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https://nrich.maths.org/public/leg.php?code=5039&cl=2&cldcmpid=215 | 1,435,658,493,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375093400.45/warc/CC-MAIN-20150627031813-00109-ip-10-179-60-89.ec2.internal.warc.gz | 890,569,255 | 9,941 | # Search by Topic
#### Resources tagged with Interactivities similar to A Dotty Problem:
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### There are 221 results
Broad Topics > Information and Communications Technology > Interactivities
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### Cycling Squares
##### Stage: 2 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
### Which Symbol?
##### Stage: 2 Challenge Level:
Choose a symbol to put into the number sentence.
##### Stage: 2 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Multiplication Square Jigsaw
##### Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Times Tables Shifts
##### Stage: 2 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### Colour Wheels
##### Stage: 2 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
### Counters
##### Stage: 2 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Tubular Path
##### Stage: 2 Challenge Level:
Can you make the green spot travel through the tube by moving the yellow spot? Could you draw a tube that both spots would follow?
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Target Archery
##### Stage: 2 Challenge Level:
A simulation of target archery practice
##### Stage: 1 and 2 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Train
##### Stage: 2 Challenge Level:
A train building game for 2 players.
### Factor Lines
##### Stage: 2 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Coordinate Tan
##### Stage: 2 Challenge Level:
What are the coordinates of the coloured dots that mark out the tangram? Try changing the position of the origin. What happens to the coordinates now?
### Board Block Challenge
##### Stage: 2 Challenge Level:
Choose the size of your pegboard and the shapes you can make. Can you work out the strategies needed to block your opponent?
### Tetrafit
##### Stage: 2 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
### Sorting Symmetries
##### Stage: 2 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
##### Stage: 2 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### Ratio Pairs 2
##### Stage: 2 Challenge Level:
A card pairing game involving knowledge of simple ratio.
### A Maze of Directions
##### Stage: 2 Challenge Level:
Use the blue spot to help you move the yellow spot from one star to the other. How are the trails of the blue and yellow spots related?
### Twice as Big?
##### Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
##### Stage: 1 and 2 Challenge Level:
Our 2008 Advent Calendar has a 'Making Maths' activity for every day in the run-up to Christmas.
### Rod Ratios
##### Stage: 2 Challenge Level:
Use the Cuisenaire rods environment to investigate ratio. Can you find pairs of rods in the ratio 3:2? How about 9:6?
### Noughts and Crosses
##### Stage: 2 Challenge Level:
A game for 2 people that everybody knows. You can play with a friend or online. If you play correctly you never lose!
### More Carroll Diagrams
##### Stage: 2 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Overlapping Circles
##### Stage: 2 Challenge Level:
What shaped overlaps can you make with two circles which are the same size? What shapes are 'left over'? What shapes can you make when the circles are different sizes?
##### Stage: 2 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
### Part the Piles
##### Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### Rabbit Run
##### Stage: 2 Challenge Level:
Ahmed has some wooden planks to use for three sides of a rabbit run against the shed. What quadrilaterals would he be able to make with the planks of different lengths?
### Fractions and Coins Game
##### Stage: 2 Challenge Level:
Work out the fractions to match the cards with the same amount of money.
### Arrangements
##### Stage: 2 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Board Block for Two
##### Stage: 1 and 2 Challenge Level:
Board Block game for two. Can you stop your partner from being able to make a shape on the board?
### Four Triangles Puzzle
##### Stage: 1 and 2 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Chocolate Bars
##### Stage: 2 Challenge Level:
An interactive game to be played on your own or with friends. Imagine you are having a party. Each person takes it in turns to stand behind the chair where they will get the most chocolate.
### 100 Percent
##### Stage: 2 Challenge Level:
An interactive game for 1 person. You are given a rectangle with 50 squares on it. Roll the dice to get a percentage between 2 and 100. How many squares is this? Keep going until you get 100. . . .
### Building Stars
##### Stage: 2 Challenge Level:
An interactive activity for one to experiment with a tricky tessellation
### Makeover
##### Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### Calculator Bingo
##### Stage: 2 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### Coordinate Cunning
##### Stage: 2 Challenge Level:
A game for 2 people that can be played on line or with pens and paper. Combine your knowledege of coordinates with your skills of strategic thinking.
### Beat the Drum Beat!
##### Stage: 2 Challenge Level:
Use the interactivity to create some steady rhythms. How could you create a rhythm which sounds the same forwards as it does backwards?
### Combining Cuisenaire
##### Stage: 2 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods? | 2,077 | 9,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2015-27 | longest | en | 0.881209 |
https://ncertmcq.com/rd-sharma-class-9-solutions-chapter-10-congruent-triangles-ex-10-4/ | 1,709,389,284,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00742.warc.gz | 417,152,150 | 14,726 | ## RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
Other Exercises
Question 1.
In the figure, AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.
Solution:
AB || CD and l is transversal ∠1 : ∠2 = 3 : 2
Let ∠1 = 3x
Then ∠2 = 2x
But ∠1 + ∠2 = 180° (Linear pair)
∴ 3x + 2x = 180° ⇒ 5x = 180°
⇒ x = $$\frac { { 180 }^{ \circ } }{ 5 }$$ = 36°
∴ ∠1 = 3x = 3 x 36° = 108°
∠2 = 2x = 2 x 36° = 72°
Now ∠1 = ∠3 and ∠2 = ∠4 (Vertically opposite angles)
∴ ∠3 = 108° and ∠4 = 72°
∠1 = ∠5 and ∠2 = ∠6 (Corresponding angles)
∴ ∠5 = 108°, ∠6 = 72°
Similarly, ∠4 = ∠8 and
∠3 = ∠7
∴ ∠8 = 72° and ∠7 = 108°
Hence, ∠1 = 108°, ∠2= 72°
∠3 = 108°, ∠4 = 72°
∠5 = 108°, ∠6 = 72°
∠7 = 108°, ∠8 = 12°
Question 2.
In the figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠l, ∠2 and ∠3.
Solution:
l || m || n and p is then transversal which intersects then at X, Y and Z respectively ∠4 = 120°
∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
But ∠3 + ∠4 = 180° (Linear pair)
⇒ ∠3 + 120° = 180°
⇒ ∠3 = 180° – 120°
∴ ∠3 = 60°
But ∠l = ∠3 (Corresponding angles)
∴ ∠l = 60°
Hence ∠l = 60°, ∠2 = 120°, ∠3 = 60°
Question 3.
In the figure, if AB || CD and CD || EF, find ∠ACE.
Solution:
Given : In the figure, AB || CD and CD || EF
∠BAC = 70°, ∠CEF = 130°
∵ EF || CD
∴ ∠ECD + ∠CEF = 180° (Co-interior angles)
⇒ ∠ECD + 130° = 180°
∴ ∠ECD = 180° – 130° = 50°
∵ BA || CD
∴ ∠BAC = ∠ACD (Alternate angles)
∴ ∠ACD = 70° (∵ ∠BAC = 70°)
∵ ∠ACE = ∠ACD – ∠ECD = 70° – 50° = 20°
Question 4.
In the figure, state which lines are parallel and why.
Solution:
In the figure,
∵ ∠ACD = ∠CDE = 100°
But they are alternate angles
∴ AC || DE
Question 5.
In the figure, if l || m,n|| p and ∠1 = 85°, find ∠2.
