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

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: (b)

Let the number x be added.

${17 + x}/{24 + x} = 1/2$

34 + 2x = 24 + x

2x - x = 24 - 34

x = - 10

Hence, 10 should be subtracted.

Answer: (a)

Original number of boys in school

= $5/9$ × 432 = 240

Number of girls

= 432 - 240 = 192

Let the new number of girls be x.

According to the question,

${240 + 12}/{192 + x} = 7/6$

$252/{192 + x} = 7/6$

192 × 7 + 7x = 252 × 6

1344 + 7x = 1512

7x = 1512 - 1344 = 168

$x = 168/7$ = 24

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)

Numbers = 3x and 5x (let)

According to question,

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

69x - 207 = 60x - 108

69x - 60x = 207 - 108

9x = 99 ⇒ $x = 99/9$ = 11

Smaller number = 3x = 3 × 11 = 33

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.