Concise Mathematics – Class 8 Selina Exercise 14B Solution by experts | 14B Selina Concise Solution Class 8 | Class 8 Selina Concise Ex 14B Page 169 Solution | RK Bansal Class 8 ICSE Book Solution Online.
Exercise 14(B)
(1) Fifteen less than 4 times a number is 9. Find the number.
Solution: Let the number be x
Therefore, 4x – 15 = 9
=> 4x = 9 + 15
=> 4x = 24
=> x = 24 / 4
=> x = 6
(2) If Megha’s age is increased by three times her age, the result is 60 years. Find her age.
Solution: Let Megha’s age be x years.
Therefore, x + 3x = 60
=> 4x = 60
=> x = 60 / 4
=> x = 15
Ans. Her age is 15 years
(3) 28 is 12 less than 4 times a number. Find the number.
Solution: Let the number be x
Therefore,
28 = 4x – 12
=> 4x – 12 = 28
=> 4x = 28 + 12
=> 4x = 40
=> x = 40 / 4
=> x = 10
(4) Five less than 3 times a number is – 20. Find the number.
Solution: Let the number be x
Therefore,
3x – 5 = -20
=> 3x = -20 + 5
=> 3x = -15
=> x = -5
Ans. The number is -5.
(5) Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she?
Solution: Let Neetu’s age be x years.
Therefore,
3x + 15 = 4x
=> 3x – 4x = – 15
=> -1x = -15
=> x = 15
Ans Her age is 15 years.
(6) A number decreased by 30 is the same as 14 decreased by 3 times the number. Find the number.
Solution: Let the number be x
Therefore,
X – 30 = 3x – 14
=> x + 3x = – 14 + 30
=> 4x = 44
=> x = 11
The number is 11
(7) A’s salary is same as 4 times B’s salary. If together they earn Rs. 3,750 a month, find the salary of each.
Ans. Let A’s salary be x
B’s salary will 4x
Therefore,
X + 4x = 3750
=> 5x = 3750
=> x = 3750 / 5
=> x = 750
A’s salary is Rs. 750
B’s salary is 4 x 750 = Rs. 3000
(8) Separate 178 in two parts so that the first part is 8 less than twice the second part.
Ans. Let the second part be ‘x’
Then the first part is 2x – 8
By the problem, x+2x-8=178
3x-8=178
3x=178+8=186
x=186/3 = 62
So the first part is 2x-8 = 2(62)-8 = 124-8=116
Second part is x = 62
=> 3y = 120
=> y = 120/3
=> y = 40
The number is 40
(10) Let the width be x.
Length = 2x – 5
Therefore,
2 x (L + B) = Perimiter
=> 2 x (2x – 5 – 5 + x + 2) = 74
=> 2 x (3x – 8) = 74
=> 6x – 16 = 74
=> 6x = 74 + 16
=> 6x = 90
=> x = 90 / 6
=> x = 15
The length of the original rectangle is 2 x 15 – 5 = 30 – 5 = 25 cm
Width is 15 cm
(12) Let the three consecutive odd number be x, (x + 2) and (x |+ 4)
Therefore,
X + (x + 2) + (x + 4) = 57
=> 3x + 6 = 57
=> 3x = 57 – 6
=> 3x = 51
=> x = 17
The three numbers are 17, 17 + 2 = 19, 17 + 4 = 21
(13) Let sons present age be x
Man’s age = 3x
In 12 years Sons age become = x + 12
In 12 years Mans age become = 3x + 12
Therefore,
3x + 12 = 2(x+12)
=> 3x + 12 = 2x + 24
=> 3x – 2x = 24 – 12
=> 1x = 12
=> x = 12
Son’s present age = 12 years
Mans present age = 3 x 12 = 36 years.
(14) Let in x years will the age will the age of the son be half the age of the man the age of the man at that time.
After x years Man’s age = 42 + x
After x years Son’s age = 12 + x
Therefore,
42 + x = 2(12+x)
=> 42 + x = 24 + 2x
=> x – 2x = 24 – 42
=> -1x = -18
=> x = 18
Ans. 18 years.
=> 5x+816-6x = 720
=> -1x = 720 – 816
=> -1x = -96
=> x = 96
Ans. 96 km covered at 18 km/hr.
(16) Let the first number be x
And the second number be x + 3
Therefore,
(x+3)2 – x2 = 69
=> x2 + 6x + 9 – x2 = 69
=> 6x = 69 – 9
=> 6x = 60
=> x = 10
The first number is 10
And the second number is = 10 + 3 = 13
(17) Let, the two consecutive number x and x+1
Therefore
(20) Let, present age of sons be x
Mans age = 2x
Eight years hence Mans age be 2x + 8
Son’s age be = x + 8
Therefore,
=> 8x + 32 = 7x + 56
=> 8x – 7x = 56 – 32
=> x = 24
The present age of man and son are = 2×24 = 48 years and 24 years respectively.
Question number 6 equation is wrong
It should be x-30=14-3x
Instead of x-30= 3x-14