Solution:
In the figure, l || m, n|| p and ∠1 = 85°
∵ n || p
∴ ∠1 = ∠3 (Corresponding anlges)
But ∠1 = 85°
∴ ∠3 = 85°
∵ m || 1
∠3 + ∠2 = 180° (Sum of co-interior angles)
⇒ 85° + ∠2 = 180°
⇒ ∠2 = 180° – 85° = 95°
Question 6.
If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
Solution:
Question 7.
Two unequal angles of a parallelogram are in the ratio 2:3. Find all its angles in degrees.
Solution:
In ||gm ABCD,
∠A and ∠B are unequal
and ∠A : ∠B = 2 : 3
Let ∠A = 2x, then
∠B = 3x
But ∠A + ∠B = 180° (Co-interior angles)
∴ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = $$\frac { { 180 }^{ \circ } }{ 5 }$$ = 36°
∴ ∠A = 2x = 2 x 36° = 72°
∠B = 3x = 3 x 36° = 108°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 72° and ∠D = 108°
Hence ∠A = 72°, ∠B = 108°, ∠C = 72°, ∠D = 108°
Question 8.
In each of the two lines is perpendicular to the same line, what kind of lines are they to each other?
Solution:
AB ⊥ line l and CD ⊥ line l
∴ ∠B = 90° and ∠D = 90°
∴ ∠B = ∠D
But there are corresponding angles
∴ AB || CD
Question 9.
In the figure, ∠1 = 60° and ∠2 = ($$\frac { 2 }{ 3 }$$)3 a right angle. Prove that l || m.
Solution:
In the figure, a transversal n intersects two lines l and m
∠1 = 60° and
∠2 = $$\frac { 2 }{ 3 }$$ rd of a right angle 2
= $$\frac { 2 }{ 3 }$$ x 90° = 60°
∴ ∠1 = ∠2
But there are corresponding angles
∴ l || m
Question 10.
In the figure, if l || m || n and ∠1 = 60°, find ∠2.
Solution:
In the figure,
l || m || n and a transversal p, intersects them at P, Q and R respectively
∠1 = 60°
∴ ∠1 = ∠3 (Corresponding angles)
∴ ∠3 = 60°
But ∠3 + ∠4 = 180° (Linear pair)
60° + ∠4 = 180° ⇒ ∠4 = 180° – 60°
∴ ∠4 = 120°
But ∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
Question 11.
Prove that the straight lines perpendicular to the same straight line are parallel to one another.
Solution:
Given : l is a line, AB ⊥ l and CD ⊥ l
Question 12.
The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60°, find the other angles.
Solution:
In quadrilateral ABCD, AB || DC and AD || BC and ∠A = 60°
∵ AD || BC and AB || DC
∴ ABCD is a parallelogram
∴ ∠A + ∠B = 180° (Co-interior angles)
60° + ∠B = 180°
⇒ ∠B = 180°-60°= 120°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 60° and ∠D = 120°
Hence ∠B = 120°, ∠C = 60° and ∠D = 120°
Question 13.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measure of ∠AOC, ∠COB, ∠BOD and ∠DOA.
Solution:
Two lines AB and CD intersect at O
and ∠AOC + ∠COB + ∠BOD = 270°
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° (Angles at a point)
∴ 270° + ∠DOA = 360°
⇒ ∠DOA = 360° – 270° = 90°
But ∠DOA = ∠BOC (Vertically opposite angles)
∴ ∠BOC = 90°
But ∠DOA + ∠BOD = 180° (Linear pair)
⇒ 90° + ∠BOD = 180°
∴ ∠BOD= 180°-90° = 90° ,
But ∠BOD = ∠AOC (Vertically opposite angles)
∴ ∠AOC = 90°
Hence ∠AOC = 90°,
∠COB = 90°,
∠BOD = 90° and ∠DOA = 90°
Question 14.
In the figure, p is a transversal to lines m and n, ∠2 = 120° and ∠5 = 60°. Prove that m || n.
Solution:
Given : p is a transversal to the lines m and n
Forming ∠l, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠2 = 120°, and ∠5 = 60°
To prove : m || n
Proof : ∠2 + ∠3 = 180° (Linear pair)
⇒ 120°+ ∠3 = 180°
⇒ ∠3 = 180°- 120° = 60°
But ∠5 = 60°
∴ ∠3 = ∠5
But there are alternate angles
∴ m || n
Question 15.
In the figure, transversal l, intersects two lines m and n, ∠4 = 110° and ∠7 = 65°. Is m || n?
Solution:
A transversal l, intersects two lines m and n, forming ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠4 = 110° and ∠7 = 65°
To prove : Whether m || n or not
Proof : ∠4 = 110° and ∠7 = 65°
∠7 = ∠5 (Vertically opposite angles)
∴ ∠5 = 65°
Now ∠4 + ∠5 = 110° + 65° = 175°
∵ Sum of co-interior angles ∠4 and ∠5 is not 180°.
∴ m is not parallel to n
Question 16.
Which pair of lines in the figure are parallel? Give reasons.
Solution:
Given : In the figure, ∠A = 115°, ∠B = 65°, ∠C = 115° and ∠D = 65°
∵ ∠A + ∠B = 115°+ 65°= 180°
But these are co-interior angles,
Similarly, ∠A + ∠D = 115° + 65° = 180°
∴ AB || DC
Question 17.
If l, m, n are three lines such that l ||m and n ⊥ l, prove that n ⊥ m.
Solution:
Given : l, m, n are three lines such that l || m and n ⊥ l
To prove : n ⊥ m
Proof : ∵ l || m and n is the transversal.
∴ ∠l = ∠2 (Corresponding angles)
But ∠1 = 90° (∵ n⊥l)
∴ ∠2 = 90°
∴ n ⊥ m
Question 18.
Which of the following statements are true (T) and which are false (F)? Give reasons.
(i) If two lines are intersected by a transversal, then corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.
(iii) Two lines perpendicular to the same line are perpendicular to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.
Solution:
(i) False. Because if lines are parallel, then it is possible.
(ii) True.
(iii) False. Not perpendicular but parallel to each other.
(iv) True.
(v) False. Sum of interior angles on the same side is 180° not are equal.
Question 19.
Fill in the blanks in each of the following to make the statement true:
(i) If two parallel lines are intersected by a transversal then each pair of corresponding angles are ……..
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are …….
(iii) Two lines perpendicular to the same line are ……… to each other.
(iv) Two lines parallel to the same line are ……… to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are …….
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are …….
Solution:
(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.
(iii) Two lines perpendicular to the same line are parallel to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are parallel.
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are parallel.
Question 20.
In the figure, AB || CD || EF and GH || KL. Find ∠HKL.
Solution:
In the figure, AB || CD || EF and KL || HG Produce LK and GH
∵ AB || CD and HK is transversal
∴ ∠1 = 25° (Alternate angles)
∠3 = 60° (Corresponding angles)
and ∠3 = ∠4 (Corresponding angles)
= 60°
But ∠4 + ∠5 = 180° (Linear pair)
⇒ 60° + ∠5 = 180°
⇒ ∠5 = 180° – 60° = 120°
∴ ∠HKL = ∠1 + ∠5 = 25° + 120° = 145°
Question 21.
In the figure, show that AB || EF.
Solution:
Given : In the figure, AB || EF
∠BAC = 57°, ∠ACE = 22°
∠ECD = 35° and ∠CEF =145°
To prove : AB || EF,
Proof : ∠ECD + ∠CEF = 35° + 145°
= 180°
But these are co-interior angles
∴ EF || CD
But AB || CD
∴ AB || EF
Question 22.
