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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D

Question 1.
If a : b = 3 : 5, find : (10a + 3b) : (5a + 2b)
Solution:

Question 2.
If 5x + 6y : 8x + 5y = 8 : 9, find x : y
Solution:

Question 3.
If k (3x – 4y) : (2x – 3y) = (5x – 6y) : (4x – 5y), find x : y.
Solution:

Question 4.
Find the :
(i) duplicate ratio of 2√2 : 3√5
(ii) triplicate ratio of 2a : 3b,
(iii) sub-duplicate ratio of 9x² a⁴ : 25y⁶ b²
(iv) sub-triplicate ratio of 216 : 343
(v) reciprocal ratio of 3 : 5
(vi) ratio compounded of the duplicate ratio of 5 : 6, the reciprocal ratio of 25 : 42 and the sub-duplicate ratio of 36 : 49.
Solution:

Question 5.
Find the value of x, if :
(i) (2x + 3) : (5x – 38) is the duplicate ratio of √5 : √6
(ii) (2x + 1) : (3x + 13) is the sub-duplicate ratio of 9 : 25.
(iii) (3x -7): (4x + 3) is the sub-triplicate ratio of 8 : 27.
Solution:

Question 6.
What quantity must be added to each term of the ratio x : y so that it may become equal to c : d ?
Solution:

Question 7.
A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 84 kg?
Solution:
Ratio in reduction of weight = 7 : 5
Originally weight of the woman = 84 kg
Reduced weight = \(\frac { 84 x 5 }{ 7 }\) = 60 kg

Question 8.
If 15 (2x² – y²) = 7xy, find x : y, if x and y both are positive.
Solution:

Question 9.
Find the :
(i) fourth proportional to 2xy, x² and y².
(ii) third proportional to a² – b² and a + b.
(iii) mean proportion to (x – y) and (x³ – x²y)
Solution:
(i) Let a be the fourth proportional
Then 2xy : x² :: y² : a

Question 10.
Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.
Solution:
Let x and y be the two numbers.
Then 14 is the mean proportional between x and y
xy = 14² => xy = 196 ….(i)
and 112 is the third proportional to x, y
y² = 112 x ….(ii)

Question 11.
If x and y be unequal and x : y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.
Solution:

Question 12.
If q is the mean proportional between p and r, prove that:

Solution:

Question 13.
If a, b and c are in continued proportion, prove that: a : c = (a² + b²) : (b² + c²).
Solution:

Question 14.

Solution:

Question 15.
If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that a : b = c : d.
Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.
There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3 : 1. How many more girls should be added to the council so that the ratio of number of boys to the number of girls may be 9 : 5 ?
Solution:
Total number of members = 36
Ratio in boys and girls = 3 : 1

Question 19.
If 7x – 15y = 4x + y, find the value of x : y. Hence, use componendo and dividendo to find the values of:

Solution:

Question 20.

Solution:

Question 21.
If x, y, z arc in continued proportion, prove that

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.
Using componendo and dividendo, find the value

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D

Question 1.
If a : b = c : d, prove that:
(i) 5a + 7b : 5a – 7b = 5c + 7d : 5c – 7d.
(ii) (9a + 13b) (9c – 13d) = (9c + 13d) (9a – 13b).
(iii) xa + yb : xc + yd = b : d.
Solution:

Question 2.
If a : b = c : d, prove that: (6a + 7b) (3c – 4d) = (6c + 7d) (3a – 4b).
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), prove that a : b = c : d.
Solution:

Question 6.

Solution:

Question 7.
If (a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d), prove that a : b = c : d.
Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
If a, b and c are in continued proportion, prove that

Solution:

Question 11.
Using properties of proportion, solve for x:

Solution:



But x = – 1 does not satisfy it
x = 1

Question 12.

Solution:

Question 13.
Using the properties of proportion, solve for x, given

Solution:

Question 14.

Solution:

Question 15.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D

Question 1.
Find the fourth proportional to :
(i) 1.5, 4.5 and 3.5
(ii) 3a, 6a² and 2ab²
Solution:
(i) Let x be the fourth proportional to 1.5, 4.5

Question 2.
Find the third proportional to :
(i) 2\(\frac { 2 }{ 3 }\) and 4
(ii) a – b and a² – b²
Solution:

Question 3.
Find the mean proportional between :
(i) 6 + 3√3 and 8 – 4√3
(ii) a – b and a³ – a²b.
Solution:

Question 4.
If x + 5 is the mean proportion between x + 2 and x + 9 ; find the value of x.
Solution:
x + 5 is the mean proportion between x + 2 and x + 9
(x + 5)² = (x + 2) (x + 9) {b² = ac}
⇒ x² + 10x + 25 = x² + 11x + 18
⇒ x² + 10x – x² – 11x = 18 – 25
⇒ -x = -7
⇒ x = 7
Hence x = 7

Question 5.
If x², 4 and 9 are in continued proportion, find x.
Solution:
x², 4 and 9 are in continued proportion

Question 6.
What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional. (2005, 2013)
Solution:
Let x to be added to each number then 6 + x, 15 + x, 20 + x and 43 + x are in proportion.

Question 7.

Solution:

Question 8.
What least number must be subtracted from each of the numbers 7,17 and 47 so that the remainders are in continued pro-portion ?
Solution:
Let x be subtracted from each of the numbers 7, 17 and 47.
Then 7 – x, 17 – x and 47 – x are in continued proportion.
7 – x : 17 – x : : 17 – x : 47 – x
⇒ (7 – x) (47 – x) = (17 – x) (17 – x)
⇒ 329 – 7x – 47x + x² = 289 – 17x – 17x + x²
⇒ -7x – 47x + x² + 17x + 17x – x² = 289 – 329
⇒ -54x + 34x = – 40
⇒ -20x = -40
⇒ x = 2
2 is to be the subtracted

Question 9.
If y is the mean proportional between x and z; show that xy + yz is the mean proportional between x² + y² and y² + z².
Solution:
y is the mean proportional between x and z.
y² = xz
Now, we have to prove that
xy + yz is the mean proportional between x² + y² and y² + z²
i.e., (xy + yz)² = (x² + y²) (y² + z²)
L.H.S. (xy + yz)² = [y(x + z)]² = y² (x + z)² = xz (x + z)²
R.H.S. (x² + y²) (y² + z²) = (x² + xz) (xz + z²) = x (x + z) z (x + z) = xz (x + z)²
L.H.S. = R.H.S.
Hence proved.

