Interview Questions.

Q1. On Dividing A Number By 999,the Quotient Is 366 And The Remainder Is 103.The Number Is?

Number (Dividend) = Divisor * quotient + the rest.

Number = 999 * 377 + 105 = 3767.

Q2. What Is The Number Of Digits In 333? Given That Log3 = 0.47712?

Let   Let x=(333) = (33)three 

 Then, log(x) = 33 log(three)  

= 27 x zero.47712 = 12.88224 

Since the function in the resultant fee of log x is 12

∴The quantity of digits in x is (12 + 1) = 13 

Hence the required variety of digits in 333is 13.

Q3. A Reduction Of 20% In The Price Of Strawberries Enables A Person To Purchase 12 More For Rs.

Price x Consumption = Expenditure

(15 / 8x) – (15 / x) = 12

x = (15 x 2) / (12 x eight)

For sixteen Strawberries = [(15 x 2) / (12 x 8)] x sixteen = five.

Q4. How Long Will A Boy Take To Run Round A Square Field Of Side 35 Meters, If He Runs At The Rate Of 9 Km/hr?

Speed = 9 km/hr = nine x (five/18) m/sec = 5/2 m/sec  

Distance = (35 x 4) m = 140 m.

Time taken = a hundred and forty x (2/five) sec= fifty six sec.

Q5. The Number Of Prime Factors Of (3 X 5)12 (2 X 7)10 (10)25 Is?

The equation can be facorize as three*five*3*2*2*2*7*2*5*2*five*five*5 or 2^5*three^2*5^five*7^1

overall no of prime thing =(5+1)*(5+1)*(2+1)*(1+1)=216.

Q6. Three Cubes Of Edges 6 Cm, 8 Cm And 10 Cm Are Meted Without Loss Of Metal Into A Single Cube. The Edge Of The New Cube Will Be?

Since the cube is melted so the volume of the brand new dice must be the equal.

Volume of latest cube = Volume of dice 1 + dice 2 + cube 3 = sixty three + 83 + 103 = 216 + 512 + one thousand

a^three = 1728, a = (1728)^(1/3) = 12.

Q7. The Smallest Number, Which Is A Perfect Square And Contains 7936 As A Factor Is:

7936 => 2^2 * 2^2 * 2^2 * 2^2 * 31^1

To make it as a great square, we ought to multiply 7936 with 31.

Hence the reqd no. Is 7936*31 = 246016.

Q8. What Least Value Must Be Assigned To * So That The Number 63576*2 Is Divisible By eight?

The test for divisibility via 8 is that the last three digits of the range in question must be divisible by way of 8.

So, 6*2 needs to be divisibile via eight.

I understand 512 is divisible via 8.

Also 592 is divisible by means of eight.

So, 632 is divisible by 8.

So * is three.

Q9. A Shopkeeper Gives Two Successive Discounts Of 20 % And 10 % On Surplus Stock. Further, He Also Gives 5 % Extra Discount On Cash Payment. If A Person Buys A Shirt From The Surplus Stock And Pays In Ca

Let the marked price of the blouse be Rs. One thousand

=> Price after first discount = Rs. A thousand – 20 % of Rs. 1000 = Rs. 1000 – 200 = Rs. 800

=> Price after 2nd cut price = Rs. 800 – 10 % of Rs. 800 = Rs. 800 – eighty = Rs. 720

=> Price after coins cut price = Rs. 720 – five % of Rs. 720 = Rs. 720 – 36 = Rs. 684

Therefore, overall cut price = Rs. One thousand – 684 = Rs. 316

=> Overall discount percent = (316 / 1000) x a hundred = 31.60 %.

Q10. What Is The Smallest Four-digit Number Which When Divided By 6, Leaves A Remainder Of five And When Divided By 5 Leaves A Remainder Of 3?

Remainder when  m is split by way of five  = 2

Smallest m is two.

Hence, N = 1001 + 6 * 2 = 1013.

Q11. If A And B Are Natural Numbers And A-b Is Divisible By three, Then A3-b3 Is Divisible By?

