Download The Mathematics Department Stage : 3rd Prep Date : / / [1] Complete

The Mathematics Department
Stage : 3rd Prep
Date : / /
1st Term
Practice Sheet
Ratio , Proportion and Continued Proportion
☻ The ratio must be between two quantities of the same kind and the same unit
a
c
=
☻If
, then a  d = b  c
b
d
a
c
=
☻If
, then the quantities a, b, c and d are proportional
b
d
a
c
=
☻If the quantities a, b, c and d are proportional, then
b
d
a
c
= , then a = c m and b = d m , where m is a constant  0
☻If
b
d
☻If a d = b c , then each of the following proportions is true:
a
c
a
b
b
d
c
d
=
=
i)
ii)
iii)
iv)
=
=
b
d
c
d
a
c
a
b
a
c
e
☻If
=
=
= ........ and m1 , m2 , m3 , ……are non-zero real numbers then:
b
d
f
m1 a + m2c + m3e + .......
= one of the given ratios
m1b + m2d + m3f + .........
a
b
=
☻a, b and c are said to be in continued proportion if
b
c
● b2 = a c
b= ±
ac
● The middle proportional between two quantities
= ± the product of the twoquantities
Notice that: a & c should be positive together or negative together
[1] Complete each of the following:
y
7
1] If 7x = 10y , then = ……
…..
10
x
3
15
=
2]
4
...20..
3] If a = 3b , then
a
= …3……
b
4] 12 : …15…. = 4 : 5
5] 3a + 5b = 0 , then
a
-5
=…
…..
3
b
1
6]
..6 x2..
2k
=
15x2
5k
3x + 4y
x
= 2 , then = …- 2…..
2x + 5y
y
6x + 10y = 3x + 4y
3x = - 6y
8] Two numbers with ratio 7 : 12 one of them exceeds the other by 275 then the
two numbers are …385…. , …660…
7] If
9] The number c is added to both terms of the ratio 5 : 37 to become
c = …11….
[
5+c
1
=
37 + c
3
15 + 3c = 37 + c 2c = 22
10] If
7-4
3
a
a- b
7
= , then
=…
…..
=
7+4
11
b
a+ b
4
11] If
5
5a - 7b
b
= zero, then = … ….
7
a
8a + 11b
[5a – 7b = 0
4
a
= … ..
3
b
3a = 4b (a & b ∈ R+)]
12] If 9a2 – 16b2 = 0 where a & b  R+, then
[(3a + 4b)(3a – 4b) = 0
a
3
a 2
13] If = and = , then a : b : c = …6... : …10.. : …21..
b
5
c
7
14] If x : y = 3 : 1 and x + y = 28, then x = …21…. and y = …7….
15] If
2a + ..5a...
a
a c
=
, then
=
..2b... + 5d
b
b
d
16] If
8
a
3
a 3
c
= and =
, then = … …..
5
b
5
c
8
b
17] If
6
x
2
, then x = … …..
=
5
3 5
18] If x ,
8 , 7 , 14 2 are proportional, then x = …1…
19] The fourth proportional of 2 , 4 and 6 is ……12…..
2
1
, then
3
c = 11]
5a = 7b]
20] The fourth proportional of x , y and (x + y) is ……
y (x + y)
…
x
21] If 3 , 4 , x and 11 are four proportional quantities , then x = …
22] If 5a , 2y , 3b and 7y are proportional , then
23] If a , b , 2 and 3 are proportional , then
33
…..
4
6
a
=…
……
35
b
3
b
= … …..
2
a
24] The middle proportional between 8 and 18 is …± 12….
25] The third proportional of the numbers -2 and 6 is …- 18…..
26] 9x2 – 25y2 : ……1…… = (3x – 5y) (3x + 5y)
27] If
3
1
3
a
b
2
a
=
and =
, then = … =
….
6
2
c
3
b
c
4
28] The third proportional of the numbers 4x + 5y , 4x – 5y and 16x2 – 25y2 is
……(4x + 5y)2……
29] If
5
x
z
5
x+z
, then
= … ……
= =
4
y
e 4
y+ e
30] If
3
a - 2c + e
a c
e 3
, then
= … …….
= = =
5
b d
f
5
b - 2d + f
31] If
x- e
y- x
x+ y
y+ e e+ x
=
then
=
=
2
....- 2...
2
3
2
32] If 7 , x and
1
are in continued proportion, then x2 y = …7…..
y
33] If 2 , 4 + x and 18 are proportional and x  R- then x = …- 10…
2
4+x
=
(4 + x)2 = 36
x2 + 8x + 16 – 36 = 0
4+x
18
2
∴ x + 8x – 20 = 0
x = - 10 & x = 2
3
2ac + bd
a
1
c
7
then find the ratio
= and =
b 3
d 2
bc - 3ad
a = m & b = 3m
c = 7k & d = 2k
2 (m)(7k ) + (3m)(2k )
2ac + bd
2mk + 6mk
8mk
8
=
=
=
=
bc - 3ad
(3m)(7k ) - 3 (m)(2k ) 21mk - 6mk 15mk 15
[2] If
[3] If x2 – 6xy + 9y2 = 0 then find x : y
∵ x2 – 6xy + 9y2 = 0
(x – 3y)2 = 0
∴ x = 3y
x:y=3:1
x – 3y = 0
[4] If 3x2 – 10xy + 7y2 = 0 then find x : y
(3x – 7y)(x – y) = 0
3x = 7y
x=y
x
7
x
=
=1
y
3
y
[5] The ratio between two positive real numbers is 4 : 7 if we subtract 16 from these
two numbers then the ratio between them become 2 : 5. Find these two numbers.
x
4
→
x = 4m
→
y = 7m
=
y
7
x - 16
2
4m - 16
2
→
→
20m – 80 = 14m - 32
=
=
7m - 16
5
y - 16
5
20m – 14m = 80 – 32
6m = 48
m=8
x = 4 × 8 = 32
&
y = 7 × 8 = 56
The two numbers are 32 & 56
[6] The ratio between two real numbers is 8 : 15, find these two numbers if the
difference between them is 40.6
x
8
→
15x = 8y
→
15x – 8y = 0
(1)
=
y
15
y – x = 40.6 →
x – y = 40.6
(2)
From (1) & (2) x = 46.4 & y = 87
a
b
c
a + 2b
7
, prove that
=
=
=
2x + y 3y - x
4x + 5y
4b + c 17



