S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15 B

  Exercise 15 B

Question 1 

Ans: (i) Given,

r= 7cm, 'h=8cm

∴ Volume = $\pi r^{2} h$

$=\frac{22}{7} \times 7 \times 7+8$

$=22 \times 56$

$=1232 \mathrm{~cm}^{3}$

(ii) Given,

$\begin{aligned}&r=7 \mathrm{~cm} \\&h=12 \mathrm{~cm} .\end{aligned}$

$\therefore$ Volume $=\pi r^{2} h$

$\begin{aligned} &=\frac{22}{7} \times 7 \times 7 \times 12 \\ &=22 \times 84 \\ &=1848 \mathrm{~cm}^{3} \end{aligned}$

(iii) Given,

$r=14 \mathrm{~cm}$

$\mathrm{h}=16 \mathrm{~cm} .$

$\therefore \quad$ Volume $=\pi r^{2} h$

$-\frac{22}{7} \times 14 \times 14 \times 6$

$=44 \times 224$

$=9856 \mathrm{~cm}^{3}$

(iv) Given, 

r= 21cm,

h=40cm

Volume = $\pi r^{2} h$,

$=\frac{22}{7} \times 21 \times 21 \times 40$

$=66 \times 840$

$=\quad 55440 \mathrm{~cm}^{3}$

Question 2

Ans: (a) Given, 

Volume, = $44 \mathrm{Cm}^{3}$

Height $=3.5 \mathrm{~cm}$

ஃ Volume = $\pi r^{2}h$

$44=\frac{22}{7}\times r^{2}\times \frac{3.5}{10}$

$ 44=11 r^{2}$

$ \frac{44}{11}=r^{2}$

$4=r^{2}$

$r^{2}=4 .$

$r=\sqrt{4} .$

$r=2 \mathrm{cm} $

∴ Diameter = 2r 

$=2 \times 2$

$=4 c \mathrm{~m}$

(b) Given, 

Volume , = $385 \mathrm{~cm}^{3}$

Height $=1 \mathrm{dm}=10 \mathrm{~cm}$

$\therefore \quad$ Volume $=\pi r^{2} h$

$385=\frac{22}{7} \times r^{2} \times 10$

$\frac{49}{4}=r^{2}$

$r^{2}=\frac{49}{4}$

$r=\sqrt{\frac{49}{4}}$

$r=\frac{7}{2}$

∴ Diameter = 2r = $2 \times \frac{7}{2}=7 \mathrm{~cm}$

 

Question 3

Ans: (a) Given,

Volume = $66 \mathrm{~cm}^{3}$

Radius = 2cm

∴ Volume = π$(r^{2}$h

$\frac{66 \times 7}{22 \times 4}=h$

$\frac{21}{4}=h$

$h=\frac{21}{4}$

$h=5.25 \mathrm{~cm}$

(b) Given,

Volume = 4litres= 4000 $\mathrm{Cm}^{3}$

Radius= 5cm

$\therefore \quad$ Volume $=\pi r^{2} h$

$4000=\frac{22}{7} \times 5 \times 5 \times h$

$\frac{4000 \times 7}{22 \times 25}=h$

$\frac{80 \times 7}{11}$

$\frac{560}{11}=h$

$h=\frac{510}{11} \mathrm{~cm}$

Question 4

Ans: Given,

Height(h) = 7m

Radius(r) = $\frac{20}{2}=100 \mathrm{~cm}$ $=\frac{10}{100}$ $=\frac{1}{10}m$

$\therefore$ Volume $=\pi r^{2} h$

$=\frac{11}{50}$

According to question,

Total weight $=\frac{11}{50} \times 225$

$=\frac{99}{2}$

$=49.5 \mathrm{~kg}$

Question 5

Ans: Giver,

Internal radius. $(r)=3 \mathrm{~cm}$

Thickness of pipe= $1 \mathrm{~cm}$

$\therefore$ Outer radius $(R)=3+1$=4cm

Length = 6cm

$\begin{aligned} \therefore \quad \text { Volume } &=\pi R^{2} h-\pi r^{2} h . \\ &=\pi h\left(R^{2}-r^{2}\right) . \\ &=\frac{22}{7} \times 6\left((4)^{2}-(3)^{2}\right) \end{aligned}$

$=\frac{132}{7}(16-9)$

$=\frac{132}{7} \times 7 $

$=132 \mathrm{cm}^{3}$

Question 6

Ans: Given,

Sum of radius, of the base and the height of a cylinder (h+r) = 37cm,

Total surface area = $1628 \mathrm{~cm}^{2}$

∴ Total surface area = $2πr h+2πr^{2}$

$1628=2 \pi r(h+r)$

$1628=$ $2 \times \frac{22}{7} \times r \times 37 .$

r=7

$\therefore r=7 \mathrm{~cm} .$

So, $h+r=37$

$h+7=37$.

$h=377$

$h=30 \mathrm{~cm}$.