Question 10.
If q is the mean proportional between p and r, show that: pqr (p + q + r)³ = (pq + qr + pr)³.
Solution:
q is the mean proportional between p and r,
q² = pr
Now L.H.S. = pqr (p + q + r)³
= qq² (p + q + r)³
= q³ (p + q + r)³
= [q(p + q + r)]³
= (pq + q² + qr)³
= (pq + pr + qr)³
= (pq + qr + pr)³
= R.H.S.

Question 11.
If three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.
Solution:
Let x, y and z are three quantities which are
in continued proportion
Then x : y : y : z ⇒ y² = zx
Now, we have to prove that
x : z = x² : y² or xy² = zx²
L.H.S. = xy² = x x zx (y² = zx)
= x² z = R.H.S.
Hence Proved.

Question 12.
If y is the mean proportional between x and z, prove that:

Solution:

Question 13.
Given four quantities a, b, c and d are in proportion. Show that:
(a – c) b² : (b – d) cd = (a² – b² – ab) : (c² – d² – cd)
Solution:
a, b, c and d are in proportion Then a : b :: c : d

Question 14.
Find two numbers such that the mean proportional between them is 12 and the third proportional to them is 96.
Solution:
Let a and b be the two numbers, whose mean proportional is 12

Question 15.

Solution:

Question 16.
If p : q = r : s ; then show that: mp + nq : q = mr + ns : s.
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C

Question 1.
Show that (x – 1) is a factor of x³ – 7x² + 14x – 8. Hence, completely factorise the given expression.
Solution:

Question 2.
Using Remainder Theorem, factorise : x³ + 10x² – 37x + 26 completely. (2014)
Solution:
f(x) = x³ + 10x² – 37x + 26
f(1) = (1)³ + 10(1)² – 37(1) + 26 = 1 + 10 – 37 + 26 = 0
x = 1
x – 1 is factor of f(x)

Question 3.
When x³ + 3x² – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
Solution:
Let f(x) = x³ + 3x² – mx + 4
and x – 2 = 0 then x = 2
f(2) = (2)³ + 3(2)² – m(2) + 4 = 8 + 12 – 2m + 4 = 24 – 2m
Remainder = 24 – 2m
But, remainder is given m + 3
m + 3 = 24 – 2m
⇒ m + 2m = 24 – 3
⇒ 3m = 21
⇒ m = 7
Hence m = 7

Question 4.
What should be subtracted from 3x³ – 8x² + 4x – 3, so that the resulting expression has x + 2 as a factor ?
Solution:
The number to be subtracted = Remainder obtained by dividing 3x³ – 8x² + 4x – 3 by x + 2
Let f(x) = 3x³ – 8x² + 4x – 3
and x + 2 = 0, then x = – 2
Remainder = f(-2) = 3 (-2)³ – 8 (-2)² + 4 (-2) – 3 = -24 – 32 – 8 – 3 = -67
Hence the number to be subtracted = – 67

Question 5.
If (x + 1) and (x – 2) are factors of x³ + (a + 1) x² – (b – 2) x – 6, find the values of a and 6. And then, factorise the given expression completely.
Solution:

Question 6.
If x – 2 is a factor of x² – ax + b and a + b = 1, find the values of a and b.
Solution:
(x – 2) is a factor of x² + ax + b
Let x – 2 = 0 ⇒ x = 2
Now x² + ax + b = (2)² + a x 2 + b = 4 + 2a + b = 2a + b + 4
x – 2 is the factor Remainder = 0 or 2a + b + 4 = 0
⇒ 2a + b = -4 …(i)
But a + b = 1 (given) …(ii)
Subtracting, we get : a = -5
Substituting the value of a in (ii)
-5 + b = 1 ⇒ b = 1 + 5 ⇒ b = 6
Hence a = -5, b = 6

Question 7.
Factorise x³ + 6x² + 11x + 6 completely using factor theorem.
Solution:

Question 8.
Find the value of ‘m’ if mx³ + 2x² – 3 and x² – mx + 4 leave the same remainder when each is divided by x – 2.
Solution:
Let f(x) = mx³ + 2x² – 3
g (x) = x² – mx + 4
Let x – 2 = 0, then x = 2
f(2) = m (2)³ + 2 (2)² – 3 = 8m + 8 – 3 = 8m + 5
g(2) = (2)² – mx² + 4 = 4 – 2m + 4 = 8 – 2m
In both cases the remainder is same
8m + 5 = 8 – 2m
⇒ 8m + 2m = 8 – 5
⇒ 10m = 3
⇒ m = \(\frac { 3 }{ 10 }\)

Question 9.
The polynomial px³ + 4x² – 3x + q is completely divisible by x² – 1; find the values of p and q. Also, for these values of p and q, factorize the given polynomial completely.
Solution:

Question 10.
Find the number which should be added to x² + x + 3 so that the resulting polynomial is completely divisible by (x + 3).
Solution:
Let k be added to f(x), then f(x) = x² + x + 3 + k
Let x + 3 = 0, then x = -3
f(-3) = (-3)2 + (-3) + 3 + k = 9 – 3 + 3 + k = 9 + k
f(x) is divisible by x + 3, then remainder will be 0.
9 + k = 0 ⇒ k = -9
-9 should be added

Question 11.
When the polynomial x³ + 2x² – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x³ + ax² – 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.
Solution:
Let f(x) = x³ + 2x² – 5ax – 1
and let x – 1 = 0, then x = 1
f(1) = (1)³ + 2(1)² – 5a x 1 – 7 = 1 + 2 – 5a – 7 = -5a – 4
-5a – 4 = A ….(i)
Let g (x) = x³ + ax² +12x + 16
and let x + 2 = 0, then x = -2
g (-2) = (-2)³ + a (-2)² – 12 (-2) + 16 = -8 + 4a + 24 + 16 = 32 + 4a
32 + 4a = B ….(ii)
2A + B = 0
2 (-5a – 4) + 32 + 4a = 0
⇒ -10a – 8 + 32 + 4a = 0
⇒ -6a + 24 = 0
⇒ 6a = 24
⇒ a = 4
a = 4

Question 12.
(3x + 5) is a factor of the polynomial (a – 1) x³ + (a + 1) x² – (2a + 1) x – 15. Find the value of ‘a’. For this value of ‘a’, factorise the given polynomial completely.
Solution:
Let f(x) = (a – 1) x³ + (a + 1) x² – (2a + 1) x – 15