If a − b is divisible with the aid of three, then a − b = 3k, for a few integer k

(a − b)² = (3k)²

a² − 2ab + b² = 9k²

a³ − b³ = (a−b) (a² + ab + b²)

= (a−b) (a² − 2ab + b² + 3ab)

= 3k (9k + 3ab)

= 3k * three (3k + ab)

= 9 ok(3k+ab)

Since okay(3k+ab) is an integer, then 9k(3k+ab) is divisible by means of nine.

Q12. If 522x Is A Three Digit Number With As A Digit X . If The Number Is Divisible By 6, What Is The Value Of The Digit X Is?

If various is Divisiable through 6 , it have to be divisible through each 2 and three

In 522x, to this range be divisible by way of 2, the price of x have to be even. So it n be 2,four or 6 from given alternatives

552x is divisible by way of 3, If sum of its digits is a multiple of three.

Five+5+2+x =12+x ,

If placed x =2 , 12+2=14 not a a couple of of three

If put x =4 , 12+6=18  is a a couple of of three

If placed x =6 , 12+2=14 not a more than one of three

The price of x is 6.

Q13. The Greatest Number That Will Divide 63, 138 And 228 So As To Leave The Same Remainder In Each Case?

The greatest number = H.C.F of (138-63), (228-138), (228-sixty three)

H.C.F of 75, 90, 165 = 15.

15 is the finest variety.

Q14. A Tap Can Fill A Bucket In 6 Hours. After Half The Bucket Is Filled, Three More Similar Taps Are Opened. What Is The Total Time Taken To Fill The Bucket Completely?

Time is taken via one tap to fill half the bucket = 3 hours.

So the element crammed four faucets in one hour = four * (1/6) = 2/3 of the bucket.

Therefore, the closing element is = (1 – half of) = 1/2

Proportionally à 2/3: half:: 1: x

=> x = 3/four hours = forty five mins. So the total time = 3 hrs forty five minutes.

Q15. If A Number Is Exactly Divisible By eighty five, Then What Will Be The Remainder When The Same Number Is Divided By 17?

Variety=divisor*quotient+the rest

so 17*five+0;

the rest is 0;

divisor is 17;

quotient is 5.

Q16. Let C Be A Positive Integer Such That C + 7 Is Divisible By

c + n^2 is divisible through five if and only if c and n^2 are both divisible by means of five.

But, if c is divisible by using 5 then c + five will not be divisible with the aid of 5.

Q17. Raju, Ramu And Razi Can Do A Piece Of Work In 20, 30 And 60 Days Respectively Depending On Their Capacity Of Doing Work. If Raju Is Assisted By Ramu And Razi On Every Third Day, Then In How Raju Will

We want t first be counted the amount of work completed in 2 days by using Raju.

Raju can do a piece of labor in 20 days.

So, in 2 days he can do = 1/20 * 2 = 1/10.

Amount of work carried out by Raju, Ramu and Razi in 1 day = 1/20 + 1/30 + 1/60 = 1/10.

Amount of labor completed in 3 days = 1/10 + 1/10 = 1/5.

So the work will be completed in three * five = 15 days.

Q18. P Is An Integer. P>88

Given P is an integer>883.

P-7 is a a couple of of 11=>there exist a fantastic integer a such that

P-7=11 a=>P=11 a+7

(P+4)(P+15)=(11 a+7+4)(eleven a+7+15)

=(eleven a+eleven)(11 a+22)

=121(a+1)(a+2)

As a is a fine integer consequently (a+1)(a+2) is divisible by 2.Hence (P+four)(P+15) is divisible through 121*2=242.

Q19. What Is The Highest Power Of 5 That Divides 90 X eighty X 70 X 60 X 50 X 40 X 30 X 20 X 10?

Take LCM of Each Number:

ninety/5=5*2*3*three——————>here we can get one 5

eighty/5=5*2*2*2*2—————>right here we are able to get one five

70/five=five*2*7————–___—->here we will get one five

60/5=5*2*2*3——————>right here we can get one five

50/5=five*five*2___——————>right here we can get Two five^2

40/five=5*2*2*2——————>here we will get one 5

30/five=five*2*three———————>here we will get one 5

20/five=five*2*2———————>right here we are able to get one five

10/five=5*2————————>here we can get one five

Here we are able to get one 5 in every number as opposed to 50(5*5*2)

So wer is five^10.

