type 7 finding sum difference product based ratio & proportion problems Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (a)

Let the numbers be $3/2$x and $8/3$x

According to question,

${3/2 x + 15}/{{8x}/3 + 15} = {5/3}/{5/2}$

${{3x + 30}/2}/{{8x + 45}/3} = 2/3$

${3(3x + 30)}/{2(8x + 45)} = 2/3$

${9x + 90}/{16x + 90} = 2/3$

27x + 270 = 32x + 180

32x - 27x = 270 - 180 = 90

5x = 90 ⇒ x = 18

The greater number

= $8/3 x = 8/3 × 18$ =48

Answer: (d)

Let the numbers be 2x and 3x.

${2x + 4}/{3x + 4} = 5/7$

15x + 20 = 14x + 28

x = 28 - 20 = 8

Required difference

Using Rule 34,

Here, a = 2, b = 3,c = 5

d = 7 and x = 4

1st Number = ${xa(c-d)}/{ad-bc}$

= ${4 ×2(5 - 7)}/{2 × 7 - 5 × 3}$

= ${8 × - 2}/{14 - 15}$ = 16

2nd Number= ${xb(c-d)}/{ad-bc}$

= ${4 ×3(5 - 7)}/{2 × 7 - 5 × 3}$

= ${4 × 3 (- 2)}/{14 - 15}$ = 24

Difference of numbers = 24 - 16 = 8

Answer: (c)

Let the original number of students be 4x , 6x and 9x.

${4x + 12}/{6x +12} = 7/9$

42x + 84 = 36x + 108

42x - 36x = 108 - 84

6x = 24 ⇒ x = 4

Required number of students

= 19x = 19 × 4 = 76

Answer: (c)

Let the numbers be 3x and 5x.

${3x - 9}/{5x - 9} = 12/23$

69x - 60x = 207 - 108

$x =99/9$ 11

The smaller number = 3x = 33

Using Rule 35,

Here, a = 3, b = 5, x= 9, c = 12, d = 23

1st Number = ${xa(d-c)}/{ad-bc}$

= ${9 ×3(23 - 12)}/{3 × 23 - 5 × 12}$

= ${27 × 11}/{69 - 60}$

=${27 × 11}/9$ = 33

2nd Number= ${xb(d-c)}/{ad-bc}$

= ${9 ×5(23 - 12)}/{3 × 23 - 5 × 12}$

= ${45 × 11}/{69 - 60}$

= ${45 × 11}/9$= 55

Smallest number = 33

Answer: (c)

Let the numbers be x, 2x and 3x.

According to the question,

$x^3 + (2x)^3 + (3x)^3$ = 4500

$x^3 + 8x^3 + 27x^3$ = 4500

$36x^3$ = 4500

$x^3 = 4500/36$ = 125

$x = √^3{125}$= 5 = smallest number