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, 4th 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 1st pipe, 12hr. fill 1 tank
Therefore, 1hr fill 1/12 part.
And in 2nd 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 2nd 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 1st 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.
it is wrong
Which one?
Hmm
Which one????
Where is 19 no. Question?????