$\begin{aligned} \text { Volume } &=\pi r^{2} h \\ &=\frac{22}{7} \times 7 \times 7 \times 30 \\ &=22 \times 210 \\ &=4620 \mathrm{~cm}^{3} \end{aligned}$

Question 7

Ans: Given

Capacity of a cylindrical tank = $6160 \mathrm{m}^{3}$

Radius $(r)  \frac{28}{2}=14 \mathrm{~m}$

$\therefore$ Volume $=\pi r^{2} h$

$6160=\frac{22}{7} \times 14 \times 14 \times h$

$\frac{6160}{22 \times 281}=h $

h= 10m

Area of curved surface of taken inner sides = $2 \pi r h$

$=2 \times \frac{22}{7} \times 14 \times 10$

$=44 \times 20$

$=880 \mathrm{m}^{2}$

Question 8

Ans: Given,

Curved surface area of cylinder = $4400 \mathrm{~cm}^{2}$

Circumference $=110 \mathrm{~cm} $

Circumference $=2 \pi r$

$110=2 \times \frac{22}{7} \times r$

$\frac{110 \times 7}{2 \times 22}=r$

$\frac{35}{2}$=r

$r=\frac{35}{2} \mathrm{~cm}$

(i) Curved Surface area $=2$ πrh

4400= $=\frac{2}\times \frac{22}{7} \times \frac{35}{2} \times h$

$\frac{4400}{22 \times 5}=h$

$40=h$

$h=40$ cm

hence the height of cylinder is 40cm

(ii) Volume = $\pi r^{2} h$

$=110 \times 350$

$=38500 \mathrm{~cm}^{3}$

Question 9

Ans: (i) Given ,

Height of the wall = 20 meter 

Radius = $\frac{2}{2}=1$

$\because$ Volume =  πr^{2}$

$\begin{aligned} &=\frac{22}{7} \times 1 \times 1 \times 20 \\=& \frac{440}{7} \\=& 62 \frac{6}{7} \mathrm{~m}^{3} . \end{aligned}$

(ii) Curved surface area = $2 π r h$

$\begin{aligned} &=2 \times \frac{22}{7} \times 1 \times 20 \\=& \frac{44 \times 20}{7} \\=& \frac{880}{7} \end{aligned}$

Rate of plastering the inner surface = Rs 5  per m².

Total cost = 880\7 x 5

$=\frac{4400}{7}$

$=7628.57 $

Question 10

Ans: Giver

Radius of cylinder $=\frac{20}{2}=10 \mathrm{~cm}

Curved surface area = $1000 \mathrm{~cm}^{2}$

(i) Curved surface area $=2 \pi r$

$1000=2 \times \frac{22}{7} \times 10 \times h$

$\frac{175}{11}=h$

$h=\frac{175}{11}$

$h=15.9 \mathrm{~cm} .$

(ii) $\begin{aligned} \text { Volume } &=\pi r^{2}h \\ &=3.14 \times 10 \times 10 \times 15-9 . \\ &=31.4 \times 159 . \\ &=4992.6 \mathrm{~cm}^{3}\end{aligned}$

Question 11

Ans: Given, 

Radius =  $\frac{35}{2} \mathrm{~cm}$

height $=1.2 \mathrm{~m}=1.2 \times 100 \mathrm{~cm}=120 \mathrm{~cm}$.

(i) Outer lateral surface area = $2 \pi r h$

$=110 \times 120$

$=13200 \mathrm{~cm}^{2} $

(ii) Capacity = $\pi r^{2} h$

 $=55 \times 2100$

$=115500 \mathrm{~cm} .$

=115.5 liters 

Question 12

Ans: Given, 

Radius of cylindrical glass = $\frac{8}{2}=4 \mathrm{~cm}$

and its height = 15cm

Radius, of cylindrical vessel = $\frac{30}{2}=15 \mathrm{~cm}$

and its height = 80 cm

Volume of cylindrical glass= $=\pi r^{2} \mathrm{~h}$

$\pi \times 4 \times 4 \times 15$

$=240 \pi$

Volume of cylindrical vessel $=\pi r^{2} h$

$=\pi \times 15 \times 15 \times 80$

$=\pi \times 225 \times 80$

$=18000 \pi$.

$\therefore$ Number of glasses $=\frac{\text { Volume of vessel }}{\text { Volume of glass}}

$=\frac{18000 \pi}{240 \pi}$ 

$=75$ glasses

Question 13

Ans: Given, 

R= 22m

r= 20 

h=$\frac{7}{100} m$

$\pi R^{2} h-\pi{r}^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times \frac{7}{100}(R+r)(R-r)$

$=\frac{22}{7}\times{7}{100}(22) \times 18=18.48 \mathrm{~mm}$

Question 14

Ans: Given

Radius of iron cylindrical block = $=\frac{0.5m}{20}$ =$\frac{1}{4} \mathrm{~m}=$0.25m= $0.25\times $100cm  =25cm

