125876 + Interview Questions in General QA

1.	Identify the wrongly matched pair. (a) Equal – == (b) Not Equal – ! = (c) Not Identical – <> (d) Identical – ===
Answer» (c) Not Identical – <>

Discussion

2.	Space is considered as a character. (a) True (b) False
Answer» Correct Answer is : (a) True

Discussion

3.	How many possible values does Boolean data type has? (a) 2 (b) 3 (c) 4 (d) 5
Answer» Correct Answer is : (a) 2

Discussion

4.	!$x is an example of …………operator. (a) And (b) Or (c) Not (d) Xor
Answer» !$x is an example of Not operator.

Discussion

5.	Which operator is mostly used during iteration in the program logics. (a) Increment (b) Decrement (c) both a & b(d) none of these
Answer» (c) both a & b

Discussion

6.	Match the following 1. And – (i) && 2. Or – (ii) \|\| 3. Not – (iii) ! 4. Not equal – (iv) image(a) 1-(i) 2-(ii) 3-(iii) 4-(iv) (b) 1-(iv) 2-(iii) 3 -(ii) 4-(iii)
Answer» (a) 1-(i) 2-(ii) 3-(iii) 4-(iv)

Discussion

7.	OOP’S stands for ……………
Answer» OOP’S stands for object oriented programming.

Discussion

8.	How many clarifications of web scripting languages are there? (a) 1 (b) 2 (c) 3 (d) 4
Answer» Correct Answer is : (b) 2

Discussion

9.	How many client server architecture models there? (a) 2 (b) 3(c) 4 (d) 5
Answer» Correct Answer is : (b) 3

Discussion

10.	Under which directory httpd.conf file is located? (a) configure (b) config (c) configuration (d) conf
Answer» Correct Answer is :(d) conf

Discussion

11.	The statements in PHP ends with (a) $ (b) = (c) ;(d) :
Answer» Correct Answer is : (c) ;

Discussion

12.	Which one of the following statement is correct related to variable? (a) data type is required (b) data type is not required
Answer» (b) data type is not required

Discussion

13.	The variable in PHP begins with a …………symbol.(a) $ (b) # (c) II (d) =
Answer» The variable in PHP begins with a $ symbol.

Discussion

14.	The ………is a separate hardware machine which is connected with server in the networks.
Answer» The Client is a separate hardware machine which is connected with server in the networks.

Discussion

15.	Match the following 1. Webservers – (a) ASP 2. Server Scripting Language – (b) Server 3. Service provider – (c) Microsoft IIS 4. Service requester – (d) Client (a) 1 -(c) 2-(a) 3-(b) 4-(d) (b) 1-(a)2 (b) 3-(c) 4-(d) (c) 1-(d)2-(c)3-(b)4-(a) (d) 1-(d) 2-(a) 2-(b) 4-(c)
Answer» (a) 1 -(c) 2-(a) 3-(b) 4-(d)

Discussion

16.	Which one of the following is the modulus operator? (a) + (b) % (c) \|\| (d) !
Answer» Correct Answer is : (b) %

Discussion

17.	Write is the purpose of Web servers?
Answer» 1. Webserver is software which is running in server hardware. 2. It takes the responsibilities for compilation and execution of server side scripting languages. 3. After receiving the request from client machine the Web server tries to compile and interpret the PHP code which is available in remote machine. 4. Next a response will be generated and sent back to the client machine over the network from Webserver.

17.

Write is the purpose of Web servers?

Answer»

1. Webserver is software which is running in server hardware.

2. It takes the responsibilities for compilation and execution of server side scripting languages.

3. After receiving the request from client machine the Web server tries to compile and interpret the PHP code which is available in remote machine.

4. Next a response will be generated and sent back to the client machine over the network from Webserver.

Discussion

18.	Write a menu driven program to add or delete stationary items. You should use dictionary to store items and the brand?
Answer» stationary = { } while((ch == 1) or (ch == 2)) print(” 1. Add Item \n 2. Delete Item") ch = int(input(“Enter your choice “)) if(ch==1): n = int(input(“Enter the number of items to be added in the stationary shop”)) for i in range(n): item = input(“Enter an item “) brand = input(“Enter the brand Name”) stationary[item] = brand print(stationary) elif(ch == 2): remitem = input(“Enter the item to be deleted from the shop”) dict.pop(remitem) print( stationary) else: print(“Invalid options. Type 1 to add items and 2 to remove items “) ch = int(input(“Enter your choice :”) Output: 1. Add item 2. Delete Item Enter your choice :1 Enter the number of items to be added in the stationary shop : 2 Enter an item : Pen Enter the brand Name : Trimax Enter an item : Eraser Enter the brand Name : Camlin Pen : Trimax Eraser : Camlin Enter your choice : 2 Enter the item to be deleted from the shop : Eraser Pen : Trimax Enter your choice : 3 Invalid options. Type 1 to add items an 2 to remove items.

