How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th
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1 Answer

Answer :

∵ Volume of one brick = (25 × 11.25 × 6) cm3    and volume of the wall = (800 × 600 × 22.5) cm3  ∴ Number of bricks = Volume of the walls / Volume of one brick    =  800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400
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Related questions

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

ormal size of brick used in India is a) 22 x 11.2 x 7 cm* b) 25 x 16 x 8 cm c) 19 x 12 x 9 cm d) 20 x 10 x 5 cm

Description : ormal size of brick used in India is a) 22 x 11.2 x 7 cm* b) 25 x 16 x 8 cm c) 19 x 12 x 9 cm d) 20 x 10 x 5 cm

Last Answer : a) 22 x 11.2 x 7 cm*

Minimum thickness of wall where single Flemish bond can be used is (A) Half brick thick (B) One brick thick (C) One and a half bricks thick (D) Two bricks thick

Description : Minimum thickness of wall where single Flemish bond can be used is (A) Half brick thick (B) One brick thick (C) One and a half bricks thick (D) Two bricks thick

Last Answer : Answer: Option C

The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Description : The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Last Answer : Total surface area of one brick = 2(lb +bh+lb) = [2(22.5 10+10 7.5+22.5 7.5)] cm2 = 2(225+75+168.75) cm2 = (2 468.75) cm2 = 937.5 cm2 Let n bricks can be painted out by the ... 93750 cm2 So, we have, 93750 = 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. Draw a line segment BC of length 4.8 cm. From B, point A is at a distance of 3.6 cm. ... at P. Joining BP, we obtain angle bisector of ∠B. Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = ½ x 139° = 19.5°

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : In conclusion, the range of data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, and 20 is 26.

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : (d) In a given data, maximum value = 32 and minimum value = 6 We know, range of the data = maximum value – minimum value = 32 – 6 = 26 Hence, the range of the given data is 26.

To construct a 10 cm thick partition wall, you will prefer (A) English bond (B) Flemish bond (C) Header bond (D) Stretcher bond

Description : To construct a 10 cm thick partition wall, you will prefer (A) English bond (B) Flemish bond (C) Header bond (D) Stretcher bond

Last Answer : Answer: Option D

Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 7 cm, 8 cm and 5 cm? Give reason for your answer. -Maths 9th

Last Answer : Solution :- Yes, because in each case sum of two sides is greater than the third side.

Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Description : Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Last Answer : 1.Draw any line l. 2.Take any point M on it and draw a line p perpendicular to l. 3.With M as centre, cut off MC = 6 cm 4.At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.

Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Description : Construct an equilateral triangle of altitude 6 cm. -Maths 9th

Last Answer : 1.Draw any line l. 2.Take any point M on it and draw a line p perpendicular to l. 3.With M as centre, cut off MC = 6 cm 4.At C, with initial line CM construct angles of measures 30° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.

Construct an equilateral triangle if its altitude is 6 cm. -Maths 9th

Description : Construct an equilateral triangle if its altitude is 6 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw a line XY. (ii) Construct perpendicular PD at any point D on the line XY. (iii) From point D, cut-off line segment AD = 6 cm. (iv) Construct ∠BAD = ∠CAD ... 30 °+ 30° = 60° and AD perpendicular BC therefore, △ABC is an equilateral triangle with altitude AD = 6 cm.

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Description : A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

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A hemispherical bowl is made of steel 0.25 cm thick. -Maths 9th

Description : A hemispherical bowl is made of steel 0.25 cm thick. -Maths 9th

Last Answer : Outer radius of hemispherical bowl = R = r + 0.25 = 5 + 0.25 = 5.25 cm Outer curved surface area of bowl = 2 πR2 = 2 x 22/7 x (5.25)2 = 2 x 22 x 5.25 x 0.75 = 173.25 cm2

Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Description : Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Last Answer : 1.Draw a line segment QR = 5 cm. 2.With Q as centre, construct an angle of 90° and let this line through Q is QX. 3. With R as centre, construct an angle of 90° and let this line through R is RY. Yes, the perpendicular lines QX and- RY are parallel.

Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Description : Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

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Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

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Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

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For 12 mm thick cement plastering 1 : 6 on 100 sq.m new brick work, the quantity of cement required, is (A) 0.200 m3 (B) 0.247 m3 (C) 0.274 m3 (D) 0.295 m

Description : For 12 mm thick cement plastering 1 : 6 on 100 sq.m new brick work, the quantity of cement required, is (A) 0.200 m3 (B) 0.247 m3 (C) 0.274 m3 (D) 0.295 m

Last Answer : (C) 0.274 m3

A godown measures 45 m x 25 m x 10 m. -Maths 9th

Description : A godown measures 45 m x 25 m x 10 m. -Maths 9th

Last Answer : Total number of crates that can be accommodated in the godown = Volume of godown/Volume of one crate = 45 x 25 x 10/ 1.5 x 1.25 x 0.5 = 12000 Number of crates already stored = 8000 Remaining crates that can be accommodated = 12000 - 8000 = 4000

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm ? -Maths 9th

Last Answer : No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 ... , the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

Construct a square of side 3 cm. -Maths 9th

Description : Construct a square of side 3 cm. -Maths 9th

Last Answer : Steps of construction : 1. Take AB = 3cm . 2. At A , draw AY ⊥ AB. 3. With A as center and radius = 3cm , describe an arc cutting AY at D. 4. With B and D as centers and radii equal to 3cm , draw arcs intersecting at C. 5. Join BC and DC . ABCD is the required square .

