Quiz-1

A variable name in C should not start with a special symbol such as $.

SIMPLE TRICK: how may number of marbles 59(take front and back value) 58,59,60(LCM) =Ans.102660

well explaination

let, speed of Jake is x kmph; hence Paul\'s is (7-x) then time taken be Jake is 24/x and Pul\'s is 24/(7-x) Hence, the formed equation will be: (24/x)+(24/(7-x))=14 so, solving it, x=3,4 According to question, as Jake has more speed than Paul, Hence Paul\'s speed is 4 kmph. (The given answer of this problem is Wrong!!)

It will be option (B)

is

Here all options are in present tense.Began only is in past tense because they asking question about that is happened in past.

15 degree. according to the foirmula (11/2 M-5H) we gwt 15 degree.

11/2*m=30h+-angle

If all the previous questions were answered correctly this is understood.

Answer is south west...

The global variable is the one whose scope is all through the program even if the variable value is changed in some function it is limited to that function itself.

weakens

if else if nested if else are called control statements or can be taken as decision-making statements

as it is signed integer

range is 2^6..i.e -32 to 31...so 32 cant be represented

strstr() is a predefined function used for string handling. strupr() is used to convert string to upercase. strcmp() is used to compare two strings. strchr() is used to Locate first occurrence of character in string. So, basically none is the right answer.

Let X be the total number of balls. 2x/3 be the number of blue balls and 1x/3 be the number of pink balls. given no of non defective balls is 146 since 5/9 of blueballs is defected then remaining non defected blue balls is 4/9 since 7/8 of pink balls is defected then remaining non defected pink balls are 1/8. no of non defective blue balls +no of non defective pink balls =146. 4/9[2x/3]+1/8[x/3]=146 8x/27+x/24=146 x=432

b

Since infant weight should be 60% i.e(3/5) therefore the weight of daughter should be a multiple of 5 hence in the options there is only one option i.e 10 hence 10 is the answer

is

Most of the python functions are anonymous functions.

By using a function whenever the statements are required they can be called easily.

yield

the loop statements will take more time than the normal statements as it is only 2*2 matrix it is enough to calculate them by 2 stmnts

firstly remove even numbers - 60 now remove multiples of 5 which are not multiple of 2 - 5,15,25,35,45,55,65,75,85,95,105,115 : Total = 12 now remove multiples of 7 which are odd and not multiple of 5 - 7,21,49,63,77,91,119 : Total = 7 Adding them up will give the number of students who have atleast opted for one subject : 60+12+7 = 79 Now students who opted for none of the three subjects = 120-79 = 41

In binary, there are only two possibilities- zero and one.  Thus, looking at the powers of 2, we realize that 32 is the closest exponential result of 2 to 28, where 2^5 = 32. Hence, to cover all the possibilities, we need 5 bits at the least. But there are seven letters. Thus, the number of bits we require is 7 * 5 = 35 bits 

The only number divisble by 9 is 1683

let\'s suppose total quantity of the mixture in both of the tank is 100 ltr now in vessel A milk : Water 4 : 1 so, 4/5*100 = 80 ltr will be milk and rest 20 ltr will be water for vessel B milk : water 3 : 2 so, 3/5*100 = 60 ltr will be milk and rest 40 ltr will be water let\'s assume we take 100 ltr mixture from vessel A and 100 ltr from vessel B i.e. vessel c = vessel A (+) vessel B c = (80ltr milk + 20 ltr water) from A (+) (60ltr milk + 40 ltr water) from B c = (80 ltr milk + 60 ltr milk) (+) (20 ltr water + 40 ltr water) c = 140 ltr milk (+) 60 ltr water so final ratio of mixture in vessel c is c = Milk : Water 140 : 60 7 : 3 (Ans.)

fact of numbers from 5 ends with 0, hence its safe to assume that the sum of the numbers from 5! also ends with a 0, hence considering the numbers from 4! , which is 24....similarly for 1,2,3,4 we have 1+2+6+24=33 as it ends with 3, as told above using 3 cyclicity,....we can find that the answer to this question is 3

xy=696*11 Let x=99 so y=77

Let us find the number of students who took at least one of the three subjects and subtract the result from the overall 120 to get the number of students who did not opt for any of the three subjects. Number of students who took at least one of the three subjects can be found by finding out n(A U B U C), where A is the set of students who took Physics, B is the set of students who took Chemistry and C is the set of students who opted for Math. Now, n(A?B?C) = n(A) + n(B) + n(C) - {n(A ? B) + n(B ? C) + n(C ? A)} + n(A ? B ? C) A is the set of those who opted for Physics = 1202 = 60 students B is the set of those who opted for Chemistry = 1205 = 24 C is the set of those who opted for Math = 1207 = 17. Number of students who opted for Physics and Chemistry Students whose numbers are multiples of 2 and 5 i.e., common multiples of 2 and 5 would have opted for both Physics and Chemistry. The LCM of 2 and 5 will be the first number that is a multiple of 2 and 5. i.e., 10 is the first number that will be a part of both the series. The 10th, 20th, 30th..... numbered students or every 10th student starting from student number 10 would have opted for both Physics and Chemistry. Therefore, n(A ? B) = 12010 = 12 Number of students who opted for Physics and Math Students whose numbers are multiples of 2 and 7 i.e., common multiples of 2 and 7 would have opted for both Physics and Math. The LCM of 2 and 7 will be the first number that is a multiple of 2 and 7. i.e., 14 is the first number that will be a part of both the series. The 14th, 28th, 42nd..... numbered students or every 14th student starting from student number 14 would have opted for Physics and Math. Therefore, n(C ? A) = 12014 = 8 Number of students who opted for Chemistry and Math Students whose numbers are multiples of 5 and 7 i.e., common multiples of 5 and 7 would have opted for both Chemistry and Math. The LCM of 5 and 7 will be the first number that is a multiple of 5 and 7. i.e., 35 is the first number that will be a part of both the series. The 35th, 70th.... numbered students or every 35th student starting with student number 35 would have opted for Chemistry and Math. Therefore, n(B ? C) = 12035 = 3 Number of students who opted for all 3 subjects Students whose numbers are multiples of 2, 5, and 7 i.e., common multiples of 2, 5, and 7 would have opted for all 3 subjects. The LCM of 2, 5, and 7 will be the first number that is a multiple of 2, 5, and 7. i.e., 70 is the first number that will be a part of all 3 series. 70 is the only multiple of 70 in the first 120 natural numbers. So, the 70th numbered student is the only one who would have opted for all three subjects. Therefore, n(A?B?C) = 60 + 24 + 17 - (12 + 8 + 3) + 1 = 79. Number of students who opted for none of the three subjects = 120 - 79 = 41.

bases are same. therefore :- 4.5×x = 5.6×1.8 x = 5.6×1.8/4.5

60+24+17-12-8-3+1=41

=> L+M+N=48*3=144 => L+M+N+O=44*4=176 144+O=176 O=176-144=32 years => P=O+3 years=32 years + 3 years= 35years => M+N+O+P=43*4=172 M+N+32+35=172 M+N=172-67=105 => L+M+N=144 L+105=144 L=144-105=39 years

simple trick 5^n-1 5^9-1=5^8

be have

tend

first one will be 60 and the next will be 140 hence they differ by 80

:Here, ++b[1] means that firstly b[1] will be incremented so, b[1]=2 then assigned to k i.e. k=2. b[1]++ means firstly b[1] will be assigned to variable l i.e. l=2, Then value stored in b[1] will be incremented i.e. b[1]=3. b[k++] means first b[k] will be assigned to m i.e. m=32, then value of k will be incremented i.e. k=3