60

Last Answer : D Let distance = x km Rate = y km/hr x/y - x / (y+6) = 30/60 (x(y+6)-xy) / y(y+6) = ½ (xy +6x - xy) / y(y+6) = ½ 12x = y(y+6) ------------>1 And x/ (y-4) - x/y = 30/60 (Xy - x(y-4 ... = (y+6) / (y-4) 3y - 12 = 2y +12 Y = 24km Put y value in 1 we get 12x = 24 (24+6) 12x = 24 *30 X=60km

A taxi driver make a trip from the plains to ooty which are 340km apart at an average speed of 60km/hr. In the return trip, he covers the speed distance at an average of 30 km/hr. the average speed of the taxi over the entire distance of 680 km is a) 30 b) 40 c) 50 d) 60

Description : A taxi driver make a trip from the plains to ooty which are 340km apart at an average speed of 60km/hr. In the return trip, he covers the speed distance at an average of 30 km/hr. the average speed of the taxi over the entire distance of 680 km is a) 30 b) 40 c) 50 d) 60 

Last Answer : B Average speed =(2xy/( x+y)) km/hr Given x =60 km/hr y = 30km/hr Average speed = (2*60*30/ (60+30)) km/hr =(3600/ 90) =40 km/hr

Tharun has to cover a distance of 84 km in 40 minutes. If he covers one-half of the distance in one-fourth of the total time, to cover the remaining distance in the remaining time, what should be his speed in km/hr? a) 42 km/hr b) 64km/hr c) 84km/hr d)76km/hr

Description : Tharun has to cover a distance of 84 km in 40 minutes. If he covers one-half of the distance in one-fourth of the total time, to cover the remaining distance in the remaining time, what should be his speed in km/hr? a) 42 km/hr b) 64km/hr c) 84km/hr d)76km/hr

Last Answer : C Tharun needs to cover 84 km in 40minutes Given that he covers one-half of the distance in one-fourth of the total time ⇒ he covers half of 84km in one-fourth of 40 minutes ⇒ He covers 42 km in ¼ * ... Distance =42km Time =30minutes =1/2 hr Required Speed = Distance / Time =42/(1/2)=84 km/hr

Suppose engineers are able to build a space elevator within the next 40 years, would you enter the elevator on the ground floor and press the button for floor 100,000?

Description : Suppose engineers are able to build a space elevator within the next 40 years, would you enter the elevator on the ground floor and press the button for floor 100,000?

Last Answer : Of course, as long as the chance of dying horribly was less than 10%. Even so, if something happened and you fell, that’d be one hell of a base jump. :) Or you know, Tower of Terror. Depending.

Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Last Answer : Length of cuboid (l) = 80 cm Breadth (b) = 40 cm Height (h) = 20 cm (i) ∴ Lateral surface area = 2h(l + b) = 2 x 20(80 + 40) cm² = 40 x 120 = 4800 cm² (ii) Total surface area = 2(lb ... x 40 + 40 x 20 + 20 x 80) cm² = 2(3200 + 800 + 1600) cm² = 5600 x 2 = 11200 cm²

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Description : A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Last Answer : Given: Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m Height of cone, h = 1m Slant height of cone is l, and l2 = (r2+h2) Using given values, l2 = (0.22+12) = (1.04) Or l ... (32.028 12) = Rs.384.336 = Rs.384.34 (approximately) Therefore, the cost of painting all these cones is Rs. 384.34.

The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Description : Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Last Answer : Since mean of 20 observations is 17 Sum of the 20 observations = 17 x 20 = 340 New sum of 20 observations = 340 – 40 + 12 = 312 New mean=312 / 20 =15.6

The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Description : The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Last Answer : Let the 8 observations are x1, x2, x3, x4, x5, x6, x7, x8 ∴ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = 40 × 8 = 320 New mean = 320 + 5 × 8 = 360 / 8 = 45

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Description : Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean. -Maths 9th

Last Answer : Since mean of 20 observations is 17 Sum of the 20 observations = 17 x 20 = 340 New sum of 20 observations = 340 – 40 + 12 = 312 New mean=312 / 20 =15.6

The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Description : The mean of 8 observations is 40. If 5 is added to each observation, then what will be the new mean ? -Maths 9th

Last Answer : Let the 8 observations are x1, x2, x3, x4, x5, x6, x7, x8 ∴ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = 40 × 8 = 320 New mean = 320 + 5 × 8 = 360 / 8 = 45