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : NEED ANSWER

Construct a square of side 3 cm. -Maths 9th

Description : Construct a square of side 3 cm. -Maths 9th

Last Answer : We know that, each angle of a square is right angle (i.e., 90°). To construct a square of side 3 cm, use the following steps. 1.Draw a line segment AS of length 3 cm. 2.Now, generate an angle of 90° ... (obtained in step iv) at C. 6.Join DC and BC. Thus, ABCD is the required square of side 3 cm.

Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Description : Construct an equilateral triangle whose altitude is 7 cm -Maths 9th

Last Answer : This answer was deleted by our moderators...

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Description : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw BC = 12 cm. (ii) Construct ÐCBY = 90°. (iii) From ray BY, cut-off line segment BD = 18 cm. (iv) Join CD. (v) Draw the perpendicular bisector of CD intersecting BD at A. (vi ... = AC Now, BD = BA + AD ⇒ BD = AB + AC Hence, △ABC is the required triangle.

A tank internally measuring 150 cm × 120 cm × 100 cm has 1281600 cm^3 of water in it. Porous bricks are placed in the water until the tank -Maths 9th

Description : A tank internally measuring 150 cm × 120 cm × 100 cm has 1281600 cm^3 of water in it. Porous bricks are placed in the water until the tank -Maths 9th

Last Answer : Volume of a brick = 20 cm 6 cm 4 cm = 480 cm3. Water absorbed by one brick = (110 480)(110 480) cm3 = 48 cm3. Let x bricks be placed in the water. Then, x bricks absorb 48x cm3 of water. ... tank ⇒ 1281600 + 480xx - 48xx = 150 x 120 x 100 ⇒ 432xx = 1800000 - 1281600 = 518400 ⇒ xx = 1200.

Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

A queen closer is a (A) Brick laid with its length parallel to the face or direction of wall (B) Brick laid with its breadth parallel to the face or direction of wall (C) Brick having the same length and depth as the other bricks but half the breadth (D) Brick with half the width at one end and full width at the other

Description : A queen closer is a (A) Brick laid with its length parallel to the face or direction of wall (B) Brick laid with its breadth parallel to the face or direction of wall (C) Brick having the same ... bricks but half the breadth (D) Brick with half the width at one end and full width at the other

Last Answer : Answer: Option C

Brick nogging type of partition wall, is constructed by (A) Laying bricks as stretchers in cement mortar (B) Laying bricks as headers in cement mortar (C) Reinforcing brick wall with iron straps (D) Constructing brick work within a wooden framework

Description : Brick nogging type of partition wall, is constructed by (A) Laying bricks as stretchers in cement mortar (B) Laying bricks as headers in cement mortar (C) Reinforcing brick wall with iron straps (D) Constructing brick work within a wooden framework

Last Answer : Answer: Option D

The size of modular bricks, is (A) 10 × 10 × 9 cm (B) 19 × 9 × 9 cm (C) 22.5 × 10 × 8.5 cm (D) 22.5 × 8.0 × 9 cm

Description : The size of modular bricks, is (A) 10 × 10 × 9 cm (B) 19 × 9 × 9 cm (C) 22.5 × 10 × 8.5 cm (D) 22.5 × 8.0 × 9 cm

Last Answer : Answer: Option B

According to Indian Standards Institute, the actual size of modular bricks is (A) 23 cm × 11.5 cm × 7.5 cm (B) 25 cm × 13 cm × 7.5 cm (C) 19 cm × 9 cm × 9 cm (D) 20 cm × 10 cm × 10 cm

Description : According to Indian Standards Institute, the actual size of modular bricks is (A) 23 cm × 11.5 cm × 7.5 cm (B) 25 cm × 13 cm × 7.5 cm (C) 19 cm × 9 cm × 9 cm (D) 20 cm × 10 cm × 10 cm

Last Answer : (C) 19 cm × 9 cm × 9 cm

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. What is the volume of steel used in making the bowl ? -Maths 9th

Description : A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. What is the volume of steel used in making the bowl ? -Maths 9th

Last Answer : answer:

A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

For one cubic metre of brick masonry, number of bricks required, is (A) 400 (B) 425 (C) 450 (D) 500

Description : For one cubic metre of brick masonry, number of bricks required, is (A) 400 (B) 425 (C) 450 (D) 500

Last Answer : Answer: Option D

Number of bricks required for one cubic metre of brick masonry is (A) 400 (B) 450 (C) 500 (D) 550

Description : Number of bricks required for one cubic metre of brick masonry is (A) 400 (B) 450 (C) 500 (D) 550

Last Answer : Answer: Option C

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Description : Number of bricks required for 1 cubic meter of brick masonry is a)400 b) 450 c) 500* d) 550

Last Answer : c) 500*

Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Description : Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Last Answer : Area of a trapezium = 1/2(sum of parallel sides) x (perpendicular diatance between them) = 1/2(25 + 13) x 8 = 152cm2

The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Description : The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Last Answer : In right-angled triangle ABC AB2 + BC2 = AC2 (By Pythagoras Theorem) ⇒ AB2 + 42 = 52 ⇒ AB2 = 25 – 16 = 9 5 cm ⇒ AB = 3 cm ∴ Area of △ABC = 1/2 BC x AB = 1/2 x 4 x 3 = 6cm2