In the figure, PQ || AB and PR || BC. If ∠QPR = 102°. Determine ∠ABC. Give reasons.
Solution:
In the figure, PQ || AB and PR || BC
∠QPR = 102°
Produce BA to meet PR at D
∵ PQ || AB or DB
∴ ∠QPR = ∠ADR (Corresponding angles)
∴∠ADR = 102° or ∠BDR = 102°
∵ PR || BC
∴ ∠BDR + ∠DBC = 180°
(Sum of co-interior angles) ⇒ 102° + ∠DBC = 180°
⇒ ∠DBC = 180° – 102° = 78°
⇒ ∠ABC = 78°
Question 23.
Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.
Solution:
Given : In two angles ∠ABC and ∠DEF AB ⊥ DE and BC ⊥ EF
To prove: ∠ABC + ∠DEF = 180° or ∠ABC = ∠DEF
Construction : Produce the sides DE and EF of ∠DEF, to meet the sides of ∠ABC at H and G.
Proof: In figure (i) BGEH is a quadrilateral
∠BHE = 90° and ∠BGE = 90°
But sum of angles of a quadrilateral is 360°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360° – 180°= 180°
∴ ∠ABC and ∠DEF are supplementary
In figure (if) in quadrilateral BGEH,
∠BHE = 90° and ∠HEG = 90°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360°- 180° = 180° …(i)
But ∠HEF + ∠HEG = 180° …(ii) (Linear pair)
From (i) and (ii)
∴ ∠HEF = ∠HBG
⇒ ∠DEF = ∠ABC
Hence ∠ABC and ∠DEF are equal or supplementary
Question 24.
In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.
Solution:
Given : In the figure, AB || CD
P is a point between AB and CD PD
and PB are joined
To prove : ∠APB + ∠CDP = ∠DPB
Construction : Through P, draw PQ || AB or CD
Proof: ∵ AB || PQ
∴ ∠ABP = BPQ …(i) (Alternate angles)
Similarly,
CD || PQ
∴ ∠CDP = ∠DPQ …(ii)
(Alternate angles)
∠ABP + ∠CDP = ∠BPQ + ∠DPQ
Hence ∠ABP + ∠CDP = ∠DPB
Question 25.
In the figure, AB || CD and P is any point shown in the figure. Prove that:
∠ABP + ∠BPD + ∠CDP = 360°
Solution:
Given : AB || CD and P is any point as shown in the figure
To prove : ∠ABP + ∠BPD + ∠CDP = 360°
Construction : Through P, draw PQ || AB and CD
Proof : ∵ AB || PQ
∴ ∠ABP+ ∠BPQ= 180° ……(i) (Sum of co-interior angles)
Similarly, CD || PQ
∴ ∠QPD + ∠CDP = 180° …(ii)
∠ABP + ∠BPQ + ∠QPD + ∠CDP
= 180°+ 180° = 360°
⇒ ∠ABP + ∠BPD + ∠CDP = 360°
Question 26.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF.
Solution:
Given : In ∠ABC and ∠DEF. Their arms are parallel such that BA || ED and BC || EF
To prove : ∠ABC = ∠DEF
Construction : Produce BC to meet DE at G
Proof: AB || DE
∴ ∠ABC = ∠DGH…(i) (Corresponding angles)
BC or BH || EF
∴ ∠DGH = ∠DEF (ii) (Corresponding angles)
From (i) and (ii)
∠ABC = ∠DEF
Question 27.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC + ∠DEF = 180°.
Solution:
Given: In ∠ABC = ∠DEF
BA || ED and BC || EF
To prove: ∠ABC = ∠DEF = 180°
Construction : Produce BC to H intersecting ED at G
Proof: ∵ AB || ED
∴ ∠ABC = ∠EGH …(i) (Corresponding angles)
∵ BC or BH || EF
∠EGH || ∠DEF = 180° (Sum of co-interior angles)
⇒ ∠ABC + ∠DEF = 180° [From (i)]
Hence proved.
Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. | 4,940 | 12,341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-10 | latest | en | 0.615475 |
https://docs.microsoft.com/en-us/archive/blogs/ericlippert/floating-point-and-benfords-law-part-two | 1,582,136,694,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144165.4/warc/CC-MAIN-20200219153707-20200219183707-00360.warc.gz | 353,303,801 | 8,794 | # Floating Point And Benford's Law, Part Two
A number of readers asked for an explanation of my offhand comment that Benford's Law can be used to show that binary is the best base for doing floating point math.
One of the desired properties of a floating point system is that the "representation error" is as small as possible. For example, suppose we want to express "one third", in ten-place fixed point decimal notation. The closest we can get is "0.3333333333", which has a representation error of 0.0000000000333333…
A useful way to get a handle on representation error is to look at the granularity of the system. By granularity I mean the smallest difference we can make between two values. For ten-place fixed point decimal notation, the smallest nonzero difference between any two numbers is 10-10. In a floating point system, the granularity changes as the exponent changes. Really large numbers have large granularity; the difference between two successive floats might be millions or billions. And really small numbers have tiny granularity. The maximum possible representation error is half the granularity. Since there is a clear relationship between granularity and representation error, I’m just going to talk about granularity from now on. Small granularity = small representation error = goodness.
Let's consider two similar systems, one which uses a binary mantissa and one which uses a hexadecimal mantissa. I’m going to continue my convention of showing binary numbers in blue, and we’ll do hex in green.
Suppose we've got on the one hand, 32 bit IEEE floats, aka "singles". That is, we've got one sign bit, nine bits of exponent biased by -255 (so exponents can be from -255 to 256), a 22 bit mantissa, and an implicit leading "1." So a number like
(0, 100000011, 1100000000000000000000) would be +1.1100 x 24 = 28
I’m getting sick of spelling out the exponents in non-biased binary. From now on, I’m just going to give them in decimal. And I’m going to give the sign bit by plus and minus, not zero and one.
We now want a hex system with roughly similar range that doesn't take up more storage. Suppose we've got one sign bit, seven bits of exponent biased by -63, and six hex digits for the mantissa. We don't get a leading "1." because not every hex number can be so expressed, so we'll have to use a leading 0. That system has roughly the same range, and if we were to "behind the scenes" express this thing as bits, we'd still be using 32 bits. One for the sign, seven for the exponent, and 24 for the mantissa.
In this system,
(+, 2, A00000) would be +0.A00000 x 162 = 160
An immediate disadvantage of this system is that numbers have multiple representations.
(+, 2, A00000) and (+, 3, 0A0000) are the same number. Let’s ignore that for now. We’ll ignore all hex mantissas with leading zeros. (And besides, the binary IEEE system wastes lots of cases for denormals and NaNs too, so it's not clear that this is any worse.)
What's the granularity of the hex system when, say, the exponent is 2? Well, consider a number like
(+, 2, A00000) – the smallest possible number higher than this is (+, 2, A00001) which is 2-16 larger. Clearly, 2-16 is the granularity for all values with an exponent of 2. More generally, if the exponent is N then the granularity is 24N-24.
How would we represent
(+, 2, A00000) in our binary system? That would be (+, 7, 0100000000000000000000) = +1.01 x 27 = 160. The next largest number that can be represented in our system is (+, 7, 0100000000000000000001) which is 2-15 larger. So the hex system has smaller granularity and hence smaller representation error, and is therefore the better system, right?
Not so fast. Let’s make a chart.