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Last Answer : S(△ABC)=60+80+402=90S(△ABC)=60+80+402=90 ar△ABDar△ABD =90(90−80)(90−60)(90−40)−−−−−−−−−−−−−−−−−−−−−−−√=90(90−80)(90−60)(90−40) =90×10×30×50−−−−−−−−−−−−−−√=90×10×30×50 =30015−−√m2=30015m2 ar□ABCE=2×ar△ABDar◻ABCE=2×ar△ABD =60015−−√m2

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2

The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Last Answer : Length of the plank=4m=400cm Breadth=50cm Height=20cm Volume of the plank=L*B*H =400*50*20 =400000cm^3 Length of the pit=16m=1600cm Breadth=12m=1200cm Height=4m=400cm Volume of the pit= L ... *1200*400 =768000000cm^3 Number of planks that can be fitted= 768000000/400000 =1920 planks is the answer.

The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Description : The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Last Answer : NEED ANSWER

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : NEED ANSWER

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Last Answer : Area of the parallelogram

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2.

The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Last Answer : Solution of this question

The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Description : The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Last Answer : (b) Given, mean of 25 observations = 36 ∴ Sum of 25 observations = 36 x 25 = 900 Now, the mean of first 13 observations = 32 ∴ Sum of first 13 observations = 13 x 32 = 416 and the mean of last 13 ... - (Sum of 25 observations) = (520 + 416)-900 = 936 - 900 = 36 Hence, the 13th observation is 36.

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : Median will be a good representative of the data, because 1.each value occurs once. 2.the data is influenced by extreme values.

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

The ratio of girls and boys in a class is 1: 3. Set up an equation between the students of a class and boys and then draw its graph. Also find the number of boys in a class of 40 students from the graph. -Maths 9th

Description : The ratio of girls and boys in a class is 1: 3. Set up an equation between the students of a class and boys and then draw its graph. Also find the number of boys in a class of 40 students from the graph. -Maths 9th

Last Answer : Total number of boys and girl = 40, Ratio = 1 : 3 Number of girls be A and Number of boys be B. Ratio of number of girls and boys is 1 : 3, so Therefore 3A=B To find number of boys we ... the number 30 represents the number of girls. 40 as total on the line A = 10, which is the common equation.

In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Description : In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Last Answer : Solution :-

In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Last Answer : Solution :-

The sides of a triangular field are 41 m, 40 m and 9 m. -Maths 9th

Description : The sides of a triangular field are 41 m, 40 m and 9 m. -Maths 9th

Last Answer : Let a = 41m, b = 40 m, c = 9 m. s = (a + b + c)/2 = (41 + 40 +9)/2 = 90/2 ⇒ s = 45 m Area of the triangular field = root under( √s(s - a)(s - b)(s -c)) = root under( ... x 5 x 36 ) = 180 m2 = 1800000 cm2 Number of rose beds = Total area / Area needed for one rose bed = 1800000/900 = 2000

A river 3 m deep and 40 m wide -Maths 9th

Description : A river 3 m deep and 40 m wide -Maths 9th

Last Answer : Length of water canal in one minute = 2 x 1000 m/60 = 100/3 m Volume of water flowing into the sea in one minute = l x b x h = (100 x 40 x 3)/3 = 4000 m 3 = 4000 x 1000 L = 4000000 L ( Since,1 m3 = 1000 L)

A field is 70 m long and 40 m broad. -Maths 9th

Description : A field is 70 m long and 40 m broad. -Maths 9th

Last Answer : Area of the field on which earth taken out is to be spread = 70 X 40 m2 - 10 X 8 m2 = 2800 m2 - 80 m2 = 2720 m2 Volume of the earth dug out = 10 X 8 X 5 m3 = 400 m3 Rise in level of the field = ... earth dugout/ Area on which earth taken out is to be spread = 400/ 2720 = 0.147 m = 14.7 cm

The length of 40 leaves of a plant are -Maths 9th

Description : The length of 40 leaves of a plant are -Maths 9th

Last Answer : (i) Consider the class 118 - 126 and 127 - 135 The lower limit of 127 - 135 = 127 The upper limit of 118 - 126 = 126 Half of the difference = 127 -126/2 = 0.5 So, the new class interval ... , frequency polygon. (ii) No, this frequency includes all leaves whose length are from 144.5 mm to 153.5 mm.

In a survey of 364 children aged 20-40 months, -Maths 9th

Description : In a survey of 364 children aged 20-40 months, -Maths 9th

Last Answer : Number of children = 364 Number of children not like to eat potato chips = 364 - 91 = 273. The required probability = 273/364 = 0.75