Length = 3.5m= 3.5 $\times 100$cm = 350 cm

Volume of block =  $\pi r^{2} b$

$=\frac{22}{7} \times 25 \times 25 \times 350$

$=550 \times 1250$

$=687500 \mathrm{~cm}^{3}$

So , volume of base = $687500 \mathrm{~cm}^{3}$

Area of square base $=25 \times 25$

$=625 \mathrm{~cm}^{2}$

Height of bar= $\frac {Volume of bar}{Area of square base}$

$=\frac{687500}{625}$

$=1100 \mathrm{~cm} $

$=\frac{1100}{100}$

$=11 \mathrm{~m}$

Question 15

Ans: Given, 

Length of swimming pole (l) = 70m

Breadth (b)= 44m

and depth (h) =  3m

Volume = lbh

$=70 \times 40 \times 3$

$=924012^{3}$

Radius of pipe= $\frac{14}{2} c m=7 cm=\frac{7}{160} \mathrm{~m}$

Volume = $πr^{2}h

9240= $\frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times h$ (Volume = 9240)

$\frac{9240 \times 100 \times 100}{22 \times 7}=h$

$\frac{92400000}{154}=h .$

$600000=h .$

$\therefore h=600000$

Let be the distance 

$\therefore \quad$ Distance $=$ speed \times $ Time.

$600000=2 \times$ Time. (speed = 2m)

$\frac{600000}{2}=$ Time.

$83 \frac{1}{3}$ hours

Question 16

Ans: Given, 

External radius of a hollow cylinder = $\frac{12}{2}=6 \mathrm{~cm}$

Internal radius of a hollow cylinder = 6_ 0.25

= 5.75cm

Length (h)= 15cm

Volume of hollow cylinder = $\pi R^{2} h-\pi r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 15\left((6)^{2}-\left(5.70^{2}\right)\right.$

$=\frac{330}{7}(36-33.0625) .$

$=\frac{330}{7} \times 2.9375$

$=\frac{969.375}{7 }$

Radius of solid cylinder $\frac{2}{2}=1 \mathrm{~cm}$ (given).

$\begin{aligned} \therefore \text { Volume } &=\pi r^{2} h \\ \frac{969-375}{7} &=\frac{22}{7} \times 1 \times 1 \times h \end{aligned}$

$\frac{969.375}{22}=h$ 

$\therefore h=\frac{969.375}{22}$

$h=44.0625 \mathrm{~cm} $

Question 17

Ans: Given, 

Internal radius of tube= $\frac{11.2}{20} \mathrm{~cm}=5.6 \mathrm{~cm}$

$\operatorname{Length}(h)=21 \mathrm{~cm}$.

Thickness $=0.4 \mathrm{~cm}$.

∴ Outer radius 5.6+0.4

=6cm

Volume of Metal = $\pi R^{2} h - r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$\left.=\frac{22}{7}\times 21 \times(6)^{2}-(5-6)^{2}\right)$

$=66 \times(36-31.36)$

$=66 \times 464$

$=306.24 \mathrm{cm}^{3}$

$=306.2 \mathrm{~cm}^{3}$

Question 18

Ans: Given, 

Volume of water = 1000 lit = $1000 \times 1000 \mathrm{~cm}^{3}=1000000 \mathrm{~cm}^{3}$

Radius of pip = 0.6cm

∴ Volume = $\pi r^{2} h$

$1000000=\frac{22}{7} \times 0-6 \times 0.6 \times h .$

$\frac{1000000 \times 7}{22 \times 0-6 \times 0.6}=h .$

$\frac{700000000}{7.92}=h .$

$883838.38=h .$

$h=883838-38 \mathrm{~cm}$

Let height (h) be the distance, 

Distance = speed $\times$ times 

$883838.38=8 \times$ time.

$\frac{88383838}{800}=$ time

$110479.7975=$ Time

Time = 110479.7975sec

Time = $\frac{110479.7975}{60 \times 60}$ hours 

$=\frac{11047967975}{36000060}$ hours

$=30.69$ hours 

Question 19

Ans: Given,

Radius =$\frac{28}{2} \mathrm{~cm}=14 \mathrm{~cm}$

Height = 72cm

$\therefore$ Volume $\pi r 2 h$

$=\frac{22}{7} \times 14 \times 14 \times 72$

$=44 \times 1008$

$=44352 \mathrm{~cm}^{3}$

Length f tank = 66cm

Breadth of tank = 28cm

Volume = $=l \times b \times h$

$\begin{aligned} 44352 &=66 \times 28 \times h . \\ 44352 &=1848 h . \\ \frac{44352}{1848} &=h . \\ 24 &=h \\ \therefore h &=24 \mathrm{~cm} . \end{aligned}$

