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Q.1. An investigator interviewed 100 students to determine their preferences for the three drinks : milk (M), coffee (C) and tea (T). He reported the following : 10 students had all the three drinks M, C and T; 20 had M and C; 30 had C and T; 25 had M and T; 12 had M only; 5 had C only; and 8 had T only.
Using a Venn diagram find how many did not take any of the three drinks. (1978)
Ans. Sol. We have
n (U) = 100, where U stands for universal set
n (M ∩ C ∩ T) = 10; n (M ∩ C) = 20;.
n (C ∩ T) = 30; n (M ∩ T) = 25;
n (M only) = 12; n (only C) = 5;
n (only T) = 8
Filling all the entries we obtain the Venn diagram as shown :
∴ n (M ∩ C ∪ T) = 12 + 10 + 5 + 15 + 10 + 20 + 8 = 80
∴ n (M ∪ C ∪ T)' = 100 – 80 = 20
Q.2. (a) Construct a triangle with base 9 cm and altitude 4 cm, the ratio of the other two sides being 2 : 1.
(b) Construct a triangle in which the sum of the three sides is 15 cm with base angles 60° and 45°. Justify your steps.
(1979)
Ans. Sol. (a) To construct a Δ with base = 9 cm, altitude = 4 cm and ratio of the other two sides as 2 : 1.
Steps of Construction : 1. Draw BC = 9 cm
2. Divide BC internally at P and externally at Q in the ratio 2 : 1.
3. Draw a semicircle on PQ.
4. Draw a line  BQ at a distance of 4 cm from intersecting semicircle at A and A'.
5. ABC and A'BC are the required Δ's.
(b) To construct a Δ with perimeter = 15 cm, base angles 60° and 45°.
Steps of Construction :
1. Draw PQ = 15 cm
2. At P draw ∠RPQ = 60° and
and at Q draw ∠SQP = 45° and ∠YQP =
3. PX and QY meet each other at A.
4. Through A draw AB  PR and AC  QS.
5. ABC is the required Δ.
Justification: Q AB  PR and PA transversal
∴∠PAB = ∠RPA = x 60° = 30°
∠APB = ∠BAP = 30° ⇒ AB = PB
Similarly AC = CQ
∴ AB + BC + CA = PB + BC + CQ = 15 cm
Also ∠ABC = ∠RPB = 60° and ∠ACB = ∠SQS = 45°
Q.3. A tent is made in the form of a frustrum A of a right circular cone surmounted by another right circular cone B. The diameter of the ends of the frustrum A are 8 m and 4 m, its height is 3 m and the height of the cone B is 2 m. Find the area of the canvas required. (1979)
Ans. Sol. Slant height of frustum
∴ Curved surface area of frustum =
Also slant height of cone
∴ Curved surface area of cone =
∴ Area of canvas required =
Q.4. In calculating the mean and variance of 10 readings, a student wrongly used the figure 52 for the correct figure of 25. He obtained the mean and variance as 45.0 and 16.0 respectively. Determine the correct mean and variance. (1979)
Ans. Sol. Let the remaining 9 readings be x_{1}, x_{2}, x_{3} ..... x_{9} and tenth is taken by the student as 52.
∴ Incorrect mean
⇒ x_{1 }+ x_{2} + ....+ x_{9} = 450 – 52 = 398
∴ Correct mean
⇒ Correct mean = 42.3
Incorrect variance
= 111 + 90 × 398 – 9 × (45)^{2} = 17706 ...(2)
Correct variance
= 43.81
Q.5. The diameter PQ of a semicircle is 6 cm. Construct a square ABCD with points A, B on the circumference, and the side CD on the diameter PQ. Describe briefly the method of con struction. (1980)
Ans. Sol. The rough figure is as shown, let 2x be the side of square.
In ΔOBC r^{2} = x^{2} + 4x^{2}
Now on number line we locate M such that OM =
Divide OM in five equal parts and take a point C on it such that OC : CM = 3 : 2. So, that
Now, draw a line segment PQ = 6 cm whose midpoint is O.
From this cut a line segment and on opposite sides of O.
At C and B draw perpendiculars to PQ. With O as centre and OP as radius, draw a semicircle intersecting the perpendiculars draw at C & D at A and P respectively. Join AB. ABCD is the required square.
Q.6. C and D are any two points on the same side of a line L.
Show how to find a point P on the line L such that PC and PD are equally inclined to the line L. Justify your steps. (1980)
Ans. Sol. Here we are given two points C and D on the same side of line L. To find a point P on L such that PC and PD are equally inclined to L.
Steps of Construction : 1. From C draw CE ⊥ L and produce it to C ' such that EC ' = EC.
