CBSE Set Qa3 Maths Class X 2000 Sample Test Papers For Class 10th for students online

Q14) 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)
Ans14) 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 

Q15) 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

Q16) Determine graphically the co-ordinates of the vertices of the triangle, the equation of whose sides are:
y = x, 3y = x, x + y = 8.  (Marks 4)
Ans16)  

x = y   x 0 2 4 y 0 2 4 y = x/3 x 0 3 6 y 0 1 2 y = 8 - x x 0 4 6 y 8 4 2

Q17) 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)
Ans17) 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 

Q18) 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)
Ans18) 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

Q19) 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)
Ans19)  

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

Q20) 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 :

Ans20) 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

Q21) Show that 2sec² - sec⁴ - 2cosec² + cosec⁴ = cot⁴ - tan⁴   (Marks 4)
Ans21)  RHS = cot⁴ - tan⁴
= (cot² )² - (tan² )²
= (cosec² - 1)² - (sec² - 1)²   (^..^. 1 + tan² = sec² , 1 + cot² = cosec² )
= cosec⁴ + 1 - 2cosec² - sec⁴ - 1 + 2sec²
= 2sec² - sec⁴ - 2cosec² + cosec⁴

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