Hence, the height of the water level in the tank is 24 cm

Question 20

Ans: Given,

Radius of cylindrical vessel = $\frac{14}{2} c m=7 cm$ 

Height of water = $8 \frac{9}{14}(cm )=\frac{121}{14}$

Volume of water= $\pi r^{2} h$

$=1331 \mathrm{~cm}^{3}$

Volume of cube = $=1331 \mathrm{~cm}^{3}$

Volume of cube= $a^{3}$

$1331=a^{3}$

$\sqrt[3]{1331}=a$

$\sqrt{11 \times 11 \times 11}=a$

$a=11$

Hence , the length of the edge is 11cm

Question 21

Ans: Let the radius of the cylinder = r

And height = h

Volume = $\pi r^{2} h$

If the radius is halved 

$\therefore \quad r=\frac{r}{2}$

Height = h

$\therefore Volume=\pi\left(\frac{r}{2}\right)^{2}h $

$=\pi \frac{r^{2}}{4} h$

According to question 

$=\frac{\pi r^{2}}{4} h=\pi r^{2} h$

$=\frac{1}{4}$= 1:4

Question 22

Ans: Given,

Length of sheet = 22cm

and breadth = 12cm

If it is folded breadth wire, 

∴ Circumference = 12cm

and height = 22cm

Circumference = 2πr

$12=2 \times \frac{22}{7} \times r$

$\frac{21}{11}=r$

$y=\frac{21}{11} \mathrm{~cm}$

∴ Volume =πz $=r^{2} h$

$=\frac{22}{7} \times \frac{21}{11} \times \frac{21}{11} \times 22$

$=\frac{462 \times 42}{77}$

$=\frac{19404}{72}$

$=269.5 \mathrm{~cm}^{3}$

IF it is folded length wire 

$\therefore$ circumference  $=22 \mathrm{~cm}$. 

and Height: $12 \mathrm{~cm}$.

So circumference =$2 \pi r$

$22=2 \times \frac{22}{7} \times r$

$\frac{22 \times 7}{2 \times 22}=r$

$\frac{7}{2}=r$

$r=\frac{7}{2} cm$

Volume =$\left.\pi r^{2}\right)h$

$\frac{27}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12$

$=22 \times 21$

$=482cm^{3}$

According to question, 

$=462-269-5$

$=192.5 \mathrm{~cm}^{3}$

Question 23

Ans: Given, 

Depth of a wall = 20m

Radius = $\frac{7}{2} \mathrm{~m}$

Volume of earth = $\pi r^{2} h$

$=22 \times 35$

$=770 \mathrm{m}^{3}$

Length of platform= 22m,

Breadth = 14m

Volume of platform = $770 \mathrm{~m} 3$

Volume = lbh 

770= $22 \times 14 \times h$

$\frac{770}{22 \times 14}=h$

$\frac{770}{308}=h$

$2-5=h$

$h=2.5 \mathrm{~m}$

Hence , the height of the platform is 2.5m

Question 24

Ans: Given,

Height of cylindrical barrel of pen = 7cm

Radius = $\frac{5}{2} \cdot mm =$ $\frac{5}{2} \times \frac{1}{10} \mathrm{~cm}$=1\4cm

Volume of ink in it =  $\pi r^{2} h$

$-\frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times 7$

$=\frac{11}{8} \mathrm{~cm}^{3}$

Volume of link in bottle =  $\frac{1}{5}l=\frac{1}{5} \times 1000 \mathrm{~cm}=200 \mathrm{cm}^{3}$

$\therefore$ Total number of barrels $=200 \div \frac{11}{8}$

$=200 \times \frac{8}{11}$

$=\frac{1600}{11}$

Word written in one barrel = 310 words 

Total number of words = $\frac{1600}{11} \times 320$

$=\frac{496000}{11 .}$

$=45080.90 .$

$=45090$ words 

Question 25

 

Ans: Given , 

Length of a rectangular box= 40cm,

Breadth = 30cm and 

Height = 25cm

Volume = lbh 

$=40 \times 30 \times 25$

$=1200 \times 25$

$=30000 \mathrm{~cm}^{3}$

 Radius of cylindrical tin= 17.5cm

Volume = $30000 \mathrm{~cm}^{3}$

Volume $=\pi r^{2} h$

$30000=3-14 \times 17.5 \times 17.5 \times h $

$30000=3-14 \times 306.25 \dot{x h}$

$30000=961.625h $

$\frac{30000000}{961625}=h$

$31 \cdot 2$

$h=31-2 \mathrm{~cm}$

Hence the height of the cylindrical tin is 31.2cm

Question 26

 

Ans: Given, 

Diameter of a circular tank = 17.5m

So, the radius = $\frac{175}{20}=\frac{35}{4}=$ 8.75m

Outer radius = 8.75+4

= 12.75m

Height= 2m 

Volume of the embankment = $\pi R^{2} h-\pi r^{2} h$

$\begin{aligned} & \pi h\left(R^{2}-r^{2}\right) \\=& \frac{22}{7} \times 2\left((12-75)^{2}-(8.75)^{2}\right) \\=& \frac{44}{7}(162.5625-76.5625) \\=& \frac{44}{7} \times 86 . \\=& \frac{3784}{7} \\ 540.57 \mathrm{~m}^{3} . \end{aligned}$