18.

Write a menu driven program to add or delete stationary items. You should use dictionary to store items and the brand?

Answer»

stationary = { }

while((ch == 1) or (ch == 2))

print(” 1. Add Item \n 2. Delete Item")

ch = int(input(“Enter your choice “))

if(ch==1):

n = int(input(“Enter the number of items to be added in the stationary shop”))

for i in range(n):

item = input(“Enter an item “)

brand = input(“Enter the brand Name”)

stationary[item] = brand

print(stationary)

elif(ch == 2):

remitem = input(“Enter the item to be deleted from the shop”)

dict.pop(remitem)

print( stationary)

else:

print(“Invalid options. Type 1 to add items and 2 to remove items “)

ch = int(input(“Enter your choice :”)

Output:

1. Add item

2. Delete Item Enter your choice :1

Enter the number of items to be added in the

stationary shop : 2

Enter an item : Pen

Enter the brand Name : Trimax

Enter an item : Eraser

Enter the brand Name : Camlin

Pen : Trimax

Eraser : Camlin

Enter your choice : 2

Enter the item to be deleted from the shop : Eraser Pen : Trimax

Enter your choice : 3

Invalid options. Type 1 to add items an 2 to remove items.

Discussion

19.	How will you delete a string in Python?
Answer» Python will not allow deleting a particular character in a string. Whereas you can remove entire string variable using del command. Example: Code lines to delete a string variable >>> str1= ”How about you” >>> print (str1) How about you >>> del str1 >>> print (str1) Name Error: name ‘str1’ is not defined

19.

How will you delete a string in Python?

Answer»

Python will not allow deleting a particular character in a string. Whereas you can remove entire string variable using del command.

Example:

Code lines to delete a string variable

>>> str1= ”How about you”

>>> print (str1)

How about you

>>> del str1

>>> print (str1)

Name Error: name ‘str1’ is not defined

Discussion

20.	Write a C++ program to print multiplication table of a given number.
Answer» # include using namespace std; int main () { int num; cout << “Enter Number to find its multiplication table”; cin >>num; for (int a = 1; a < = 10; a++) { cout<<num <<“” << a << “ = ” << numa << endl; } return( ); }

20.

Write a C++ program to print multiplication table of a given number.

Answer»

# include

using namespace std;

int main ()

{

int num;

cout << “Enter Number to find its

multiplication table”; cin >>num;

for (int a = 1; a < = 10; a++)

{

cout<<num <<“*” << a << “ = ” <<

num*a << endl;

}

return( );

}

Discussion

21.	Deepika wants to remove all rows from the table BANK. But he needs to maintain the structure of the table. Which command is used to implement the same?
Answer» DELETE FROM BANK.

Discussion

22.	Consider the following tables CABHUB and CUSTOMER and answer (b) and (c) parts of this question :1. Give a suitable example of a table with sample data and illustrate Primary and Candidate Keys in it.2. Write SQL commands for the following statements:(i) To display the names of all the white coloured vehicles.(ii) To display name of vehicle name and capacity of vehicles in ascending order of their sitting capacity.(iii) To display the highest charges at which a vehicle can be hired from CABHUB.(iv) To display the customer name and the corresponding name of the vehicle hired by them.3. Give the output of the following SQL queries :(i) SELECT COUNT(DISTINCT Make) FROM CABHUB;(ii) SELECT MAX(Charges), MIN(Charges)(iii) FROM CABHUB;(iv) SELECT COUNT (*) Make FROM CABHUB;(v) SELECT Vehicle FROM CABHUB WHERE Capacity=4;
Answer» 1. Primary key of CABHUB = Vcode alternate key of CABHUB = Vehicle Name. Primary key of Customer = Ccode Alternate Key of CUSTOMER = Cname. 2. (i) SELECT VehicleName FROM CABHUB WHERE Colour = “WHITE”; (ii) SELECT VehicleName, capacity From CABHUB ORDER BY Capacity ASC; (iii). SELECT MAX(Charges) FROM CABHUB; (iv). SELECT Cname,VehicleName FROM CABHUB, CUSTOMER WHERE CUSTOMER. Vcode=CABHUB. Vcode; 3. (i) 4 (ii) MAX(Charges) MIN (Charges) 35 12 (iii) 5 (iv) SX4 (v) C Class

22.

