Selina Concise Class 10 Math Chapter 7 Ratio and Proportion Solutions
Exercise – 7 (A)
(Q1) If a:b = 5:3, find; 5a – 3b/5a + 3b
= Solution:
Given that a:b = 5:3
Means, a/b = 5/3 = a = 5/3 b.
Find 5a – 3b/5a + 3b = 5a – 3b/5a + 3b
From equation (1) we get.
= 5 (5/3 b) – 3b/5(5/3 b) + 3b
= b (25/3 – 3)/b (25/3 + 3)
= (25/3 – 3)/(25/3 + 3)
(Q2) If x:y = 4:7, find the value of (3x + 2y) : (5x + y)
Given: x:y = 4:7
Means: x/y = 4/7
x = 4/7 y —– (1)
(Q3) If a:b = 3:b, find the value of 4a + 3b/6a – b
= Solution:
Given that a:b = 3:8
Means: a/b = 3/8
a = 3/8 b ——— (1)
Find 4a + 3b/6a – b = (4 × 3/8 b + 3b)/(6 × 3/8 b – b)
(∵ from (1))
= b (12/8 + 3)/b (18/8 – 1)
= (12/8 + 3)/(18/8 – 1)
(Q4) If (a – b) : (a + b) = 1:11 find the ratio (5a + 4b + 15) : (5a – 4b + 3)
= Solution:
Given that a-b/a+b = 1/11
Find (5a + 4b + 15/(5a – 4b + 3)
∴ a – b/a + b= 1/11
11 (a – b) = 1 (a + b)
11a – 11b = a + b
11a – a = b + 11b
10a = 12b
a/b = 12/10
a/b = 6/5
Assume that, a = 6p and b = 5p
(5a + 4b +15)/(5a – 4b + 3) = (5×6p + 4(5p) +15)/(5× 6p – 4(5P) + 3)
= (30p + 20p + 15)/(30p – 20p + 3)
= (30p + 20p + 15)/(10p + 3)
= 5 (5p + 4p + 3)/(10p + 3)
= 5 (10p + 3)/(10p + 3)
= 5
∴ (5a + 4b + 15) : (5a – 4b+ 3) = 5:1
(Q5) Find the number which bears the same ratio to 7/33 that 8/21 does to 4/9
= Solution:
Let, the number be x.
∴ x: 7/33 = 8/21 : 4/9
x/(7/33) = (8/21)/(4/9)
33x/7 = 8/21 × 9/4
33x/7 = 4/21 × 9/2
33x/7 = 2/21 × 9/1
x = 2×9×7/21×33×1
x = 2×9/3×33
x = 2×3/33
x = 2/11
∴ The number is 2/11 or 2:11
(Q6) If m+n/m+3n = 2/3; find 2n2/3m2 + mn
= Solution:
Given that the ratio is m+n/m+3n = 2/3
Find 2n2/3m2 + mn
∴ m+n/m+3n = 2/3
3 (m+n) = 2 (m + 3n)
3m + 3n = 2m+6n
3m – 2m = 6n – 3n
m/n = 3/1
Let, m = 3t and n = 1t
2n2/3m2 + mn = 2×t2/3 (3t)2 + (3t) (t)
= 2t2 / 3 (9t2) + 3t2
= 2t2/27t2 + 3t2
2n2/3m2 + mn = 2t2/30t2
2n2/3m2 + mn = 1/15
(Q7) Find x/y; when x2 + 6y2 = 5xy.
Solution:
Given that x2 + 6y2 = 5xy
∴ x2 – 5xy + 6y2 = 0
x2 – (2 + 3) xy + 6y2 = 0
x2 – 2xy – 3xy + 6y2 = 0
x (x – 2y) – 3y (x – 2y) = 0
(x – 2y) (x – 3y) = 0
x – 2y = 0 or x – 3y = 0
x = 2y or x = 3y
x/y = 2/1 or x/y = 3/1
x/y = 2 or x/y = 3
(Q8) If the ratio between 8 and 11 is the same as the ratio of 2x – y to x + 2y find the value of 7x/9y
= Solution:
Given that the ratio is 2x – y/x + 2y = 8/11
11 (2x – y) = 8 (x + 2y)
22x – 11y = 8x + 16y
22x – 8x = 16y + 11y
14x = 27y
x/7 = 27/14
But find the value of 7x/9y
Multiply by 7 on denominator and multiply by 9 on numerator.
