Key Ratio Proportion Grade 10

7/18/2019 Key Ratio Proportion Grade 10

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1. Arrange the following ratios in descending order.

2 : 3, 3 : 4, 5 : 6, 1 : 5

Solution:

Given ratios are 2/3, 3/4, 5/6, 1/5

The L.C.M. of 3, 4, 6, 5 is 2 × 2 × 3 × 5 = 60

No, 2/3 = !2 × 20"/!3 × 20" = 40/60

3/4 = !3 × 15"/!4 × 15" = 45/60

5/6 = !5 × 10"/!6 × 10" = 50/60

1/5 = !1 × 12"/!5 × 12" = 12/60

C#ear#$, 50/60 % 45/60 % 40/60 % 12/60

Therefore, 5/6 % 3/4 % 2/3 % 1/5

&o, 5 : 6 % 3 : 4 % 2 : 3 % 1 : 5

2. To n'()ers are in the ratio 3 : 4. *f the s'( of n'()ers is 63, fin+ the n'()ers.

Solution:

&'( of the ter(s of the ratio = 3 4 = -

&'( of n'()ers = 63

Therefore, first n'()er = 3/- × 63 = 2-

&eon+ n'()er = 4/- × 63 = 36

Therefore, the to n'()ers are 2- an+ 36.

3. *f : $ = 1 : 2, fin+ the va#'e of !2 3$" : ! 4$"

Solution:

: $ = 1 : 2 (eans /$ = 1/2

No, !2 3$" : ! 4$" = !2 3$"/! 4$" [Divide numerator and denominator by y.]

= !2 3$"/$/! 4$"/2 = 2!/$" 3/!/$" 4, 't /$ = 1/2

e et = 2 !1/2" 3"/!1/2 4" = !1 3"/!1 "/2 = 4/!/2" = 4/1 × 2/ = /



Therefore the va#'e of !2 3$" : ! 4$" = :

More solved problems on ratio and proportion are explained here with full description.

. 7 )a ontains 8510 in the for( of 50 , 25 an+ 20 oins in the ratio 2 : 3 : 4. 9in+ the

n'()er of oins of eah t$e.

Solution:

Let the n'()er of 50 , 25 an+ 20 oins )e 2, 3 an+ 4.

Then 2 × 50/100 3 × 25/100 4 × 20/100 = 510

/1 3/4 4/5 = 510

!20 15 16"/20 = 510

⇒ 51/20 = 510

= !510 × 20"/51

= 200

2 = 2 × 200 = 400

3 = 3 × 200 = 600

4 = 4 × 200 = 00.

Therefore, n'()er of 50 oins, 25 oins an+ 20 oins are 400, 600, 00 resetive#$.

!. *f 27 = 3 = 4C, fin+ 7 : : C

Solution:

Let 27 = 3 = 4C =

&o, 7 = /2 = /3 C = /4

The L.C.M of 2, 3 an+ 4 is 12

Therefore, 7 : : C = /2 × 12 : /3 × 12 : /4 = 12

= 6 : 4 : 3

= 6 : 4 : 3

Therefore, 7 : : C = 6 : 4 : 3

". hat ('st )e a++e+ to eah ter( of the ratio 2 : 3, so that it (a$ )eo(e e;'a# to 4 : 5<



Solution:

Let the n'()er to )e a++e+ )e , then !2 " : !3 " = 4 : 5

⇒ !2 "/!5 " = 4/5

5!2 " = 4!3 "

10 5 = 12 4

5 4 = 12 10

= 2

#. The #enth of the ri))on as oriina##$ 30 (. *t as re+'e+ in the ratio 5 : 3. hat is its

#enth no<

Solution:

Lenth of ri))on oriina##$ = 30 (

Let the oriina# #enth )e 5 an+ re+'e+ #enth )e 3.