Decimal Hex Binary Binary Granularity Exponent
16 (+, 2,
100000) (+, 4, 0000000000000000000000) -18
32 (+, 2, 200000) (+, 5, 0000000000000000000000) -17
48 (+, 2, 300000) (+, 5, 1000000000000000000000) -17
64 (+, 2, 400000) (+, 6, 0000000000000000000000) -16
80 (+, 2, 500000) (+, 6, 0100000000000000000000) -16
96 (+, 2, 600000) (+, 6, 1000000000000000000000) -16
112 (+, 2, 700000) (+, 6, 1100000000000000000000) -16
128 (+, 2, 800000) (+, 7, 0000000000000000000000) -15
144 (+, 2, 900000) (+, 7, 0010000000000000000000) -15
160 (+, 2, A00000) (+, 7, 0100000000000000000000) -15
176 (+, 2, B00000) (+, 7, 0110000000000000000000) -15
192 (+, 2, C00000) (+, 7, 1000000000000000000000) -15
208 (+, 2, D00000) (+, 7, 1010000000000000000000) -15
224 (+, 2, E00000) (+, 7, 1100000000000000000000) -15
240 (+, 2, F00000) (+, 7, 1110000000000000000000) -15
The hex system always has a granularity exponent of -16. In 8/15ths of the cases, the binary system has worse granularity than the hex system. In 4/15ths of the cases, they have the same granularity, and in only 3/15ths of the cases, the binary system has better granularity. Therefore the hex system is clearly the same or better most of the time. We should use this system instead of binary floats.
Not so fast! We’ve forgotten Benford’s Law! That's true if any number is as likely as any other, but that's not realistic.
Suppose the numbers that we are manipulating obey Benford’s Law. In that case, we would expect that fully a quarter of them would begin with 1 when encoded in hex. We’d expect another quarter of them to begin with 2 or 3, another quarter to begin with 4, 5, 6 or 7, and the remaining quarter to begin with 8, 9, A, B, C, D, E and F. If we make that assumption then we must conclude that the binary system is better half the time, equal a quarter of the time, and worse a quarter of the time.
Clearly this isn't the case just for the exponent 2. For any hex exponent N, the hex system will have a granularity exponent of 4N-24. For numbers in the range expressible by the hex system with that exponent, a quarter of the time, the binary system will have a granularity exponent of 4N-26, a quarter of the time it'll be 4N-25, a quarter of the time it'll be 4N-24, and the remaining quarter it'll be 4N-23, so three-quarters of the time it'll be as good or better. On average, the binary system is considerably better if data are distributed according to Benford's Law.
This is just one example, not a proof. But we could generalize this example and show that for any system with a given number of bits, binary mantissas yield smaller representation errors on average than mantissas in any other base. | 1,789 | 6,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-10 | latest | en | 0.906218 |
https://forum.effectivealtruism.org/posts/6iChzc36Bjpsc2xAK/estimating-the-value-of-mobile-money | 1,702,247,414,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00186.warc.gz | 289,758,347 | 141,881 | # Estimating the Value of Mobile Money
Dec 21 20164 min read 18
# 8
In the US we have many options for sending money around: there's the traditional cash, check, credit card, bank transfer, etc., and then there's PayPal, Venmo, Square, Wallet, etc. In poorer countries, however, if you want to send money to someone in another city your best option is often to send cash on a bus. In 2007 Safaricom introduced M-Pesa in Kenya, where it has been very successful. Can we get an estimate of how valuable M-Pesa has been?
In The long-run poverty and gender impacts of mobile money (Suri and Jack 2016) they try to figure out the effect of the M-Pesa rollout on per-capita consumption. They argue that from 2008 to 2010, when M-Pesa was just getting started, new agents were distributed effectively randomly. [1] Then they look at how areas that were close to M-Pesa agents in 2008-2010 compare to ones that weren't, and argue that this difference represents the effect of having access to M-Pesa.
They find that having access to M-Pesa raised log [2] per-capita consumption by 0.012 (n=1593, standard error=0.005, Sidak-Holm p=0.04). (EDIT: this is an underestimate; see update below.) The reason to care about log consumption is that the more money you have the less valuable additional money is, and this relationship is approximately logarithmic (though it might fall off even more sharply). So by averaging log consumption over all the people in their sample we're getting a rough approximation of how much better M-Pesa access made people's lives.
What does a 0.012 increase in log per-capita consumption look like? Here's a table:
per-capita consumption change
$50,000$603
$25,000$302
$10,000$121
$5,000$60
$2,500$30
$1,000$12
$500$6
$250$3
$100$1
To get a comparision, another way to increase someone's consumption would be to just give them money. GiveDirectly is currently the best option for doing this, and on average GiveDirectly gives $288 per-person to recipients with an annual consumption of$238, a log-increase of 0.79. This is 66x the 0.012 log-increase for M-Pesa availability. GiveDirectly is about 80% efficient, so getting someone $288 costs about$360. So, very roughly, giving someone access to M-Pesa one year earlier is about as valuable as donating $5.45 to GiveDirectly. [3] While M-Pesa was very successful in Kenya, its expansion to other markets has been slow. Improving M-Pesa's expansion, or starting a competitor that can expand faster, could be very valuable. For example, the population of Ethiopia is 94M, so getting mobile money available in Ethiopia a year earlier would be about as good as donating$500M to GiveDirectly. [4]
This is a very rough analysis, and as it is essentially a multiplicative chain errors anywhere along the chain are magnified. For example, two errors in the first version of this post moved the final estimate by 2.3x and 1.5x respectively. This estimate is nowhere near as robust as, for example, GiveWell's estimates for GiveDirectly or the AMF, and those are still very uncertain. Still, it suggests deploying mobile money is an extremely promising intervention, competitive with the top global poverty interventions.
Update 2017-01-01: Tavneet Suri, corresponding author of the paper I'm drawing on here, responded to my question about the interpretation of cell 1 in table 1. Above I interpret it as the effect of having an M-Pesa agent within 1km, but it's actually the effect of having one additional M-Pesa agent within 1km. Since the average number of M-Pesa agents within 1km was ~9, naively we'd expect this to mean the effect of M-Pesa is 9x what I estimated above. But since the first M-Pesa agent is probably much more valuable than the tenth, and their sample mostly didn't include this no-access-to-access transition, the true effect is probably more than 9x. On the other hand, this is kind of big, maybe implausibly so. Instead of a ~1% increase, we're talking about more like 20%. Could M-Pesa really have made Kenya that much richer?
Cross-posted from jefftk.com.
[1] On the other hand, after 2010 they believe there was a systematic effort to expand in places with higher demand. Overall, this is one of the weakest parts of this analysis: if places that got agents earlier were more actually more economically dynamic then it wouldn't be at all surprising to see greater consumption growth there.
[2] The paper doesn't say what base they're using for the log, but I'm assuming here that it's e: natural log is standard in economics. Additionally, in Table S1 of the supplement they give both "Daily per Capita Consumption" and "Log Daily Per Capita Consumption", and the particular numbers they have rule out a base as large as 10. In a draft of this post I had initially assumed log10: thanks to Rob Wiblin for catching this.
[3] My model of this is that you can either wait and get the consumption boost from mobile money eventually, or you can do something to make it happen sooner. If we take a window long enough to include the eventual deployment of mobile money I would expect to see something like:
The solid line represents the path the average log(per-capita consumption) would take, with a jump at the eventual introduction of mobile money, while the dashed line represents moving that jump forward in time. So the benefit is the area between the two curves: how much better mobile money is for people, multiplied by how much you can accelerate it.