Question 13.
When divided by x – 3 the polynomials x³ – px² + x + 6 and 2x³ – x² – (p + 3) x – 6 leave the same remainder. Find the value of ‘p.’
Solution:
When (x – 3) divides x³ – px² + x + 6,
then Remainder = p(3) = (3)³ – p(3)² + (3) + 6 = 27 – 9p + 9 = 36 – 9p
When (x – 3) divides 2x³ – x² – (p + 3) x – 6,
then Remainder = p(3) = 2(3)³ – (3)² – (p + 3) (3) – 6
= 54 – 9 – 3p – 9 – 6 = 30 – 3p
A.T.Q. both remainders are equal
⇒ 36 – 9p = 30 – 3p
⇒ 36 – 30 = -3p + 9p
⇒ 6 = 6p
⇒ p = 1

Question 14.
Use the Remainder Theorem to factorise the following expression : 2x³ + x² – 13x + 6
Solution:
(a) By hit and trial, putting x = 2, we have
2 (8) + 4 – 26 + 6 = 0
⇒ (x – 2) is the factor of 2x³ + x² – 13x + 6

Question 15.
Using remainder theorem, find the value of k if on dividing 2x³ + 3x² – kx + 5 by x – 2, leaves a remainder 7. (2016)
Solution:
Let f(x) = 2x³ + 3x² – kx + 5 By the remainder theorem,
f(2) = 7
⇒ 2(2)³ + 3(2)² – k(2) + 5 = 7
⇒ 2(8) + 3(4) – k(2) + 5 = 7
⇒ 16 + 12 – 2k + 5 = 7
⇒ 2k = 16 + 12 + 5 – 7
⇒ 2k = 26
⇒ k = 13
The value of k is 13.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 8 Remainder and Factor Theorems Ex 8C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ex 7A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7D

Question 1.
If a : b = 5 : 3; find: \(\frac { 5a – 3b }{ 5a + 3b }\)
Solution:

Question 2.
If x : y = 4 : 7; find the value of (3x + 2y) : (5x + y).
Solution:

Question 3.
If a : b = 3 : 8, find the value of \(\frac { 4a + 3b }{ 6a – b }\)
Solution:

Question 4.
If (a – b): (a + b) = 1 : 11, find the ratio (5a + 4b + 15) : (5a – 4b + 3).
Solution:
(a – b) : (a + b) = 1 : 11
Let a – b = x, then a + b = 11x
Adding we get, 2a = 12x ⇒ a = 6x
Subtracting, -2b = -10x ⇒ b = 5x

Question 5.
Find the number which bears the same ratio to \(\frac { 7 }{ 33 }\) that \(\frac { 8 }{ 21 }\) does to \(\frac { 4 }{ 9 }\).
Solution:
Let x be the required number, then

Question 6.

Solution:

Question 7.
Find \(\frac { x }{ y }\) ; when x² + 6y² = 5xy.
Solution:

Question 8.
If the ratio between 8 and 11 is the same as the ratio of 2x – y to x + 2y, find the value of \(\frac { 7x }{ 9y }\)
Solution:

Question 9.
Divide ₹ 290 into A, B and C such that A is \(\frac { 2 }{ 5 }\) of B and B : C = 4 : 3.
Solution:
Total amount = ₹ 1290
A = \(\frac { 2 }{ 5 }\) B and B : C = 4 : 3
⇒ A : B = 2 : 5 and B : C = 4 : 3
LCM of 5, 4 = 20
A : B = 2 x 4 : 5 x 4 = 8 : 20
and B : C = 4 x 5 : 3 x 5 = 20 : 15
A : B : C = 8 : 20 : 15
Sum of ratios = 8 + 20 + 15 = 43

Question 10.
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:
Number of students = 630
Ratio in boys and girls = 3 : 2

Question 11.
What quantity must be subtracted from each term of the ratio 9 : 17, to make it equal to 1 : 3?
Solution:
Let x be subtracted from each term such that
\(\frac { 9 – x }{ 17 – x }\) = \(\frac { 1 }{ 3 }\)
⇒ 17 – x = 27 – 3x
⇒ -x + 3x = 21 – 17
⇒ 2x = 10
⇒ x = 5

Question 12.
The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each save Rs. 80 every month, find their monthly pocket money. [2012]
Solution:
Let the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.
Also, let their expenditure be 3y and 5y respectively. So,
5x – 3y = 80 …(i)
and 7x – 5y = 80 …(ii)
Multiplying (i) by 7 and (ii) by 5 and subtracting, we get
4y = 160 ⇒ y = 40
From (i), 5x = 80 + 3 x 40 = 200 ⇒ x = 40
So, monthly pocket money of Ravi = Rs. 5 x 40 = Rs. 200
and monthly pocket money of Sanjeev = Rs. 7 x 40 = Rs. 280

Question 13.
The work done by (x – 2) men in (4x + 1) days and the work done by (4x + 1) men in (2x – 3) days are in the ratio of 3 : 8. Find the value of x.
Solution:
(x – 2) men can do a work in = (4x +1) days
1 man can do a work in = (4x +1) (x-2) days ….(i)
Again (4x + 1) men can do work in = 2x – 3 days
1 man can do the work in = (2x – 3) (4x + 1) days ….(ii)
From (i) and (ii), we get
(4x + 1) (x – 2) : (2x – 3) (4x + 1) = 3 : 8
⇒ (4x + 1) (x – 2) x 8 = (2x – 3) (4x + 1) x 3
⇒ 8 (4x² – 8x + x – 2) – 3 (8x² + 2x – 12x – 3)
⇒ 32x² – 64x + 8x – 16 = 24x² + 6x – 36x – 9
⇒ 32x² – 64x + 8x – 24x² – 6x + 36x – 16 + 9 = 0
⇒ 8x² – 70x + 44x – 7 = 0
⇒ 8x² – 26x – 7 = 0
⇒ 8x² – 28x + 2x – 7 = 0
⇒ 4x ( 2x – 7) + 1 (2x – 7) = 0
⇒ (2x – 7) (4x + 1) = 0
Either 2x – 7 = 0, then x = \(\frac { 7 }{ 2 }\)
or 4x + 1 = 0, then x = \(\frac { -1 }{ 4 }\) but it is not possible.
x = \(\frac { 7 }{ 2 }\) or 3.5 Ans.

