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

Answer: (a)

Let the number of ladies and gents be 3x and 2x respectively.

According to the question,

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

9x = 4x + 40

5x = 40 ⇒ x = 8

Number of ladies = 3x

= 3 × 8 = 24

Answer: (b)

Numbers = x, 2x and 3x

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

4x + 10 = 3x + 15

x = 5

Numbers = 5, 10 and 15

Answer: (a)

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

Due to increase of 40 students in each class,

we have ${2x + 40}/{3x + 40} = 4/5$

10x + 200 = 12x + 160

2x = 200 - 160

2x = 40 ⇒ x = 20

Original strength

= 10x = 10 × 20 = 200

Answer: (b)

According to the question,

Sum of remaining two numbers

= 11 × 36 - 9 × 34

= 396 - 306 = 90

Ratio of the remaining two numbers = 2 : 3

Smaller number

= $2/5$ × 90 = 36

Answer: (c)

Let the numbers be 5x and 7x.

Now, ${5x - 9}/{7x - 9} = 7/11$

11 (5x - 9) = 7 (7x - 9)

55x - 99 = 49x - 63

55x - 49x = 99 - 63

6x = 36 ⇒ x = 6

Required difference

= 7x - 5x = 2x = 2 × 6 = 12

Here, a = 5, b = 7, x = 9, c = 7, d = 11

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

= ${9 ×5(11 - 7)}/{5 ×11 - 7 × 7}$

= ${45 × 4}/{55 - 49}$

= ${45 × 4}/6$= 30

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

= ${9 ×7(11 - 7)}/{5 ×11 - 7 × 7}$

= ${63 × 4}/{55 - 49}$

= ${63 × 4}/6$= 42

Their difference = 42 - 30 = 12

Answer: (b)

Using Rule 21,

Initially number of boys

= $8/{8 + 5} × 286 = 8/13 × 286$ = 176

Number of girls

= $5/13 × 286$ = 110

22 more girls get admitted.

∴ Required ratio

= $176/{110 + 22} = 176/132 = 4/3$ = 4 : 3