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

Answer: (a)

Let the original number of boys and girls be x and y respectively.

Then $x/{y - 15} = 2/1$

x = 2y - 30 ....(i)

Again, ${x - 45}/{y - 15} =1/5$

5x - 225 = y - 15

5x = y - 15 + 225

5 (2y - 30) = y + 210

[From equation (i)]

10y - 150 = y + 210

10y - y = 210 + 150

9y = 360 ⇒ $y = 360/9$ = 40

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: (d)

Number of boys

= $13/{13 + 11} × 504$

= $13/24 × 504$ = 273

Number of girls

= 504 - 273 = 231

3 girls are admitted.

Required ratio

= 273 : 234 = 7 : 6

Answer: (a)

Let the original number of students in three classes be 2x, 3x and 5x respectively.

As given,

${2x + 20}/{3x + 20} = 4/5$

10x + 100 = 12x + 80

12x - 10x = 100 - 80

2x = 20 ⇒ $x = 20/2$ = 10

Total number of students originally

= 2x + 3x + 5x = 10x

= 10 × 10 = 100

Answer: (d)

Let two positive numbers be 3x and 4x.

According to the question,

$(3x)^2 + (4x)^2$ = 400

$9x^2 + 16x^2$ = 400

$25x^2$ = 400

$x^2 = 400/25$ = 16

$x = √16$ = 4

Sum of numbers

= 3x + 4x = 7x

= 7 × 4 = 28