Maths RD Sharma Solutions Class 10

RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
Find two consecutive numbers whose squares have the sum 85. (C.B.S.E. 2000)
Solution:
Let first number = x
Then second number = x + 1
According to the condition
x² + (x + 1)² = 85
=> x² + x² + 2x + 1 = 85
=> 2x² + 2x + 1 – 85 = 0
=> 2x² + 2x – 84 = 0
=> x² + x – 42 = 0
=> x² + 7x – 6x – 42 = 0
=> x (x + 7) – 6 (x + 7) = 0
=> (x + 7) (x – 6) = 0
Either x + 7 = 0, then x = -7 or x – 6 = 0, then x = 6
(i) If x = -7, then the first number = -7 and second number = -7 + 1 = -6
(ii) If x = 6, then the first number = 6 and second number = 6 + 1 = 7
Hence numbers are -7, -6 or 6, 7

Question 2.
Divide 29 into two parts so that the sum of the squares of the parts is 425.
Solution:
Total = 29
Let first part = x
Then second part = 29 – x
According to the condition
x² + (29 – x)² = 425
=> x² + 841 + x² – 58x = 425
=> 2x² – 58x + 841 – 425 = 0
=> 2x² – 58x + 416 = 0
=> x² – 29x + 208 = 0 (Dividing by 2)
=> x² – 13x – 16x + 208 = 0
=> x(x – 13) – 16 (x – 13) = 0
=> (x – 13) (x – 16) = 0
Either x – 13 = 0, then x = 13 or x – 16 = 0, then x = 16
(i) If x = 13, then First part =13 and second part = 29 – 13 = 16
(ii) If x = 16, then First part =16 and second part = 29 – 16 = 13
Parts are 13, 16

Question 3.
Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm². Find the sides of the squares. (C.B.S.E. 1997)
Solution:
Side of the first square = x cm
Its area = (side)² = x² cm²
Side of the second square = (x + 4) cm
Its area = (x + 4)² cm²
According to the condition,
x² + (x + 4)² = 656
=> x² + x² + 8x + 16 = 656
=> 2x² + 8x + 16 – 656 = 0
=> 2x² + 8x – 640 = 0
=> x² + 4x – 320 = 0 (Dividing by 2)
=> x² + 20x – 16x – 320 = 0
=> x (x + 20) – 16 (x + 20) = 0
=> (x + 20) (x – 16) = 0
Either x + 20 = 0, then x = -20 Which is not possible being negative
or x – 16 = 0, then x = 16
Side of the first square = 16 cm
and side of the second square = 16 + 4 = 20 cm

Question 4.
The sum of two numbers is 48 and their product is 432. Find the numbers.
Solution:
Sum of two numbers = 48
Let first number = x
The second number = 48 – x
According to the condition,
x (48 – x) = 432
=> 48x – x² = 432
=> – x² + 48x – 432 = 0
=> x² – 48x + 432 = 0
=> x² – 12x – 36x + 432 = 0
=> x (x – 12) – 36 (x – 12) = 0
=> (x – 12) (x – 36) = 0
Either x – 12 = 0, then x = 12 or x – 36 = 0, then x = 36
(i) If x = 12, then First number = 12 and second number = 48 – 12 = 36
(ii) If x = 36, then First number = 36 and second number = 48 – 36 = 12
Numbers are 12, 36

Question 5.
If an integer is added to its square, the sum is 90. Find the integer with the help of quadratic equation.
Solution:
Let the given integer be = x
According to the condition
x² + x = 90
=> x² + x – 90 = 0
=> x² + 10x – 9x – 90 = 0
=> x (x + 10) – 9 (x + 10) = 0
=> (x + 10) (x – 9) = 0
Either x + 10 = 0, then x = -10 or x – 9 = 0, then x = 9.
The integer will be -10 or 9

Question 6.
Find the whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.
Solution:
Let the given whole number = x
Then its reciprocal = \(\frac { 1 }{ x }\)
According to the condition,
x – 20 = 69 x \(\frac { 1 }{ x }\)
=> x – 20 = \(\frac { 69 }{ x }\)
=> x² – 20x = 69
=> x² – 20x – 69 = 0
=> x² – 23x + 3x – 69 = 0
=> x (x – 23) + 3 (x – 23) = 0
=> (x – 23) (x + 3) = 0
Either x – 23 = 0, then x = 23
or x + 3 = 0, then x = -3, but it is not a whole number
Required whole number = 23

Question 7.
Find two consecutive natural numbers whose product is 20.
Solution:
Let first natural number = x
Then second number = x + 1
According to the condition,
x (x + 1) = 20
=> x² + x – 20 = 0
=> x² + 5x – 4x – 20 = 0
=> x (x + 5) – 4 (x + 5) = 0
=> (x + 5) (x – 4) = 0
Either x + 5 = 0, then x = -5 which is not a natural number
or x – 4 = 0, then x = 4
First natural number = 4 and second number = 4 + 1=5

Question 8.
The sum of the squares of two consecutive odd positive integers is 394. Find them.
Solution:
Let first odd number = 2x + 1
Then second odd number = 2x + 3
According to the condition
(2x + 1)² + (2x + 3)² = 394
=> 4x² + 4x + 1 + 4x² + 12x + 9 = 394
=> 8x² + 16x + 10 = 394
=> 8x² + 16x + 10 – 394 = 0
=> 8x² + 16x – 384 = 0
=> x² + 2x – 48 = 0 (Dividingby8)
=> x² + 8x – 6x – 48 = 0
=> x(x + 8) – 6(x + 8) = 0
=> (x + 8) (x – 6) = 0
Either x + 8 = 0, then x = 8 but it is not possible as it is negative
or x – 6 = 0, then x = 6
First odd number = 2x + 1 = 2 x 6 + 1 = 13
and second odd number = 13 + 2 = 15