2. Join C ' D intersecting L at P. Also join CP.
3. By simple geometry ∠1 = ∠2 and ∠2 = ∠3 ⇒ ∠1 = ∠3
∴ PC and PD are the required lines inclined equally to L.
Q.7. (i) Set A has 3 elements, and set B has 6 elements. What can be the minimum number of elements in the set A ∪ B? (1980) (ii) P, Q, R are subsets of a set A. Is the following equality true ?
R × (P^{c} ∪ Q^{c})^{c} = (R × P) ∩ (R × Q)? (1980)
(iii) For any two subset X and Y of a set A define X o Y = (X^{c} ∩ Y) ∪ (X ∩Y^{c}) Then for any three subsets X, Y and Z of the set A, is the following equality true. (X o Y) o Z = X o (Y o Z)? (1980)
Ans. Sol. (i) n (A) = 3, n (B) = 6
We know that n (A ∪ B) ≥ max (n (A), n (B)) ⇒ n (A ∪ B) ≥ 6
∴ Min number of element that A ∪ B can have is 6.
(ii) Here R × (P^{c} ∪ Q^{c})^{c} = R × (P ∩ Q) = (R × P) ∩ (R × Q)
∴ Given equality is true.
(iii) Yes
From (1) and (2) (X o Y) o Z = X o (Y o Z)
Q.8. Suppose A_{1}, A_{2}, ........ A_{30} are thirty sets each with five elements and B_{1}, B_{2}, ....... B_{n} are n sets each with three elements. Let Assume that each element of S belongs to exactly ten of the Ai’ s and to exactly nine of the Bj’ s. Find n. (1981  2 Marks)
Ans. Sol. We are given that ....(1)
Each A'_{i}^{s} contain 5 elements, so contains 5 × 30 = 150 elements (with repetition) out of which each element is repeated 10 times, (as given that each element of S belongs to 10 A'_{i}^{s})
∴ Number of different elements in
∴ From eqn (1) we can say S contains 15 elements...(2) Again each B'_{j}^{s} contains 3 elements, so contains 3 × n = 3_{n} elements (with repetition), out of which each element is repeated 9 times (as each element of S belongs to 9 B'_{j}^{s})
∴ No. of different elements in
∴ From eqn (1) we can say S contain n/3 elements....(3) From (2) and (3) we get =15 ⇒ n = 45
Q.9. The mean square deviations of a set of observations x_{1}, x_{2}, ....., x_{n} about a points c is defined to be
The mean suqare deviations about –1 and +1 of a set of observatons are 7 and 3 respectively. Find the standard deviation of this set of observations. (1981  2 Marks)
Ans. Sol. Given that : Mean square deviation for the observations x_{1}, x_{2}, .... x_{n}, about a point. c is given by Also given that mean square deviations about – 1 and + 1 are 7 and 3 for a particular set of observations.
and
and
and NOTE THIS STEP
...(1)
and ...(2)
Subtracting (2) from (1), we get
Now standard deviation for same set of observations
{using the given value}
Q.10. The marks obtained by 40 students are grouped in a frequency table in class intervals of 10 maks each. The mean and the variance obtained from this distribution are found to be 40 and 49 respectively. It was later discovered that two observations belonging to the class interval (21–30) were included in the class interval (31–40) by mistake. Find the mean and the variance after correcting the error. (1982  3 Marks)
Ans. Sol. n = 40
....(1)
Also Var. = 49 =
2 x 1600 x 40 x 40
...(2)
Let 21 – 30 and 31 – 40 denote the k^{th} and (k + 1)^{th }class intervals respectively.
Then if before correction f_{k} and f_{k+1} are the frequencies of these intervals then after correction (2 observations are shifted from 31 – 40 to 21 – 30), frequency of k^{th} intervals becomes f_{k} + 2 and frequency of (k + 1)^{th} interval becomes f_{k+1} – 2.
Then, we get
= 40 – 0.5 = 39.5
= 1649 – 3160 + 1560.25 = 49.25 [Using eqn. (1) and (2)]
Q.11. A relation R on the set of complex numbers is defined by z_{1} R z_{2} if and only if
is real. Show that R is an equivalence relation. (1982  2 Marks)
Ans. Sol. Given that is real.
To show that R is an equivalence relation.
Reflexivity : For z_{1 }= z_{2} = z (say)
0 which is real
∴ R is reflexive.
Symmetric : Let z_{1} R z_{2} is real also real
is real ⇒ z_{2} R z_{1} ⇒ R is symmetric.
Transitivity : Let z_{1} R z_{2} and z_{2} R z_{3}
is real and is also real
Now, is real ⇒
...(1)
Similarly,...(2)
From (1) and (2) we get
⇒ = 0 ⇒ is real
⇒ z_{1}R z_{3} ∴ R is transitive.
Thus R is reflexive, symmetric and transitive.
Hence R is an equivalence relation.
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