Sun	Mon	Tue	Wed	Thu	Fri	Sat
27	28	29	30	1	2	3
4	5	6	7	8	9	10
11	12	13	14	15	16	17
18	19	20	21	22	23	24
25	26	27	28	29	30	31
1	2	3	4	5	6	7

Back to Main Page

Multiple Choice Questions

The number of 39 Gold 39 coins in bag 39 A 39 is what percent more less than the number of 39 copper...........

Quantitative Aptitude(IBPS RRB PO) Case Studies

Directions: Answer the questions based on the information given below.

There are three types of coins i.e. 'Gold', 'Silver', and 'Copper' in each of three bags A, B, and C. The total number of coins in all the bags is 1510. The number of coins in bag 'A' is 580 and the total number of coins in bag 'B' is 510. The ratio between the number of copper coins in bag 'A', the number of copper coins in bag 'B', and the number of 'Gold' coins in bag 'C' is 3 : 2 : 4 respectively. The number of silver coins in bag 'B' is 60 more than the number of copper coins in bag 'B'. The number of 'Gold' coins in bag 'B' is 25% less than the number of silver coins in bag 'A'. The total number of 'gold' coins in all three bags is 570. The ratio of the total number of copper coins in all three bags and the total number of silver coins in all three bags is 20 : 27.

The number of 'Gold' coins in bag 'A' is what percent more/less than the number of 'copper' coins in bag 'C'?

This questions was previously asked in

RRB Officer Scale-I Mains (1st Oct 2022)

View All RRBA Previous Year Papers

10%

20%

25%

40%

50%

छोटे Courses बड़े Results

Master Case Studies

Prepare through Micro-courses

Try Now

Explanation:

Let the total number of copper coins in all three bags = 20k And, the total number of silver coins in all three bags = 27k The total number of 'gold' coins in all three bags = 570
The total number of coins in all three bags = 1510
Therefore, 570 + 20k + 27k = 1510
⇒ 47k = 940
⇒ k = 20
So, the total number of copper coins in all three bags = 20k = 400
And, the total number of silver coins in all three bags = 27k = 540
Let the number of copper coins in bag 'A' = 3x
The number of copper coins in bag 'B' = 2x
Therefore, the number of copper coins in bag 'C' = 400 - 5x
Let the number of silver coins in bag 'A' = 4y
The number of gold coins in bag 'B' = 4yx (3/4) = 3y
The number of silver coins in bag 'B' = 2x + 60
The number of silver coins in bag 'C' = 540-(2x+60 + 3y) = 480 - 2x-4y
The number of gold coins in bag 'C' = 4x
The number of gold coins in bag 'A' = 570-(4x + 3y) = 570-4x-3y
The number of coins in bag 'A' = 580
Therefore, 3x + 4y + 570 - 4x-3y = 580
⇒y - x = 10 ...(a)
The number of coins in bag 'B' = 510
Therefore, 2x + 3y + 2x + 60 = 510
=4x+3y=450...(b)
Using (a) and (b)
4x (y-10) + 3y = 450
⇒ 7y = 490
⇒ y = 70
x = y-10 = 60
Putting all the values in a table we get