Consider the following tables CABHUB and CUSTOMER and answer (b) and (c) parts of this question :1. Give a suitable example of a table with sample data and illustrate Primary and Candidate Keys in it.2. Write SQL commands for the following statements:(i) To display the names of all the white coloured vehicles.(ii) To display name of vehicle name and capacity of vehicles in ascending order of their sitting capacity.(iii) To display the highest charges at which a vehicle can be hired from CABHUB.(iv) To display the customer name and the corresponding name of the vehicle hired by them.3. Give the output of the following SQL queries :(i) SELECT COUNT(DISTINCT Make) FROM CABHUB;(ii) SELECT MAX(Charges), MIN(Charges)(iii) FROM CABHUB;(iv) SELECT COUNT (*) Make FROM CABHUB;(v) SELECT Vehicle FROM CABHUB WHERE Capacity=4;

Answer»

1. Primary key of CABHUB = Vcode

alternate key of CABHUB = Vehicle Name.

Primary key of Customer = Ccode

Alternate Key of CUSTOMER = Cname.

2. (i) SELECT VehicleName FROM CABHUB WHERE Colour = “WHITE”;

(ii) SELECT VehicleName, capacity From CABHUB ORDER BY Capacity ASC;

(iii). SELECT MAX(Charges) FROM CABHUB;

(iv). SELECT Cname,VehicleName FROM CABHUB, CUSTOMER WHERE CUSTOMER. Vcode=CABHUB. Vcode;

3. (i) 4

(ii) MAX(Charges) MIN (Charges) 35 12

(iii) 5

(iv) SX4

(v) C Class

Discussion

23.	Consider the following tables STOCK and DEALERS and answer (a) and (b) parts of this question:(a) Write SQL commands for the following statements:1. To display the details of all Items in the STOCK table in ascending order of StockDate.2. To display ItemNo and Item name of those items from STOCK table whose UnitPrice is more than Rupees 10.3. To display the details of those items whose dealer code (Dcode) is 102 or quantity in STOCK (Qty) is more than 100 from the table Stock.4. To display maximum UnitPrice of items for each dealer individually as per Dcode from the table STOCK.(b) Give the output of the following SQL queries:1. SELECT COUNT(DISTINCT Dcode) FROM STOCK;2. SELECT Qty* UnitPrice FROM STOCK WHERE ItemNo=5006;3. SELECT Item, Dname FROM STOCK S, Dealers D WHERE S.Dcode=D.Dcode AND ItemNo = 5004;4. SELECT MIN (StockDate) FROM STOCK;
Answer» (a) 1. SELECTFROM STOCK ORDER BY StockDate; 2. SELECT Item No, Item FROM STOCK WHERE UnitPrice >10; 3. SELECT FROM DEALERS, STOCK WHERE (DEALERS.Dcode=”102″OR STOCK.Qty >100 and DEALERS. DCODE = STOCK.DCODE); 4. SELECT MAX (Unitprice) FROM DEALERS, STOCK ORDER BY STOCK. Dcode WHERE DEALERS.Dcode = STOCK.Dcode; (b) 1. 3 2. 4400 3. Item Dname Eraser Big Clear Deals 4. 01-Jan-09

23.

Consider the following tables STOCK and DEALERS and answer (a) and (b) parts of this question:(a) Write SQL commands for the following statements:1. To display the details of all Items in the STOCK table in ascending order of StockDate.2. To display ItemNo and Item name of those items from STOCK table whose UnitPrice is more than Rupees 10.3. To display the details of those items whose dealer code (Dcode) is 102 or quantity in STOCK (Qty) is more than 100 from the table Stock.4. To display maximum UnitPrice of items for each dealer individually as per Dcode from the table STOCK.(b) Give the output of the following SQL queries:1. SELECT COUNT(DISTINCT Dcode) FROM STOCK;2. SELECT Qty* UnitPrice FROM STOCK WHERE ItemNo=5006;3. SELECT Item, Dname FROM STOCK S, Dealers D WHERE S.Dcode=D.Dcode AND ItemNo = 5004;4. SELECT MIN (StockDate) FROM STOCK;

Answer»

(a)

1. SELECT*FROM STOCK ORDER BY StockDate;

2. SELECT Item No, Item FROM STOCK WHERE UnitPrice >10;

3. SELECT *FROM DEALERS, STOCK WHERE (DEALERS.Dcode=”102″OR STOCK.Qty >100 and DEALERS. DCODE = STOCK.DCODE);

4. SELECT MAX (Unitprice) FROM DEALERS, STOCK ORDER BY STOCK. Dcode WHERE DEALERS.Dcode = STOCK.Dcode;

(b)