Question 14.
The bus fare between two cities is increased in the ratio 7 : 9. Find the increase in the fare, if :
(i) the original fare is Rs. 245 ;
(ii) the increased fare is Rs. 207.
Solution:
The increase in bus fare between two cities is in the ratio = 7 : 9.
(i) If the original fare is Rs. 245
then increase fare = Rs. 245 x \(\frac { 9 }{ 7 }\) = Rs. 315
Increase = Rs. 315 – Rs. 245 = Rs. 70
(ii) The increased fare is Rs. 207
Original fare = Rs. \(\frac { 207 x 7 }{ 9 }\) = Rs. 161
Increase = Rs. 207 – Rs. 161 = Rs. 46

Question 15.
By increasing the cost of entry ticket to a fair in the ratio 10 : 13; the number of visitors to the fair has decreased in the ratio 6 : 5. In what ratio has the total collection increased or decreased ?
Solution:
Increase in the entry tickets = 10 : 13.
But decrease in visitors = 6 : 5
Let original price of per ticket = Rs. 10
Then increased in price will be = Rs. 13
Collection in first case = Rs. 10 x 6 = Rs. 60
and collection in second case = Rs. 13 x 5 = Rs. 65
Hence increase in collection will be = Rs. 60 : Rs. 65 = 12 : 13

Question 16.
In a basket, the ratio between the number of oranges and the number of apples is 7 : 13. If 8 oranges and 11 apples are eaten, the ratio between the number of oranges and the number of apples becomes 1 : 2. Find the original number of oranges and the original number of apples in the basket.
Solution:
The ratio in number of oranges and apples = 7 : 13
Let number of oranges = 7x
Then number of apples = 13x
When 8 oranges and 11 apples are eaten, then
According to the sum,

Original number of oranges = 7x = 7 x 5 = 35
and number of apples = 13x = 13 x 5 = 65

Question 17.
In a mixture of 126 kg of milk and water, milk and water are in ratio 5 : 2. How much water must be added to the mixture to make this ratio 3 : 2?
Solution:
Mixture of milk and water = 126 kg
Ratio in milk and water = 5 : 2
Quantity of milk = \(\frac { 126 }{ 5 + 2 }\) x 5

Question 18.
(a) If A : B = 3 : 4 and B : C = 6 : 7, find:
(i) A : B : C
(ii) A : C
(b) If A : B = 2 : 5 and A : C = 3 : 4, find: A : B : C
Solution:


A : B : C = 6 : 15 : 8

Question 19.
(i) If 3A = 4B = 6C ; find A : B : C.
(ii) If 2a = 3b and 4b = 5c, find a : c
Solution:

Question 20.
Find the compound ratio of:
(i) 2 : 3, 9 : 14 and 14 : 27.
(ii) 2a : 3b, mn : x² and x : n.
(iii) √2 : 1, 3 : √5 and √20 = 9.
Solution:
(i) Compound ratio of 2 : 3, 9 : 14 and 14 : 27

Question 21.
Find the duplicate ratio of:
(i) 3 : 4
(ii) 3√3 : 2√5
Solution:

Question 22.
Find triplicate ratio of:
(i) 1 : 3
(ii) \(\frac { m }{ 2 }\) : \(\frac { n }{ 3 }\)
Solution:

Question 23.
Find sub-duplicate ratio of:
(i) 9 : 16
(ii) (x – y)⁴ : (x + y)⁶
Solution:

Question 24.
Find sub-triplicate ratio of:
(i) 64 : 27
(ii) x³ : 125y³
Solution:

Question 25.
Find the reciprocal ratio of :
(i) 5 : 8
(ii) \(\frac { x }{ 3 }\) : \(\frac { y }{ 7 }\)
Solution:

Question 26.
If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.
Solution:
(x + 3) : (4x + 1) is the duplicate ratio of 3 : 5

Question 27.
If m : n is the duplicate ratio of m + x : n + x; show that x² = mn.
Solution:

Question 28.
If (3x – 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.
Solution:
(3x – 9) : (5x + 4) is the triplicate ratio of 3 : 4

⇒ 27 (5x + 4) = 64 (3x – 9)
⇒ 135x + 108 = 192x – 576
⇒ 192x – 135x = 108 + 576
⇒ 57x = 684
⇒ x = 12

Question 29.
Find the ratio compounded of the reciprocal ratio of 15 : 28, the sub-duplicate ratio of 36 : 49 and the triplicate ratio of 5 : 4.
Solution:
Reciprocal ratio of 15 : 28

Question 30.
(a) If r² = pq, show that p : q is the duplicate ratio of (p + r) : (q + r).
(b) If (p – x) : (q – x) be the duplicate ratio of p : q then show that : \(\frac { 1 }{ p }\) + \(\frac { 1 }{ q }\) = \(\frac { 1 }{ x }\)
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems (Based on Quadratic Equations) Ex 6E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E

Question 1.
The distance by road between two towns A and B is 216 km., and by rail it is 208 km. A car travels at a speed of x km/hr. and the train travels at a speed which is 16 km/hr faster than the car. Calculate:
(i) the time taken by the car to reach town B from A, in terms of x ;
(ii) the time taken by the train, to reach town B from A, in terms of x.
(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x, and solve it.
(iv) Hence, find the speed of the train. [1998]
Solution:
Distance between two stations by road = 216 km and by rail = 208 km.
Speed of car = x km/hr.
and speed of train = (x + 16) km/ hr.

⇒ 8x + 3456 = 2×2 + 32x
⇒ 2x² + 32x – 8x – 3456 = 0
⇒ 2x² + 24x – 3456 = 0
⇒ x² + 12x – 1728 = 0 (Dividing by 2)
⇒ x² + 48x – 36x – 1728 = 0
⇒ x (x + 48) – 36 (x + 48) = 0
⇒ (x + 48 ) (x – 36) = 0
Either x + 48 =0, then x = – 48 which is not possible.
or x – 36 = 0, then x = 36
(iv) Speed of train = x + 16 = 36 + 16 = 52 km/hr.

Question 2.
A trader buys x articles for a total cost of Rs. 600.
(i) Write down the cost of one article in terms of x. If the cost per article were Rs. 5 more, the number of articles that can be bought for Rs. 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it for x. [1999]
Solution:
C.P. of x articles = Rs. 600

⇒ 600x = (x – 4) (600 + 5x) (By cross multiplication)
⇒ 600x = 600x + 5x² – 2400 – 20x
⇒ 5x² – 20x – 2400 = 0
(ii) x² – 4x – 480 = 0
⇒ x² – 24x + 20x – 480 = 0
⇒ x (x – 24) + 20 (x – 24) = 0
⇒ (x – 24) (x + 20) = 0 (Zero Product Rule)
Either x – 24 = 0, then x = 24
or x + 20 = 0, then x = -20 Which is not possible.
Hence no. of articles = 24

Question 3.
A hotel bill for a number of people for overnight stay is Rs. 4,800. If there were 4 people more, the bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight. [2000]
Solution:
Amount of the bill = Rs. 4800
Let the number of persons staying overnight = x
Then amount to be paid by each person


⇒ x (x + 12) – 8 (x + 12) = 0
⇒ (x + 12) (x – 8) = 0 (Zero Product Rule)
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 8 = 0, then x = 8
Hence no. of persons staying overnight = 8

Question 4.
An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression lor the time taken for :
(i) the omvard journey;
(ii) the return journey.
If the return journey took 30 minutes less than the on ward journey, write down an equation in x and find its value. |2002]
Solution:
Distance between A and B = 400 km.
Speed of aeroplane onward journey = x km/hr.
and Speed of aeroplane on return journey = (x + 40) km/hr.
Now time taken for onward journey = \(\frac { 400 }{ x }\) hrs.