 

∴ Volume of earth of the tank = $540.57 \mathrm{~m}^{3}$

$\begin{aligned} \therefore \text { Volyme } &=\pi r^{2} h \\ 540.57 &=\frac{22}{7} \times 8.75 \times 8.75 \times h . \end{aligned}$

$540.57=\frac{1684.375 \mathrm{~h}}{7}$

$2.25=h$

$h=2.25 \mathrm{~m} .$

Hence the depth of the circular tank is 2.25m

Question 27

 

Ans: Given, 

Speed of water = $7 m=700 \mathrm{~cm}$

Internal radius= $\frac{2}{2} c m=1 c m$

Radius of tank = 40cn

Time =$\frac{1}{2}$ hour $=$ $\frac{1}{2} \times 60 \times 60=\frac{3600}{2} \mathrm{sec}=$

=1800sec 

∴ Distance  (h) = Times into speed 

$=1800 \times 700$

$=1260000 \mathrm{~cm}$

Volume $=\pi r^{2} \mathrm{~h}$. 

=$\frac{22}{7}1 \times1 \times \times 1260000$

$=3960000 \mathrm{~cm}^{3}$

∴ Volume of water in the tank =$3960000 \mathrm{cm}^{3} $

Volume $=\pi r^{2} \mathrm{h}$

$3960000=\frac{22}{7} \times 40 \times 40 \times h$.

$\frac{17325}{22}=h$

$787.5=h$

$h=787.5 \mathrm{cm} $

Question 28

 

Ans: Given, 

Radius of pipe = $\frac{7}{2} cm=\frac{7}{200}$ m

Speed = 36 km\hr = $36 \times \frac{1000}{60}$ = 600m

Radius of tank = $35 \mathrm{~cm}=\frac{35}{100} \mathrm{~m}$

Height = 1m

∴ Volume of tank =π$r^{2}h$

$\frac{22}{7} \times \frac{35}{100} \times \frac{35}{100} \times 1$

$=\frac{77}{200} \mathrm{~m}^{3}$

$\because$ Volume $=\pi r^{2} h .$

$\frac{77}{200}=\frac{22}{7} \times \frac{7}{200} \times \frac{7}{200} \times h .$

$\frac{2200}{221}=h$

$100=h .$

$h=100$

Let height (h) be the distance 

Distance = speed into Time

100= $600 \times Time$

Time $=\frac{100}{600}$

Time $=0.167 \mathrm{~min} $

Question 29

 

Ans: Given, 

Radius of coin =$\frac{1.5}{20} \mathrm{~cm}=\frac{3}{4} \mathrm{~cm}$

and its thickness = 0.2 cm

Volume of one coin = $\pi r^{2} h$

$=\frac{22}{7} \times \frac{3}{4} \times \frac{3}{4} \times 0.2 .$

$=\frac{39.6}{112}$

$=0.35 \mathrm{~cm}^{3}$

Radius of cylinder = $\frac{4.5}{20}=\frac{9}{4} \mathrm{~cm}$

Height = 10cm

Volume of cylinder = $\pi r^{2} h$

$\begin{aligned} &=\frac{22}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \\=& \frac{17820}{112} \\=& 159.11 \mathrm{~cm}^{3} . \end{aligned}$

Volume of cylinder = volume of one coin X Number of coin

159.11 = $0.35 \times$ Number of coin.

$\frac{159 \cdot 11}{0.35}$ =Number of coin.

$454.6=$ Number of coin

Number of coin $=454.6$

Question 30

 

Ans: Given

Weight of  $1cm^{3}=21 g$

Length of pipe = $1 \mathrm{~m}=100 \mathrm{~cm}$

Internal radius = $\frac{3}{2} \mathrm{~cm}=1.5 \mathrm{~cm}$

External radius = 1.5+1= 2.5cm

∴ Volume = Volume of outer - Volume of inner surface 

$π R^{2} h$-$π r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 100\left((2.5)^{2}-(1.5)^{2}\right)$

$=\frac{2200}{7} \times(6.25-2.25) .$

$=\frac{2200}{7} \times 4 $

$=\frac{8800}{7} \mathrm{~cm}^{3} $

∴ Total Weight of metal = $\frac{8800}{7} \times {21}$

$=26400 \mathrm{~g}$

S.chand books class 8 maths solution chapter 15 Linear Equations exercise 15 B

EXERCISE 15 B

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Q1 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 1

The sum of three consecutive odd natural numbers is 87. What are the three numbers ?

Sol :
Let the three consecutive odd numbers be (2x+1) , (2x+3) and(2x+5)
A.T.Q
⇒(2x+1)+(2x+3)+(2x+5)=87

⇒2x+1+2x+3+2x+5=87

⇒6x+9=87

⇒6x=87-9

⇒6x=78

⇒$x=\dfrac{78}{6}$

⇒x=13

Hence , the consecutive numbers are

⇒2x+1=2×13+1=27 ,

⇒2x+3=2×13+3=29 and

⇒2x+5=2×13+5=31

 

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Q2 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 2

The sum of two numbers is 80. If the larger number exceeds four times the smaller by 5 , what is the smaller number ?

