NCTB Class 7 Math Chapter Two Exercise 2.3 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 2″Proportion, Profit & Loss” Exercise 2.3 Solution.

Board |
NCTB |

Class |
7 |

Subject |
Math |

Chapter |
2 |

Chapter Name |
“Proportion, Profit & Loss” |

Exercise |
2.3 Solution |

**Exercise:- 2.3**

**(1) (a) Solution:- **Twice divided ratio of 4:9

= √4: √9

(a) = 2:3

**(2) Solution:- **

**(3) Solution:-** 4:3= 4×5: 3×5= 20:15

5:6= 5×3: 6×3= 15:18

Therefore, 20:15:18.

Second quantity is = 15 (b).

**(4) Solution:-** Ratio = 5:3:2

Therefore, (5+3+2)

= 10

Therefore, Maisha get = (5/10×30)

= 15m.

**(5) Solution:-** Tania get = 30x 2/10

= 6m.

Maria get = 30x 3/10

= 9m.

Therefore, (9-6) = 3m.

Maria get 3m more than Tania.

**(6) Solution:- **

**(7) Solution:-** Let, 4^{th} proportional is ‘x’

Therefore, 3:5 : : 15: x

=> 3/5 = 15/x

=> x = 5×15/3

(b) = 25

**(8) Solution:- ** C P= 1.50 TK. S P= 2.00

Profit = (2-1.50)

= 0.50TK.

Percentage profit= (0.5/1.50x 100)%

= 33 1/3%

**(9) Solution:-** C P of banana= 25TK. 1 bunch= 4pc

SP of banana = 27TK.

2TK profit in 1 bunch

Therefore, 1TK profit in ½ bunch

Therefore, 50TK profit in 50×1/2 bunch

= 25 bunch

**(10) (a) Solution:-** If the cost price is more than the selling price — (d) Loss oriented.

**(10) (b) Solution:-** If the cost price is less than the selling price— (b) Profitable.

**(10) (c) Solution:-** Time along with current—(a) Less time.

**(10) (d) Solution:-** Time against the current—(c) More time.

**(11) Solution:–** 5 workers can reap crops land 8 bigha in 6 days.

Therefore, 1 workers can reap crops land 8 bigha in 6×5/8 days.

Therefore, 25 workers can reap crops land 20 bigha in 6x5x20/25×8 days.

= 3 days.

**(12) Solution:-** Swapan 1 days work = 1/24

Ratan 1 days work = 1/16

Together in 1 day total = (1/24+1/10)

= 2+3/48

= 5/48

Therefore, They work 5/48 in 1 day.

They work 1 in 48/5 day.

= 3 3/5 day.

**(13) Solution:-** Hibiba 8 Aalima together can do,

20 day—- 1 part

1 day = 1/20part

Therefore, 8 day = 8/20 part.

Rest part = (1- 8/20)

= 20-8/20

= 12/20 = 6/10= 3/5 part

Therefore, Halima do, 3/5 part in 21 day

Therefore, 1 part in 21×5/3 day

= 35 days.

**(16) Solution:-** In 1^{st} pipe, 12hr. fill 1 tank

Therefore, 1hr fill 1/12 part.

And in 2^{nd} pipe,

18hr fill 1 tank.

Therefore, 1hr fill 1/18 part

If two pipes are open then in 1hr fill

= (1/12+ 1/18)

= 3+2/36

= 5/36

Therefore, 2 pipe fill 5/36 part in 1 hr.

Therefore, 2 pipe fill 1 part in 1/5/36 hr.

= 7 1/5hr.

**(17) Solution:-** Along with current,

Boat cross, 4hr-36km.

Therefore, 1hr= 36/4 km.

= 9km.

Therefore, Speed of beat +speed of current = 9km/hr.

Where, speed of current = 3km/hr.

Therefore, (9-3)= 6km/hr.

Speed of boat in still water is 6km/hr.