Which is not possible being negative
or x – 160 = 0, then x = 160
x = 160

Question 5.
Rs. 6,500 were divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs. 30 less. Find the original number of persons.
Solution:
Let original number of persons = x
Amount = Rs. 6,500

⇒ 30x² + 450x – 97500 = 0 (Dividing by 30)
⇒ x² + 15x – 3250 = 0
⇒ x² + 65x – 50x – 3250 = 0
⇒ x (x + 65) – 50 (x + 65) = 0
⇒ (x + 65) (x – 50) = 0
Either x + 65 = 0, then x = -65 which is not possible.
or x – 50 = 0, then x = 50
Original number of persons = 50

Question 6.
A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km. away in time, it has to increase its speed by 250 km./hr. from its usual speed. Find its usual speed.
Solution:
Let the usual speed of plane = x km/hr.
Distance = 1500 km.
Increased speed = (x + 250) km./hr.
Now, according to the condition,

⇒ x² + 250x – 750000 = 0
⇒ x² + 1000x – 750x – 750000 = 0
⇒ x (x + 1000) – 750 (x + 1000) = 0
⇒ (x + 1000) (x – 750) = 0
Either x + 1000 = 0, then x = – 1000 But it is not possible.
or x – 750 = 0, then x = 750
Usual speed of plane = 750 km/hr

Question 7.
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the average speed of each train.
Solution:
Let the speed of first train = x km/hr.
Then speed of second train = (x – 5) km/ hr.
In 2 hours, first train will travel = 2x km.
and second train will travel = 2 (x – 5) km.
According to the condition,
2x + 2 (x – 5) = 50
⇒ 2x + 2x – 10 = 50
⇒ 4x = 50 + 10 = 60
x = 15 km/hr.
and speed of second train = 15 – 5 = 10 km/hr.

Question 8.
The sum S of first n even natural numbers is given by the relation S = n (n + 1). Find n if the sum is 420.
Solution:
S = n (n + 1) and x = 420
⇒ n (n + 1) = 420
⇒ n² + n – 420 = 0
⇒ n² + 21n – 20n – 420 = 0
⇒ n (n + 21) – 20 (n + 21) = 0
⇒ (n + 21) (n – 20) = 0
Either n + 21 =0, then n = -21 which is not possible as it is negative
or n – 20 = 0 then n = 20

Question 9.
The Sum of the ages of a father and his son is 45 years. Five year ago, the product of their ages (in years) was 124. Determine their present ages.
Solution:
Let age of father = x years
Then age of his son = (45 – x) years (sum = 45 years)
5 years ago,
The age of father = (x – 5) years
and age of son = 45 – x – 5 = (40 – x) years
According to the condition,
(x – 5) (40 – x) = 124
⇒ 40x – x² – 200 + 5x = 124
⇒ -x² + 45x – 200 – 124 = 0
⇒ -x² + 45x – 324 = 0
⇒ x² – 45x + 324 = 0
⇒ x² – 36x – 9x + 324 = 0
⇒ x (x – 36) – 9 (x – 36) = 0
⇒ (x – 36) (x – 9) = 0
Either x – 36 = 0, then x = 36
or x – 9 = 0, then x = 9, but it is not possible as age of father cannot be less than his son.
Age of father = 36 years
and age of son = 45 – 36 = 9 years

Question 10.
In an auditorium, seats were arranged in rows and columns. The number of rows w as equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement. [2003]
Solution:
Let the number of rows in the auditorium = x
No. of seats in each row = x
and no. of total seats in the auditorium =
x x x = x²
In second case,
No. of rows = 2x
and no. of seats in each row = x – 10
Then the total seats will = x² + 300
Now, according to the condition,
2x (x – 10) = x² + 300
⇒ 2x² – 20x = x² + 300
⇒ 2x² – x² – 20x – 300 = 0
⇒ x² – 30x + 10x – 300 = 0
⇒ x (x – 30) + 10 (x – 30) = 0
⇒ (x – 30) (x + 10) = 0 (Zero Product Rule)
Either x – 30 = 0, then x = 30
or x + 10 = 0, then x = -10 Which not possible.
(i) No. of rows in original arrangement = 30
(ii) and no. of seats after re-arrangements = x² + 300 = (30)2 + 300 = 900 + 300 = 1200

Question 11.
Mohan takes 16 days less than Manoj to do a piece of work. If both working together can do it in 15 days, how many days will Mohan alone complete the work?
Solution:
Let time taken by Mohan = x days
Time taken by Manoj = (x + 16) days

⇒ 30x + 240 = x² + 16x
⇒ x² + 16x – 30x – 240 = 0
⇒ x² – 14x – 240 = 0
⇒ x² – 24x + 10x – 240 = 0
⇒ x (x – 24) + 10 (x – 24) = 0
⇒ (x + 10) (x – 24) = 0
Either x + 10 = 0, then .x = -10 But it is not possible
or x – 24 = 0, then x = 24
Mohan can do the work in = 24 days

Question 12.
Two years ago. a man’s age was three times the square of his son’s age. In three years time, his age will be four times his son’s age. Find their present ages.
Solution:
2 years ago,
let son’s age = x
Man’s age = 3x
Son’s present age = x + 2
and man’s age = 3x² + 2
and 3 years after,
Son’s age = x + 2 + 3 = x + 5
Man’s age 3x² + 2 + 3 = 3x² + 5
According to condition,
3x² + 5 = 4 (x + 5)
⇒ 3x² + 5 = 4x + 20
⇒ 3x² – 4x + 5 – 20 = 0
⇒ 3x² – 4x – 15 = 0
⇒ 3x² – 9x + 5x – 15 = 0
⇒ 3x (x – 3) + 5 (x – 3) = 0
⇒ (x – 3) (3x + 5) = 0
Either x – 3 = 0, then x = 3
or 3x + 5 = 0, then 3x = -5 ⇒ x = \(\frac { -5 }{ 3 }\)
But it is not possible.
x = 3
Son’s present age = x + 2 = 3 + 2 = 5 years
and man’s present age = 3x² + 2 = 3(3)² + 2 = 27 + 2 = 29 years