Question 9.
The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.
Solution:
Sum of two numbers = 8
Let first number = x
Then second number = 8 – x
According to the condition,

(ii) If x = 5, then First number = 5 and second number = 8 – 5 = 3
Numbers are 3, 5

Question 10.
The sum of a number and its positive square root is \(\frac { 6 }{ 25 }\). Find the number.
Solution:

Question 11.
The sum of a number and its square is \(\frac { 63 }{ 4 }\) , find the numbers.
Solution:

Question 12.
There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers ?
Solution:
Let first integer = x
Then second integer = x + 1
and third integer = x + 2
According to the condition,
x² + (x + 1) (x + 2) = 154
=> x² + x² + 3x + 2 = 154
=> 2x² + 3x + 2 – 154 = 0
=> 2x² – 16x + 19x – 152 = 0
=> 2x(x – 8) + 19 (x – 8) = 0
=> (x – 8) (2x + 19) = 0
Either x – 8 = 0, then x = 8
or 2x + 19 = 0, then 2x = -19 => x = \(\frac { -19 }{ 2 }\) But it is not an integer
First number = 8
Second number = 8 + 1=9
and third number = 8 + 2 = 10

Question 13.
The product of two successive integral multiple of 5 is 300. Determine the multiplies.
Solution:
Let first multiplie of 5 = 5x
Then second multiple = 5x + 5
According to the condition,
5x (5x + 5) = 300
=> 25 x² + 25x – 300 = 0
=> x² + x – 12 = 0 (Dividing by 25)
=> x² + 4x – 3x – 12 = 0
=> x (x + 4) – 3 (x + 4) = 0
=> (x – 4) (x – 3) = 0
Either x + 4 = 0, then x = -4
or x – 3 = 0, then x = 3
(i) When x = -4, then
Required multiples of 5 will be
5 (-4) = -20, -20 + 5 = -15
or when x = 3, then
Required multiples will be
5 x 3 = 15, 15 + 5 = 20
Required number are -20, -15 or 15, 20

Question 14.
The sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number. Find the nqmbers.
Solution:
Let first number = x
Then second number = 2x – 3
According to the condition,
x² + (2x – 3)² = 233
=> x² + 4x² – 12x + 9 = 233
=> 5x² – 12x + 9 – 233 = 0
=> 5x² – 12x – 224 = 0
=> 5x² – 40x + 28x – 224 = 0
=> 5x (x – 8) + 28 (x – 8) = 0
=> (x – 8) (5x + 28) = 0
Either x – 8 = 0, then x = 8
or 5x + 28 = 0, then 5x = -28 => x = \(\frac { -28 }{ 5 }\) But it is not possible
x = 8
First number = 8
Second number = 2x – 3 = 2 x 8 – 3 = 16 – 3 = 13
Number are 8, 13

Question 15.
Find two consecutive even integers whose squares have the sum 340.
Solution:
Let first even integer = x
The second even integer = x + 2
According to the condition,
x² + (x + 2)² = 340
x² + x² + 4x + 4 = 340
=> 2x² + 4x + 4 – 340 = 0
=> 2x² + 4x – 336 = 0
=> x² + 2x – 168 = 0
=> x² + 14x – 12x – 168 = 0
=> x (x + 14) – 12 (x + 14) = 0
=> (x + 14) (x – 12) = 0
Either x + 14 = 0, then x = -14
or x – 12 = 0, the x = 12
(i) If x = -14, then
First number = -14
and second number = -14 + 2 = -12
(ii) If x = 12, then
First number =12
and second number =12 + 2 = 14
Hence even numbers are 12, 14 or -14, -12

Question 16.
The difference of two numbers is 4. If the difference of their reciprocals is \(\frac { 4 }{ 21 }\), find the numbers. (C.B.S.E. 2008)
Solution:
Let first number = x
Then second number = x – 4
According to the condition,

Either x – 7 = 0, then x = 7
or x + 3 = 0, then x = -3
(i) If x = 7, then
First number = 7
and second number = 7 – 4 = 3
(ii) If x = -3, then
First number = -3
and second number = -3 – 4 = -7
Number are 7, 3 or -3, -7

Question 17.
Find two natural numbers which differ by 3 and whose squared have the sum 117.
Solution:
Let first number = x
Then second number = x – 3
According to the condition,
x² + (x – 3)² = 117
=> x² + x² – 6x + 9 = 117
=> 2x² – 6x + 9 – 117 = 0
=> 2x² – 6x – 108 = 0
=> x² – 3x – 54 = 0 (Dividing by 2)
=> x² – 9x + 6x – 54 = 0
=> x (x – 9) + 6 (x – 9) = 0
=> (x- 9) (x + 6) = 0
Either x – 9 = 0, then x = 9
or x + 6 = 0, then x = -6 which is not a natural number
First natural number = 9
and second number = 9 – 3 = 6

Question 18.
The sum of squares of three consecutive natural numbers is 149. Find the numbers.
Solution:
Let first number = x
Then second number = x + 1
and third number = x + 2
According to the condition,
x² + (x + 1)² + (x + 2)² = 149
=> x² + x² + 2x + 1 + x2 + 4x + 4 = 149
=> 3x² + 6x + 5 – 149 = 0
=> 3x² + 6x – 144 = 0
=> x² + 2x – 48 = 0 (Dividing by 3)
=> x² + 8x – 6x – 48 = 0
=> x (x + 8) – 6 (x + 8) = 0
=> (x + 8) (x – 6) = 0 .
Either x + 8 = 0, then x = -8, But it is not a natural number
or x – 6 = 0, then x = 6
Numbers are 6, 6 + 1 = 7, 6 + 2 = 8 or 6, 7, 8

