Board Previous Year Questions (Board PYQs)

Board Previous Year Questions

PYQ 1 (1 mark): The mean of a grouped frequency distribution is found by $\bar{x}=\dfrac{\sum f_ix_i}{?}$. The denominator is: (A) $\sum x_i$ (B) $\sum f_i$ (C) $n^2$ (D) $\sum f_ix_i$. [CBSE] Solution: $\sum f_i$. Answer: (B).

PYQ 2 (1 mark): For a symmetrical distribution, mean $=$ median $=$ mode. The empirical relation $3\,\text{Median}=\text{Mode}+2\,\text{Mean}$ then gives mode $=$: (A) mean (B) $2\times$ mean (C) $3\times$ mean (D) 0. [CBSE] Solution: If all equal to $m$: $3m=m+2m$ ✓, mode $=$ mean. Answer: (A).

PYQ 3 (2 marks): Find the mean: $0\text{-}10,10\text{-}20,20\text{-}30,30\text{-}40,40\text{-}50$ with frequencies $8,12,10,6,4$. [CBSE] Solution: marks $5,15,25,35,45$; $\sum f_ix_i=860$; $\sum f=40$; mean $=21.5$. Answer: 21.5.

PYQ 4 (3 marks): Find the mode: $0\text{-}20,20\text{-}40,40\text{-}60,60\text{-}80,80\text{-}100$ with frequencies $10,35,52,61,38$. [CBSE] Solution: modal class $60\text{-}80$ ($f_1=61$); $l=60,f_0=52,f_2=38,h=20$; mode $=60+\dfrac{61-52}{122-52-38}\times20=60+\dfrac{9}{32}\times20=60+5.625=65.625$. Answer: ~65.6.

PYQ 5 (3 marks): Find the median: $0\text{-}10,10\text{-}20,20\text{-}30,30\text{-}40,40\text{-}50$ with frequencies $5,8,12,10,5$. [CBSE] Solution: cf $5,13,25,35,40$; $\tfrac n2=20$; median class $20\text{-}30$; $l=20,cf=13,f=12,h=10$; median $=20+\dfrac{20-13}{12}\times10=25.83$. Answer: ~25.83.

PYQ 6 (3 marks): The mean of the following is 18. Find $f$: $11\text{-}13,13\text{-}15,15\text{-}17,17\text{-}19,19\text{-}21$ (marks $12,14,16,18,20$) with frequencies $3,6,9,13,f$. [CBSE] Solution: $\dfrac{498+20f}{31+f}=18\Rightarrow f=30$. Answer: 30.

PYQ 7 (3 marks): Estimate the mode from the empirical relation given mean $=45$ and median $=48$. [CBSE] Solution: mode $=3(48)-2(45)=54$. Answer: 54.

PYQ 8 (3 marks): The median of the data is 525 with $n=100$. Find $x$ and $y$: classes $0\text{-}100,\dots,900\text{-}1000$ with frequencies $2,5,x,12,17,20,y,9,7,4$. Median class $500\text{-}600$. [CBSE] Solution: $\sum f=100\Rightarrow x+y+76=100\Rightarrow x+y=24$. cf up to $400\text{-}500=2+5+x+12+17=36+x$. Median: $525=500+\dfrac{50-(36+x)}{20}\times100\Rightarrow25=\dfrac{14-x}{20}\times100\Rightarrow25=5(14-x)\Rightarrow5=14-x\Rightarrow x=9$, $y=15$. Answer: $x=9,y=15$.

PYQ 9 (2 marks): Find the median class: frequencies $4,4,8,10,12,8,4$ ($n=50$). [CBSE] Solution: cf $4,8,16,26,38,46,50$; $\tfrac n2=25$; median class is the 4th (cf 26). Answer: 4th class.

PYQ 10 (3 marks): Find the mean by step-deviation: $0\text{-}20,20\text{-}40,40\text{-}60,60\text{-}80,80\text{-}100$ (marks $10,30,50,70,90$) with frequencies $7,10,15,8,10$; $a=50,h=20$. [CBSE] Solution: $u_i=-2,-1,0,1,2$; $f_iu_i=-14,-10,0,8,20=4$; $\sum f=50$; mean $=50+20\times\dfrac{4}{50}=50+1.6=51.6$. Answer: 51.6.

PYQ 11 (3 marks): For frequencies $5,10,20,7,8$ (classes $0\text{-}10,\dots,40\text{-}50$), find the median. [CBSE] Solution: cf $5,15,35,42,50$; $\tfrac n2=25$; median class $20\text{-}30$; median $=20+\dfrac{25-15}{20}\times10=25$. Answer: 25.

PYQ 12 (4 marks): For the same data as PYQ 11, find the mean and mode, and verify the empirical relation approximately. [CBSE] Solution: mean $=25.6$, mode $=24.35$; check $3(25)=75$ vs mode$+2$mean$=24.35+51.2=75.55$ — approximately equal. Answer: mean 25.6, mode ~24.35.

PYQ 13 (1 mark): In a less-than ogive, cumulative frequency is plotted against the: (A) lower limits (B) upper limits (C) class marks (D) frequencies. [CBSE] Solution: upper limits. Answer: (B).

PYQ 14 (2 marks): The x-coordinate of the point where the less-than and more-than ogives intersect gives the: (A) mean (B) mode (C) median (D) range. [CBSE] Solution: median. Answer: (C).

PYQ 15 (3 marks): Convert to a frequency distribution and find the median: 'less than' cf — below 10 → 3, below 20 → 12, below 30 → 27, below 40 → 57, below 50 → 75, below 60 → 80. [CBSE] Solution: frequencies $3,9,15,30,18,5$; $n=80$, $\tfrac n2=40$; median class $30\text{-}40$ (cf 57); $l=30,cf=27,f=30,h=10$; median $=30+\dfrac{40-27}{30}\times10=30+4.33=34.33$. Answer: ~34.33.

PYQ 16 (3 marks): Find the mode: $10\text{-}25,25\text{-}40,40\text{-}55,55\text{-}70,70\text{-}85,85\text{-}100$ with frequencies $2,3,7,6,6,6$. [CBSE] Solution: modal class $40\text{-}55$; mode $=40+\dfrac{7-3}{14-3-6}\times15=40+12=52$. Answer: 52.

PYQ 17 (3 marks): Given mode $=65$ and mean $=61.4$, estimate the median. [CBSE] Solution: median $=\dfrac{65+2(61.4)}{3}=\dfrac{65+122.8}{3}=\dfrac{187.8}{3}=62.6$. Answer: 62.6.

PYQ 18 (4 marks): The distribution below has mean 62.8 and total frequency 50. Find the missing frequencies $f_1,f_2$: classes $0\text{-}20,\dots,100\text{-}120$ (marks $10,30,50,70,90,110$) with frequencies $5,f_1,10,f_2,7,8$. [CBSE] Solution: $\sum f=50\Rightarrow f_1+f_2=20$. $\sum f_ix_i=50+30f_1+500+70f_2+630+880=2060+30f_1+70f_2$; mean $62.8\Rightarrow 2060+30f_1+70f_2=3140\Rightarrow30f_1+70f_2=1080\Rightarrow3f_1+7f_2=108$. With $f_1+f_2=20$: $3f_1+7f_2=108$ and $3f_1+3f_2=60\Rightarrow4f_2=48\Rightarrow f_2=12$, $f_1=8$. Answer: $f_1=8,f_2=12$.