Question 13.
In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both. the fraction reduces by \(\frac { 1 }{ 14 }\) Find the fraction.
Solution:
In first case.
Let numerator of a fraction = x
then, its denominator = x + 3

Question 14.
In a two digit number, the ten’s digit is bigger. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Solution:
Difference of digits = 6
Let one’s digit = x
Then ten’s digit = x + 6
and number = x + 10 (x + 6) = x + 10x + 60 = 11x + 60
But product of digits = 27
x (x + 6) = 27
⇒ x² + 6x – 27 = 0
⇒ x² + 9x – 3x – 27 = 0
⇒ x (x + 9) – 3 (x + 9) = 0
⇒ (x + 9) (x – 3) = 0
Either x + 9 = 0, then x = – 9 But it is not possible
or x – 3 = 0, then x = 3
Number = 11x + 60 = 11 x 3 + 60 = 33 + 60 = 93

Question 15.
Some school children on an excursion by a bus to a picnic spot at a distance of 300 km. While returning, it was raining and the bus had to reduce its speed by 5 km/ hr and it took two hours longer for returning. Find the time taken to return.
Solution:
Distance = 300 km.
Let speed of the bus = x km/hr.

⇒ x² – 5x = 750
⇒ x² – 5x – 750 = 0
⇒ x² – 30x + 25x – 750 = 0
⇒ x (x – 30) + 25 (x – 30) = 0
⇒ (x – 30) (x + 25) = 0
Either x – 30 = 0, then x = 30
or x + 25 = 0, then x = -25, but it is not possible as it is negative
Speed of the bus = 30 km/hr
and time taken while returning = \(\frac { 300 }{ x }\) = \(\frac { 300 }{ 25 }\) = 12 hours

Question 16.
Rs. 480 is divided equally among ‘x’ children. If the number of children were 20 more then each would have got Rs. 12 less. Find ‘x’.
Solution:
Total amount = Rs. 480
Number of children = x

⇒ x² + 20x – 800 = 0
⇒ x² + 40x – 20x – 800 – 0
⇒ x (x + 40) – 20 (x + 40) = 0
⇒ (x + 40) (x – 20) = 0
Either x + 40 = 0, then x = – 40 which is not possible being negative
or x – 20 = 0 then x = 20
Number of children = 20

Question 17.
A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to covers the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x’ (2016)
Solution:
Let the original speed be x km/hr
Time taken by the bus with moving at speed x km/h = \(\frac { 240 }{ x }\)
Time taken by the bus with moving at speed (x – 10) km/h = \(\frac { 240 }{ x – 10 }\)
According to the given condition,

⇒ x (x – 10) = 10 x 120
⇒ x² – 10x = 1200
⇒ x² – 10x – 1200 = 0
⇒ x² – 40x + 30x – 1200 = 0
⇒ x (x – 40) + 30 (x – 40) = 0
⇒ (x – 40) (x + 30) = 0
⇒ x – 40 = 0 or x + 30 = 0
⇒ x = 40 or x = -30
Since, the speed cannot be negative, the uniform speed is 40 km/h

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems (Based on Quadratic Equations) Ex 6D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E

Question 1.
The sum S of n successive odd numbers starting from 3 is given by the relation :
S = n (n + 2). Determine n, if the sum is 168.
Solution:
S = n (n + 2) and S = 168
⇒ n (n + 2) = 168
⇒ n² + 2n – 168 = 0
⇒ n² + 14n – 12n – 168 = 0
⇒ n (n + 14) – 12 (n + 14) = 0
⇒ (n + 14) (n – 12) = 0
Either n + 14 = 0, then n = -14 which is not possible as n is positive.
or n – 12 = 0, then n = 12
Hence n = 12

Question 2.
A stone is thrown vertically downwards and the formula d = 16t² + 4t gives the distance, d metres, that it falls in t seconds. How long does it take to fall 420 metres ?
Solution:
d = 16t² + 4t, d = 420 m
Distance = 420 m.
6t² + 4t = 420
⇒ 16t² + 4t – 420 = 0
⇒ 4t² + t – 105= 0 (Dividing by 4)
⇒ 4t² + 21t – 20t – 105 = 0
⇒ t (4t + 21) – 5 (4t + 21) = 0
⇒ (4t + 21) (t – 5) = 0
Either 4t + 21 = 0, then 4t = -21 ⇒ t = \(\frac { -21 }{ 4 }\)
But it is not possible as time can not be negative.
or t – 5 = 0 , then t = 5
t = 5 seconds

Question 3.
The product of the digits of two digit number is 24. If its unit’s digit exceeds twice its ten’s digit by 2; find the number.
Solution:
Let ten’s digit = x
then unit’s digit = 2x + 2
According to the condition,
x (2x + 2) = 24
⇒ 2x² + 2x – 24 = 0
⇒ x² + x – 12 = 0 (Dividing by 2)
⇒ x² + 4x – 3x – 12 = 0
⇒ x (x + 4) – 3 (x + 4) = 0
⇒ (x + 4) (x – 3) = 0
Either x + 4 = 0, then x = – 4, which is not possible.
x – 3 = 0, then x = 3.
Ten’s digit = 3
and unit’s digit = 3 x 2 + 2 = 6 + 2 = 8
Number = 8 + 10 x 3 = 8 + 30 = 38

Question 4.
The ages of two sisters are 11 years and 14 years. In how many years time will the product of their ages be 304 ?
Solution:
Let the number of years = x
Age of first sister = 11 + x
and of second sister = 14 + x
Now according to the condition,
(11 + x) ( 14 + x) = 304
⇒ 154 + 11x + 14x + x² = 304
⇒ x² + 25x – 150 = 0
⇒ x² + 30x – 5x – 150 = 0
⇒ x (x + 30) – 5 (x + 30 ) = 0
⇒ (x + 30) (x – 5) = 0
Either x + 30 = 0 , then x = -30 But it is not possible as can’t be in negative
or x – 5 = 0 , then x = 5
Number of years = 5

