Maths Class X Question Bank

Question 1 of 41

1. Question
1 point(s)
Find the mode of following frequency distribution :
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Hint

Modal class $=30-40$
Mode $=l+\left(\dfrac{f-f_1}{2f-f_1-f_2}\right)\times h$
$=30+\left(\dfrac{20-15}{40-15-14}\right)\times10\\ \\ =30+\dfrac{5}{11}\times10\\ \\ =30+4.54=34.54$
Question 2 of 41

2. Question
1 point(s)
Find the median of the following data :
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Hint

$\dfrac{N}{2}=\dfrac{50}{2}=25$
Median class $=120-140$
Median $=l+\left(\dfrac{\dfrac{N}{2}-cf}{f}\right)\times h$
$=120+\left(\dfrac{25-12}{14}\right)\times20=138.57$
Question 3 of 41

3. Question
1 point(s)
Find mode of the given data :
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Hint

Modal class $=40-50$
Mode $=l+\left(\dfrac{f-f_1}{2f-f_1-f_2}\right)\times h$
$=40+\left(\dfrac{27-11}{54-11-18}\right)\times10\\ \\ =40+\dfrac{16}{25}\times 10=46.4$
Question 4 of 41

4. Question
1 point(s)
The mean of the following frequency distribution is 57.6 and the sum of observations is 50. Find the missing frequencies $ f_1$ and $f_2 $.
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Hint

$x=\dfrac{\sum f_ix_i}{\sum f_i}\\ \\ \Rightarrow 57.6=\dfrac{2060+30f_1+70f_2}{50}\\ \\ \Rightarrow 2880=1940+30f_1+70f_2\\ \\ \Rightarrow 94=3f_1+7f_2$ ….(1)
And $32+f_1+f_2=50$
$f_1+f_2=18\Rightarrow f_1=18-f_2$
Solving both the equations
$f_1=8,\ f_2=10$
Question 5 of 41

5. Question
1 point(s)
Find the mode from the following frequency distribution :
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Hint

Modal class $=100-125$
Mode $=l+\left(\dfrac{f-f_1}{2f-f_1-f_2}\right)\times h$
$=100+\left(\dfrac{30-20}{60-20-25}\right)\times 25\\ \\ =100+\dfrac{10}{15}\times 25\\ \\ =100+16.67=116.67$
Question 6 of 41

6. Question
1 point(s)
Find the modal marks from the following table :
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Hint

Modal marks $=30-40$ as frequency is high on this class.
Question 7 of 41

7. Question
1 point(s)
Find the mode from the following data :
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Hint

Modal class $=90-100$
Mode $=l+\left(\dfrac{f-f_1}{2f-f_1-f_2}\right)\times h$
$=90+\left(\dfrac{11-7}{22-7-5}\right)\times10\\ \\ =90+\dfrac{4}{14}\times10=94$
Question 8 of 41

8. Question
1 point(s)
Find the missing frequencies $f_1 $ and $f_2 $ in the following frequency distribution, if it is known that the mean of the distribution is 50 and the total frequency is 150.
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Hint

$\sum f_1=150\\ \\ \Rightarrow 68+f_1+f_2=150\\ \\ \Rightarrow f_1+f_2=82$ …(1)
$\dfrac{\sum f_ix_i}{\sum f_i}=50\\ \\ \Rightarrow \dfrac{3480+30f_1+70f_2}{150}=750\\ \\ \Rightarrow 3f_1+7f_2=402$ ….(2)
Solving both the eqn we get
$f_2=39,\ f_1=43$
Question 9 of 41

9. Question
1 point(s)
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.
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Hint

Median $=l+\dfrac{\dfrac{N}{2}-cf}{f} \times h$
$=35+\dfrac{50-45}{33}\times5=35.76$
Question 10 of 41

10. Question
1 point(s)
Find the mean marks of the students from the following frequency distribution.
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Hint