a + 2b 4b + c
R.T.P. :
=
7
17
A.R. =  + 2 
&
A.R. = 4 + 
a + 2b
a + 2b
4a + c
a + 2b
=
=
A.R.=
(1)
A.R. =
(2)
2x + y + 6y - 2x
7y
12y - 4x + 4x + 5y
17y
[7] Let
4
a + 2b 4b + c
=
7
17
[8] If a, b, c and d are in proportional prove that:
a2 - c 2
ac
5a - 2c
4a + 3c
=
(i)
(ii) 2
=
bd
5b - 2d
4b + 3d
b - d2
a
c
→
a=bm
&
c = dm
=
=m
b
d
5bm - 2dm m (5b - 2d)
(i) L.H.S. =
=m
=
5b - 2d
5b - 2d
4bm + 3dm
m (4b + 3d)
R.H.S. =
=m
L.H.S. = R.H.S. = m
=
4b + 3d
4b + 3d
From (1) & (2)
b2m2 - d2m2
m2 (b2 - d2 )
(ii) L.H.S. =
= m2
=
2
2
2
2
b - d
b - d
R.H.S. =
bm . dm
= m2
bd
L.H.S. = R.H.S. = m2
[9] If a, b, c, d, e and f are in proportional prove that:
c 2 + e2
2ac - 5e2
4a + c - 3e
=
(i)
= one of the ratio
(ii) 2
4b + d - 3f
d + f2
2bd - 5f 2
a
c
e
→
a=bm
,
c=mc
=
=
=m
b
d
f

 
4a + c - 3e
(i)
= one of the ratio
4b + d - 3f
A.R. = 4 +  - 3
4a + c - 3e
=
4b + d - 3f
c 2 + e2
2ac - 5e2
(ii) 2
=
d + f2
2bd - 5f 2
d2m2 + f 2m2
m2 (d2 + f 2 )
L.H.S. =
=
= m2
2
2
2
2
d + f
d +f
2(bm)(dm) - 5f 2m2
m2 (2bd - 5f 2 )
=
= m2
2
2
2bd - 5f
2bd - 5f
L.H.S. = R.H.S. = m2
R.H.S. =
5
&
e=fm
[10] If a, b and c are in proportional prove that:
2
æb - c ö
c
a
b
a- b
÷
=
(i)
(ii) çç
=
=
÷
a
a+ b b+ c a- c
èça - b ÷
ø
a
b
→
a = c m2 & b = c m
=
=m
b
c
a
b
a- b
(i)
=
=
a+ b b+ c a- c
cm2
cm2
m
L.H.S. =
=
=
2
m+1
cm + cm cm (m + 1)
cm
cm
m
M.H.S. =
=
=
cm + c c(m + 1)
m+1
R.H.S. =
m (m - 1)
cm2 - cm
cm(m - 1)
m
=
=
=
m+1
(m + 1) (m - 1)
cm2 - c
c(m2 - 1)
2
æb - c ö
c
÷
ç
(ii) ç
÷ =
çèa - b ÷
a
ø
2
ö
æ cm - c ÷
ö2 æ
c
(m
1)
1
÷
÷
÷
L.H.S. = ççç 2
= ççç
= 2
÷
÷
çècm - cm ÷
çè c m (m - 1) ÷
m
ø
ø
c
1
1
R.H.S. =
L.H.S. = R.H.S. = 2
= 2
2
cm
m
m
[11] If a, b, c and d are in continued proportional prove that:
a2 + b2 + c 2
b
a
b+ d
=
(i) 3 =
(ii) 2
2
2
2
2
d
c
d(c + d )
b + c + d
a
b
c
→
a = f m3 & b = f m2 & c = fm
=
=
=m
b
c
f
a
b+ d
(i) 3 =
c
d(c 2 + d2 )
L.H.S. =
dm3
1
=
d3m3
d2
R.H.S. =
dm2 + d
d(m2 + 1)
1
=
=
d(d2m2 + d2 )
d3 (m2 + 1)
d2
a2 + b2 + c 2
b
=
(ii) 2
d
b + c 2 + d2
L.H.S. =
d2m6 + d2m4 + d2m2
d2m2 (m4 + m2 + 1)
=
= m2
2 4
2 2
2
2
4
2
dm + dm + d
d (m + m + 1)
6
b
dm2
R.H.S. = =
= m2
d
d
7