1. 3

2. 4400

3. Item Dname

Eraser Big Clear Deals

4. 01-Jan-09

Discussion

24.	Consider the following tables STORE and SUPPLIERS and answer (a) and (b) parts of this question:(a) Write SQL commands for the following statements:1. To display details of all the items in the STORE table in ascending order of BestBuy.2. To display ItemNo and Item name of those items from STORE table whose Rate is more than 15 Rupees.3. To display the details of those items whose supplier code (Scode) is 22 or Quantity in Store (Qty) is more than 110 from the table Store.4. To display the minimum Rate of items for each supplier individually as per Scode from the table STORE.(b) Give the output of the following SQL queries:1. SELECT COUNT(DISTINCT Scode) FROM STORE;2. SELECT Rate* Qty FROM STORE WHERE ItemNo=2004;3. SELECT Item, Sname FROM STORE S, Suppliers P4. WHERE S.Scode=PScode AND ItemNo=2006;5. SELECT MAX(LastBuy) FROM STORE;
Answer» (a) 1. SELECT * FROM STORE ORDER BY LastBuy ASC; 2. SELECT ItemNo, Item FROM STORE WHERE Rate > 15; 3. SELECT * FROM STORE WHERE (Scode = ’22’ OR Qty >’110′); 4. SELECT Sname, MIN(Rate) FROM STORE, SUPPLIERS WHERE STORE. Scode = SUPPLIERS.Scode GROUP BY Sname; (b) 1. 3 2. 880 3. Item Sname Gel Pen Classic Premium Stationers 4. 24-Feb-10

24.

Consider the following tables STORE and SUPPLIERS and answer (a) and (b) parts of this question:(a) Write SQL commands for the following statements:1. To display details of all the items in the STORE table in ascending order of BestBuy.2. To display ItemNo and Item name of those items from STORE table whose Rate is more than 15 Rupees.3. To display the details of those items whose supplier code (Scode) is 22 or Quantity in Store (Qty) is more than 110 from the table Store.4. To display the minimum Rate of items for each supplier individually as per Scode from the table STORE.(b) Give the output of the following SQL queries:1. SELECT COUNT(DISTINCT Scode) FROM STORE;2. SELECT Rate* Qty FROM STORE WHERE ItemNo=2004;3. SELECT Item, Sname FROM STORE S, Suppliers P4. WHERE S.Scode=PScode AND ItemNo=2006;5. SELECT MAX(LastBuy) FROM STORE;

Answer»

(a)

1. SELECT * FROM STORE ORDER BY LastBuy ASC;

2. SELECT ItemNo, Item FROM STORE WHERE Rate > 15;

3. SELECT * FROM STORE WHERE (Scode = ’22’ OR Qty >’110′);

4. SELECT Sname, MIN(Rate) FROM STORE, SUPPLIERS WHERE STORE. Scode = SUPPLIERS.Scode GROUP BY Sname;

(b)

1. 3

2. 880

3. Item Sname

Gel Pen Classic Premium Stationers

4. 24-Feb-10

Discussion

25.	Sonal needs to display name of teachers, who have “0” as the third character in their name. She wrote the following query. SELECT NAME FROM TEACHER WHERE NAME = “$$0?”; But the query is’nt producing the result. Identify the problem.
Answer» The wildcards are incorrect. The corrected query is SELECT NAME FROM TEACHER WHERE NAME LIKE ‘_ _0%’.

Discussion

26.	Consider the following tables SCHOOL and ADMIN and answer this question :Write SQL statements for the following:1. To display TEACHERNAME, PERIODS of all teachers whose periods are more than 25.2. To display all the information from the table SCHOOL in descending order of experience.3. To display DESIGNATION without dupli¬cate entries from the table ADMIN.4. To display TEACHERNAME, CODE and corresponding DESIGNATION from tables SCHOOL and ADMIN of Male teachers.
Answer» 1. SELECT TEACHERNAME, PERIODS FROM SCHOOL WHERE PERIODS>25: 2. SELECT * FROM SCHOOL; 3. SELECT DISTINCT DESIGNATION FROM ADMIN; 4. SELECT TEACHERNAME.CODE DESIGNATION FROM SCHOOL.CODE = ADMIN.CODE WHERE GENDER = MALE;

26.

Consider the following tables SCHOOL and ADMIN and answer this question :Write SQL statements for the following:1. To display TEACHERNAME, PERIODS of all teachers whose periods are more than 25.2. To display all the information from the table SCHOOL in descending order of experience.3. To display DESIGNATION without dupli¬cate entries from the table ADMIN.4. To display TEACHERNAME, CODE and corresponding DESIGNATION from tables SCHOOL and ADMIN of Male teachers.

