Let x be the smaller of the two consecutive odd positive intgers. Then the other odd integer is x + 2.
It is given that both the integers are smaller than 10 and their sum is more than 11.
∴ x + 2 < 10 and, x + (x + 2) > 11
⟹ x < 10 - 2 and 2x + 2 > 11
⟹ x < 8 and 2x > 9
⟹ x < 8 and x > 9/2
⟹ 9/2 < x < 8
⟹ x = 5, 7 [ ∵ x is an odd integer]
Hence, the reuired pairs of odd integers are (5, 7) and (7, 9).
Let x be the smaller of the two consecutive odd natural numbers. Then the other odd integer is x + 2.
It is given that both the natural number are greater than 10 and their sum is less than 40.
∴ x > 10 and, x + x + 2 < 40
⟹ x > 10 and 2x < 38
⟹ x > 10 and x < 19
⟹ 10 < x < 19
⟹ x = 11, 13, 15, 17 [͙∵ x is an odd number]
Hence, the required pairs of odd natural number are (11, 13), (13, 15), (15, 17) and (17,19).
Let x be the smaller of the two consecutive even positive integers.
Then the other even integer is x + 2.
It is given that both the even integers are greater than 5 and their sum is less than 23.
∴ x > 5 and, x + x + 2 < 23
⟹ x > 5 and 2x < 21
⟹x > 5 and x < 21/2
⟹ 5 < x < 21/2 = 10.5
⟹ x = 6, 8, 10 [∵ x is an even integer]
Hence, the required pairs of even positive integer are (6, 8), (8, 10) and (10, 12).
Suppose Rohit scores x marks in the third test then,
⟹ 195 ≤ 135 + x
⟹ 195 – 135 ≤ x
⟹ 60 ≤ x
Hence, the minimum marks rohit should score in the thrid tes is 60.
We have,
F1 = 86°F
⟹ C2 = 7 × 5 = 35°C
∴ The range of temperature of the solution is from 30°C to 35°C.
We have,
C1 = 30°C
⟹ F2 = 9 × 7 + 32
⟹ F2 = 63 + 32
⟹ F2 = 95°F
∴ Hence, the temperature of the solution lies between 86°F to 95°F.
Suppose Shikha scores x marks in the fifth paper. Then,
⟹ 90 × 5 ≤ 182 + 186 + x
⟹ 450 ≤ 368 + x
⟹ 450 - 368 ≤ x
⟹ 82 ≤ x
Hence, the minimum marks is required in the last paper is 82.
We have,
Profit = Revenue - cost
Therefore, to earn some profit, we must have
Revenue > Cost
Hence, the manufacturer must sell more than 600 cassettes to realize some profit.
Let the length of the shortest side be x.
Then, the length of the longest side and third side of the triangle are 3x and 3x - 2 respectively.
According to question,
Perimeter of triangle ≥ 61
⟹ x + 3x – 2 + 3 x ≥ 61
⟹ 7x ≥ 61 + 2
⟹ 7x ≥ 63
⟹ x ≥ 63/7
⟹ x ≥ 9
∴ The minimum length of the shortest side is 9 cm.
Let the quantity of water to be added to solution = x liters.
∴ 25 % (1125 + x) < 45% of 1125
⟹ 1.5 × 1125 < 1125 + x
⟹ 1687.5 < 1125 + x
⟹ 1687.5 – 1125 < x
⟹ 562.5 < x ……… (ii)
Using (i) and (ii), we get 562.5 < x < 900
Hence, quantity of water lies between 562.5 litres and 900 litres.
Let x liters of 2% solution will have to be added to 640 liters of the 8% solution of acid.
Total quantity of mixture = (640 + x)
Total acid in the (640 + x) liters of mixture
It is given that acid content in the resulting mixture must be more than 4% but less than 6%.
⟹ 4[640 + x] < (2x + 8640) < 6[640 + x]
⟹ 2560 + 4x < 2x + 8640 and 2x + 8640 < 3840 + 6x
⟹ 2560 – 8640 < 2x - 4x and 2x - 6x < 3840 - 8640
⟹ x < 1280 and x > 320
More than 320 litres but less than 1280 liters of 2% is to be added.