$\bar x=a+h\left(\dfrac{\sum f_iu_i}{\sum f}\right)\\ \\ =55+\left(\dfrac{-56}{85}\right)\times10=48.41$
Question 11 of 41

11. Question
1 point(s)
Calculate the mode from the following data :
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Hint

Mode $=l+\left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$
$=5000+\dfrac{60}{110}\times 5000=7727.27$
Question 12 of 41

12. Question
1 point(s)
Compute the median for the following data :
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Hint

Median $=l+\left(\dfrac{\dfrac{N}{2}-cf}{f}\right)\times$
$=110+\left(\dfrac{75-45}{45}\right)\times 10=116.67$
Question 13 of 41

13. Question
1 point(s)
The median value for the following frequency distribution is 35 and the sum of all the frequencies is 170. Find the values of x and y.
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Hint

Median $=l+\left(\dfrac{\dfrac{n}{2}-cf}{f}\right)\times h$
$\Rightarrow 35=30+\left(\dfrac{85-30-f_1}{40}\right)\times 1\\ \\ \Rightarrow f_1=35\Rightarrow x=35$
Also $110+f_1+f_2=170\Rightarrow f_2=60-35=23\\ \\ \Rightarrow y=23$
Question 14 of 41

14. Question
1 point(s)
Find the mode of the marks obtained by 80 students in a class test in mathematics as given below :
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Hint

Modal class $=50-60$
Mode $=l+\left(\dfrac{f-f_1}{2f-f_1-f_2}\right)\times h$
$=50+\left(\dfrac{20+13}{40-13-6}\right)\times 10\\ \\ =50+\dfrac{7}{21}\times10\\ \\ =50+3.33=53.33$
Question 15 of 41

15. Question
1 point(s)
The following table shows the ages of the patients admitted in a hospital during a year.
Find the mode and mean of the data given above. Compare and interpret the two measures of central tendency.
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Hint

$\bar x=a+\left(\dfrac{\sum f_iu_i}{\sum f_i}\right)\times h$
$=30+\dfrac{430}{80}=35.38$
Modal class $=35-45$
Mode $=35+\left(\dfrac{23-21}{46-21-14}\right)\times 10$
$=35+\dfrac{20}{11}=36.8$
Question 16 of 41

16. Question
1 point(s)
The mode of the following distribution is 65.625 hours. Find the value of p.
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Hint

$65.625=l+\dfrac{f-f_1}{2f-f_1-f_2}\\ \\ \Rightarrow 65.625=60+\dfrac{61-52}{122-52-P}\times 20\\ \\ \Rightarrow 65.625=60+\dfrac{9}{70-P}\times20\\ \\ \Rightarrow 65.625=60+\dfrac{180}{70-P}\\ \\ \Rightarrow P=38$
Question 17 of 41

17. Question
1 point(s)
Find the median for the following data :
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Hint

$\dfrac{N}{2}=100$
Median $=50+10\times \left(\dfrac{100-92}{24}\right)$
$=50+3.33=53.33$
Question 18 of 41

18. Question
1 point(s)
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained.
Find the median height.
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Hint

$N=51\ \Rightarrow \dfrac{N}{2}=25.5$
Median $=145+\dfrac{25.5-11\times5}{18}$
$=145+4.03=149.03$
Question 19 of 41

19. Question
1 point(s)
The following table gives the distribution of the lifetime of neon lamps. Find the median lifetime of a lamp.
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Hint

$N=400\ \Rightarrow \dfrac{N}{2}=200$
Median $=3000+\dfrac{200-130}{86}\times500=3406.98$
Question 20 of 41

20. Question
1 point(s)
The annual profits earned by 30 shops in a market is represented in following table.
Draw both the ogives for the above data and hence obtain the median.
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Hint

