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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to Updated on: 21 May 2019, 11:24
Originally posted by nick1816 on 21 May 2019, 04:45.
Last edited by nick1816 on 21 May 2019, 11:24, edited 1 time in total.
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57% (02:51) correct 43% (03:02) wrong based on 277 sessions

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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to boys in the ratio 5 : 3. If the number of students in Class A is at least twice as many as the number of students in Class B, what is the minimum percentage of boys possible, when both classes are considered together?

A. 30%
B. 33.33%
C. 37.5%
D. 39.17%
E. 40%
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D
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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to 21 May 2019, 05:29
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nick1816
Class A has boys to girls in the ratio 2 : 3, and Class B has girls to boys in the ratio 5 : 3. If the number of students in Class A is at least twice as many as the number of students in Class B, what is the minimum percentage of boys possible, when both classes are considered together?

A. 30%
B. 33.33%
C. 37.5%
D. 39.17%
E. 40%
Show more

You can do by weighted average method or algebraically or logically..

1) LOGICALLY - a 20 seconds answer
% of boys in A is \(\frac{2}{5}\)or 40% and
% of boys in B is \(\frac{3}{8}\) or 37.5%
So, our answer has to be between 37.5 and 40, irrespective of the strength of each class. As none of the class has a strength of 0, the extremes 37.5 and 40 are not possible.
ONLy D left

2) Weighted average method.
Class A B
strength 2 1
% 40% 37.5%
As the strength of A is more, the answer should be closer to 40 than 37.5, again only D possible.
\(40-(40-37.5)*\frac{1}{1+2}=40-0.8=39.2%\)
D

3) Substitution ..
Let strength of B be 40, so boys = \(40*\frac{3}{8}=15\).
Strength of A is at least 2*40=80, so boys = \(80*\frac{2}{5}=32\).
Total boys = 15+32=47 out of 40+80=120.
As a % = \(\frac{47}{120}*100=39.2%\)
We could have taken strength as x and 2x and done it algebraically.
D

Choose what suits you.
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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to 21 May 2019, 06:00
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chetan2u
nick1816
Class A has boys to girls in the ratio 2 : 3, and Class B has girls to boys in the ratio 5 : 3. If the number of students in Class A is at least twice as many as the number of students in Class B, what is the minimum percentage of boys possible, when both classes are considered together?

A. 30%
B. 33.33%
C. 37.5%
D. 39.17%
E. 40%

You can do by weighted average method or algebraically or logically..

1) LOGICALLY - a 20 seconds answer
% of boys in A is \(\frac{2}{5}\)or 40% and
% of boys in B is \(\frac{3}{8}\) or 37.5%
So, our answer has to be between 37.5 and 40, irrespective of the strength of each class. As none of the class has a strength of 0, the extremes 37.5 and 40 are not possible.
ONLy E left

2) Weighted average method.
Class A B
strength 2 1
% 40% 37.5%
As the strength of A is more, the answer should be closer to 40 than 37.5, again only E possible.
\(40-(40-37.5)*\frac{1}{1+2}=40-0.8=39.2%\)
E

3) Substitution ..
Let strength of B be 40, so boys = \(40*\frac{3}{8}=15\).
Strength of A is at least 2*40=80, so boys = \(80*\frac{2}{5}=32\).
Total boys = 15+32=47 out of 40+80=120.
As a % = \(\frac{47}{120}*100=39.2%\)
We could have taken strength as x and 2x and done it algebraically.
E

Choose what suits you.
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Hi ,

If you solve (47 ×100)÷120 it comes out to be 39.166 which is equuivalent to 39.17 thus answer should be D

Posted from my mobile device
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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to 21 May 2019, 06:08
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nick1816
Class A has boys to girls in the ratio 2 : 3, and Class B has girls to boys in the ratio 5 : 3. If the number of students in Class A is at least twice as many as the number of students in Class B, what is the minimum percentage of boys possible, when both classes are considered together?

A. 30%
B. 33.33%
C. 37.5%
D. 39.17%
E. 40%

You can do by weighted average method or algebraically or logically..