**(18) Solution:-** A ship travel 11 hrs — 77km against the current

Therefore, a ship travel 1 hrs — 77/11 km against the current.

= 7 km.

In still water, speed of ship 9km/hr.

Therefore, against the stream,

Speed of ship = (9-7) km/hr.

= 2km/hr.

**(20) Solution:-** Farmers, cultivate,

With 5 pair cows in 8 days cultivate 40 hectors land

Therefore, with 1 pair cows in 8 days cultivate 40/5 hectors land.

Therefore, with 7 pair cows in 1 days cultivate 40×7/5×8 hectors land.

Therefore, with 7 pair cows in 12 days cultivate 40x7x12/8×5 hectors land

= 84 hectors land.

**(21) Solution:-** Lily can do, a work in 10 hr.

Therefore, 10hr— 1 part

Therefore, 1hr — 1/10 part.

And, mily can do that work in 8hr.

Therefore, 8hr — 1 part.

Therefore, 1 hr —- 1/8 part.

They work together totally

= (1/10+ 1/8)

= 4+5/40

= 9/40

Therefore, Together, 9/40 part do in 1hr.

Therefore, 1 part do in 40/9 hr.

= 4 4/9 hr.

**(22) Solution:-** By 2^{nd} pipe,

In 30 min fill 1 tank 1/30 part tank full.

Therefore, 18 min fill 1×18/30 part tank full.

= 3/5 Part.

Therefore, Rest part = (1- 3/5)= 2/5 part

And, by 1^{st} pipe, 1 part tank full in 20 min

Therefore, 2/5 part tank full in 20x 2/5 min

= 8min.

**(23) Solution:-** 1hr= 3600 sec.

Therefore, In 3600 sec, train cover 48/3600 km.

Therefore 30 sec train 48×30/3600 km

= 2/5km.

= 2/5x 1000

= 400m.

Therefore, length of the bridge

= (400-100)m.

= 300m.

**(24) Solution:-** If train cover total bridge, then total distance cover by train

= (120+330)m

= 450m, ; 30km= 30,000m.

1hr= 3600sec.

30,000m distance cover in 36000sec

Therefore, 1m distance cover in 3600/30,000 sec.

Therefore, 450m distance cover in 3600×450/30,000 sec

= 54sec.

**(25) (a) Solution:- **

**(25) (b) Solution:-** Ratio given= 3:6:10

Total= 19

Given weight = 190gm.

Therefore, weight of bronze = (3/19×190)

= 30gm.

Therefore, weight of zinc = (6/19×190)

= 60gm.

Therefore, weight of silver= (190×10/19) gm

= 100 gm.

**(25) (c) Solution:-** Weight bronze in ornament = 30gm

Therefore, weight of zinc in ornament = 60gm

Let, x gm zinc to be added,

Therefore, 30: (60+x) = 1:3

=> 30/60+x = 1/3

=> 60+x = 90

= x= 90-60

= 30gm.

Therefore, 30 gm to be added to get ratio 1:3.

**(26) (a) Solution:-** Let, C P = 100TK.

(a) 10% loss, S P= (100-10) = 90TK.

Therefore, if S P 90, than C P = 100TK.

Therefore, 1 TK then C P 100/90 TK.

Therefore, 625 TK. Then C P 100×625/90

= 69 -4 -4/9 TK

Therefore, Cost price of watch 69 – 4 -4/9 TK

Therefore, Loss (C P- S P)

= (694- 4/9 – 625)

= (6250/9- 625)

= 6250-625/9

= 625/9 TK.

**(26) (b) Solution:- **buying price of watch 694- 4/9.TK.

**(26) (c) Solution:-** 10% profit, S P of watch (100+10)

= 110TK.

Therefore,. If C P 100 then S P = 110

Therefore, C P 1 than S P = 110/6250/9×10

= 6875/9 TK.

= 763- 8/9.