[4] Or donating $105M to the AMF, according to GiveWell's cost-effectiveness estimates. # 8 # Reactions # More posts like this Comments18 Sorted by Click to highlight new comments since: Because it is helpful to think about exactly what intervention is needed to help mobile money expand (which may differ by country), I'm throwing here a few potential barriers (mostly based on my own experience in Kenya and Myanmar): • Regulatory barriers (India allowed it only recently because of this; in Myanmar it's still ongoing) • Network effects: in Kenya I heard that an important reason it took off was that Safaricom had a very high market share (maybe near 70%?); in Nigeria I heard that the fragmentation of the telecommunication market is one reason it didn't take off. I'm not sure if more countries are similar to Kenya or Nigeria, also these are all anecdotal. One interesting thing though is that lack of competition in Kenya may have contributed to the high charges (though there is more competition now including from mobile carriers and banks). • Lack of trust: people may not trust mobile carriers or mobile money agents. Probably less of a problem in a close knit community where agents are shopkeepers. Also, lack of trust in banks is a common problem in developing countries but I have no idea about trust on mobile carriers/agents. • Existing alternatives already good enough: this has been mentioned to me in Myanmar, that the traditional "hundi" system of money transfer works well and is cheap which may dampen adoption of mobile money. If that's true then mobile money wouldn't contribute much anyway, but I'm skeptical since mobile money is really much more convenient. (Also it can be used as a savings tool like a checking account, and the poor often face savings constraint too, but I'm not sure how effective that will be; interventions that tackle "self-control" have worked well on this so such elements might need to be bundled in order for mobile money to help with saving) The intervention I had in mind when writing this post was joining a start-up that has been working on this and has been seeing good results so far: http://www.jefftk.com/p/leaving-google-joining-wave Wave is really good! (I use it) Another thing one can do is to work for some mobile money company in a developing country to design products that benefit the poor (e.g. saving, credit, that I mention in the other post), like the American guys I met in Myanmar's Wave Money (but they are still early stage and has many challenges before having an impact). (Not suggesting you should do it though -- involves moving to a developing country etc., and could be much less likely to succeed due to regulations etc.). BTW this is the mobile credit scoring company I had in mind: http://tala.co/. Thanks for this Jeff - a very informative post. The study doesn't appear to control for cash transfers received through access to M-Pesa. I was thinking about how much of the 0.012 increase in ln(consumption) was due to GiveDirectly cash transfers. Back of the envelope: • M-Pesa access raises ln(consumption) by 0.012 for 45% of population (c.20m people). • 0.012 * 20m = 234,000 unit increases in ln(consumption) • GiveDirectly gave c.$9.5m in cash transfers between 2012-14 to people with access to M-Pesa. [1]
• GiveWell estimate each $to GiveDirectly raises ln(consumption) by 0.0049 • 9.5m * 0.0049 = 46,000 unit increases in ln(consumption) So GiveDirectly accounted for (very roughly) a fifth of the 0.012 increase in ln(consumption) due to M-Pesa. [1] this is an overestimate as it assumes all transfers went to Kenya and none to Uganda (Done in haste - may have got my sums / methodology wrong) The study doesn't appear to control for cash transfers received through access to M-Pesa. Good point! I hadn't thought about this at all. GiveDirectly's cash transfers were very large, enough that$9.5m would go to 33k people ($288/person). The population was 43M, so 1 in 1300 people received money from GiveDirectly. Their sample size is just 1593, so you expect 0-2 GiveDirectly recipients. I think they should be pretty visible in the data? Might be worth writing to the authors. It seems like you're assuming that the GiveDirectly money would have gone only to the M-Pesa-access side of the (natural) experiment, but they categorized areas based on whether they had M-Pesa access in 2008-2010, not 2012-2014 when access was much higher. GiveWell estimate each$ to GiveDirectly raises ln(consumption) by 0.0049
I didn't notice that GiveWell had an estimate for this, and checking now I still don't see it. Where's this estimate from?
(In my post I just took their average amount transferred, figured out what effect that had on the average recipient's income, and then discounted by .8 for GiveDirectly's overhead.)
(My method gives an estimate of 0.0022 per dollar to GiveDirectly, so if GiveWell is estimating 0.0049 then my bottom line numbers are roughly 2x too high.)
It seems like you're assuming that the GiveDirectly money would have gone only to the M-Pesa-access side of the (natural) experiment, but they categorized areas based on whether they had M-Pesa access in 2008-2010, not 2012-2014 when access was much higher.
Ah yes - that kind of invalidates what I was trying to do here.
I didn't notice that GiveWell had an estimate for this, and checking now I still don't see it. Where's this estimate from?
It came from the old GiveWell cost-effectiveness analysis excel sheet (2015). "Medians - cell V14". Actually looking at the new one the equivalent figures seems to be 0.26% so you're right! (Although this is the present value of total increases in current and future consumption).
I spent about two hours looking at this in further depth and made an initial stab at modeling out the impact. I estimate an effectiveness of $200/hr (95% interval:$50/hr to $511/hr), not taking into account the value of donating the salary earned from working at Wave. Some places where I notice we disagree or I am confused: 1.) I disagree with you here (footnote 1) that there is a 50% chance of failure (or success). I think the chance of failure could be significantly higher. From https://en.wikipedia.org/wiki/M-Pesa: M-Pesa expanded to Kenya (>10M subscribers), Tanzania (5M), South Africa (100K in a year, 1M in five years), India (???), Mozambique (???), and Lesotho (???). Also, a 2016 Vodaphone press-release suggested M-Pesa seems to have 25M customers worldwide after 10 years of effort. Based on this, I model that a Kenya-level success (>10M subscribers) thus looks like it would have a less than 1/10 chance and a South Africa-level success (1-10M subscribers) looks like it would have a ~3/10 chance. However, I think this success figure could be lower due to diminishing marginal returns since M-Pesa has already plucked low hanging fruit. It's possible better technology could increase this chance. I'd have to know more specifically about what problems M-Pesa runs into and how these are addressed. - 2.) I think your estimate that getting M-Pesa a year earlier is only 66x worse than getting a$288 transfer from GiveDirectly is an overestimate because I expect future roll-outs will take place in countries with higher base consumption. However, as you point out, that estimate is also already an underestimate due to misunderstanding the study. I don't know how to correct for this either way, so I used the 66x number literally in my calculation.
-
3.) I disagree with you here (footnote 1a) that marginal ETG donations are at GiveDirectly levels of cost-effectiveness. I expect them to continue at AMF levels (or greater) for at least a few more years, for reasons OpenPhil mentioned and Carl mentioned. I did an AMF-adjustment in my model for this reason.
-
4.) I really don't know how many staff years it would take to either complete a roll-out or know that it's not going to happen and this is an important part of the model. I currently guess 5-10 full-time staff for 2-5 years, or 10-50 total staff years. This does not count field agents or other hired locals. I couldn't find any information on M-Pesa's total staff count anywhere. I note that Wave has at least 44 staff (from counting faces on the about page), but I don't know if they're all full-time or all focused on expanding cash transfers.
-
5.) I'm confused about why GiveDirectly is stated to be 5x more cost-effective than AMF when from GiveWell's cost-effectiveness estimate, GiveDirectly has a median of $7702 per life saved, ranging from$2200 to $16000, excluding outliers. AMF has a median of$3282 per life saved, ranging from $2200 to$4800, excluding outliers. Together, this implies a comparison centered around 2.35x but ranging from 1x to 3.3x. Maybe I misread the sheet -- I haven't invested that much time in making sure I fully understand it yet.
