CPhill

0.000001728  = 1.728 x 10^-6 in scientific notation.......

6.955871851049422469085290325852817460147947559359959249... × 10^-89

Mr Raju had some $2 notes and $5 notes in his wallet. The number of $2 notes is 3 times the number of $5 notes. After he spent 2/3 of his $2 notes, he had $70 left in his wallet. How much did he spend?

Let x = the number of   $5 notes

And 3x = the number of $2 notes.........and after he spends 2/3 of these he has (1/3)(3x )   = x left

So

2x + 5x = 70

7x = 70

x = 10

So he spent    (2/3)(3x)($2) = (2/3)(30)*$2  = (20)*$2   = $40

Check;

He started  with  $5(10) + $2(30)  = $110 .......  and he spent $40  = $70 left

Given: A = {a, b, c}; B = {a, e, i, o, u}; C = {c, d, f, g, h, i}
Find (A ∩ C) ∪ (B ∩ C)

(A ∩ C}  =  {a}

(B ∩ C)  =  { i }

So

(A ∩ C) ∪ (B ∩ C)  =

{a} ∪ {i}  =

{a, i }

Josephine had a total of 540 Singapore and Malaysian stamps in the ratio 11:7. After she had given away an equal number of each type of stamps, the number of Singapore and Malaysia stamps left was in the ratio 7:1. How many Singapore stamps did she give away?

Let x be the original number of Singapore stamps and 540 - x  be the original number of Malaysian stamps....and we're told that

x / [540 - x]  = 11/7    cross-multiply

7x = 11[540 - x]  simplify

7x = 5940 - 11x    add 11x to both sides

18x = 5940     divide both sides by 18

x = 330........this is the original number of Singapore stamps

And 540 - 330  =  210  = the original number of Malaysian stamps

Let n be the number of each that she gave away.....and we're told that........

[330 - n] / [210 - n] =  7/1       cross-multiply, again

330 - n = 7[210 - n]   simplify

330 - n = 1470 - 7n

6n = 1140   divide both sides by 6

n = 190  = the number of Singapore [and Malaysian] stamps she gave away

In a Mathematics test, Bala scored 5/6 of Calvin's score and Calvin scored 8 more marks than Dan. If the average score of the 3 boys is 88, what is Calvin's score?

Here's a way to solve this in one variable :

Let x be Calvin's score

Let x - 8  - Dan's score

Let (5/6)x = Bala's score      ....... and we have

[ (5/6)x + x + x - 8] / 3 =  88     nultiply both sides by 3

(5/6)x + 2x - 8  =  264    add 8 to both sides, simplify the left side    ....2x =  (12/6)x

(17/6)x = 272     nultiply both sides by  6/17

x= 272(6/17)  =  96  = Calvin's score

Daisy and Cindy have some stickers. If Daisy gives away 6 stickers, the ratio of the number of stickers Daisy has to the number of stickers Cindy has will be 3:5. If Cindy gives away 6 stickers, the ratio of the number of stickers Daisy has to the number of stickers Cindy has will be 11:13. What is the total number of stickers they have ?

Let D be the original number htat Daisy had and C be the original number that Cindy had....and we have

[D - 6] / [C + 6] = 3/5   

[D + 6] / [ C - 6]  = 11/13

Cross-multiply  in both equations

5[D - 6 ] =  3[C + 6]                                       13[D + 6]  = 11[C - 6]

5D - 30   = 3C + 18                                       13D + 78  = 11C - 66

5D - 3C  = 48                                                13D - 11C   =  -144

So we have this system

  5D - 3C  = 48

13D - 11C = -`144      multiply the top equation by 11 and the bottom equation by -3

 55D - 33C =   528

-39D + 33C  = 432     add both equations

16D = 960     divide both sides by 16

D = 60

And using 5D - 3C = 48  →  5(60) - 48 = 3C  →  100 - 16  = C = 84

So there are 60 + 84 =  144 total stickers

EDIT

Given: A = {2, 3, 4}; B = {1, 3, 5}; C = {5, 6, 7, 8} Find A U B

A U B  = {1, 2, 3, 4, 5 }

[ This is a "union"  rather than an intersection ]

The total length of the wire used = 144  cm

The height of each isosceles triangle = 24cm and the two equal sides = 26 cm each.  So....using the Pythagorean Theorem, 1/2 of the base  = sqrt(26^2 - 24^2)  = sqrt(100)  = 10cm

So.......the whole base = 20 and the perimeter of each triangle = 20 + 2(26)   = 72 cm

And the area of the isosoceles triangles = 1/2(base) x ht =    10cm x 24cm  = 240cm^2

(4/11)x 55 are nickels  = 20 are nickes  =          $1.00

(2/11) x 55 are dimes =  10 are dimes =            $1.00

(5/11) x 55 are quartes  = 25 are quarters =      $6.25

                                                                          --------

                                                                           $8.25  

                                                              =  total value of all the coins