- Costs of two watches were in the ratio of 16 : 23. The cost of first watch increases by 10% and that of second by Rs. 477. Now the costs of two watches are in a ratio of 11 : 20. The price of the second watch (in Rs.) in the beginning was
Correct Option: B
Given in question , The ratio of cost price of two watches = 16 : 23Let C.P. of first watch = Rs.16x
After 2nd temple visit, he is left with Rs (4x-300) After 3rd temple visit, he is left with Rs (8x-700) After 4th temple visit, he is left with Rs (16x-1500), which is zero (as he is left with empty pockets) Hence he started of with x i.e. On his way out he donates another Rs. 100 and he is left with Rs. According to the question, after his last visit, he completely runs out of money.
C.P. of second watch = Rs. 23x
According to the question,| Ratio after corresponding increases = | |
| 20 |
| ⇒ | = | ||
| 100(23x + 477) | 20 |
| ⇒ | = 1 | |
| 5(23x + 477) |
⇒ 160x = 115x + 2385
⇒ 160x – 115x = 2385⇒ 45x = 2385
| ⇒ x = | = 53 | |
| 45 |
∴ Original C.P of second watch = Rs. 23x = Rs. (23 × 53 )
Original C.P of second watch = Rs.1219
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EXERCISE 20A
(1) Perimeter = 2 × (135 + 65)m
= 2 × 200 m = 400m
Rate of fencing = Rs 60/m
∴Total cost of fencing = Rs (400 × 60) = Rs 24000
(2) Let the length will be 7x m and breadth will be 4x m.
Then, 2 × (7x + 4x) = 440
⇒ 2 × 11x = 440
⇒ 22x = 440
⇒ x = 440/22
⇒ x = 20
Therefore, length = (7 × 20) = 140 m and breadth = (4 × 20) = 80m
Rate of fencing = Rs 150/m
∴ Total cost of fencing = Rs (150 × 440) = Rs 66,000
(3) Let the breadth will be x m.
Applying Pythagoras theorem,
AC2 = AB2 + BC2
⇒ (34)2 = (30)2 + x2
⇒ x2 = (34)2 – (30)2
⇒ x2 = 1156 – 900
⇒ x2 = 256
Rs-16x Drum-machine
⇒ x = √256 = 16
Therefore, breadth = 16 m.
∴ Perimeter = 2 × (30 + 16) m
= 2 × 46 m = 92 m
(4) Let the each side will be x m.
Applying Pythagoras theorem,
AC2 = AB2 + BC2
⇒ (12√2)2 = x2 + x2
⇒ (144 × 2) = 2x2
⇒ 2x2 = 288
⇒ x2 = 288/2
⇒ x = √144 = 12
Therefore, length of each side = 12 m
∴ Perimeter = (4 × 12) m = 48 m.
(5) Here, 16 dm = 1.6 m
∴ Perimeter = 2 × (22.5 + 1.6) m
= 2 × 24.1 m = 48.2 m
(6) Let the breadth will be x cm.
Applying Pythagoras theorem,
AC2 = AB2 + BC2
⇒ (25)2 = (24)2 + x2
⇒ x2 = (25)2 – (24)2
⇒ x2 = 625 – 576
⇒ x = √49 = 7
Breadth = 7 cm
∴ Perimeter = 2 × (24 + 7) cm
= 2 × 31 cm = 62 cm
(7) Let the length and the breadth will be 5x and 3x m respectively.
Perimeter = 2 × (5x + 3x) m = (2 × 8x) m = 16x m
Rate of fencing = Rs 48/m
Total cost of fencing = Rs (16x × 48) = Rs 768x
But given = Rs 19200
∴ 768x = 19200
⇒ x = 19200/768
⇒ x = 25
Therefore, Length = (5 × 25) = 125m
And breadth = (3 × 25) = 75 m
(8) Perimeter of square shape = (4 × 20) cm = 80 cm
Let the breadth of rectangular shape be x cm.
Then, Perimeter of rectangular shape = 2 × (24 + x) cm
Then, 2 × (24 + x) = 80
⇒24 + x = 80/2
⇒ x = 40 – 24 = 16
Therefore, breadth = 16 cm.
(9) P = perimeter, Length = l and breadth = b
(i) P = 2 × (l + b)
Lcd 16x2 Rs Pin
⇒ P = 2 × (38 + 27)
⇒ P = 2 × 65 = 130 cm
(ii) 2 × (l + b) = P
⇒ (24 + b) = 88/2
⇒ b = 44 – 24
⇒ b = 20
(iii) 2 × (l + b) = P
⇒ l + 28 = 96/2
⇒ l = 48 – 28
⇒ l = 20
(10) Let the length of each side will be x m.
Perimeter = 4x m
Rate of fencing = Rs 75/m
Rs-16x Os X
Total cost of fencing = Rs (4x × 75)
But given = Rs 67500
∴ 4x × 75 = 67500
⇒ 4x = 67500/75
⇒ 4x = 900
⇒ x = 900/4
⇒ x = 225
Hence, the side of square = 225m.
(11) Perimeter of rectangular field = 2 × (36 + 28) cm
= 2 × 64 cm = 128 cm
Given, its perimeter = perimeter of a square
Let the length of each side will be x cm.
Then, its perimeter = 4x cm
∴ 4x = 128
⇒ x = 128/4
⇒ x = 32