Re 1, it's worth noting that M-Pesa was administered by very different teams in different countries. In Kenya it was allowed to operate mostly as a startup with limited oversight from Safaricom (or anyone else), whereas in other countries (to varying degrees) the people running M-Pesa were constrained by stricter management from the telecom's country head. This means that there are obvious ways in which M-Pesa was executed less well in other countries. For instance, Wave integrates with M-Pesa in both Kenya and Tanzania, and despite running on exactly the same technology, the Tanzanian system's uptime is substantially worse. Similarly, the quality of their agents and their customer support staff in Tanzania is noticeably lower.
Since Wave isn't hamstrung by oversight from a stodgy and risk-averse telecom, I think you should give less weight to examples from countries other than Kenya as a base rate.
1) ... I think the chance of failure could be significantly higher.
Possibly, but they are already starting to operate in the country in question, and my understanding is that's been going pretty well. My impression is that they're much more competent than Safaricom. My inside view is much higher than 50%, and getting down to 50% was a discount from there.
2) ... I expect future roll-outs will take place in countries with higher base consumption
I'm confused. I was trying to talk about the counterfactual for a specific very poor country if Wave were not working there. So if future mobile money rollouts by other organizations happen first in countries with higher base consumption then that increases the counterfactual impact of Wave choosing to come into a country with very low consumption.
3) ... I expect them to continue at AMF levels (or greater) for at least a few more years
4) ... I really don't know how many staff years it would take
That, combined with estimating marginal impact, makes this pretty awkward. I figure something like 40 person years?
5) ... I'm confused about why GiveDirectly is stated to be 5x more cost-effective than AMF
This comes from cell F31 of the "Results" tab. I haven't put time into understanding how that's calculated, but it looked like the relevant bottom line number.
Thanks Jeff!
-
1) ... I think the chance of failure could be significantly higher.
Possibly, but they are already starting to operate in the country in question, and my understanding is that's been going pretty well. My impression is that they're much more competent than Safaricom. My inside view is much higher than 50%, and getting down to 50% was a discount from there.
Sounds like you definitely have inside info that I don't have, so for now I'd have to rely on my outside view, but I can work to acquire that inside info if I look into this more.
-
2) ... I expect future roll-outs will take place in countries with higher base consumption
I'm confused. I was trying to talk about the counterfactual for a specific very poor country if Wave were not working there. So if future mobile money rollouts by other organizations happen first in countries with higher base consumption then that increases the counterfactual impact of Wave choosing to come into a country with very low consumption.
I don't know what country Wave is looking at or how they are doing what they do because you have inside info that I don't have. If it has consumption comparable to Kenya than my point is invalid. I just was concerned that it wouldn't.
-
3) ... I expect them to continue at AMF levels (or greater) for at least a few more years
Cool. Sounds like this isn't a disagreement between us then.
-
4) ... I really don't know how many staff years it would take
That, combined with estimating marginal impact, makes this pretty awkward. I figure something like 40 person years?
Agreed that it is pretty awkward to estimate. I modified my model to use some of your inputs -- such as a 40% chance of 1-10M subscribers and a 10% chance of >10M subscribers and 40 person years -- and it comes out to $383/hr (95%:$145/hr to \$834/hr). The new mean is still in my old 95% interval which is about the best I can hope for with this level of uncertainty.
-
5) ... I'm confused about why GiveDirectly is stated to be 5x more cost-effective than AMF
This comes from cell F31 of the "Results" tab. I haven't put time into understanding how that's calculated, but it looked like the relevant bottom line number.
Oh, I see that now. I suppose this is a question for GiveWell and not you. I'll ask them.
Sounds like you definitely have inside info that I don't have, so for now I'd have to rely on my outside view, but I can work to acquire that inside info if I look into this more.
If you're interested in working for Wave, or are advising other people on whether it's a good idea for them, I could imagine they'd be quite interested in talking to you!
if it has consumption comparable to Kenya than my point is invalid. I just was concerned that it wouldn't.
It's poorer than Kenya.
That sounds pretty awesome, who do you think would be a good person to reach out to when I'm ready?
Ben Kuhn maybe?
Very cool post.
Just saw that the transaction costs for m-pesa are quite high - the company makes ~20% profit... so there might be something that a Wave-like startup could do: https://en.wikipedia.org/wiki/M-Pesa#Cost.2C_transaction_charges.2C_statistics
maybe using crypocurrency - see here:
http://phys.org/news/2016-10-cryptocurrency-bottom-billion.html
Just saw that the transaction costs for m-pesa are quite high - the company makes ~20% profit
The transaction costs listed on the wikipedia page you cite aren't trivial, but would average well less than 20% unless most transactions are (a) very small and (b) to unregistered users. I'm missing something.
EDIT: could it just be that their profit is 20% of expenses, as opposed to 20% of the money that flows through the M-Pesa network?
maybe using crypocurrency
That article doesn't really show that cryptocurrency helps here. Mostly they're unhappy with transaction fees on international remittances, but you can have low transaction fees just by automating interactions with the money transfer organization, without going to cryptocurrency. And with cryptocurrency generally you pay someone a fee to get your money into the cryptocurrency and then your recipient pays someone else a fee to get it into their local currency.
I think their profit is 20% of their revenue (for a money transfer company, revenue is the total fees brought in, not total money paid into the network).
I just got back from Myanmar and I talked with some people running Wave Money (one of the mobile money companies in Myanmar, and the only licensed one so far; not related to the Wave that Jeff mentioned which sends money to Africa).
• Getting people to adopt could be a big challenge, depending on the country. In Kenya, the anecdotal story of why mobile money took off so quickly is 1) the need to send remittances, 2) preexisting methods for this being not very good for various reasons (insecurity is one); some also argue that Safaricom's unusually high market share in the country played a role in speeding up adoption through bundling of services + network effects (telecommunication markets in other countries seem more fragmented). Mobile money has not taken off in some countries, e.g. Nigeria (this article argues it's due to regulation https://iea.org.uk/blog/why-mobile-money-transformed-kenya-failed-to-take-in-nigeria). In Myanmar it remains an open question: the traditional hundi system for remittances works well for most purposes (being cheap, reliable, and not too slow), which may hinder adoption of mobile money.
• Other potential functions of mobile money: other than through remittances (which is what Suri's paper is estimating), it can also help poor populations by
• Saving: providing a safe place to save. This may be important as many poor people seem "savings constrained" (e.g. see https://www.econ.uzh.ch/dam/jcr:5f6e818a-ad07-466a-8962-fdc77bb1dfc2/casaburi_macchiavello_dairy_20160731.pdf, http://www.simonrquinn.com/research/TwoSidesOfTheSameRupee.pdf) and bank are either scarce or expensive in many rural places -- although it might be hard to convince people to adopt mobile money just for saving purposes.
• Credit: e.g. the Mpesa-based mobile loan Mshwari. There are some startups (including one in Kenya, whose name I forgot) that creates credit scores for people based on their mobile money transaction history, mobile phone records etc. (Mshwari does a version of this but doesn't seem very sophisticated; the startups probably use more "big data") which may help the poor access credit in a way that's much cheaper and sustainable than the traditional microfinance model. (In Myanmar I know one startup trying to do this and giving out small amounts of loan for shorter periods, basically competing with money lenders in slums -- seems hard but could be really good if they succeed; they are quite early stage now.)