Question 19.
The sum of two numbers is 16. The sum of their reciprocals is \(\frac { 1 }{ 3 }\). Find the numbers. (C.B.S.E. 2005)
Solution:
Sum of two numbers = 16
Let first number = x
Then second number = 16 – x
According to the condition,

Either x – 12 = 0, then x = 12
or x – 4 = 0, then x = 4
(i) If x = 12, then
First number = 12
and second number = 16 – 12 = 4
(ii) If x = 4, then First number = 4
and second number = 16 – 4 = 12
Hence numbers are 4, 12

Question 20.
Determine two consecutive multiples of 3 whose product is 270.
Solution:
Let first multiple of 3 = 3x
Then second multiple of 3 = 3x + 3
According to the condition,
3x (3x + 3) = 270
=> 9x² + 9x – 270 = 0
=> x² + x – 30 = 0 (Dividing by 9)
=> x² + 6x – 5x – 30 = 0
=> x (x + 6) – 5 (x + 6) = 0
=> (x + 6) (x – 5) = 0
Either x + 6 = 0, then x = -6
or x – 5 = 0, then x = 5
(i) When x = -6, then
First number = 3x = 3 x (-6) = -18 and second number = -18 + 3 = -15
(ii) If x = 5, then
First number = 3x = 3 x 5 = 15 and second number =15 + 3 = 18
Hence numbers are 15, 18 or -18, -15

Question 21.
The sum of a number and its reciprocal is \(\frac { 17 }{ 4 }\) , Find the number.
Solution:

Question 22.
A two-digit number is such that the product of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number.
Solution:
Product of two digits = 8
Let units digit = x
Then tens digit = \(\frac { 8 }{ x }\)
Number = x + 10 x \(\frac { 8 }{ x }\) = x + \(\frac { 80 }{ x }\)

Question 23.
A two-digit number is such that the product of the digits is 12. When 36 is added to the number the digits interchange their places. Determine the number.
Solution:

Question 24.
A two-digit number is such that the product of the digits is 16. When 54 is subtracted from the number, the digits are interchanged. Find the number.
Solution:

Question 25.
Two numbers differ by 3 and their product is 504. Find the numbers. (C.B.S.E. 2002C)
Solution:
Difference of two numbers = 3
Let first number = x
Then second number = x – 3
According to the condition,
x (x – 3) = 504
=> x² – 3x – 504 = 0
=> x² – 24x + 21x – 504 = 0
=> x (x – 24) + 21 (x – 24) = 0
=> (x – 24) (x + 21) = 0
Either x – 24 = 0, then x = 24
or x + 21 = 0, then x =-21
(i) If x = 24, then
First number = 24
and second number = 24 – 3 = 21
(ii) If x =-21, then
First number = -21
and second number = -21 – 3 = -24
Hence numbers are 24, 21 or -21, -24

Question 26.
Two numbers differ by 4 and their product is 192. Find the numbers. (C.B.S.E. 2000C)
Solution:
Let first number = x
Then second number = x – 4
According to the condition,
x (x – 4) = 192
=> x² – 4x – 192 = 0
=> x² – 16x + 12x – 192 = 0
=> x (x – 16) + 12 (x – 16) = 0
=> (x – 16) (x + 12) = 0
Either x – 16 = 0, then x = 16
or x + 12 = 0, then x = -12
(i) If x = 16, then
First number = 16
and second number = 16 – 4 = 12
(ii) If x = -12, then
First number = -12
and second number = -12 – 4 = -16
Hence numbers are 16, 12 or -12, -16

Question 27.
A two digit number is 4 times the sum of its digits and twice the product of its digits. Find the number. (C.B.S.E. 1999C)
Solution:
Let units digit of the number = x
and tens digit = y
Number = x + 10y
According to the given conditions,
x + 10y = 4 (x + y)
=> x + 10y = 4x + 4y
=> 10y – 4y = 4x – x
=> 3x = 6y
=>x = 2y …(i)
and x + 10y = 2xy ….(ii)
Substituting the value of x in (i)
2y + 10y = 2 x 2y x y
=> 12y = 4y2
=> 4y2 – 12y = 0
=> y2 – 3y = 0
=> y (y – 3) = o
Either y = 0, but it is not possible because y is tens digit number
or y – 3 = 0, then y = 3
x = 2y = 2 x 3 = 6
and number = x + 10y = 6 + 10 x 3 = 6 + 30 = 36

Question 28.
The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers. [CBSE 2014]
Solution:
Let first large number = x
and smaller number = y
According to the condition,
x² – y² = 180 …(i)
and y² = 8x
From (i) and (ii),
x² – 8x – 180 = 0
=> x² – 18x + 10x – 180 = 0
=> x (x – 18)+ 10 (x – 18) = 0
=> (x – 18) (x + 10) = 0
Either x – 18 = 0, then x = 18
or x + 10 = 0, then x = -10 But it is not possible being negative
x = 18
First number =18
Then second number y² = 8x
y² = 8 x 18 = 144 = (12)²
=> y = 12
Numbers are 18, 12

Question 29.
The sum of two numbers is 18. The sum of their reciprocals is \(\frac { 1 }{ 4 }\). Find the numbers. (C.B.S.E. 2005)
Solution:
Sum of two numbers = 18
Let one number = x
Then second number = 18 – x
According to the condition,