7×x/9×y = 27×7/14×9
7x/9y = 3/2
(Q9) Divide RS 1290 into A, B and C such that A is 2/5 of B and B:C = 4:3
Solution:
Given that B:C = 4:3
Means B/C = 4/3
Also, given that A is 2 of B
A = 2/5 B
A/B = 2/5
∴ A/B = 2/5 —— (1) B/C = 4/3 —– (2)
Multiply and divide by 4, in equation (1)
A/B = 2×4/5×4
A/B = 8/20
Multiply and divide by 5 in equation (2)
B/C = 4/3 × 5/3 = 20/15
A/B = 8/20, B/C = 20/15
∴ A:B:C = 8:20:15
Let, A = 8x, B = 20x, C = 15x
Now, A + B + C = 1290
8x + 20x + 15x = 1290
43x = 1290
x = 1290/43
x = 30
A = 8x = 8 (30) = 240
B = 20x = 20 (30) = 600
C = 15x = 15 (30) = 450
(Q10) A school has 630 students. The ratio of the number of boys to the number of girls is 3:2. This ratio changes to 7:5 after the admission of 90 new students. Find the number of newly admitted boys.
= Solution:
Let us consider the number of newly admitted boys is x.
∴ Let us assume that the number of boys be 3x and the number of girls be 2x.
∴ Given that total number of students be 630.
∴ boys + girls = 630
= 3x + 2x = 630
= 5x = 630
= x = 630/5
= x = 126
∴ The number of boys = 3x = 3 × (126) = 378
And the number of girls = 2x = 2 × (126) = 252
After the 90 new admission students.
Now, The total number of students
= 630 + 90
= 720
Let us consider the number of boys is 7x and the number of girls is 5x
= 7x + 5x = 720
= 12x = 720
= x = 720/12
= x = 60
∴ The number of boys = 7x
= 7×60
= 420
And the number of girls = 5x
= 5×60
= 300
∴ The number of newly admitted boys = 420 – 378
= 42
(Q11) What quantity must be subtracted from each term of the ratio 9:17 to make it equal to 1:3 ?
= Solution:
Let we subtract x.
9:17 = 1:3
9/17 = 1/3
∴ 9-x/17-x = 1/3
3 (9 – x) = 1 (17 – x)
27 – 3x = 17 – x
– 3x + x = 17 – 27
– 2x = – 10
x = 10/2
x = 5
∴ x = 5 must be subtracted from each term of the ratio 9:17 to make it equal to 1:3
(Q12) The monthly pocket money of Ravi and Sanjeev are in the ratio 5:7. Their expenditures are in the ratio 3:5. If the each saves RS 80 every month, find their monthly pocket money.
= Solution:
Let monthly pocket money Ravi = 5x and the monthly pocket money of Sajeev = 7x
Let monthly expenditure of Ravi = 3y and monthly expenditure of Sanjeev = 5y
5x – 3y = 80 —– (1)
7x – 5y = 80 ——- (2)
We have to find the value of x and y.
Equation (1) × 7 – equation (2) × 5,
7×5x – 7×3y = 80×7
5×7x – 5×5y = 80×5
35x – 21y = 560
35x – 25y = 400
(-) (+) (-)
__________________
4y = 160
Y = 160/4
Y = 40
Substitute y = 40 in equation (i),
5x – 3(40) = 80
5x – 120 = 80
5x = 80 + 120
5x = 200
x = 200/5
x = 40
∴ Ravi = 5x
= 5 × (40)
= ₹200
And Sanjeev = 7x
= 7×40
= ₹280
Here is your solution of Selina Concise Class 10 Math Chapter 7 Ratio and Proportion Exercise 7A Solutions
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