Ratio, Proportion
- The ratio of the number of boys and that of girls in a school having 504 students is 13 :11. What will be the new ratio if 3 more girls are admitted?
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Number of boys
= 13 × 504 13 + 11 = 13 × 504 = 273 24
Number of girls
= 504–273 = 231
3 girls are admitted.
∴ Required ratio = 273 : 234
= 7 : 6Correct Option: A
Number of boys
= 13 × 504 13 + 11 = 13 × 504 = 273 24
Number of girls
= 504–273 = 231
3 girls are admitted.
∴ Required ratio = 273 : 234
= 7 : 6
- The students in three classes are in the ratio 2 : 3 : 5. If 20 students are increased in each
class, the ratio changes to 4 : 5 : 7. Originally the total number of students was :
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Let the original number of students in three classes be 2x, 3x and 5x respectively.
As given,2x + 20 = 4 3x + 20 5
⇒ 10x + 100 = 12x + 80
⇒ 12x – 10x = 100 – 80
⇒ 2x = 20⇒ x = 20 = 10 2
∴ Total number of students originally
= 2x + 3x + 5x = 10x
= 10 × 10 = 100Correct Option: C
Let the original number of students in three classes be 2x, 3x and 5x respectively.
As given,2x + 20 = 4 3x + 20 5
⇒ 10x + 100 = 12x + 80
⇒ 12x – 10x = 100 – 80
⇒ 2x = 20⇒ x = 20 = 10 2
∴ Total number of students originally
= 2x + 3x + 5x = 10x
= 10 × 10 = 100
- The ratio of number of boys to that of girls in a group becomes 2:1 when 15 girls leave. But, afterwards, when 45 boys also leave, the ratio becomes 1 : 5. Originally the number of girls in
the group was
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Let the original number of boys and girls be x and y respectively.
Thenx = 2 y − 15 1
⇒ x = 2y – 30 ....(i)Again, x − 45 = 1 y − 15 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 = 40 9 Correct Option: C
Let the original number of boys and girls be x and y respectively.
Thenx = 2 y − 15 1
⇒ x = 2y – 30 ....(i)Again, x − 45 = 1 y − 15 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 = 40 9
- Two numbers are in the ratio 3 : 5. If 9 is subtracted from each, then they are in the ratio 12 : 23. Find the smaller number.
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Let the numbers be 3x and 5x.
∴ 3x − 9 = 12 5x − 9 23
⇒ 69x – 60x = 207 – 108⇒ x = 99 = 11 9
∴ The smaller number
= 3x = 33
Second Method :
Here, a = 3, b = 5, x= 9, c = 12, d = 231st Number = xa(d − c) ad − bc = 9 × 3(23 − 12) 3 × 23 − 5 × 12 = 27 × 11 69 − 60 = 27 × 11 = 33 9 2nd Number= xb(d − c) ad − bc = 9 × 5(23 − 12) 3 × 23 − 5 × 12 = 45 × 11 69 − 60 = 45 × 11 = 55 9
∴ Smallest number = 33Correct Option: B
Let the numbers be 3x and 5x.
∴ 3x − 9 = 12 5x − 9 23
⇒ 69x – 60x = 207 – 108⇒ x = 99 = 11 9
∴ The smaller number
= 3x = 33
Second Method :
Here, a = 3, b = 5, x= 9, c = 12, d = 231st Number = xa(d − c) ad − bc = 9 × 3(23 − 12) 3 × 23 − 5 × 12 = 27 × 11 69 − 60 = 27 × 11 = 33 9 2nd Number= xb(d − c) ad − bc = 9 × 5(23 − 12) 3 × 23 − 5 × 12 = 45 × 11 69 − 60 = 45 × 11 = 55 9
∴ Smallest number = 33
- Two numbers are in the ratio 5 : 7. If 9 is subtracted from each of them, their ratio becomes 7 : 11. The difference of the numbers is
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Let the numbers be 5x and 7x.
Now, 5x − 9 = 7 7x − 9 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
Second Method :
Here, a = 5, b = 7, x = 9, c = 7, d = 111st Number = xa(d − c) ad − bc = 9 × 5(11 − 7) 5 × 11 − 7 × 7 = 45 × 4 55 − 49 = 45 × 4 = 30 6 2nd Number
=xb(d − c) ad − bc = 9 × 7(11 − 7) 5 × 11 − 7 × 7 = 63 × 4 55 − 49 = 63 × 4 = 42 6 Correct Option: B
Let the numbers be 5x and 7x.
Now, 5x − 9 = 7 7x − 9 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
Second Method :
Here, a = 5, b = 7, x = 9, c = 7, d = 111st Number = xa(d − c) ad − bc = 9 × 5(11 − 7) 5 × 11 − 7 × 7 = 45 × 4 55 − 49 = 45 × 4 = 30 6 2nd Number
=xb(d − c) ad − bc = 9 × 7(11 − 7) 5 × 11 − 7 × 7 = 63 × 4 55 − 49 = 63 × 4 = 42 6