Question 5.
One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.
Solution:
One year ago, let the age of son = x years
and age of his father = 8x.
But present age of father is = (8x + 1) years
8x + 1 = (x + 1)²
⇒ x² + 2x + 1 = 8x + 1
⇒ x² + 2x + 1 – 8x – 1 = 0
⇒ x² – 6x = 0
⇒ x (x – 6) = 0
Either x = 0, which is not possible.
or x – 6 = 0, then x = 6
Present age of father = 8x + 1 = 8 x 6 + 1 = 48 + 1 = 49 years.
and age of son = x + 1 = 6 + 1 = 7 years

Question 6.
The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.
Solution:
Let age of son = x
Then age of father will be = 2x²
8 years hence,
age of son = x + 8
and age of father = 2x² + 8
According to the condition,
2x² + 8 = 3 (x + 8) + 4
⇒ 2x² + 8 = 3x + 24 + 4
⇒ 2x² + 8 – 3x – 28 = 0
⇒ 2x² – 3x – 20 = 0
⇒ 2x² – 8x + 5x – 20 = 0
⇒ 2x (x – 4) + 5 (x – 4) = 0
⇒ (x – 4) (2x + 5) = 0
Either x – 4 = 0, then x = 4
or 2x + 5 = 0, then 2x – 5 ⇒ x = \(\frac { -5 }{ 2 }\)
Which is not possible being negative
x = 4
Present age of son = 4 years
and age of father = 2x² = 2 (4)² = 2 x 16 = 32 years

Question 7.
The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return down-stream to the original point in 4 hours 30 minutes, find the speed of the stream.
Solution:
Let the speed of stream = x km/hr.
Distance = 30 km.
Speed of boat in still water = 15 km/hr.

⇒ 9x² – 2025 + 1800
⇒ 9x² – 225 = 0
⇒ x² – 25 = 0
⇒ (x)² – (5)² = 0
⇒ (x + 5) (x – 5) = 0
Either x + 5 = 0, then x = -5 which is not possible.
or x – 5 = 0, then x = 5
Speed of stream = 5 km/hr.

Question 8.
Mr. Mehra sends his servant to the market to buy oranges worth Rs. 15. The servant having eaten three oranges on the way, Mr. Mehra pays 25 paise per orange more than the market price. Taking x to be the number of oranges which Mr. Mehra receives, form a quadratic equation in x. Hence, find the value of x.
Solution:
No. of oranges received by Mr. Mehra = x
No. of oranges eaten by the servant = 3
Total no. of oranges bought = x + 3
Total cost = Rs. 15
Price of one orange = Rs. \(\frac { 15 }{ x + 3 }\)
Now according to the sum,

⇒ x² + 3x = 180
⇒ x² + 3x – 180 = 0
⇒ x² + 15x – 12x – 180 = 0
⇒ x (x + 15) – 12 (x + 15) = 0
⇒ (x + 15) (x – 12) = 0
Either x + 15 = 0, then x = – 15 which is not possible
or x – 12 = 0, then x = 12
x = 12

Question 9.
Rs. 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.
Solution:
Let the number of children = x
Amount to be divided = Rs. 250

⇒ 6250 x 2 = x² + 25x
⇒ x² + 25x – 12500 = 0
⇒ x² + 125x – 100x – 12500 = 0
⇒ x (x + 125) – 100 (x + 125) = 0
⇒ (x + 125) (x – 100) = 0
Either x + 125 = 0 then x = -125 which is not possible.
or x – 100 = 0, then x = 100
No. of children = 100

Question 10.
An employer finds that if he increases the weekly wages of each worker by Rs. 5 and employs five workers less, he increases his weekly wage bill from Rs. 3,150 to Rs. 3,250. Taking the original weekly wage of each worker as Rs. x; obtain an equation in* and then solve it to find the weekly wages of each worker.
Solution:
In first case,
Let weekly wages of each employee = Rs. x
and number of employees = y
and weekly wages = 3150
xy = 3150 ⇒ y = \(\frac { 3150 }{ x }\) ….(i)
In second case,
Weekly wages = x + 5
and number of employees = y – 5
and weekly wages = 3250
(x + 5) (y – 5) = 3250
⇒ xy + 5y – 5x – 25 = 3200

⇒x (x + 70) – 45 (x + 70) = 0
⇒ (x + 70) (x – 45) = 0
Either x + 70 = 0, then x = -70 which is not possible being negative
or x – 45 = 0, then x = 45
Weekly wages per worker = Rs. 45

Question 11.
A trader bought a number of articles for Rs. 1,200. Ten were damaged and he sold each of the remaining articles at Rs. 2 more than what he paid for it, thus getting a profit of Rs. 60 on the whole transaction. Taking the number of articles he bought as x, form an equation in x and solve it.
Solution:
Let number of articles = x
C.P. = Rs. 1200
Profit = Rs. 60
S.P. = Rs. 1200 + 60 = Rs. 1260
No. of articles damaged = 10
Remaining articles = x – 10

⇒ 2x² – 80x – 12000 = 0
⇒ x² – 40x – 6000 = 0 (Dividing by 2)
⇒ x² – 100x + 60x – 6000 = 0
⇒ x (x – 100) + 60 (x – 100) = 0
⇒ (x – 100) (x + 60) = 0
Either x – 100 = 0, then x = 100
or x + 60 = 0, then x = – 60 which is not possible.
Number of articles = 100

Question 12.
The total cost price of a certain number of identical articles is Rs. 4,800. By selling the article at Rs. 100 each, a profit equal to the cost price of 15 articles is made. Find the number of articles bought.
Solution:
Total cost of some articles = Rs. 4800
Let number of articles = x
S.P. of one article = Rs. 100
S.P. of x articles = Rs. 100x
Profit = Cost price of 15 articles

⇒ x² = 48x + 720 (Dividing by 100)
⇒ x² – 48x – 720 = 0
⇒ x² – 60x + 12x – 720 = 0
⇒ x (x – 60) + 12 (x – 60) = 0
⇒ (x – 60) (x + 12) = 0
Either x – 60 = 0, then x = 60
or x + 12 = 0, then x = -12 Which is not possible.
x = 60
Number of articles = 60

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems (Based on Quadratic Equations) Ex 6C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6D
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6E

Question 1.
The speed of an ordinary train is x km/hr and that of an express train is (x + 25) km per hour.
(i) Find the time taken by each train to cover 300 km.
(ii) If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.
Solution:
Speed of an ordinary train = x km/ hr
and speed of an express train = (x + 25) km/hr.
(i) Time taken by ordinary train to cover 300 km = \(\frac { 300 }{ x }\) hr
(ii) Time taken by express train to cover 300 km = \(\frac { 300 }{ x + 25 }\) hr