Sol :

Let the two numbers be x and y and lets assume x>y.
Also, x=4y+5..(i)
A.T.Q
x+y=80 ..(ii)
substituting value of (i) in (ii)
⇒(4y+5)+(y)=80

⇒4y+5+y=80

⇒5y+5=80

⇒5y=80-5

⇒5y=75

⇒$y=\dfrac{75}{5}$

⇒y=15

 

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Q3 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 3

A girl was asked to multiply a number by $\dfrac{7}{8}$ . Instead she divided the number by $\dfrac{7}{8}$ and got the result 15 more than the correct result . What is the number ?

Sol :
Let the number be x
On dividing x by $\dfrac{7}{8}$

⇒$x \div \dfrac{7}{8}$

⇒$x \times \dfrac{8}{7}$

⇒$\dfrac{8x}{7}$

A.T.Q
⇒$x\times \dfrac{7}{8}=\dfrac{8x}{7}-15$

⇒$\dfrac{7x}{8}=\dfrac{8x}{7}-15$

⇒$15=\dfrac{8x}{7}-\dfrac{7x}{8}$

⇒$15=\dfrac{8(8x)-7(7x)}{56}$

⇒$15=\dfrac{64x-49x}{56}$

⇒$15=\dfrac{15x}{56}$

⇒$15\times 56=15x$

⇒$\dfrac{15\times 56}{15}=x$

⇒x=56

 

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Q4 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 4

What is the sum of two consecutive even numbers , the difference of whose square is 84 ?

Sol :
Let the two consecutive even numbers be (2x) and (2x+2). Then ,
A.T.Q
⇒(2x+2)²-(2x)²=84

⇒{(2x)²+2²+2(2x)(2)}-4x²=84 [using(a+b)²=a²+b²+2ab]

⇒4x²+4+8x-4x²=84

⇒8x+4=84

⇒8x=84-4

⇒8x=80

⇒$x=\dfrac{80}{8}$

⇒x=10

So , the numbers are 2x=2×10=20 and 2x+2=2×10+2=22
Then, sum is 20+22=42

 

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Q5 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 5

One number is 7 more than another and its square is 77 more than the square of the smaller number. What are the numbers ?

Sol :
Let the numbers are x and y.Then ,
A.T.Q

⇒x=y+7

⇒y=x-7..(i)

Also,

⇒(x)²=77+y²

⇒x²=77+(x-7)² [from (i)]

⇒x²=77+x²+49-2(x)(7) [(a-b)²=a²+b²-2ab]

⇒x²=77+x²+49-14x

⇒x²-x²+14x=77+49

⇒14x=126

⇒$x=\dfrac{126}{14}$

⇒x=9

Putting value of x in (i)

⇒y=9-7

⇒y=2

The numbers are 2 and 9

 

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Q6 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 6

The numerator of a fraction is 4 less than its denominator if the numerator is decreased by 2 and the denominator is increased by 1 , then the fraction becomes $\dfrac{1}{8}$ . Find the fraction .

Sol :
Let the numerator be x and denominator be y
A.T.Q
x=y-4..(i)

Also,
⇒$\dfrac{x-2}{y+1}=\dfrac{1}{8}$

⇒8(x-2)=1(y+1)

⇒8x-16=y+1

⇒8x=y+1+16

⇒8x=y+17 ..(ii)

Substituting (i) in (ii)

⇒8(y-4)=y+17

⇒8y-32=y+17

⇒7y=17+32

⇒$y=\dfrac{49}{7}$

⇒y=7

Putting value of y in (i) , we get

⇒x=(7)-4

⇒x=3

Hence , the fraction is $\dfrac{x}{y}=\dfrac{3}{7}$

 

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Q7 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 7

The digits of a two-digit number are in the ratio 2:3 and the number obtained by interchanging the digits is greater than the original number by27 . What is the original number ?

Sol:

Let the common ration be x . Then , the tens digit be 2x and once digit be 3x.

Original number be 10(tens)+(once) which can be written as 10(2x)+3x=23x

Original number 23x ..(i)

On interchanging digits 10(3x)+2x=32x

And 32x is greater than original number by 27 , which means

⇒32x=23x+27

⇒32x-23x=27

⇒9x=27

⇒$x=\dfrac{27}{9}$

⇒x=3

So , the tens digit be 2x=2(3)=6

The once digit(unit digit) be 3x=3(3)=9

Hence , the original number be 10(tens)+(once)=10(6)+9=60+9=69

 

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Q8 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 8

The sum of the digits of a two- digit number is 10. The number formed by reversing the digits is 18 less than the original number. Find the original number .