Answer»

1. SELECT TEACHERNAME, PERIODS FROM SCHOOL WHERE PERIODS>25:

2. SELECT * FROM SCHOOL;

3. SELECT DISTINCT DESIGNATION FROM ADMIN;

4. SELECT TEACHERNAME.CODE

DESIGNATION FROM

SCHOOL.CODE = ADMIN.CODE

WHERE GENDER = MALE;

Discussion

27.	Write SQL query to add a column total price with datatype numeric and size 10, 2 in a table product.
Answer» ALTER TABLE PRODUCT ADD TOTAL PRICE NUMBER (10,2).

Discussion

28.

Consider the following tables SCHOOL and ADMIN and answer this question : Give the output the following SQL queries :1. Select Designation Count () From Admin Group By Designation Having Count () <2;2. SELECT max (EXPERIENCE) FROM SCHOOL;3. SELECT TEACHER FROM SCHOOL WHERE EXPERIENCE >12 ORDER BY TEACHER;4. SELECT COUNT (*), GENDER FROM ADMIN GROUP BY GENDER;

Answer»

(i)

VICE PRINCIPAL

01

(ii)

16

(iii)

UMESH

YASH RAJ

(iv)

5 MALE

2 FEMALE

Discussion

29.	Which types of debenture are purchases or sales in the stock exchange?
Answer» Listed debentures are sold and purchased in stock exchange.

Discussion

30.	What is the second mortgage debenture?
Answer» Second mortgage (charge) debentures are those having a second claim(after first mortgage debentures) on the assets charged.

Discussion

31.	What is the first mortgage debenture?
Answer» First mortgage (charge) debentures are those that have a first claim on the assets charged.

Discussion

32.	In Fig, lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.
Answer» Given that, ∠x = 45° From the figure we can write as ∠x = ∠z = 45° Also from the figure, we have ∠y = ∠u From the property of linear pair we can write as ∠x + ∠y + ∠z + ∠u = 360° 45° + 45° + ∠y + ∠u = 360° 90° + ∠y + ∠u = 360° ∠y + ∠u = 360° – 90° ∠y + ∠u = 270° ∠y + ∠z = 270° 2∠z = 270° ∠z = 135° Therefore, ∠y = ∠u = 135° So, ∠x = 45°, ∠y = 135°, ∠z = 45° and ∠u = 135°

32.

In Fig, lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.

Answer»

Given that, ∠x = 45°

From the figure we can write as

∠x = ∠z = 45°

Also from the figure, we have

∠y = ∠u

From the property of linear pair we can write as

∠x + ∠y + ∠z + ∠u = 360°

45° + 45° + ∠y + ∠u = 360°

90° + ∠y + ∠u = 360°

∠y + ∠u = 360° – 90°

∠y + ∠u = 270°

∠y + ∠z = 270°

2∠z = 270°

∠z = 135°

Therefore, ∠y = ∠u = 135°

So, ∠x = 45°, ∠y = 135°, ∠z = 45° and ∠u = 135°

Discussion

33.	In Fig., POS is a line, find x.
Answer» ∠POQ + ∠QOS = 180°(Linear pair) ∠POQ + ∠QOR + ∠SOR = 180° 60° + 4x + 40° = 180° 4x = 80° x = 20°

33.

In Fig., POS is a line, find x.

Answer»

∠POQ + ∠QOS = 180°(Linear pair)

∠POQ + ∠QOR + ∠SOR = 180°

60° + 4x + 40° = 180°

4x = 80°

x = 20°

Discussion

34.	In Fig, POS is a line, find x.
Answer» From the figure we can write as angles of a straight line, ∠QOP + ∠QOR + ∠ROS = 180° 60° + 4x + 40° = 180° On rearranging we get, 100° + 4x = 180° 4x = 180° – 100° 4x = 80° x = 80°/4 x = 20°

34.

In Fig, POS is a line, find x.

Answer»

From the figure we can write as angles of a straight line,

∠QOP + ∠QOR + ∠ROS = 180°

60° + 4x + 40° = 180°

On rearranging we get, 100° + 4x = 180°

4x = 180° – 100°

4x = 80°

x = 80°/4

x = 20°

Discussion

35.	In figure, POS is a line, find x.
Answer» From figure, ∠POQ and ∠QOS are linear pairs. Therefore, ∠POQ + ∠QOS = 180° ∠POQ + ∠QOR +∠SOR = 180° 60° + 4x + 40° = 180° 4x = 180° -100° 4x = 80° x = 20° Hence, the value of x is 20°.