1) LOGICALLY - a 20 seconds answer
% of boys in A is \(\frac{2}{5}\)or 40% and
% of boys in B is \(\frac{3}{8}\) or 37.5%
So, our answer has to be between 37.5 and 40, irrespective of the strength of each class. As none of the class has a strength of 0, the extremes 37.5 and 40 are not possible.
ONLy E left

2) Weighted average method.
Class A B
strength 2 1
% 40% 37.5%
As the strength of A is more, the answer should be closer to 40 than 37.5, again only E possible.
\(40-(40-37.5)*\frac{1}{1+2}=40-0.8=39.2%\)
E

3) Substitution ..
Let strength of B be 40, so boys = \(40*\frac{3}{8}=15\).
Strength of A is at least 2*40=80, so boys = \(80*\frac{2}{5}=32\).
Total boys = 15+32=47 out of 40+80=120.
As a % = \(\frac{47}{120}*100=39.2%\)
We could have taken strength as x and 2x and done it algebraically.
E

Choose what suits you.

Hi ,

If you solve (47 ×100)÷120 it comes out to be 39.166 which is equuivalent to 39.17 thus answer should be D

Posted from my mobile device
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All the answers are 39.2 only. So answer is D in all methods. E is a typo.
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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to 21 May 2019, 08:25
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nick1816
Class A has boys to girls in the ratio 2 : 3, and Class B has girls to boys in the ratio 5 : 3. If the number of students in Class A is at least twice as many as the number of students in Class B, what is the minimum percentage of boys possible, when both classes are considered together?

A. 30%
B. 33.33%so
C. 37.5%
D. 39.17%
E. 40%
Show more

let 3 boys plus 5 girls=8 minimum students in Class B
total students in Class A must be multiple of 5
least multiple of 5>2*8=20
so 8 boys plus 12 girls=20 minimum students in Class A
(3+8) total boys/(8+20) total students=11/28=39.17%
D
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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to 26 May 2019, 10:46
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nick1816
Class A has boys to girls in the ratio 2 : 3, and Class B has girls to boys in the ratio 5 : 3. If the number of students in Class A is at least twice as many as the number of students in Class B, what is the minimum percentage of boys possible, when both classes are considered together?

A. 30%
B. 33.33%
C. 37.5%
D. 39.17%
E. 40%
Show more



let class b total students = 40
so class a total students= 80
total = 120
so class a boys ; 2/5 * 80 = 32
and class b boys; 3/8*40 ; 15
total boys % ; 47/120 * 100 = 39.17
IMO D
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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to 06 Jan 2021, 00:04
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Given, Class A: B/G=2/3 (multiplier factor= x, say) and Class B: G/B=5/3 (multiplier factor= y, say)

Further, 5*x = 8*y *2 or, 5x= 16y

% of boys = (2*x +3*y) / (5*x +8*y) *100 = 9.4/24 *100 = 39.17%

So, I think D. :)
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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to 12 Jul 2022, 23:21
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In Class A, boys are 2/5 or 40% of the class, while in Class B, the percentage is 37.5%. So to minimise the percentage of boys, the ratio of the total number of students in A and B (A: B) should be the least possible.

Let us assume that the number of students in Class A is 80 and the number of students in Class B is 40.

The number of boys in Class A = (2/5)*80 = 32.
The number of boys in Class B = (3/8)*40 = 15

Thus, the total number of boys = 47.
Total number of students = 120

Thus, the minimum percentage of boys = (47/120)*100 = 39.17%.

Thus, the correct option is D.
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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to 12 Jul 2022, 23:39
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Given: Class A has boys to girls in the ratio 2 : 3, and Class B has girls to boys in the ratio 5 : 3.

Asked: If the number of students in Class A is at least twice as many as the number of students in Class B, what is the minimum percentage of boys possible, when both classes are considered together?

Percentage of boys in class A = 2/5 = 40%
Percentage of boys in class B = 3/8 = 37.5%

To minimize percentage of boys posssible, we should maximize students in class B.
Minimum percentage of boys in classes together = (2*40% + 37.5%)/3 = 117.5%/3 = 39.17%

IMO D
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Class A has boys to girls in the ratio 2 : 3, and Class B has girls to 12 Jun 2024, 01:02
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chetan2u

2) Weighted average method.
Class A B
strength 2 1
% 40% 37.5%
As the strength of A is more, the answer should be closer to 40 than 37.5, again only D possible.
\(40-(40-37.5)*\frac{1}{1+2}=40-0.8=39.2%\)
D

 
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­how did you \(40-(40-37.5)*\frac{1}{1+2}=40-0.8=39.2%\) get this equation could you eloborate i know weighted average method as well but i am finding it difficult like how did you get to this formula ?
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