't 5 = 30 (

= 30/5 ( = 6 (

Therefore, re+'e+ #enth = 3 (

= 3 × 6 ( = 1 (

More worked out problems on ratio and proportion are explained here step-by-step.

$. Mother +ivi+e+ the (one$ a(on >on, &a( an+ Maria in the ratio 2 : 3 : 5. *f Maria ot 8150,

fin+ the tota# a(o'nt an+ the (one$ reeive+ )$ >on an+ &a(.

Solution:

Let the (one$ reeive+ )$ >on, &a( an+ Maria )e 2, 3, 5 resetive#$.

Given that Maria has ot 8 150.

Therefore, 5 = 150

or, = 150/5

or, = 30

&o, >on ot = 2



= 8 2 × 30 = 860

&a( ot = 3

= 3 × 60 = 80

Therefore, the tota# a(o'nt 8!30 60 0" = 810

%. ?ivi+e 83-0 into three arts s'h that seon+ art is 1/4 of the thir+ art an+ the ratio

)eteen the first an+ the thir+ art is 3 : 5. 9in+ eah art.

Solution:

Let the first an+ the thir+ arts )e 3 an+ 5.

&eon+ art = 1/4 of thir+ art.

= !1/4" × 5

= 5/4

Therefore, 3 !5/4" 5 = 3-0

!12 5 20"/4 = 3-0

3-/4 = 3-0

= !3-0 × 4"/3-

= 10 × 4

= 40

Therefore, first art = 3

= 3 × 40

= 8120

&eon+ art = 5/4

= 5 × 40/4

= 850

Thir+ art = 5

= 5 × 40

= 8 200



1&. The first, seon+ an+ thir+ ter(s of the roortion are 42, 36, 35. 9in+ the fo'rth ter(.

Solution:

Let the fo'rth ter( )e .

Th's 42, 36, 35, are in roortion.

@ro+'t of etre(e ter(s = 42 ×

@ro+'t of (ean ter(s = 36 A 35

&ine, the n'()ers (aBe ' a roortion

Therefore, 42 × = 36 × 35

or, = !36 × 35"/42

or, = 30

Therefore, the fo'rth ter( of the roortion is 30.

More worked out problems on ratio and proportion using step-by-step explanation.

11. &et ' a## ossi)#e roortions fro( the n'()ers , 12, 20, 30.

Solution:

e note that × 30 = 240 an+ 12 × 20 = 240

Th's, × 30 = 12 × 20 ..!*"

Dene, : 12 = 20 : 30 .. !i"

e a#so note that, × 30 = 20 × 12

Dene, : 20 = 12 : 30 .. !ii"

!*" an a#so )e ritten as 12 × 20 = × 30

Dene, 12 : = 30 : 20 .. !iii"

Last !*" an a#so )e ritten as

12 : 30 = : 20 .. !iv"

Th's, the re;'ire+ roortions are : 12 = 20 : 30



: 20 = 12 : 30 12 : = 30 : 20 12 : 30 = : 20

12. The ratio of n'()er of )o$s an+ ir#s is 4 : 3. *f there are 1 ir#s in a #ass, fin+ the n'()er

of )o$s in the #ass an+ the tota# n'()er of st'+ents in the #ass.

Solution:

N'()er of ir#s in the #ass = 1

>atio of )o$s an+ ir#s = 4 : 3

7or+in to the ;'estion,

o$s/Gir#s = 4/5

o$s/1 = 4/5

o$s = !4 × 1"/3 = 24

Therefore, tota# n'()er of st'+ents = 24 1 = 42.

13. 9in+ the thir+ roortiona# of 16 an+ 20.

Solution:

Let the thir+ roortiona# of 16 an+ 20 )e .

Then 16, 20, are in roortion.

This (eans 16 : 20 = 20 :

&o, 16 × = 20 × 20

= !20 × 20"/16 = 25

Therefore, the thir+ roortiona# of 16 an+ 20 is 25.

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Key Ratio Proportion Grade 10