Sol :

Let the original number be 10x+y . Where x is tens and y is unit digit

A.T.Q

⇒x+y=10 [ sum of the digit of number ]

⇒x=10-y ..(i)

On reversing the number we get 10y+x which is equal to (10x+y)-18

⇒10y+x=(10x+y)-18

⇒10y+x=10x+y-18

⇒10y-y-10x+x=-18

⇒9y-9x=-18

⇒-9(-y+x)=-18

⇒$(-y+x)=\dfrac{-18}{-9}$

⇒x-y=2

⇒(10-y)-y=2

⇒10-y-y=2

⇒-2y=2-10

⇒-2y=-8

⇒$y=\dfrac{-8}{-2}$

⇒y=4

Putting value of y in (i) , we get

⇒x=10-y

⇒x=10-4

⇒x=6

Hence , the original number be 10x+y=10(6)+4=64

 

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Q9 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 9.

A man ate 100 grapes in 5 days. Each day he ate 6 more grapes than those he ate on the previous day. How many grapes did he eat the first day .

Sol :

Let the grapes eaten at first day be x . Also , each day he ate 6 more grapes

A.T.Q

1st day + 2nd day + 3rd day + 4th day + 5th day = 100 grapes

⇒(x)+(x+6)+{(x+6)+6}+{(x+6+6)+6}+{(x+6+6+6)+6}=100

⇒x+(x+6)+(x+12)+(x+18)+(x+24)=100

⇒x+x+6+x+12+x+18+x+24=100

⇒5x+60=100

⇒5x=100-60

⇒5x=40

⇒$x=\dfrac{40}{5}$

⇒x=8

Hence , grapes eaten by man on first day is 8

 

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Q10 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 10

In an examination, a student scores 4 marks for every correct answer and losses 1 marks for every incorrect answer. If he attempts all 75 questions and secures 125 marks . What are the number of questions he attempted correctly ?

Sol :

+4 Marks for every correct answer is given

-1 for every incorrect answer is given

Let the correct attempt be x

then , incorrect attempt be (75-x)

A.T.Q

⇒+4(x)-1(75-x)=125

⇒4x-75+x=125

⇒5x=125+75

⇒5x=200

⇒$x=\dfrac{200}{5}$

⇒x=40

Which means 40 attempts are correct

 

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Q11 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 11

Arun's age is four years more than Prashant's age 4 years ago , the ratio of their ages was 5 : 4 . Find their present ages.

Sol:

Let the Prashant's age be x (4 years ago from present)

Then Arun's age be (x+4) (4 years ago from present)

A.T.Q

$\dfrac{(x+4)}{x}=\dfrac{5}{4}$ [cross multiply]

⇒4(x+4)=5x

⇒4x+16=5x

⇒5x-4x=16

⇒x=16

Prashant's age 4 years ago be 16 And for present age (x+4) = 16+4 = 20 years

Arun's age 4 years ago be (x+4) which is 20 and now at present {(x+4)+4} ={(16+4)+4}=24 years

 

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Q12 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 12

A father is three times as old as his son. 15 years hence, he will be twice as old as his son. What is the sum of their present ages ?

Sol :

Let the son age be x and father's age be 3x

In 15 years

Son age be x + 15 and Father age be 2(x+15) “OR” 3x + 15 because both equation represent fathers present age

A.T.Q

⇒2(x + 15) = 3x + 15

⇒2x + 30 = 3x + 15

⇒30-15=3x-2x

⇒15=x

Therefore, Present age of son is x=15 and

fathers age is 3x = 3(15) = 45

And sum of their present age is 45+15 = 60 years

 

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Q13 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 13

4 years ago, the ratio of their ages of A and B was 2 : 3 and after 4 years it becomes 5 : 7 . Find their present ages .

Sol:

Given that, 4 years ago from present , the ratio of the ages of A and B = 2 :3

Hence we can assume that
age of A 4 years ago = 2x
age of B 4 years ago = 3x

After 4 years from present, the ratio of their ages = 5 : 7

⇒{(2x+4)+4} : {(3x+4)+4} = 5 : 7

⇒$\dfrac{2x+8}{3x+8}=\dfrac{5}{7}$

⇒7(2x+8)=5(3x+8) [cross multiplication]

⇒14x+56=15x+40

⇒56-40=15x-14x

⇒x=16

A's present age = (2x+4) = 2(16)+4 = 36

B's present age = (3x+4) = 3(16)+4 = 52

 

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Q14 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 14

Length of a rectangular blackboards is 8 m more than its breadth. If its length is increased by 7 m and its breadth is decreased by 4 m , its area remains unchanged. What is the length of the blackboard ?

