**Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics”.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB ,KVS,STET Exam.**

**Q1. Calculate the mode of a series the mean and median values of which are 16 cm and 20 cm respectively.**

(a) 32 cm

(b) 56 cm

(c) 28 cm

(d) 22 cm

**Q2. The median and mean values of the marks obtained by the students of a class are 46.67 and 45.5 respectively. Find out the mode of the marks.**

(a) 60.00

(b) 58.58

(c) 49.01

(d) 40

**Q3. The median and mean weight of the students of a class are 35.83 kg and 37.06 kg respectively. Calculate the mode.**

(a) 33.37 kg

(b) 26.58 kg

(c) 43.22 kg

(d) 35 kg

**Q5. If the median and mean of a distribution are 18.8 and 20.2 respectively, what will be its mode?**

(a) 24.4

(b) 12.9

(c) 16.0

(d) 14.45

**Q6. If in a symmetrical distribution, the median is 280 and mean is 310, what will be the mode?**

(a) 265

(b) 220

(c) 156

(d) 245

**Q7. The mean of the median, mode and large of the observations 6, 6, 9, 14, 8, 9, 9, 8 is**

(a) 8.5

(b) 8.8

(c) 10.3

(d) 10.5

**Q8. When the mode is 400 and median is 550, the mean will be**

(a) 625

(b) 540

(c) 442

(d) 225

**Q9. When the mode is 90 and mean is 120, the median will be**

(a) 98

(b) 105

(c) 110

(d) 135

**Q10. Calculate the median of an asymmetrical distribution if the mode is 78 and arithmetic mean is 87.**

(a) 84

(b) 56

(c) 94

(d) 96

Solutions

S1. Ans.(c)

Sol. Mode = 3 Median – 2 Mena

Here, median = 20 cm, mean = 16 cm

Mode (Z) = 3 × 20 – 2 × 16

= 60 – 32 = 28

∴ Z = 28 cm

S2. Ans.(c)

Sol. Mode = 3 Median – 2 Mean

Here, median = 46.67, mean = 45.5

Mode (Z) = 3 × 46.67 – 2 × 45.5

= 140.01 – 91 = 49.01

∴ Z = 49.01

S3. Ans.(a)

Sol. Mode = 3 Median – 2 Mean

Here, median = 35.83, mean = 37.06

Mode (Z) = 3 × 35.83 – 2 × 37.06

∴ Z = 33.37 kg

S4. Ans.(c)

Sol. Sales in January = Rs. 3.75 million

Sales in February = Rs. 3.5 million

Sales in March = Rs. 4.25 million

Sales in April = Rs. 4.75 million

Sales in May = Rs. 3.75 million

Sales in June = Rs. 4.25 million

∴ From the given data, it can be observed that the change in sales is the highest from April to May.

S5. Ans.(c)

Sol. Here, median = 18.8, mean = 20.2

Mode (Z) = 3 × 18.8 – 2 × 20.2

= 56.4 – 40.4 = 16

∴ Z = 16

S6. Ans.(b)

Sol. Mode = 3 Median – 2 Mean

Here, mode (Z) = ?, mean = 310,

M = 280

Z = 3 × 280 – 2 × 310

Z = 840 – 620

Z = 220

S7. Ans.(a)

Sol. Arrange the given values in ascending order as:

6, 6, 8, 8, 9, 9, 9, 14

As the given sequence has an even number of values, the median is the average of two middle values.

Here, the two middle values are 8 and 9.

∴ Median is (8 + 9)/2=17/2 = 8.5

The mode is the value that appears the maximum number of times, and here, this value is 9.

Therefore, mode = 9

Range = Maximum value – Minimum value

= 14 – 6 = 8

Required mean = (8.5 + 9 + 8)/3=25.5/3=8.5

S8. Ans.(a)

Sol. Mode = 3 Median – 2 Mean

Here, mode (Z) = 400, Median = 550

400 = 3 × 550 – 2 × Mean

2 Mean = 1650 – 400

∴ Mean = 1250/2 = 625

S9. Ans.(c)

Sol.Mode = 3 Median – 2 Mean

Here, mode (Z) = 90, mean = 120

∴ 90 = 3 × Median – 2 × 120

⇒ 3 × Median = 90 + 2 × 120

⇒ Median = 330/3 = 110

S10. Ans.(a)

Sol. Mode = 3 Median – 2 Mean

Here, mode (Z) = 78, M = ?, mean = 87

∴ 78 = 3 Median – 2 × 87

3 Median = 78 + 174

Median = 252/3 = 84