Math's Class X (CBSE)
Sample Paper Set - (SOLVED) II

Time allowed : 3 hours

Q1) For what value of k will the following system of linear equations have an infinite number of solutions: 2x + 3y = 2; (k + 2)x + (2k + 1)y = 2(k - 1) ?  
(Marks 2)

Ans1)  a₁/a₂ = b₁/b₂ = c₁/c₂
=> 2/(k + 2) = 3/(2k + 1) = 2/(2(k - 1))
From (i) & (ii) and From (iii) & (iv)

.^.. k = 4

Q2) Reduce the following relation expression into lowest form:
(x⁴ - 10x² + 9)/(x³ + 4x² + 3x)  (Marks 2)

Ans2) x⁴ - 10x² + 9 = (x + 3)(x - 3)(x + 1)(x - 1)
x³ + 4x² + 3x = x(x + 3)(x + 1)
... (x⁴ - 10x² + 9)/(x³ + 4x² + 3x) = ((x + 3)(x - 3)(x + 1)(x - 1))/(x(x + 3)(x + 1))
=> (x² - 4x + 3)/x

Q3) Solve for x and y
ax + by = a - b  - (i)
bx - ay = a + b  - (ii)  (Marks 2)

Ans3) Multiply (i) by a and (ii) by b and add
a²x + aby = a² - ab
b²x - aby = ab + b²
(a² + b²)x = a² + b²
=> x = 1
subs x = 1 in (i)
a + by = a - b
y = -1
.^.. solution is x = 1, y = -1

Q4) Find the value of k such that sum of the roots of the quadratic equation
3x² + (2k + 1)x - (k + 5) = 0 is equal to the product of its roots.  (Marks 2)
Ans4)  Sum of the roots = -(2k + 1)/3
Product of the roots = -(k + 5)/3
.^.. -(2k + 1)/3 = -(k + 5)/3
=> -2k - 1 = -k - 5
=> 4 = k

Q5) Find two consecutive numbers, whose square have sum 85.  (Marks 2)

Ans5) Let the two consecutive number be x and x + 1
.^.. x² + (x + 1)² = 85
=> x² + x² + 2x + 1 = 85
=> 2x² + 2x - 84 = 0
x² + x - 42 = 0
(x + 7)(x - 6) = 0
x = -7 or x = 6
Two numbers are -7, -6 or 6, 7

Q6) Rita purchased a car, with a marked price of Rs. 2,10,000 at a discount of 5%. If the sales tax charged at 10%, find the amount Rita had to pay for purchasing the car.  (Marks 2)
Ans6) Marked price of the car = Rs. 2,10,000
Discount = 5/100 x 2,10,000
= Rs. 10,500
Price of the car after the discount = 1,99,500
Sales tax = 10/100 x 1,99,500
= Rs. 19,950
Rita had to pay fpr the car = Rs. (1,99,500 + 19,950)
= Rs. 2,19,450 

Q7) The mean weight of 21 students of a class is 52 kg. If the mean weight of first 11 students of the class is 50 kg and that of last 11 students is 54 kg, find the weight of the 11th student.  (Marks 2)
Ans7)  Total weight of 21 student = 52 x 21
= 1092kg
Total weight of first 11 students = 50 x 11
= 550 kg
Total weight of last 11 students = 11 x 54
= 594 kg
Total weight of 22 students = 550 + 594
= 1144 kg
.^.. Weight of the 11th student = 1144 - 1092
= 52 kg

Q8) The following data has been arranged in ascending order: 12, 14, 17, 21, x, 26, 28, 32, 36. If the median of the data is 23, find x. If 32 is changed to 23, find the new median.  (Marks 2)

Ans8) Here n = no. of observations = 10
.^.. Median = (5^th observation + 6^th observation)/2
23 = (22 + x)/2
=> x = 24
If 32 is changed to 23
Median = (22 + 23)/2
= 25/2 = 22.5 

Q9) For what value of x, in the mode of the following data 5 ?
2, 4, 3, 5, 4, 5, 6, 4, x, 7, 5.  (Marks 2)

Ans15) The data is 2, 3, 4, 4, 4, 5, 5, 5, 6, 7, ,x
So, if mode is 5 then x = 5

Q9) A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, he has to pay Rs. 1000 as hostel charges where as a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charge and the cost of food per day.  (Marks 4)

Ans10) Let the fixed charge be Rs. x
Let the fixed cost for food per day be Rs. y
x + 20y = 1000
x + 26y = 1180
Solving we get,
x = 400, y = 30
.^.. Fixed charges = Rs. 400
charge of food per day = Rs. 30 

Q10) Find the value of a and b so that the polynomials p(x) and q(x) have (x + 1)(x - 2) as their HCF.
p(x) = (x² + 3x + 2)(x² + x + a)
q(x) = (x² - 3x + 2)(x² - 3x + b)  (Marks 4)

Ans10) p(x) = (x² + 3x + 2)(x² + x + a)
q(x) = (x² - 3x + 2)(x² - 3x + b)
.^.. x - 2 is a factor of x² + x + a => 4 + 2 + a = 0
=> a = -6
x + 1 is a factor of x² - 3x + b => 1 + 3 + b = 0
=> b = -4
.^.. a = -6, b = -4

Q11) A page of pass book of Ved is given below :-

Find the interest Ved gets for the period March, 98 to Dec.' 98 at 5% per annum simple interest.  (Marks 4)
Ans11)  

Total Principal = Rs. 71,300
Interest = Rs. (71,300 x 5 x 1)/(100 x 12)
= Rs. 297.08

Q12) The annual income of Seema (excluding HRA) is Rs. 1,60,000. She contributes Rs. 5000 per month to her provident fund and pays a half yearly insurance premium of Rs. 5000. Calculate the income tax along with surcharge Seema has to pay in the last month of the year if her earlier deductions as income tax for the first 11 months were at the rate of Rs. 400 per month.  (Marks 4)

Assume the following for calculating income tax :

An12) Annual Income = Rs. 1,60,000, Standard Deduction = Rs. 20,000
.^.. Taxable Income = Rs. 1,60,000 - Rs. 20,000
= Rs. 1,40,000
Income Tax = Rs. 1,000 + Rs. 16000
= Rs. 17,000
Total Saving = 5000 x 12 + 5000 x 2
= 60,000 + 10,000
= Rs. 70,000
Rebate = Rs. 12,000
Income Tax payable = Rs. 17,000 - Rs. 12,000
= Rs. 5000
Total Income tax, including surcharge = 5000 + 10/100 x 5000
= Rs. 5500
Advance tax paid = 11 x 400  = Rs. 4400
.^.. Income Tax payable in the last month = Rs. 1100

Q13) Solve for x : 9^{x + 2} - 6 x 3^{x + 1} + 1 = 0  (Marks 6)

Ans13) 3^2x . 3⁴ - 6 . 3^x . 3 + 1 = 0
=> 81 x (3^x)² - 18 x 3^x + 1 = 0
Put 3^x = y
=> 81y² - 18 y + 1 = 0
=> (9y - 1)² = 0
=> y = 1/9
 at y = 1/9
3^x = y
=> 3^x = 1/9
=> 3^x = 1/3²
=> 3^x = 3^-2
=> x = -2