Now, we mark on x-axis lower class limits, y-axis cumulative frequency
Thus, we plot the points (5,30),(10,28),(15,16),(20,14),(25,10),(30,7) and(35,3)
Now we mark the upper class limits along x-axis and cumulative frequency along y-axis.
Thus we plot the points (10,2),(15,14),(20,16),(25,20),(30,23),(35,27),(40,30)
We find that the two types of curves intersect of P from point L it is drawn on x-axis. The value of a profit corresponding to M is 17.5. Hence median is 17.5 lakh
Question 21 of 41

21. Question
1 point(s)
From the following data, draw the two types of cumulative frequency curves and determine the median.
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Hint
Question 22 of 41

22. Question
1 point(s)
Convert the following data to a less than type distribution and draw its ogive. Also find the median from the graph.
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Hint

Median $=138.5$ (approx)
Question 23 of 41

23. Question
1 point(s)
Change the given distribution to more than type distribution and draw its ogive.
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Question 24 of 41

24. Question
1 point(s)
Draw more than ogive for the following distributions. Find the median from the curve.
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Hint

$\dfrac{N}{2}=\dfrac{100}{2}=50$
Median $=25$
Question 25 of 41

25. Question
1 point(s)
Draw ‘less than’ ogive and ‘more than’ ogive for the following distribution and hence find its median.
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Hint
Question 26 of 41

26. Question
1 point(s)
If the median of the following data is 525. Find the values of x and y if the sum of the frequencies is 100.
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Hint

$76+x+y=100\\ \\ \Rightarrow x+y=24$
Median $=l+\dfrac{\dfrac{N}{2}cf}{f}\times h\\ \\ \Rightarrow 525=500+\left(\dfrac{50-36-x}{20}\right)\times 100\\ \\ \Rightarrow 25=70-5x\\ \\ \Rightarrow x=9\\ \\ x+y=24\Rightarrow y=15$
Question 27 of 41

27. Question
1 point(s)
Calculate the mode of the following frequency distribution table.
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Hint

Mode $=45+\left(\dfrac{20-10}{40-10-9}\right)\times 10$
$=45+4.7=49.7$
Question 28 of 41

28. Question
1 point(s)
Find the median wage of a worker engaged at a construction site whose data are given below :
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Hint

Total number of workers = 100
Median wage $=\left(\dfrac{n}{2}\right)^{th}$
$=\dfrac{100}{2}=50$
4100 is the median
Question 29 of 41

29. Question
1 point(s)
Change the following data into less than type distribution and draw its ogive. Hence find the median of the data.
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Hint
Question 30 of 41

30. Question
1 point(s)
Draw less than and more than ogive for the following distribution and hence find its median.
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Question 31 of 41

31. Question
1 point(s)
The following distribution gives the annual profit earned by 30 shops of a shopping complex.
Change the above distribution to more than type distribution and draw its ogive.
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Hint

Now, we mark on x-axis lower class limits, y-axis cumulative frequency
Thus, we plot the points (5,30),(10,28),(15,16),(20,14),(25,10),(30,7) and(35,3)
Now we mark the upper class limits along x-axis and cumulative frequency along y-axis.
Thus we plot the points (10,2),(15,14),(20,16),(25,20),(30,23),(35,27),(40,30)
We find that the two types of curves intersect of P from point L it is drawn on x-axis. The value of a profit corresponding to M is 17.5. Hence median is 17.5 lakh
Question 32 of 41

32. Question
1 point(s)
Following distribution shows the marks obtained by the class of 100 students. Draw less than ogive for the above data. Find median graphically and verify the result by actual method.
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Hint

$N=100,\dfrac{N}{2},\dfrac{N}{2}=50$
From the graph median is 39 (approx)
Question 33 of 41

33. Question
1 point(s)
Draw both types of cumulative frequency curve on the same graph paper and then determine the median.
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Hint

Median $=80+10\dfrac{\left(\dfrac{35}{2}-17\right)}{12}=80.42$
Question 34 of 41

34. Question
1 point(s)
If the median of the distribution given below is 28.5, find the values of x and y.
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Hint