35.

In figure, POS is a line, find x.

Answer»

From figure, ∠POQ and ∠QOS are linear pairs.

Therefore, ∠POQ + ∠QOS = 180°

∠POQ + ∠QOR +∠SOR = 180°

60° + 4x + 40° = 180°

4x = 180° -100°

4x = 80°

x = 20°

Hence, the value of x is 20°.

Discussion

36.	In figure, determine the value of x.
Answer» The sum of all the angles around a point O is equal to 360°. Therefore, 3x + 3x + 150 + x = 360° 7x = 360° – 150° 7x = 210° x = 210/7 x = 30° Hence, the value of x is 30°.

36.

In figure, determine the value of x.

Answer»

The sum of all the angles around a point O is equal to 360°.

Therefore, 3x + 3x + 150 + x = 360°

7x = 360° – 150°

7x = 210°

x = 210/7

x = 30°

Hence, the value of x is 30°.

Discussion

37.	In Fig, AOC is a line, find x.
Answer» From the figure we can write as ∠AOB + ∠BOC = 180° [linear pair] Linear pair 2x + 70° = 180° 2x = 180° – 70° 2x = 110° x = 110°/2 x = 55°

37.

In Fig, AOC is a line, find x.

Answer»

From the figure we can write as

∠AOB + ∠BOC = 180° [linear pair]

Linear pair

2x + 70° = 180°

2x = 180° – 70°

2x = 110°

x = 110°/2

x = 55°

Discussion

38.	In figure, AOC is a line, find x.
Answer» From the figure, ∠AOB and ∠BOC are linear pairs, ∠AOB + ∠BOC = 180° 70 + 2x = 180° 2x = 180° – 70° 2x = 110° x = 110/2 x = 55° Therefore, the value of x is 55°.

38.

In figure, AOC is a line, find x.

Answer»

From the figure, ∠AOB and ∠BOC are linear pairs,

∠AOB + ∠BOC = 180°

70 + 2x = 180°

2x = 180° – 70°

2x = 110°

x = 110/2

x = 55°

Therefore, the value of x is 55°.

Discussion

39.	In the given figure p, q, r and s are parallel lines and t is a transversal. Find x, y and z.
Answer» p ∥ q So, 80° + x = 180° (co-interior angles) 80° + x 80° = 180° 80° x = 100° q ∥ r x + y = 180° (co-interior angles) 100° + y = 180° (we know x – 100°) 100° + y – 100° = 180°- 100° y = 80° r∥s y = z (Alternate exterior angles) y = z = 80° (we know y = 80°) x = 100°, y = 80° and z = 80°

39.

In the given figure p, q, r and s are parallel lines and t is a transversal. Find x, y and z.

Answer»

p ∥ q So, 80° + x = 180° (co-interior angles)

80° + x 80° = 180° 80°

x = 100°

q ∥ r

x + y = 180° (co-interior angles)

100° + y = 180° (we know x – 100°)

100° + y – 100° = 180°- 100°

y = 80°

r∥s

y = z (Alternate exterior angles)

y = z = 80° (we know y = 80°)

x = 100°, y = 80° and z = 80°

Discussion

40.	Two lines \(\overleftrightarrow{PS}\) and \(\overleftrightarrow{QT}\) intersect at M. Observe the figure and find x.
Answer» In the given figure \(\overleftrightarrow{PS}\) and \(\overleftrightarrow{QT}\) intersecting at M. ∠PMQ = ∠TMS (Vertically opposite angles) ∠QMS = ∠PMT (Vertically opposite angles) ∠QMR + ∠RMS = ∠PMT (we know ∠QMS = ∠QMR + ∠RMS) But, given, ∠QMR = 40°, ∠RMS = x° and ∠PMT = 105° ⇒ 40° + x° – 40° = 105° – 40° ∴ x° = 65° According to question, PS and QT intersect at M. So, angles QMS and PMT are equal ( Vertically opposite angles) Angle QMS =40+x° Angle PMT= 105 ° According to question, 40+x°=105° x=105-40= 65° Thus, the value of x=65°

40.

Two lines $\overleftrightarrow{PS}$ and $\overleftrightarrow{QT}$ intersect at M. Observe the figure and find x.

Answer»

In the given figure $\overleftrightarrow{PS}$ and $\overleftrightarrow{QT}$ intersecting at M.