Sol :

Case 1:

Let the breadth be x

Length of blackboard = x+8

Area of rectangular blackboard = l × b =(x) × (x+8)

= x² + 8x..(i)

 

Case 2:

Breadth is decreased by 4m then breadth = x-4

Length is increased by 7m then length = x+8+7 = x+15

Area of rectangular blackboard = l × b = (x-4) × (x+15)

= x(x+15) - 4(x+15)

= x²+15x-4x-60

= x²+11x-60..(ii)

Both (i) and (ii) are equal because it is given that area remains unchanged(equal)

⇒x² + 8x = x²+11x-60

⇒x²-x²+8x-11x=-60

⇒-3x=-60

⇒$x=\dfrac{-60}{-3}$

⇒x=20

Length of blackboard = x+8 = 20+8 = 28

 

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Q15 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 15

Two cars A and B leave Mumbai at the same time, travelling in opposite directions . If the average speed of car A is 8 km/h more than that of B,and they are 300 km apart at the end of 6 hours. What is the average speed of car A ?

Sol :

Let x be average speed of B

Then Average speed of A = (x+8) km/hr

Distance travelled by A in 6 hrs = 6(x+8) km [Distance=speed×time]

Distance travelled by B in 6 hrs= 6(x) km

Distance travelled by A + Distance travelled by B = Total distance

⇒6(x+8)+6x=300

⇒6x+48+6x=300

⇒12x=300-48

⇒$x=\dfrac{252}{12}$

⇒x=21

Hence , average speed of A is (x+8)=21+8=29km/hrs

 

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Q16 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 16

A boat takes 9 hours to travel a distance upstream and takes 3 hours to travel the same distance downstream. If the speed of the boat in still water is 4 km/h. What is the speed of the stream ?

Sol :

Let the distance covered by the boat each way be d km. And let the velocity of the stream be v km/hr.

Case 1 : (Downstream)

While going downstream the speed of the boat is (4+v) km/hr and time taken is 3 hours

[Distance=velocity×time]

d=(4+v)3..(i) [here , v is +ve because direction of stream and boat is same]

 

Case 2 : (Upstream)

While going upstream the speed of the boat is (4-v) km/hr and time taken is 9hours

[Distance=velocity×time]

d=(4-v)9..(ii) [here , v is -ve because direction of boat and stream is opposite]

(i)=(ii) , As the distance travelled upstream and downstream is the same,

⇒(4+v)3=(4-v)9

⇒12+3v=36-9v

⇒3v+9v=36-12

⇒12v=24

⇒$v=\dfrac{24}{12}$

⇒v=2km/hr [here , + minus sign shows direction of boat and stream is same]

Hence , the velocity of stream is 2km/hr

 

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Q17 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 |Linear Equations | myhelper

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Question 17

Kartik buys some apples at the rate of 10 rupees per apple . He also buys an equal number of oranges at the rate of 6 rupees per orange. He makes a15% profit on apples and 8% profit on oranges. The total profit earned by Kartik at the end of the day is 396. When all fruit is sold out , find the number of apples purchased .

Sol :

Number of apples be x and Number of oranges also be x (equal)

Cost of apples 10x and cost of oranges be 6x

⇒(15% of 10x )+(8% of 6x)=396

⇒$\left(\dfrac{15}{100}\times 10x\right)+\left(\dfrac{8}{100}\times6x\right)=396$

⇒$\dfrac{150x}{100}+\dfrac{48x}{100}=396$

⇒$\dfrac{150x+48x}{100}=396$

⇒198x=396×100

⇒$x=\dfrac{39600}{198}$

⇒x=200

Hence, Total number of apples purchased is 200

 

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Q18 | Ex-15B | Class 8 | S.Chand | Composite maths | chapter 15 | Linear Equations | myhelper

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Question 18

Bittu spends $\dfrac{1}{3}$ of the money with him on books , $\dfrac{1}{4}$of the remaining on food and $\dfrac{2}{5}$ of the remaining on travel. Now, he is left with 450 rupees . How much money did he have in the beginning?

Sol :

Total money at the beginning be x

Spending on books [1/3 of total] $=\dfrac{1}{3}\times x=\dfrac{1x}{3}$

Remaining $=x-\dfrac{1x}{3}$ $=\dfrac{3x-1x}{3}=\dfrac{2x}{3}$

 

Spending on foods [1/4 of remaining ]$=\dfrac{1}{4}\times\left(x-\dfrac{1x}{3}\right)$ or $=\dfrac{1}{4}\times \dfrac{2x}{3}$$=\dfrac{2x}{12}=\dfrac{x}{6}$

Remaining $=\dfrac{2x}{3}-\dfrac{x}{6}$ $=\dfrac{2(2x)-x}{6}=\dfrac{4x-x}{6}$$=\dfrac{3x}{6}=\dfrac{x}{2}$

 

Spending on travel , [2/5 of remaining] $=\dfrac{2}{5}\times\dfrac{x}{2}=\dfrac{x}{5}$

Remaining $=\dfrac{x}{2}-\dfrac{x}{5}$ $=\dfrac{5x-2x}{10}=\dfrac{3x}{10}$

 

A.T.Q

Money left after spending on travel must be equal to 450

$\dfrac{3x}{10}=450$

3x=450×10

$x=\dfrac{4500}{3}$

x=1500

Hence , total money at beginning is 1500

 

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