Qualcomm Averages Quiz-1

The answer is \" four\"As we know that the 29th February came after every four years. So according to that 26 total birthday will be celebrated in 101 years. Also after every years date are moved by 1 so in every birthday date will be moved by 5 days . This means wednesday will came on every 7th birthday of the john . So by simple calculation total 4 birthdays will be on the wednesday. That day will be on 1st , 8th , 15th and 22th birthday.

must have

As they have given new price is not less than half the original price therefore 164.90/2= 82.45 And new price is more than 82.45 Considering the options it would be 97.

m+n=8....(1) m-n=6.....(2) solving equation 1 and 2 we get m=7 and n=1 according to given condition mn 7*1=7 7%6=1 so the rem. is 1

option b is the correct answer.

c+1 means c is base address i.e. address of 1st one Dimensional array and on incrementing it by 1 means it points to 2nd one 2 Dimensional array. So, c+1=65474 + ( 4 * 2)= 65482 But, when we are writing &c, that means address of this whole array i.e. address of next new array. So, &c+1=65474 + (12 * 2)=65498

for cout the operator overloaded is <<

x++ is not executed inside sizeof function

n(n-3)/2=77 n^2-3n=154 n^2-3n-154=0 n=-11,14

if side of a polygon=n then diagonal=n(n-3)/2

x++ is not executed inside sizeof function

Set P = {4, 8, 12, ….496} ? 124 elements {all elements from 1 * 4 to 124 * 4} Set Q = {7, 21, 35, 49,……497} ? {7 * 1, 7 * 3, 7 * 5, ….. 7 * 71} ? 36 elements. Set R = {6, 12, 18, 24, …..498} ? {6 * 1, 6 * 2, 6 * 3, ….. 6 * 83} ? 83 elements. Sets P and R have only even numbers; set Q has only odd numbers. So, P ? Q = Null set Q ? R = Null set P ? Q ? R = Null set So, If we find P ? R , we can plug into the formula and get P ? Q ? R P ? R = Set of all multiples of 12 less than 500 = {12, 24, 36,…..492} = {12 * 1, 12 * 2 , 12 * 3, …12 * 41} ? This has 41 elements P ? Q ? R = P + Q + R – (P ? Q) – ( Q ? R) – (R ? P) + (P ? Q ? R) P ? Q ? R = 124 + 36 + 83 – 0 – 0 – 41 + 0 = 202 The question is \"How many elements are present in P ? Q ? R?\" Hence, the answer is \"202\".

decided

There is no return type given for the MyDisplay we need to consider it as null or we can take it as a string hence the integer cannot be returned.

Only if n==1, then sum is equal to one. So , if n is 1 then it should directly return 1.

check the recursive function it fails for the n=1 hence the terminating condition will be the 1

reached

The only thing to observe here is that every term beyond and including 4! is a multiple of 24. So, their contribution to the remainder will be 0. Simply adding up the first 3 terms, we get 1+2+6=9.

18*6 and 7 and 8 and 9 and 10 and 11. Answer is 6

0.1 by default is of type double which has different representation than float resulting in inequality even after conversion.

0.0056=56/10000=7/1250

the speed of first car must be 50 kmph.

x+X+20+x-20=120 3x=120 x=40 x is A\'s share so B\'s share is 20

x+X+20+x-20=120 3x=120 x=40 x is A\'s share so B\'s share is 20

It is given as an unsigned integer hence it can store only -32 to 31 but not 32

received

use the recursion tree

given 58 not 50

ratio of eff. of a & c=5:3 so, eff. of a is more than c and a takes 6 day less than c so,day takes by a & c=9 day and 15 day respectively ratio of no. of day b & c=2:3 means day required =10 & 15 day respectively therefore, no of day taken by a ,b,c=9,10,15 respectively so, total work=90( LCM of no of days) a completed 10 units of work in one day same b is 9 unit and c is 6 unit in a day total work done by all in 1 day=(10+9+6)=25 in two days +25*2=50(in two days remaining work is 40 unit completed by a ) so a takes 4 days to complete the remaining work.

ratio of eff. of a & c=5:3 so, eff. of a is more than c and a takes 6 day less than c so,day takes by a & c=9 day and 15 day respectively ratio of no. of day b & c=2:3 means day required =10 & 15 day respectively therefore, no of day taken by a ,b,c=9,10,15 respectively so, total work=90( LCM of no of days) a completed 10 units of work in one day same b is 9 unit and c is 6 unit in a day total work done by all in 1 day=(10+9+6)=25 in two days +25*2=50(in two days remaining work is 40 unit completed by a ) so a takes 4 days to complete the remaining work.

Answer: b Explanation: signed char will be a negative number. Output: $ cc pgm2.c $ a.out -128

answer is wrong ,correct is 155 Let Amith or his friend alone can complete the work in \'x\' days, then 1 day work of Amith=1 day work of his friend=1/x Their 30 day\'s work=1/x + 3/x + 6/x + 10/x .... =1 => 1/x (1 + 3 + 6 + 10 +....upto 30 terms)=1 Sequence 1,3,6,10,15,... is called triangle number series with sum=n(n+1)(n+2)/6 => (1/x) (30*31*32/2) =1 => 1/x=1/4960 As girl is of twice capacity of a boy, so 1 girl\'s 1 day work = 2*(1/x)=2*1/4960=1/2480 So 16 girl\'s 1 day work=16*(1/2480)=1/155 Hence required days=155

told

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