Inequalities Class 11 OP Malhotra Exe-11C ISC Maths Solutions Ch-11 Solutions. In this article you would learn about Word Problems on Linear Inequations in One Variables. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Inequalities Class 11 OP Malhotra Exe-11C ISC Maths Solutions Ch-11
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-11 | Inequalities |
| Writer | O.P. Malhotra |
| Exe-11(C) | Word Problems on Linear Inequations in One Variables. |
Exercise- 11C
Inequalities Class 11 OP Malhotra Exe-11C Solution.
Que-1: Find all pairs of consecutive even positive integers, both of which are larger than 8, such that their sum is less than 25.
Sol: Let x be the smaller of the two consecutive even positive integers, then the other even integer is x + 2.
Given x > 8 and x + (x + 2) < 25.
⇒ x > 8, and 2x + 2 < 25.
⇒ x > 8, 2x < 23 ⇒ x > 8, x < 23/2
⇒ 8 < x < 23/2 ⇒ x = 10,
∴ the required parity even integers is (10, 12)
Que-2: A solution is to be kept between 68°F and 77°F. What is the range in temperature in degree Celsius (°C) if the Celsius, Fahrenheit conversion formula is given by F = (9/5)C + 32 ?
Sol: Since the solution is to be kept between 68° F and 77° F, 68° < F < 77° F
Putting F = 9/5 C + 32, we obtain:
68 < 9/5 C + 32 < 77
⇒ 68 – 32 < 9/5 C < 77 – 32
⇒ 36 < 9/5 C < 45
⇒ 36 x 5/9 < C < 45 x 5/9
⇒ 20 < C < 25
Thus, the required range of temperature in degrees Celsius is between 20° C and 25° C
Que-3: In the first four papers each of 100 marks, Rishi got 95, 72, 73, 83 marks If he wants an average of greater than or equal to 75 marks and less than 80 marks, find the range of marks he should score in the fifth paper.
Sol: Let score be x in the fifth paper, then
75 ≤ {(95+72+73+83+x)/5} ≤ 80
75 ≤ {(323+x)/5} ≤ 80
375 ≤ 323+x ≤ 400
52 ≤ x ≤ 77
Hence Ravi must score between 52 and 77 marks.
Que-4: In drilling world’s deepest hole it was found that the temperature T in degree celsius, x km below the earth’s surface was given by T = 30+25(x-3), 3≤x≤15. At what depth will the temperature be between 155°C and 205°C ?
Sol: T = 30 + 25(x – 3), 3 ≤ x ≤ 15
Where, T = temperature and x = depth inside the earth
The Temperature should be between 155°C and 205°C.
So, we get,
⇒ 155 < T < 205
⇒ 155 < 30 + 25(x – 3) < 205
⇒ 155 < 30 + 25x – 75 < 205
⇒ 155 < 25x – 45 < 205
Adding 45 to each term, we get
⇒ 200 < 25x < 250
Dividing each term by 25, we get.
⇒ 8 < x < 10
Hence, temperature varies from 155° C to 205° C at a depth of 8 km to 10 km.
Que-5: A solution of 9% acid is to be diluted by adding 3% acid solution to it. The resulting mixture is to be more than 5% but less than 7% acid. If there is 460 L of the 9% solution, how many litres of 3%solution will have to be added ?
Sol: Let x litres of 3% solution be added to 460 litres of 9%.
∴ Total amount of mixture = (460 + x) litres
Given that the acid contents in the resulting mixture is more than 5% but less than 7% acid.
∴ 5% of (460 + x) < (9/100) + (3/100) × x < 7% of (460+x)
(5/100) (460+x) < {(4140+3x)/100} < (7/100) (460+x)
⇒ 5(460 + x) < 4140 + 3x < 3220 + 7x
⇒ 2300 + 5x < 4140 + 3x < 3220 + 7x
⇒ 2300 + 5x < 4140 + 3x and 4140 + 3x < 3220 + 7x
⇒ 5x – 3x < 4140 – 2300 and 3x – 7x < 3220 – 4140
⇒ 2x < 1840 and –4x < –920
⇒ x < 1840/2 and 4x > 920
⇒ x < 920
∴ x > 920/4 and x > 230
Hence the required amount of acid solution is more than 230 litres and less than 920 litres.
Que-6: A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and third length is to be twice as long as the shortest. What are the possible lengths for the shortest board, if third piece is to be at least 5 cm longer than the second ?
Sol: Let the length of the shortest piece in cm be x.
Then, the length of second and third piece in cm are (x + 3) and 2x respectively.
Since the three lengths are to be cut from a single piece of board of length 91 cm.
x + (x + 3) + 2x ≤ 91
⇒ 4x + 3 ≤ 91
⇒ 4x ≤ 91 – 3
⇒ 4x ≤ 88
⇒ x ≤ 88/4
⇒ x ≤ 22 …. (1)
Also, the third piece is at least 5 cm longer than the second piece
2x ≥ (x + 3) + 5
⇒ 2x ≥ x + 8
⇒ 2x – x ≥ 8
⇒ x ≥ 8 ….(2)
From (1) and (2), we obtain
8 ≤ x ≤ 22
Thus, the possible length of the shortest board is greater than or equal to 8 cm but less than
or equal to 22 cm
–: End of Inequalities Class 11 OP Malhotra Exe-11C ISC Maths Ch-11 Solutions. :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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