# Moderate Bar Charts Solved QuestionData Interpretation Discussion

Common Information

There are ten boxes namely Box 1, Box 2, Box 3,….. Box 9 and Box 10 with Mr. Zero. Each of these ten boxes is colored with one out of the four colors namely Black, White, Yellow and Pink. The number of coins in each of these mentioned ten boxes is one of the five numbers 12, 15, 20, 25 and 30.The following bar – graphs provides information about the number of boxes that are colored Black, White, Yellow and Pink and also about the number of boxes that have different number of coins.

No two boxes that are colored with the same color have equal number of coins.

 Q. Common Information Question: 1/5 The number of boxes that have exactly 20 coins is:
 ✖ A. 4 ✖ B. 3 ✖ C. 2 ✔ D. 1

Solution:
Option(D) is correct

The number of boxes that are colored Pink, Yellow, White and Black is 1, 2, 2 and 5 respectively.
It is also given that no two boxes that are colored with the same color have equal number of coins.
It is also given that the number of coins in each of the ten boxes is 12, 15, 20, 25 or 30.

Also, since there are five boxes that are colored, the number of coins in the boxes are 12, 15, 20, 25 and 30.

Table below can be scrolled horizontally

Number of Coins Number of Boxes

12

1 - 3

15

1 - 3

20

1

25

3

30

2

So the number of boxes that have exactly 20 coins is 1

Edit: Thank you, Shashank Dave and Akhtar, updated image in the given question.

Edit 2: For a detailed solution explaining calculation, check comment by Deepak.

## (6) Comment(s)

Vejayanantham TR
()

In the image,

Atmost 15 coins - 4 boxes

Atmost 25 coins - 8 boxes

How it is possible, if already 4 boxes has atmost 15, then it is possible that only 6 boxes can contain atmost 25 ?

Lakshman
()

Can someone explain how coins for 25 and 30 are distributed?

Deepak
()

It's easy, we will go backwards (first no. of boxes containing 30 coins then 25 coins then 20 coins then 15 coins and lastly 12 coins).

Let's have some notation to refer bags with coins.

Bags containing 12 coins $=B_{12}$

similarly, bags containing 15, 20, 25 and 30 coins are respectively $=B_{15}, B_{20}, B_{25}$ and $B_{30}$

Let's start our calculations from the last bar of the second chart (the blue one).

Given that boxes containing 'at most 25 coins ' = 8

So, Total no. of boxes containing 30 coins = Total no. of Boxes - boxes containing at most 25 coins

$B_{30}$ $=(B_{12}+B_{15}+ B_{20}+B_{25}+B_{30})$ $-(B_{12}+B_{15}+ B_{20}+B_{25})$

$=10-8=2 \text{ boxes}$

Now,

Total no. of boxes containing 25 coins = (No. of boxes containing at most 25 coins + No. of boxes containing at least 25 coins) - Total no. of Boxes

$B_{25}$ $= [(B_{25} + B_{30})+(B_{12}+B_{15}+B_{20}+B_{25})]$ $-(B_{12}+B_{15}+ B_{20}+B_{25}+B_{30})$

$=[8+5]-10$

$=3 \text{ boxes}$

Now,

Total no. of boxes containing 20 coins = No. of boxes containing at least 20 coins - No. of boxes containing at least 25 coins

$B_{20}=(B_{20}+B_{25}+B_{30})- (B_{25}+B_{30})$

$=6-5$

$= 1 \text{ box}$

This accounts 6 of total 10 boxes.

Also, it is given in the blue chart that, no. of boxes containing at most 15 coins = 4

$\Rightarrow B_{12}+B_{15}=4$

If neither $B_{12}$ or $B_{15}$ is zero then they both can have 1 to 3 coins in such a way that $B_{12}+B_{15}=4$

Ram
()

Yes me too. Little confused about the calculation.

Deepak
()