CBSE [ Delhi ]_X_Mathematics_2008_Set II
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Q1
Find the value of k, so that the following system of equations has no solution: 3x – y –5 = 0; 6x – 2y – k = 0.
Marks:1Answer:
The given system of equations is:
3x – y –5 = 0; 6x – 2y – k = 0
The system of equations has no solution,
i.e.,
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Q2
The nth term of an AP is 6n + 2. Find its common difference.
Marks:1Answer:
nth term of an A.P. tn = 6n + 2.
First term t1 = 6.1 + 2 = 8
Second term t2 = 6.2 + 2 = 14
Third term t3 = 6.3 + 2 = 20
Common difference = 14 – 8 = 20 – 14 = 6. -
Q3
In Figure, AD = 4 cm, BD = 3 cm and CB = 12 cm, find cot
.
Marks:1Answer:
Here, AD = 4 cm, BD = 3 cm and CB = 12 cm.
In right angled triangle ADB, -
Q4
Write the zeroes of the polynomial x2 – x – 6.
Marks:1Answer:
x2 – x – 6
= x2 – (3 - 2)x – 6
= x2 – 3x + 2x – 6
= x(x – 3)(x + 2)
The zeroes of the given polynomial are 3, -2. -
Q5
If (p/q) is a rational number (q
0), what is the condition on q so that the decimal representation of (p/q) is terminating?
Marks:1Answer:
If q is of the form of 2m x 5n, where m and n are positive integers, then the decimal representation of the rational number (p/q) is terminating.
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Q6
From a well shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen.
Marks:1Answer:
Favorable cases = 2
Total cases = 52
Probability of getting a black queen = 2/52 = 1/26. -
Q7
Which measure of central tendency is given by the x-coordinate of the point of intersection of the “more than ogive” and “less than ogive”?
Marks:1Answer:
Median.
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Q8
In figure, O is the centre of a circle. The area of sector OAPB is 5/18 of the area of the circle. Find x.
Marks:1Answer:
Let radius of the circle be r.
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Q9Marks:1
Answer:
In right angled triangle PAR,
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Q10
In figure, P and Q are the points on the sides AB and AC respectively of
ABC such that AP = 3.5 cm, AQ = 3 cm, PB = 7 cm and QC = 6 cm. If PQ = 4.5 cm, find BC.
Marks:1Answer: