type 7 finding sum difference product based ratio & proportion problems Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

According to the question,

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

(Numbers = 3x and 5x )

69x - 207 = 60x - 108

9x = 207 - 108 = 99

x = 11

Required numbers

3 × 11 = 33 and 5 × 11 = 55

Using Rule 35,

Here, a = 3, b = 5, c = 12,

d = 23, x = 9

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

Numbers are 33, 55

Answer: (c)

Required number = x

${11 - x}/{15 - x} = 2/3$

33 - 3x = 30 - 2x

3x - 2x = 33 - 30

x = 3

Answer: (a)

Let the numbers be x and 5x.

According to the question,

x × 5x = 320

$5x^2$ = 320

$x^2 = 320/5$ = 64

$x = √64$ = 8

Required difference

= $(5x)^2 - x^2$

= $25x^2 - x^2 = 24x^2$

= 24 × 8 × 8 = 1536

Answer: (a)

Let A and B have Rs.2x and Rs.x initially.

2x - 2 = x + 2

x = 4

Initial amount with A = Rs.8

Initial amount with B = Rs.4.

Answer: (d)

Tricky Approach

Required ratio

= 15 × 22 : 11 × 25 = 6 : 5

Answer: (d)

Total numbers of girls in the school

= $504 ×11/{13 + 11}$

= $504 × 11/24$ = 231

Total numbers of boys in the school

= $504 × 13/{13 + 11}$

= $504 × 13/29 = 273$

Now, total no. of girls when 12 more girls are admitted

= 231 + 12 = 243

Required ratio

= 273 : 243 = 91 : 81