=> x² – 12x – 6x + 72 = 0
=> x (x – 12) – 6 (x – 12) = 0
=> (x – 12) (x – 6) = 0
Either x – 12 = 0, then x = 12
or x – 6 = 0, then x = 6
(i) If x = 12, then
First number = 12
Second number =18 – 12 = 6
(ii) If x = 6, then
First number = 6
Then second number = 18 – 6 = 12
Numbers are 6, 12

Question 30.
The sum of two numbers a and b is 15, and the sum of their reciprocals \(\frac { 1 }{ a }\) and \(\frac { 1 }{ b }\) is \(\frac { 3 }{ 10 }\). Find the numbers a and b. (C.B.S.E. 2005)
Solution:

or b – 5 = 0, then b = 5
(i) a = 15 – 10 = 5
(ii) or a = 15 – 5 = 10
Numbers are 5, 10 or 10, 5

Question 31.
The sum of two numbers is 9. The sum of their reciprocals is \(\frac { 1 }{ 2 }\). Find the numbers. [CBSE 2012]
Solution:
Sum of two numbers = 9
Let first number = x
Then second number = 9 – x
According to the condition,

By cross multiplication
18 = 9x – x²
=> x² – 9x + 18 = 0
=> x² – 6x – 3x + 18 = 0
=> x (x – 6) – 3 (x – 6) = 0
=> (x – 6) (x – 3) = 0
Either x – 6 = 0, then x = 6
or x – 3 = 0, then x = 3
Numbers are 6 and (9 – 6) = 3, or 3 and (9 – 3) = 6
Numbers are 3, 6

Question 32.
Three consecutive positive integers are such that the sum of the square of the’ first and the product of other two is 46, find the integers. [CBSE 2010]
Solution:
Let first number = x
Then second number = x + 1
and third number = x + 2
According co the condition,
(x)² + (x+ 1) (x + 2) = 46
x² + x² + 3x + 2 = 46
=> 2x² + 3x + 2 – 46 = 0
=> 2x² + 3x – 44 = 0
=> 2x² + 11x – 8x – 44 = 0
=> x (2x + 11) – 4 (2x + 11) = 0
=> (2x + 11) (x – 4) = 0
Either 2x + 11 = 0, then x = \(\frac { -11 }{ 2 }\) which is not possible being fraction
or x – 4 = 0, then x = 4
Numbers are 4, 5, 6

Question 33.
The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers. [CBSE 2010]
Solution:
Let smaller number = x
Then larger number = 2x – 5
According to the condition,
(2x – 5)² – x² = 88
=> 4x² – 20x + 25 – x² – 88 = 0
=> 3x² – 20x – 63 = 0
=> 3x² – 27x + 7x – 63 = 0
=> 3x (x – 9) + 7 (x – 9) = 0
=> (x – 9) (3x + 7) = 0
Either x – 9 = 0, then x = 9
or 3x + 7 = 0, then x = \(\frac { -7 }{ 3 }\) which is not possible
Smaller number = 9
and greater number = 2x – 5 = 2 x 9 – 5 = 18 – 5 = 13
Hence numbers are 13, 9

Question 34.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find two numbers. [NCERT]
Solution:

Question 35.
Find two consecutive odd positive integers, sum of whose squares is 970.
Solution:
Let two consecutive positive integers be x and x + 2
A.T.Q.,
(x)² + (x + 2)² = 970
=> x² + x² + 4x + 4 – 970 = 0
=> 2x² + 4x – 966 = 0
=> x² + 2x – 483 = 0 (Dividing by 2)
=> x² + 23x – 21x – 483 = 0
=> x (x + 23) – 21 (x + 23) = 0
=> (x – 21) (x + 23) = 0
Either x – 21 = 0 or x + 23 = 0
x = 21 or x = – 23 (rejected being -ve)
As integers should be +Ve
x = 21 and x + 2 = 21 + 2 = 23
Hence integers are 21, 23

Question 36.
The difference of two natural numbers is 3 and the difference of their reciprocals is \(\frac { 3 }{ 28 }\). Find the numbers.
Solution:

y(y + 7) – 4(y + 7) = 0
(y – 4) (y + 7) = 0
y – 4 = 0 or y + 7 = 0
y = 4 or y = -7 (rejected being natural no.)
When y = 4, x = 3 + 4 = 7 [From (ii)]
Number are 7, 4

Question 37.
The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Solution:
Let two consecutive positive integers be x and x + 2
A.T.Q.,
(x)² + (x + 2)² = 394
x² + x² + 4x + 4 – 394 = 0
2x² + 4x – 390 = 0
x² + 2x – 195 = 0 (Dividing by 2)
x² + 15x – 13x – 195 = 0
x (x + 15) – 13 (x + 15) = 0
(x – 13) (x + 15) = 0
Either x – 13 = 0 or x + 15 = 0
x = 13 or x = -15 (rejected)
Number should be x = 13 and x = 13 + 2 = 15
or x = -15 and x = -15 + 2 = -13
Hence odd numbers are 13, 15 or -15, -13

Question 38.
The sum of the squares of two consecutive multiple of 7 is 637. Find the multiples. [ICSE 2014]
Solution:
Let first multiple of 7 = 7x
Then second = 7x + 7
(7x)² + (7x + 7) = 637
49x² + 49x² + 98x + 49 = 637
98x² + 98x + 49 – 637 = 0
98x² + 98x – 588 = 0
x² + x – 6 = 0 (dividing by 98)
x² + 3x – 2x – 6 = 0
x (x + 3) – 2 (x + 3) = 0
(x + 3) (x – 2) = 0
Either x + 3 = 0, then x = -3, but not possible being negative
or x – 2 = 0, then x = 2
Numbers will be 14, 21