∠PMQ = ∠TMS (Vertically opposite angles)

∠QMS = ∠PMT (Vertically opposite angles)

∠QMR + ∠RMS = ∠PMT (we know ∠QMS = ∠QMR + ∠RMS)

But, given, ∠QMR = 40°, ∠RMS = x° and

∠PMT = 105°

⇒ 40° + x° – 40° = 105° – 40°

∴ x° = 65°

According to question,

PS and QT intersect at M.

So, angles QMS and PMT are equal ( Vertically opposite angles)

Angle QMS =40+x°

Angle PMT= 105 °

According to question,

40+x°=105°

x=105-40= 65°

Thus, the value of x=65°

Discussion

41.	In the given figure, two lines \(\overleftrightarrow{AD}\) and \(\overleftrightarrow{EC}\) intersects at O. Name two pairs of vertically opposite angles in the given figure.
Answer» In the given figure, two lines \(\overleftrightarrow{AD}\) and \(\overleftrightarrow{EC}\) intersecting at O. ∠AOE = ∠COD (Vertically opposite angles) ∠1 = ∠3 ∠EOD = ∠AOC (Vertically opposite angles) ∠EOD = ∠AOB + ∠BOC (we know ∠AOC = ∠AOB + ∠BOC) ∠2 = ∠5 + ∠4

41.

In the given figure, two lines $\overleftrightarrow{AD}$ and $\overleftrightarrow{EC}$ intersects at O. Name two pairs of vertically opposite angles in the given figure.

Answer»

In the given figure, two lines $\overleftrightarrow{AD}$ and $\overleftrightarrow{EC}$ intersecting at O.

∠AOE = ∠COD (Vertically opposite angles)

∠1 = ∠3

∠EOD = ∠AOC (Vertically opposite angles)

∠EOD = ∠AOB + ∠BOC (we know ∠AOC = ∠AOB + ∠BOC)

∠2 = ∠5 + ∠4

Discussion

42.	Treatment of bills payable and bills receivable in dissolution of a firm.
Answer» In dissolution bills payable will be transferred to credit of realisation account and bills receivable to the debit of realisation account and if anything recovers from bills receivable, realisation account will be credited with the cash amount received and the amount of bills payable paid is to be recorded on the debit side of realisation account.

Discussion

43.	Find the general solution of the following equation:cot4x = −1
Answer» cot 4x = -1 cot 4x = cot(π - π/4) cot 4x = cot(3π/4) 4x = nπ + 3π/4 x = \(\frac{n\pi}4+\frac{3\pi}{16}, n\in z\)

43.

Find the general solution of the following equation:cot4x = −1

Answer»

cot 4x = -1

cot 4x = cot(π - π/4)

cot 4x = cot(3π/4)

4x = nπ + 3π/4

x = $\frac{n\pi}4+\frac{3\pi}{16}, n\in z$

Discussion

44.	cos−1(cos7π/6) is equal to(A) 7π/6 (B) 5π/6 (C) π/3 (D) π/6
Answer» Answer: (b) \(\frac{5\pi}6\) Given that \(cos^{-1}(cos\frac{7\pi}6)\) We know that \(cos^{-1}(cos\,x)=x\,if\,x\in[0,\pi],\) which is the principle value branch \(\therefore cos^{-1}(cos\frac{7\pi}6)=cos^{-1}[cos(2\pi-\frac{5\pi}6)]\) \(cos(2\pi-\frac{5\pi}6)=-cos\,\frac{5\pi}6\) \(=cos^{-1}(cos(2\pi-\frac{5\pi}6))=cos^{-1}(-cos\frac{5\pi}6)\) \(=\pi-cos^{-1}(cos\frac{5\pi}6)\) \(=\pi-cos^{-1}(cos(\pi-\frac{\pi}6))\) \(=\pi-cos^{-1}(-cos\frac{5\pi}6)\) \(=\pi-(\pi-cos^{-1}cos\frac{5\pi}6)\) \(=cos^{-1}(cos(\frac{5\pi}6))\) \(=\frac{5\pi}6\)

44.

cos−1(cos7π/6) is equal to(A) 7π/6 (B) 5π/6 (C) π/3 (D) π/6

Answer»

Answer: (b) $\frac{5\pi}6$

Given that $cos^{-1}(cos\frac{7\pi}6)$

We know that $cos^{-1}(cos\,x)=x\,if\,x\in[0,\pi],$ which is the principle value branch