⇒ 7500 = 2x² + 50x
⇒ 2x² + 50x – 7500 = 0
⇒ x² + 25x – 3750 = 0 (Dividing by 2)
⇒ x² + 75x – 50x – 3750 = 0
⇒ x (x + 75) -50 (x + 75) = 0
⇒ (x + 75) (x – 50) = 0
Either x + 75 = 0 then x = -75 But it is not possible,
or x – 50 = 0 then x = 50
Speed of an ordinary train = 50 km/ hr
and speed of express train = 50 + 25 = 75 km/ hr

Question 2.
If the speed of a car is increased by 10 km per hr, it takes 18 minutes less to cover a distance of 36 km. Find the speed of the car.
Solution:
Let the speed of ear = x km/hr
time taken to cover 36 km = \(\frac { 36 }{ x }\) hr
In second case,
Speed of car = (x + 10) km/hr

⇒ 3x² + 30 x = 3600
⇒ 3x² + 30x – 3600 = 0 .
⇒ x² + 10x – 1200 = 0 (Dividing by 3)
⇒ x² + 40x – 30x – 1200 = 0
⇒ x (x + 40) – 30 (x + 40) = 0
⇒ (x + 40) (x – 30) = 0
Either x + 40 = 0 then x = – 40 But it is not possible,
or x – 30 = 0, then x = 30
Speed of car = 30 km/hr.

Question 3.
If the speed of an aeroplane is reduced by 40 km/per hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.
Solution:
Let the speed of aeroplane = x km/hr.

⇒ x² – 40x = 144000
⇒ x² – 40x – 144000 = 0
⇒ x² – 400x + 360x – 144000 = 0
⇒ x (x – 400) + 360 (x – 400) = 0
Either x – 400 = 0, then x = 400
or x + 360 = 0, then x = – 360, But it is not possible.
x = 400
Hence speed of aeroplane = 400 km/hr.

Question 4.
A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Solution:
Let the original speed of the car be x km/h.

⇒ 1200x + 14400 – 1200x = 5x² + 60x
⇒ 14400 = 5x² + 60x
⇒ x² + 12x – 2880 = 0 (Dividing by 5)
⇒ x² + 60x – 48x – 2880 = 0
⇒ x (x + 60) – 48 (x + 60) = 0
⇒ (x + 60) (x – 48) = 0
⇒ x = 48 or x = -60
⇒ x = 48 (Rejecting x = -60, being speed)
Hence, original speed of the car = 48 km/h.

Question 5.
A girl goes to her friend’s house, which is at a distance of 12 km. She covers half of the distance at a speed of ‘x’ km/hr. and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs. 30 minutes to cover the whole distance; find ‘x’
Solution:
Distance = 12 km.
Speed for the first half distance = x km/hr.
and for the second half distance = (x + 2) km/hr.
Total time taken = 2 hrs. 30 minutes.

But it is not possible.
Speed for first half distance (x) = 4 km/hr

Question 6.
A car made a run of 390 km in ‘x’ hours. If the speed had been 4 km per hour more, it would have taken 2 hours less for the journey. Find ‘x’.
Solution:
Distance = 390 km
time = x hours

⇒ (x – 15) (x + 13) = 0
Either x – 15 = 0, then x = 15
or (x + 13) = 0 then x = -13 Which is not possible.
Value of x = 15

Question 7.
A goods train leaves a station at 6 p.m., followed by an express train which leaves at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the trains remain constant between the two stations; calculate their speeds.
Solution:
Departure of goods train = 6 p.m.
and departure of express train = 8 p.m.
Speed of express train is more than goods trains by 20 km/hr
Total distance = 1040 km
Let speed of goods train = x km/hr
Then speed of express train = (x + 20) km/hr
Difference of time taken = 8 p.m. – 6 p.m. + 36 minutes = 2 hours 36 minutes

⇒ x – (x + 100) – 80 (x + 100) = 0
⇒ (x + 100) (x – 80) = 0
Either x + 100 = 0, then x = -100 but it is not possible being negative
or x – 80 = 0, then x = 80
Speed of goods train = 80 km/hr
and speed of express train = 80 + 20 = 100 km/hr
⇒ 13x² + 507x – 35100 = 0
⇒ x² + 39x – 2700 = 0 (Dividing by 13)
⇒ x² + 75x – 36x – 2700 = 0
⇒ x (x + 75) – 36(x + 75) = 0
⇒ (x + 75) (x – 36) = 0
Either x + 75 = 0, then x = – 75 Which is not possible,
or x – 36 = 0 then x = 36
Speed of goods train = 36 km/hr
and speed of express train = 36 + 39 = 75 km/hr.

Question 8.
A man bought an article for Rs. x and sold it for Rs. 16. If his loss was x percent, find the cost price of the article.
Solution:
C.P. of article = Rs. x
S.R = Rs. 16
Loss = C.P. – S.P. = Rs. (x – 16)

⇒ x² – 100x + 1600 = 0
⇒ x² – 20x – 80x + 1600 = 0
⇒ x (x – 20) – 80 (x – 20) = 0
⇒ (x – 20) (x – 80) = 0
Either x – 20 = 0, then x = 20
or x – 80 = 0, then x = 80
Cost Price = Rs. 20 or Rs. 80

Question 9.
A trader bought an article for Rs. x and sold it for Rs. 52, thereby making a profit of (x – 10) per cent on his outlay. Calculate the cost price.
Solution:
Let C.P. = Rs. x
S.R = Rs. 52
Profit = S.P – C.P. = Rs. 52 – x

x² – 10x = 5200 – 100x
⇒ x² – 10x + 100x – 5200 = 0
⇒ x² + 90x – 5200 = 0
⇒ x2 + 130x – 40x – 5200 = 0
⇒ x (x + 130) – 40(x + 130) = 0
⇒ (x + 130) (x – 40 ) = 0
Either x + 130 = 0 , then x = – 130 which is not possible.
or x – 40 = 0 then x = 40
Cost price = Rs. 40

Question 10.
By selling a chair for Rs. 75, Mohan gained as much per cent as its cost. Calculate the cost of the chair.
Solution:
Let. C.P of chair = Rs. x
Profit = x %
S.P. = Rs. 75
Total profit = Rs. (75 – x)

⇒ x² = 7500 – 100x
⇒ x² + 100x – 7500 = 0
⇒ x²+ 150x – 50 – 7500 = 0
⇒ x (x + 150) – 50 (x + 150) = 0
⇒ (x + 150) (x – 50) = 0
Either x + 150 = 0, then x = -150 which is not possible
or x – 50 = 0, Then x = 50
Cost price of chair = Rs. 50

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.