Question 39.
The sum of the squares of two consecutive even numbers is 340. Find the numbers. [CBSE 2014]
Solution:
Let first even number = 2x
Then second number = 2x + 2
(2x)² + (2x + 2)² = 340
4x² + 4x² + 8x + 4 – 340 = 0
8x² + 8x – 336 = 0
x² + x – 42 = 0 (Dividing by 8)
x² + 7x – 6x – 42 = 0
x (x + 7) – 6 (x + 7) = 0
=> (x + 7) (x – 6) = 0
Either x + 7 = 0, then x = -7 but not possible being negative
or x – 6 = 0, then x = 6
Numbers are 12, 14

Question 40.
The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is \(\frac { 29 }{ 20 }\). Find the original fraction.
Solution:

Question 41.
Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. [NCERT Exemplar]
Solution:
Let n be a required natural number.
Square of a natural number diminished by 84 = n² – 84
and thrice of 8 more than the natural number = 3 (n + 8)
Now, by given condition,
n² – 84 = 3 (n + 8)
=> n² – 84 = 3n + 24
=> n² – 3n – 108 = 0
=> n² – 12n + 9n – 108 = 0 [by splitting the middle term]
=> n (n – 12) + 9 (n – 12) = 0
=> (n – 12) (n + 9) = 0
=> n = 12 [n ≠ – 9 because n is a natural number]
Hence, the required natural number is 12.

Question 42.
A natural number when increased by 12 equals 160 times its reciprocal. Find the number. [NCERT Exemplar]
Solution:
Let the natural number be x.
According to the question,
x + 12 = \(\frac { 160 }{ x }\)
On multiplying by x on both sides, we get
=> x² + 12x – 160 = 0
=> x² + (20x – 8x) – 160 = 0
=> x² + 20x – 8x – 160 = 0 [by factorisation method]
=> x (x + 20) – 8 (x + 20) = 0
=> (x + 20) (x – 8) = 0
Now, x + 20 = 0 => x = -20 which is not possible because natural number is always greater than zero
and x – 8 = 0 => x = 8.
Hence, the required natural number is 8.

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
Determine the nature of the roots of following quadratic equations :
(i) 2x² – 3x + 5 = 0 [NCERT]
(ii) 2x² – 6x + 3 = 0 [NCERT]
(iii) \(\frac { 3 }{ 5 }\) x² – \(\frac { 2 }{ 3 }\) x + 1 = 0
(iv) 3x² – 4√3 x + 4 = 0 [NCERT]
(v) 3x² – 2√6 x + 2 = 0
Solution:

Question 2.
Find the values of k for which the roots are real and equal in each of the following equations :
(i) kx² + 4x + 1 = 0
(ii) kx² – 2√5 x + 4 = 0
(iii) 3x² – 5x + 2k = 0
(iv) 4x²+ kx + 9 = 0
(v) 2kx² – 40x + 25 = 0
(vi) 9x² – 24x + k = 0
(vii) 4x² – 3kx +1 = 0
(viii) x² – 2 (5 + 2k) x + 3 (7 + 10k) = 0
(ix) (3k + 1) x² + 2(k + 1) x + k = 0
(x) kx² + kx + 1 = – 4x² – x
(xi) (k + 1) x² + 2 (k + 3) x + (k + 8) = 0
(xii) x² – 2kx + 7k – 12 = 0
(xiii) (k + 1) x² – 2 (3k + 1) x + 8k + 1 = 0
(xiv) 5x² – 4x + 2 + k (4x² – 2x – 1) = 0
(xv) (4 – k) x² + (2k + 4) x (8k + 1) = 0
(xvi) (2k + 1) x² + 2 (k + 3) x (k + 5) = 0
(xvii) 4x² – 2 (k + 1) x + (k + 4) = 0
(xviii) 4x² (k + 1) x + (k + 1) = 0
Solution:

Question 3.
In the following, determine the set of values of k for which the given quadratic equation has real roots :
(i) 2x² + 3x + k = 0
(ii) 2x² + x + k = 0
(iii) 2x² – 5x – k = 0
(iv) kx² + 6x + 1 = 0
(v) 3x² + 2x + k = 0
Solution:

Question 4.
Find the values of k for which the following equations have real and equal roots :
(i) x²- 2(k + 1) x + k² = 0 [CBSE 2001C, 2013]
(ii) k²x² – 2 (2k – 1) x + 4 = 0 [CBSE 2001C]
(iii) (k + 1) x² – 2(k – 1) x + 1 = 0 [CBSE 2002C]
(iv) x² + k(2x + k – 1) + 2 = 0 [CBSE 2017]
Solution:

Question 5.
Find the values of k for which the following equations have real roots
(i) 2x² + kx + 3 = 0 [NCERT]
(ii) kx (x – 2) + 6 = 0 [NCERT]
(iii) x² – 4kx + k = 0 [CBSE 2012]
(iv) kx(x – 2√5 ) + 10 = 0 [CBSE 2013]
(v) kx (x – 3) + 9 = 0 [CBSE 2014]
(vi) 4x² + kx + 3 = 0 [CBSE 2014]
Solution:
(i) 2x² + kx + 3 = 0
Here a = 2, b = k, c = 3

Question 6.
Find the values of k for which the given quadratic equation has real and distinct roots :
(i) kx² + 2x + 1 = 0
(ii) kx² + 6x + 1 = 0
Solution:

Question 7.
For what value of k, (4 – k) x² + (2k + 4) x + (8k + 1) = 0, is a perfect square.
Solution:
(4 – k) x² + (2k + 4) x + (8k + 1) = 0
Here, a = 4 – k, b = 2k + 4, c = 8k + 1

Question 8.
Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.
Solution:

Question 9.
Find the value of k for which the quadratic equation (3k + 1) x² + 2(k + 1) x + 1 = 0 has equal roots. Also, find the roots.
[CBSE 2014]
Solution:

Question 10.
Find the values of p for which the quadratic equation (2p + 1) x² – (7p + 2) x + (7p – 3) = 0 has equal roots. Also, find these roots.
Solution:

Question 11.
If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal-roots, find the value of k. [CBSE 2014]
Solution:

Question 12.
If 2 is a root of the quadratic equation 3x² + px – 8 = 0 and the quadratic equation 4x² – 2px + k = 0 has equal roots, find the value of k. [CBSE 2014]
Solution:

=> 16k = 16
k = 16

Question 13.
If 1 is a root of the quadratic equation 3x² + ax – 2 = 0 and the quadratic equation a(x² + 6x) – b=0 has equal roots, find the value of b.
Solution:

Question 14.
Find the value of p for which the quadratic equation (p + 1) x² – 6 (p + 1) x + 3 (p + q) = 0, p ≠ -1 has equal roots. Hence, find the roots of the equation. [CBSE 2015]
Solution:

Question 15.
Determine the nature of the roots of following quadratic equations :
(i) (x – 2a) (x – 2b) = 4ab
(ii) 9a²b²x² – 24abcdx + 16c²d² = 0, a ≠ 0, b ≠ 0
(iii) 2 (a² + b²) x² + 2 (a + b) x + 1 = 0
(iv) (b + c) x² – (a + b + c) x + a = 0
Solution:

Question 16.
Determine the set of values of k for which the given following quadratic equation has real roots :
(i) x² – kx + 9 = 0
(ii) 2x² + kx + 2 = 0
(iii) 4x² – 3kx +1=0
(iv) 2x² + kx – 4 = 0
Solution:

Question 17.
If the roots of the equation (b – c) x² + (c – a) x + (a – b) = 0 are equal, then prove that 2b = a + c. [CBSE 2002C]
Solution:

=> a + c = 2b
=> 2b = a + c
Hence proved.

Question 18.
If the roots of the equation (a² + b²) x² – 2 (ac + bd) x + (c² + d²) = 0 are equal. prove that \(\frac { a }{ b }\) = \(\frac { c }{ d }\)
Solution:

Question 19.
If the roots of the equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 are simultaneously real, then prove that b² = ac
Solution:

Question 20.
If p, q are real and p ≠ q, then show that the roots of the equation (p – q) x² + 5(p + q) x – 2(p – q) = 0 are real and unequal.
Solution:

Question 21.
If the roots of the equation (c² – ab) x² – 2 (a² – bc) x + b² – ac = 0 are equal, prove that either a = 0 or a³ + b³ + c³ = 3abc.
Solution:

Question 22.
Show that the equation 2 (a² + b²) x² + 2 (a + b) x + 1 = 0 has no real roots, when a ≠ b.
Solution:

Question 23.
Prove that both the roots of the equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are real but they are equal only when a = b = c.
Solution:

Question 24.
If a, b, c are real numbers such that ac ≠ 0, then show that at least one of the equations ax² + bx + c = 0 and – ax² + bx + c = 0 has real roots.
Solution:

Question 25.
If the equation (1 + m²) x² + 2mcx + (c² – a²) = 0 has equal roots, prove that c² = a² (1 + m²). (C.B.S.E. 1999)
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
Write the discriminant of the following quadratic equations :
(i) 2x² – 5x + 3 = 0
(ii) x² + 2x + 4 = 0
(iii) (x – 1) (2x – 1) = 0
(iv) x² – 2x + k = 0, k ∈ R
(v) √3 x² + 2√2 x – 2√3 = 0
(vi) x² – x + 1 = 0
Solution:

Question 2.
In the following, determine whether the given quadratic equations have real roots and if so, And the roots :
(i) 16x² = 24x + 1
(ii) x² + x + 2 = 0
(iii) √3 x² + 10x – 8√3 = 0
(iv) 3x² – 2x + 2 = 0
(v) 2x² – 2√6 x + 3 = 0
(vi) 3a²x² + 8abx + 4b² = 0, a ≠ 0
(vii) 3x² + 2√5 x – 5 = 0
(viii) x² – 2x + 1 = 0
(ix) 2x² + 5√3 x + 6 = 0
(x) √2 x² + 7x + 5√2= 0 [NCERT]
(xi) 2x² – 2√2 x + 1 = 0 [NCERT]
(xii) 3x² – 5x + 2 = 0 [NCERT]
Solution:

Question 3.
Solve for x :

Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
Question 1.
x² – 4 √2x + 6 = 0
Solution:

Question 2.
2x² – 7x + 3 = 0
Solution:

Question 3.
3x² + 11x + 10 = 0
Solution:

Question 4.
2x² + x – 4 = 0
Solution:

Question 5.
2x² + x + 4 = 0
Solution:

Question 6.
4x² + 4√3x + 3 = 0
Solution:

Question 7.
√2 x² – 3x – 2√2 = 0
Solution:

Question 8.
√3 x² + 10x + 7√3 = 0
Solution:

Question 9.
x² – (√2 + 1)x + √2 = 0
Solution:

Question 10.
x² – 4ax + 4a² – b² = 0
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.1
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.2
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.3
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.4
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.6
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex VSAQS
• RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