• Sending government benefits, e.g. India is considering introducing universal basic income, and already have biometric identification for most citizens, but one of the remaining barriers is the scarcity of bank branches in rural places. If each village has a mobile money agent things would be much easier (and this has implications not only for poverty reduction but maybe also curbing corruption and improving local governance etc.).
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· 5d ago · 5m read | 5,520 | 23,996 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-50 | longest | en | 0.955204 |
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Exam guide questions & Answers 2012
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## ss3 obj maths questions
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FIRST TERM EXAMINATION, 2010/2011 SESSION.
SUBJECT: MATHEMATICS 1(OBJECTIVE)
CLASS: SSS 3 TIME ALLOWED: 2˝HRS
Calculate the size of an interior angle of a regular hexagon
720° b. 120° c. 270° d. 102°
The difference between 4 and 2 is greater than the sum of and1 by
b. c. d.
What is the length of an arc of a circle that subtends an angle of at the center of a circle of radius 8cm?
10cm b. 8cm c. 40cm d. 20cm
The first term of an arithmetic progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum is 81.
12 b. 27 c. 9 d. 4
Find a two digit number such that three times the tens digit is two less than twice the units digit, and twice the number is 20 greater than the number obtained by reversing the digits.
42 b. 47 c. 24 d. 74
If x varies inversely as y, y varies directly as the square of t and x = 1 when t = 3. Find the value of x when t =.
81 b. 27 c. d.
In the diagram below, O is the centre of the circle. If the reflex angle ZOY = 300°, find ?ZXY.
a. 60°
b. 50°
c. 40°
d. 30°
Express: as a single fraction.
Use the following information to answer questions 9 and 10.
A distance of 5.6km is represented by a length of 28mm on the map.
What is the scale of the map?
2 × : 1 b. 2 × : 1 c. 2 × d. 2 ×
What length on the ground does 10cm on the map represent?
2 km b. 20 km c. 200 km d. 2000 km
A rectangular lawn is (x – 2) m long and (x – 3) m wide. If the perimeter is 70 m, calculate the area of the lawn.
289 b. 306 c. 324 d. 400
In the diagram below, WX is parallel to YZ and MN is a transversal. Express p in terms of t.
p = t – 5
P = t – 15
P = t + 10
P = t + 15
Solve :
– 1 b. c. d. 1
By selling an article for ?450:00, Abu lost 10% of the cost price. Find the cost price.
?550.00 b. ?525.00 c. ?500.00 d. ?475.00
A point on the ground is 8 m away from the foot of a vertical wall 6 m high. Calculate correct to the nearest degree, the angle of depression of the point from the top of the wall.
53° b. 45° c. 37° d. 30°
In the diagram below, POQ is a quadrant of a circle centre O and radius 14 cm. Use this information to answer questions 16 and 17. Take to be .
Calculate the length of the minor arc PQ.
11 cm b. 22 cm c. 44 cm d. 66 cm
Calculate the perimeter of the minor sector POQ.
94 cm b. 80 cm c. 50 cm d. 39 cm
To locate Musa’s house from his house, Okey has to move 7 km due east and 7 km due north. Find the bearing of Musa’s house from Okey’s house.
225° b. 135° c. 090° d. 045°
Find the value of x for which the expression is undefined.
x = -3 or 3
x = -1 or 4
x = 1 or 4
x = 1 or 3
If is a perfect square, what is the value of k?
25 b. c. -25 d.
If two straight line segments, PQ and RS intersect at a point T such that ? PTR = 2x + 2 and ? PTS = x + 4, calculate the value of ? PTR.
60° b. 62° c. 118° d. 120°
In the diagram, ? MNL = 30°, ? JKL = 73° and ? MLK = x°. Find x.
77°
81°
103°
107°
Solve the inequality:
x = 9 b. x = 9 c. x = -9 d. x = -9
In the diagram, PQTV is a parallelogram. ? TRS = 110° and ? PVT = 80°. Calculate the size of ? QTR.
50°
40°
20°
10°
A coin is thrown three times. What is the probability of having three heads or three tails in this experiment?
b. c. d.
Calculate the total surface area of a cube of side 5 cm.
150 b. 120 c. 100 d. 75
Obi has four times as many sweets as Olu. If he gives 21 sweets to Olu, they will have equal number of sweets. How many sweets have Obi?
56 b. 52 c. 48 d. 44
In the figure below, ? PQT = ? PSR = 65° and ? PTQ = ? PRS = 45°. Which of the following statements about triangles PTQ and PRS is\are CORRECT?
I, II and III
II and III only
I and III only
I and II only.
The probability that a boy passes an examination is 0.3 and that of a girl is 0.4. What is the probability that a boy and a girl will pass?
0.10 b. 0.12 c. 0.42 d. 0.70
Each interior angle of a regular polygon is 144°. How many sides have the polygon?
6 b. 9 c. 10 d. 12
From the figure below calculate ? QRS
280°
80°
70°
40°
Find the sum to infinity of the following series 3 + 2 + + + + …
1270 b. 190 c. 18 d. 9
Simplify;
b. c. 1 d.
The number converted to base ten is equal to
4 b. 23 c. 49 d. 53
A man spent of his monthly salary on rent and of the remainder on the household needs. If ?135.00 is left, how much is his monthly salary?
?450.00 b. ?225.00 c. ?192.85 d. ?125.00
Simplify;
Log 2 b. log 8 c. 1 d. 0
In the diagram, if PS = SR and PQ is parallel to SR, what is the size of angle PQR?
25°
50°
55°
65°
Find the area of the shaded portion in the figure below, if the triangle is an equilateral triangle of area 8 and the circle center O is of diameter 6 cm.
4.714
3.286
7.999
9.437
An open rectangular box externally measures 4m x 3m x 4m. Find the total cost of painting the box externally if it costs ?4.00 to paint one square meter.
?136.00 b. ?272.00 c. ?408.00 d. ?544.00
The sum of the interior angles of a regular polygon is 30 right angles. How many sides have the polygon?
17 sides b. 26 sides c. 30 sides d. 34 sides
If = 1.732, find the value of without using tables.
0.732 b. 1.346 c. 1.155 d. 3.464
The nth term of a sequence is. Which term of the sequence is?
3rd b. 4th c. 5th d.6th
A boy estimated his transport fare for a journey as ?190.00 instead of ?200.00. Find the percentage error in his estimate.
95% b. 47.5% c. 5.26% d. 45%
A man is four times as old as his son. The difference between their ages is 36 years. Find the sum of their ages.
45 years b. 60 years c. 48 years d. 74 years
In the diagram below, /AD/ = 10 cm, /DC/ = 8 cm, and /CF/ = 15 cm. Use the information to answer questions 45 and 46
If the area of triangle DCF = 24, find the area of Quadrilateral ABCD.
24 b. 48 c. 80 d. 96
Which of the following is correct?
Area of BCF = Area of DCF
Area of DABO = Area of CFEO
Area of ADE = Area of DCFE
The table below shows the scores of a group of 40m students in a mathematics test. Use the table to answer questions 47 and 48.
SCORES
1
2
3
4
5
6
7
8
9
FREQUENCY
2
3
6
7
9
6
2
2
3
If the mode of m and the median is n, then (m,n) is
(5,5) b. (5,6) c. (6,5) d.(9,5)
What is the mean of the distribution?
5.2 b. 5.0 c. 4.8 d. 4.2
The diagram below shows a triangular prism of length 7cm. The right angled triangle PQR is a cross section of the prism. /PR/ = 5cm and /RQ/ = 3cm. Use the information to answer questions 49 and 50
What is the area of the cross-section?
