Arithmetic

Central Tendency

Medium

The average of marks of 15 papers is 45. The average of marks of the first 8 papers is 48 and that of the last 8 papers is 40. The marks of the eight paper will be:

✅ Correct Option: 1

The total papers = 15

Average marks of all 15 papers = 45

Average marks of first 8 papers (papers 1 to 8) = 48

Average marks of last 8 papers (papers 8 to 15) = 40

Total marks of all 15 papers:

$\text{Total} = 45 \times 15$

$\text{Total} = 675$

Total marks of first 8 papers:

$\text{Total} = 48 \times 8$

$\text{Total} = 384$

Total marks of last 8 papers:

$\text{Total} = 40 \times 8$

$\text{Total} = 320$

The 8th paper appears in both the first 8 papers and the last 8 papers.

Adding the totals of first 8 and last 8 papers:

$384 + 320 = 704$

The difference between this sum and the actual total:

$704 - 675 = 29$

This difference represents the marks of the 8th paper, which was counted twice.

Therefore, the marks of the 8th paper = 29

2025: 28 May Shift 2

In the first 9 matches of a cricket tournament, the wickets taken by Jumpo are as follows:

3 2 0 1 3 0 5 1 4

Find the median of the number of wickets taken by Jumpo?

2023: 28 May Shift 1

Thirty six students of class IX contributed on an average Rs 42 towards flood relief fund whereas some students of class X contributed on an average Rs 54.50. If the overall average contribution per student remained Rs 47, then how many students of class X contributed ?

2026: 19 May Shift 1

The median of class frequency distribution is:

Class:	0 - 25	25 - 50	50 - 75	75 - 100	Total
Frequency:	25	32	15	16	88

2025: 3 June Shift 1

Dittee has 10 purses and each of which having an average of 25 lipsticks in each purse. Further, it is known that each purse has a minimum of 8 lipsticks and none of the purse has identical number of lipsticks.

Determine that what can be the maximum number of lipsticks that Dittee shall have in any of her purse?

2026: 25 May Shift 2

The median of numbers 14, 7, 18, 21 and 27 is:

2023: 1 June Shift 1

If mean of the numbers 2, $(2p + 2)$ , 7, 13, 17, 4 and $(p - 1)$ is 8, then find their median.

The average of marks of 15 papers is 45. The average of marks of the first 8 papers is 48 and that of the last 8 papers is 40. The marks of the eight paper will be:

Solution

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