Mark the correct alternative in each of the following:
Question 1.
Which of the following is not a measure of central tendency :
(a) Mean
(b) Median
(c) Mode
(d) Standard deviation
Solution:
Standard deviation is not a measure of central tendency. Only mean, median and mode are measures. (d)

Question 2.
The algebraic sum of the deviations of a frequency distribution from its mean is
(a) always positive
(b) always negative
(c) 0
(d) a non-zero number
Solution:
The algebraic sum of the deviations of a frequency distribution from its mean is zero
Let x₁, x₂, x₃, …… x_n are observations and \(\overline { X }\) is the mean

Question 3.
The arithmetic mean of 1, 2, 3, ….. , n is

Solution:
Arithmetic mean of 1, 2, 3, …… n is

Question 4.
For a frequency distribution, mean, median and mode are connected by the relation
(a) Mode = 3 Mean – 2 Median
(b) Mode = 2 Median – 3 Mean
(c) Mode = 3 Median – 2 Mean
(d) Mode = 3 Median + 2 Mean
Solution:
The relation between mean, median and mode is: Mode = 3 Median – 2 Mean (c)

Question 5.
Which of the following cannot be determined graphically ?
(a) Mean
(b) Median
(c) Mode
(d) None of these
Solution:
Mean cannot be determind graphically, (a)

Question 6.
The median of a given frequency distribution is found graphically with the help of
(a) Histogram
(b) Frequency curve
(c) Frequency polygon
(d) Ogive
Solution:
Median of a given frequency can be found graphically by an ogive, (d)

Question 7.
The mode of a frequency distribution can be determined graphically from
(a) Histogram
(b) Frequency polygon
(c) Ogive
(d) Frequency curve
Solution:
Mode of frequency can be found graphically by an ogive, (c)

Question 8.
Mode is
(a) least frequent value
(b) middle most value
(c) most frequent value
(d) None of these
Solution:
Mode is the most frequency value of observation or a class, (c)

Question 9.
The mean of n observations is \(\overline { X }\) . If the first item is increased by 1, second by 2 and so on,
then the new mean is

Solution:
Mean of n observations = \(\overline { X }\)
By adding 1 to the first item, 2 to second item and so on, the new mean will be
Let x₁, x₂, x_3,…..  x_n are the items whose mean is \(\overline { X }\) , then mean of
(x₁+ 1) + (x₂ + 2) + (x₃ + 3) + …… (x_n + n)

Question 10.
One of the methods of determining mode is
(a) Mode = 2 Median – 3 Mean
(b) Mode = 2 Median + 3 Mean
(c) Mode = 3 Median – 2 Mean
(d) Mode = 3 Median + 2 Mean
Solution:
Mode = 3 Median – 2 Mean (c)

Question 11.
If the mean of the following distribution is 2.6, then the value of y is

(a) 3
(b) 8
(c) 13
(d) 24
Solution:

Question 12.
The relationship between mean, median and mode for a moderately skewed distribution is
(a) Mode = 2 Median – 3 Mean
(b) Mode = Median – 2 Mean
(c) Mode = 2 Median – Mean
(d) Mode = 3 Median – 2 Mean
Solution:
The relationship between mean, median and mode is Mode = 3 Median – 2 Mean, (d)

Question 13.
The mean of a discrete frequency distribution x_i /f_i ; i= 1, 2, …… n is given by

Solution:

Question 14.
If the arithmetic mean of x, x + 3, x + 6, x + 9, and x + 12 is 10, then x =
(a) 1
(b) 2
(c) 6
(d) 4
Solution:

Question 15.
If the median of the data : 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5, then x =
(a) 27
(b) 25
(c) 28
(d) 30
Solution:

Question 16.
If the median of the data : 6, 7, x – 2, x, 17, 20, written in ascending order, is 16. Then x =
(a) 15
(b) 16
(c) 17
(d) 18
Solution:

Question 17.
The median of first 10 prime numbers is
(a) 11
(b) 12
(c) 13
(d) 14
Solution:

Question 18.
If the mode of the data : 64,60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =
(a) 44
(b) 45
(c) 46
(d) 48
Solution:
Mode of 64, 60, 48, x, 43, 48, 43, 34 is 43
∵ By definition mode is a number which has maximum frequency which is 43
∴ x = 43
∴ x + 3 = 43 + 3 = 46 (c)

Question 19.
If the mode of the data : 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =
(a) 15
(b) 16
(c) 17
(d) 19
Solution:
Mode of 16, 15, 17, 16, 15, x, 19, 17, 14 is 15
∵By definition mode of a number which has maximum frequency which is 15
∴ x = 15 (a)

Question 20.
The mean of 1, 3, 4, 5, 7, 4 is m. The number 3, 2, 2, 4, 3, 3, p have mean m – 1 and median q. Then p + q =
(a) 4
(b) 5
(c) 6
(d) 7
Solution:

Question 21.
If the mean of a frequency distribution is 8.1 and Σf_ix_i = 132 + 5k, Σf_i = 20, then k =
(a) 3
(b) 4
(c) 5
(d) 6
Solution:

Question 22.
If the mean of 6, 7, x, 8, y, 14 is 9, then
(a)x+y = 21
(b)x+y = 19
(c) x -y = 19
(d) v -y = 21
Solution:

Question 23.
The mean of n observations is \(\overline { x }\) If the first observation is increased by 1, the second by 2, the third by 3, and so on, then the new mean is

Solution:

Question 24.
If the mean of first n natural numbers is \(\frac { 5n }{ 9 }\) then n =
(a) 5
(b) 4
(c) 9
(d) 10
Solution:

Question 25.
The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is
(a) 25
(b) 18
(c) 20
(d) 22
Solution:
Arithmetic mean = 24
Mode = 12
∴ But mode = 3 median – 2 mean
⇒ 12 = 3 median – 2 x 24
⇒ 12 = 3 median =-48
⇒ 12 + 48 = 3 median
⇒ 3 median = 60
Median = \(\frac { 60 }{ 3 }\) = 20 (c) 

Question 26.
The mean of first n odd natural number is

Solution:

Question 27.
The mean of first n odd natural numbers is \(\frac { n2 }{ 81 }\) , then n = 81
(a) 9
(b) 81
(c) 27
(d) 18
Solution:

Question 28.
If the difference of mode and median of a data is 24, then the difference of median and mean is
(a) 12
(b) 24
(c) 8
(d) 36
Solution:
Difference of mode and median = 24
Mode = 3 median – 2 mean
⇒ Mode – median = 2 median – 2 mean
⇒ 24 = 2 (median – mean)
⇒ Median – mean = \(\frac { 24 }{ 2 }\) = 12 (a)

Question 29.
If the arithmetic mean of 7, 8, x, 11, 14 is x, then x =
(a) 9
(b) 9.5
(c) 10
(d) 10.5
Solution:

Question 30.
If mode of a series exceeds its mean by 12, then mode exceeds the median by
(a) 4
(b) 8
(c) 6
(d) 10
Solution:
Mode of a series = Its mean + 12
Mean = mode – 12
Mode = 3 median – 2 mean
Mode = 3 median – 2 (mode -12)
⇒ Mode = 3 median – 2 mode + 24
⇒ Mode + 2 mode – 3 median = 24
⇒ 3 mode – 3 median = 24
⇒ 3 (mode – median) = 24
⇒ Mode – medain = \(\frac { 24 }{ 3 }\) = 8 (b)

Question 31.
If the mean of first n natural number is 15, then n =
(a) 15
(b) 30
(c) 14
(d) 29
Solution:

Question 32.
If the mean of observations x₁, x₂, …, x_n is \(\overline { x }\) , then the mean of x₁ + a, x₂ + a,…, x_n + a is

Solution:

Question 33.
Mean of a certain number of observations is \(\overline { x }\) If each observation is divided by m (m ≠ 0) and increased by n, then the mean of new observation is

Solution:
Mean of some observations = \(\overline { x }\)
If each observation is divided by m and increased by n
Then mean will be = \(\frac { \overline { x } }{ m }\) +n

Question 34.
If u_i= \(\frac { xi-25\quad }{ 10 }\) Σf_iu_i = 20, Σf_i = 100, then \(\overline { x }\)
(a) 23
(b) 24
(c) 27
(d) 25
Solution:

Question 35.
If 35 is removed from the data : 30, 34, 35, 36, 37, 38, 39, 40, then the median increases by
(a) 2
(b) 1.5
(c) 1
(d) 0.5
Solution:

Question 36.
While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the classes.
(b) centred at the class marks of the classes.
(c) centred at the upper limit of the classes.
(d) centred at the lower limit of the classes.
Solution:
In computing the mean of grouped data, the frequencies are centred at the class marks of the classes. (b)

Question 37.

Solution:

Question 38.
For the following distribution:

the sum of the lower limits of the median and modal class is
(a) 15
(b) 25
(c) 30
(d) 35
Solution:

Now, \(\frac { N }{ 2 }\) = \(\frac { 66 }{ 2 }\) = 33, which lies in the interval 10-15.
Therefore, lower limit of the median class is 10.
The highest frequency is 20, which lies in the interval 15-20.
Therefore, lower limit of modal class is 15.
Hence, required sum is 10 + 15 = 25. (b)

Question 39.
For the following distribution:

the modal class is
(a) 10-20
(b) 20-30
(c) 30-40
(d) 50-60
Solution:

Here, we see that the highest frequency is 30, which lies in the interval 30-40. (c)

Question 40.
Consider the following frequency distribution:

The difference of the upper limit of the median class and the lower limit of the modal class is
(a) 0
(b) 19
(c) 20
(d) 38
Solution:

Here, \(\frac { N }{ 2 }\) = \(\frac { 67 }{ 2 }\) = 33.5 which lies in the interval 125-145.
Hence, upper limit of median class is 145.
Here, we see that the highest frequency is 20 which lies in 125-145.
Hence, the lower limit of modal class is 125.
∴ Required difference = Upper limit of median class – Lower limit of modal class
= 145 – 125 = 2 (C)

Question 41.
In the formula \(\overline { X }\) = a + \(\frac { \Sigma fidi }{ \Sigma fi }\) for finding the mean of grouped data di’s are deviations from a of
(a) lower limits of classes
(b) upper limits of classes
(c) mid-points of classes
(d) frequency of the class marks
Solution:
We know that, d_i = x_i – a
i .e , d_i‘s are the deviation from a mid-points of the classes. (c)

Question 42.
The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data gives its
(a) mean
(b) median
(c) mode
(d) all the three above
Solution:
Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa. (b)

Question 43.
Consider the following frequency distribution:

The upper limit of the median class is
(a) 17
(b) 17.5
(c) 18
(d) 18.5
Solution:
Given, classes are not continuous, so we make continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.

Here, \(\frac { N }{ 2 }\) = \(\frac { 57 }{ 2 }\) = 28.5, which lies in the interval 11.5-17.5.
Hence, the upper limit is 17.5.

Hope given RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS are helpful to complete your math homework.

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