4cm2 b. 6cm2 c. 15cm2 d. 20cm2
What is volume of the prism?
28cm3 b. 42cm3 c. 70cm3 d. 84cm3
```
### ss3 obj maths guide
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Htaccess html Php javascript Asp css maths Past Questions Practice Tests Online | 6,302 | 17,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-21 | longest | en | 0.82496 |
https://stemandleafeducation.com/2019/01/29/how-long-is-that-mountain-path/ | 1,550,783,465,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247508363.74/warc/CC-MAIN-20190221193026-20190221215026-00526.warc.gz | 708,845,394 | 35,634 | How long is that mountain path?
We have continued talking about using integrals for geometric purposes in Calculus. Yesterday, we talked about arc-length. While doing so, my mind wandered back to the trip Paula and I took to visit my sister when she got married. How are these connected? Well, we went up and down many mountains and the distance traveled through these mountain paths provides a great way to visualize arc-length.
The path
I want to walk up, down, up and down the mountains in the picture below. That is, I want to follow the path outlined in black.
If I want to determine the total distance I would have to walk in order to cover this path, I would need to know some additional information. In particular, we would need to know the height and the width of the given mountains. If we knew either one of these, or the distance to the mountains, we could find the other using Paula’s height as a reference. However, I don’t actually know any of these since I just chose a picture that looked pretty. Therefore, I am going to use arbitrary units that seem to fit the picture. Doing this, we have the following picture.
From the graph given above, we see that our path appears to go through the following points,
x y
-5 0
-3 1
-1 0
0 $$-\frac{1}{2}$$
1 0
4 3
9 0
Choosing a function type
Now that we know what we want our path to look like and what points we want it to go through, we can now determine a function to model the given path. We do have multiple choices for types of functions, so let’s look at some.
• Polynomial
• Looks like a fourth degree polynomial since it has three local extrema.
• If wanted to guarantee that all points fell on the same polynomial, we would need at least an 6th degree polynomial.
• Sinusoidal
• The function seems to be periodic, so a sinusoidal function could work.
• However, the amplitude of the crests are not the same, so this may not be the best.
• Piece-wise
• Would focus on finding the best fit for specific portions instead of the whole.
• This would require finding more than one function.
Between the options, I thought that defining the function piece-wise would end up with a good approximation and require the least amount of work possible to do so. For my pieces, I will be using quadratic functions through the points. In general, a quadratic function would be uniquely determined by three points. However, if the vertex, we can instead determine the function using two points. Therefore, I could find a function connecting each pair of points and work over each of these intervals.
Finding the function
We will be finding the quadratic through the given points. In order to do this, we can use the formula for a quadratic $$f(x)=a(x-h)^{2}+k$$. Note that, using this equation, we have that the vertex is the point $$(h,k)$$ and we would then find $$a$$ using our other point.
As we look at the first two points, we would say that the point (-3,1) is the vertex of the quadratic function and (-5,0) is on the graph. Therefore, $$h=-3$$ and $$k=1$$. We can then find that \begin{align*}0&=a(-5-(-3))^{2}+1 \\ a&=\frac{-1}{4}.\end{align*} Hence $$f(x)=\frac{-1}{4}(x+3)^{2}+1$$ on the interval $$[-5,-3)$$.
If we repeat this process for each of the consecutive pairs given above, choosing the appropriate point as the vertex, we arrive at the function $f(x)=\begin{cases} \frac{-1}{4}(x+3)^{2}+1 \text{ for } -5 \leq x < -1 \\ \frac{1}{2}x^{2}-\frac{1}{2} \text{ for } -1 \leq x < 1 \\ \frac{-1}{3}(x-4)^{2}+3 \text{ for } 1 \leq x < 4 \\ \frac{-3}{25}(x-4)^{2}+3 \text{ for } 4 \leq x \leq 9. \end{cases}$
Finding Distance
We now recall that you can find the total distance traveled by finding the arc-length of the function. If the function was linear, we could find this distance by taking $$d=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}$$. Since we do not have a linear function, we can instead use the linear approximation to approximate the distance moved over small intervals. If we then add these all up, we get the total distance traveled. That is \begin{align*}\text{distance} & \approx \sum \sqrt{(\Delta x)^{2}+(\Delta y)^{2}}. \end{align*}
As we let our $$\Delta x$$ get infinitesimally small, we would find that the approximation would be exact. In order to sum these as $$\Delta x$$ gets small, we note that this is by definition $$\int_{a}^{b}\sqrt{(dx)^{2}+(dy)^{2}}$$, where $$a$$ is the smallest $$x$$ and $$b$$ is the largest $$x$$ we are interested. By simplifying, we find that \begin{align*} \text{arc-length}&=\int_{-5}^{9}\sqrt{1+f'(x)}dx.\end{align*}
In order to find this integral, we will find the integral over the corresponding intervals where the function stays the same. We then get the first arc-length is \begin{align*}\int_{-5}^{-1}\sqrt{1+(\frac{-1}{2}(x+3))^{2}}dx&=\int_{-5}^{-1}\sqrt{1+\frac{1}{4}(x+3)^{2}}. \end{align*}
We will now need to find an anti-derivative in order to calculate this. To do so, we let $$i=\frac{1}{2}(x+3)$$. Then $$di=\frac{1}{2}dx$$, so $$dx=2di$$. We therefore get that $\int_{-5}^{-1} \sqrt{1+\frac{1}{4}(x+3)^{2}} =\int_{-1}^{1} \sqrt{1+i^{2}}2di.$ We then let $$i=\tan(\theta)$$, then $$di=\sec^{2}(\theta)d\theta$$, so \begin{align*}\int_{-1}^{1} \sqrt{1+i^{2}}2di&=2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{1+\tan^{2}(\theta)}\sec^{2}(\theta)d\theta \\ &=2 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} sec^{3}(\theta)d \theta \\ &=\sec(\theta)\tan(\theta)|_{-\frac{\pi}{4}}^{\frac{\pi}{4}}+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} sec(\theta)d\theta \\&=2\sqrt{2}+\ln|\sec(\theta)+\tan(\theta)||_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \\ &=2\sqrt{2}+\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1) \\ & \approx 4.59. \end{align*}
As I went through this, I wanted to show as much as possible in order to help show the steps involved in the integration. Here, we needed a substitution, and a trig substitution. Furthermore, in order to find the anti-derivative of $$\sec^{3}(\theta)$$, we needed a reduction formula and an integral of $$\sec(\theta)$$. While I considered showing how to get these using integration by parts and another substitution, I felt that we had already seen enough detail for one post. If you would like to see more, please let me know and I can send these details to you.
Total distance
So far, we have only found the distance traveled as we moved over the first portion of the integral. We can now follow a very similar process to find the total distance traveled over the remaining intervals. When done, we need to find the sum of these in order to find the total distance traveled. If you do this, you will find $$\text{Total Distance}=4.59+2.30+4.44+6.02=17.35$$. That is, in order to walk the path along the along the rise of the mountains, you will have to walk 17.35 units where the units correspond to the choice I made when plotting the graph.
Conclusion
What we’ve seen today is that math in general and calculus in particular can really be found anywhere. Even in a nice picture of a person fishing in front of the mountains, we were able to find a use for calculus. Furthermore, I hope that going through the example of arc-length will help those of you currently in a calculus 2 class that are working on the topic (ie those of you in my calculus class).
If you enjoyed the post and found it helpful, make sure to share with others that may also find it helpful. We really appreciate when you help spread the information we provide.
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