$\therefore cos^{-1}(cos\frac{7\pi}6)=cos^{-1}[cos(2\pi-\frac{5\pi}6)]$

$cos(2\pi-\frac{5\pi}6)=-cos\,\frac{5\pi}6$

$=cos^{-1}(cos(2\pi-\frac{5\pi}6))=cos^{-1}(-cos\frac{5\pi}6)$

$=\pi-cos^{-1}(cos\frac{5\pi}6)$

$=\pi-cos^{-1}(cos(\pi-\frac{\pi}6))$

$=\pi-cos^{-1}(-cos\frac{5\pi}6)$

$=\pi-(\pi-cos^{-1}cos\frac{5\pi}6)$

$=cos^{-1}(cos(\frac{5\pi}6))$

$=\frac{5\pi}6$

Discussion

45.	Find two numbers such that the sum of twice the first and thrice the second is 89 and four times the first exceeds five times the second by 13.
Answer» Let the two numbers be x and y. Then, equation formed are 2x + 3y = 89 ....(i) 4x - 5y = 13 ...(ii) On solving eq. (i) & (ii) we get x = 22 y = 15 Hence required numbers are 22 & 15.

45.

Find two numbers such that the sum of twice the first and thrice the second is 89 and four times the first exceeds five times the second by 13.

Answer»

Let the two numbers be x and y.

Then, equation formed are 2x + 3y = 89 ....(i)

4x - 5y = 13 ...(ii)

On solving eq. (i) & (ii) we get

x = 22

y = 15

Hence required numbers are 22 & 15.

Discussion

46.	Find the cube of : (i) 7 (ii) 11 (iii) 16 (iv) 23 (v) 31 (vi) 42 (vii) 54
Answer» (i) (7)³ = 7 x 7 x 7 = 343 (ii) (11)³ =11 x 11 x 11 = 1331 (iii) (16)³ = 16 x 16 x 16 = 4096 (iv) (23)³ = 23 x 23 x 23 = 12167 (v) (31)³ = 31 x 31 x 31 = 29791 (vi) (42)³ = 42 x 42 x 42 = 74088 (vii) (54)³ = 54 x 54 x 54 = 157464

46.

Find the cube of : (i) 7 (ii) 11 (iii) 16 (iv) 23 (v) 31 (vi) 42 (vii) 54

Answer»

(i) (7)³ = 7 x 7 x 7 = 343

(ii) (11)³ =11 x 11 x 11 = 1331

(iii) (16)³ = 16 x 16 x 16 = 4096

(iv) (23)³ = 23 x 23 x 23 = 12167

(v) (31)³ = 31 x 31 x 31 = 29791

(vi) (42)³ = 42 x 42 x 42 = 74088

(vii) (54)³ = 54 x 54 x 54 = 157464

Discussion

47.	What sort of freedom did Mandela enjoy as a boy? Was it real? Give your opinion.
Answer» Mandela was born free. He could run around in their fields. He could swim in the streams nearby. He had the freedom to roast maize under the stars. He enjoyed rides on the backs of bulls. But that freedom was very limited and purely private. Later, it turned out to be an illusion.

Discussion

48.	I was born free. a) able to act at will b) having personal rights c) not subjected to constraints d) costing nothing
Answer» c) not subjected to constraints

Discussion

49.	My freedom was curtailed. a) enhancedb) lost c) reduced d) blocked
Answer» Answer is (c) reduced

Discussion

50.	I was prevented from fulfilling my obligations. a) not able to perform b) stopped from doing c) conditioned to do d) forced to do
Answer» b) stopped from doing

Discussion

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Treatment of bills payable and bills receivable in dissolution of a firm.

Find the general solution of the following equation:cot4x = −1

cos−1(cos7π/6) is equal to(A) 7π/6 (B) 5π/6 (C) π/3 (D) π/6

Find two numbers such that the sum of twice the first and thrice the second is 89 and four times the first exceeds five times the second by 13.

Find the cube of : (i) 7 (ii) 11 (iii) 16 (iv) 23 (v) 31 (vi) 42 (vii) 54

What sort of freedom did Mandela enjoy as a boy? Was it real? Give your opinion.

I was born free. a) able to act at will b) having personal rights c) not subjected to constraints d) costing nothing

My freedom was curtailed. a) enhancedb) lost c) reduced d) blocked

I was prevented from fulfilling my obligations. a) not able to perform b) stopped from doing c) conditioned to do d) forced to do

Identify the wrongly matched pair. (a) Equal – == (b) Not Equal – ! = (c) Not Identical – <> (d) Identical – ===

Which operator is mostly used during iteration in the program logics. (a) Increment (b) Decrement (c) both a & b(d) none of these

Match the following 1. And – (i) && 2. Or – (ii) || 3. Not – (iii) ! 4. Not equal – (iv) image(a) 1-(i) 2-(ii) 3-(iii) 4-(iv) (b) 1-(iv) 2-(iii) 3 -(ii) 4-(iii)