Median $=l+\left(\dfrac{\dfrac{n}{2}-cf}{f}\right)\times h$
$\Rightarrow 28.5=20+\left(\dfrac{30-5-x}{20}\right)\times 10\\ \\ \Rightarrow 8.5\times 2=25-x\\ \\ \Rightarrow x=8\\ \\ 45+x+y=60\Rightarrow y=60-45-x\\ \\ y=7$
Question 35 of 41

35. Question
1 point(s)
The mean of the following data is 50. Find the missing frequencies $f_1 $ and $f_2 $.
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Hint

A/Q $68+f_1+f_2=120$
$\Rightarrow f_1+f_2=52$ ….(1)
Mean $=\dfrac{\sum f_ix_i}{\sum f_i}$
$\Rightarrow 50=\dfrac{3480+30f_i+70f_2}{120}\\ \\ \Rightarrow 3f_1+7f_2=252$ ….(2)
Solving two equation we get
$f_1=28,\ f_2=24$
Question 36 of 41

36. Question
1 point(s)
The following table gives the distribution of expenditures of different families on education. Find the mean expenditure on Leducation of a family.
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Hint

Mean $=275+\left(\dfrac{-35}{200}\right)\times50=266.25$
Question 37 of 41

37. Question
1 point(s)
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows :

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
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Hint

$\dfrac{n}{2}=\dfrac{100}{2}=50$
Median $=l+\left(\dfrac{\dfrac{n}{2}-cf}{f}\right)\times h$
$=7+\left(\dfrac{50-36}{40}\right)\times3\\ \\ =7+\dfrac{14}{40}\times3=8.05$
Mean $=\dfrac{832}{100}=8.32$
Mode $=7+\left(\dfrac{40-30}{80-30-16}\right)\times3$
$=7+\dfrac{10}{34}\times3\\ \\ =7+\dfrac{30}{34}=7.88$
Question 38 of 41

38. Question
1 point(s)
Find the mean, mode and median of the following data :
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Hint

Mean $=\dfrac{3560}{100}=35.60$
Median $=30+\left(\dfrac{\dfrac{100}{2}-33}{30}\right)\times10$
$=30+5.667\\ \\ =35.667$
Mode $=30+\dfrac{10\times12}{22}=35.4545$
Question 39 of 41

39. Question
1 point(s)
The following table gives the daily income of 50 workers of a factory : Find the mean, median and mode of the above data.
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Hint

Mean $=\dfrac{7260}{50}=145.2$
Median $=120+\dfrac{25-12}{14}\times20$
$=120+18.6=138.6$
Modal class $=120-140$
Mode $=120+\dfrac{14-12}{2\times14-12.8}\times20$
$=120+\dfrac{40}{8}=125$
Question 40 of 41

40. Question
1 point(s)
The median of the following data is 52.5. Find the values of x and y if the total frequency is 100.
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Hint

A/Q $76+x+y=100$
$\Rightarrow x+y=24$ ….(1)
Median $=l+\dfrac{\dfrac{n}{2}-cf}{f}\times h$
$\Rightarrow 52.5=50+\dfrac{50-36-x}{20}\times100\\ \\ \Rightarrow 52.5-50=(14-x)\times 0.5\\ \\ \Rightarrow x=9\\ \\ y=24-9=15$
Question 41 of 41

41. Question
1 point(s)
Find the mode, median and mean for the following data :
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Hint

Mean $=a+\dfrac{\sum f_id_i}{\sum f_i}$
$=60+\left(\dfrac{-1030}{100}\right)=49.7$
Modal class $=45-55$
Mode $=45+\dfrac{2}{18}\times10=46.1$
3 Median = Mode + 2 Mean
$\Rightarrow$ Median $=\dfrac{46.1+99.44}{3}=48.5$