Q20. Four Bells Begin To Toll Together And Then Each One At Intervals Of 6 S, 7 S, 8 S And nine S Respectively.The Number Of Times They Will Toll Together In The Next 2 Hr Is?

First we to find the L.C.M. Of 6, 7, eight and nine.

Prime factorization of 6 = 2*three

Prime factorization of seven = 7

Prime factorization of eight = 2*2*2

Prime factorization of nine = 3*three

L.C.M. = 2*2*2*three*3*7

= 504

The L.C.M. Of 6 seconds, 7 seconds, eight seconds and nine seconds is 504

seconds.

Now, 1 hour = 3600 seconds

So, 2 hours = 3600*2 = 7200 seconds

The quantity of instances the four bells will toll together inside the next 2 hour

= 7200/504

= 14.28 or 14 instances

They will toll collectively 14 instances in the subsequent 2 hours

Q21. The Ratio Of The No. Of White Balls In A Bag To That Of Black Balls Is 1:

Consider x black balls have been there.

After including nine grey balls the ratio is four/3.

That me, x/nine = four/3.

On fixing we are able to get x = 12.

Q22. A Hollow Iron Pipe Is 21 Cm Long And Its External Diameter Is eight Cm. If The Thickness Of The Pipe Is 1 Cm And Iron Weighs 8 G/cm3, Then The Weight Of The Pipe Is?

Given the external diameter = 8 cm. Therefore, the radius = 4 cm.

The thickness = 1 cm. Therefore the inner radius = 4 – 1 = three cm

The quantity of the iron = pi *(R^2 – r^2)*period = 22/7 *[(4^2) – (3^2)] *21 = 462 cm3>.

Therefore, the burden of iron = 462 * 8 gm = three.696 kg.

Q23. A & B Are At A Distance Of 800 M. They Start Towards Each Other At 20 & 24 Kmph. As They Start, A Bird Sitting On The Cap Of A, Starts Flying Towards B, Touches B & Then Returns Towards A & So On, Til

The chicken flies for the equal time as both A and B take to meet.

Since the time taken through A and B together and the chook is same, so the gap covered might be in the ratio of their speeds.

The ratio of the speeds is 44: 176 or 1: 4.

Hence, if A and B cowl 800 m, the bird will cowl 800*4 = 3200 m.

Q24. From A Group Of 7 Men And 6 Women, Five Persons Are To Be Selected To Form A Committee So That At Least three Men Are There In The Committee. In How Many Ways Can It Be Done?

From a set of seven guys and six girls, five men and women are to be selected with as a minimum three men.

Hence we've the subsequent three alternatives.

We can pick 5 guys à Number of approaches to do this = 7C5

ii) We can pick out four men and 1 girl à Number of approaches to do this = 7C4 × 6C1

iii)   We can pick three guys and a pair of women à Number of approaches to do this = 7C3 × 6C2

Total range of methods = 7C5+ (7C4 × 6C1) + (7C3× 6C2)

= 7C2+ (7C3× 6C1) + (7C3×6C2) —-     Expand this using nCr = nC (n – r)

= 21 + 210 + 525 = 756.

Q25. What Is The Greatest Positive Power Of 5 That Divides 30! Exactly?

Only the numbers five, 10, 15, 20, 25, and 30 have divisors of @And 25 is divisible via five^2.

So the wer is 5*5*five*5*(5^2)*five = five^7.

Q26. P Is An Integer. P Is Greater Than 883.If P -7 Is A Multiple Of 11, Then The Largest Number That Will Always Divide (p+four)(p+15) Is?

P-7= 11*a (as it's far a couple of of 11)

p=11*(a+7)

so (p+four)(p+15)= (11a+7+4)(11a+7+15);

= (11a+11)(11a+22);

=11*eleven(a+1)(a+2);

=121*2

=242.

Q27. In An Election Between Two Candidates, One Got fifty five% Of The Total Valid Votes And Got 20% Invalid Votes. At The End Of The Day When The Total Number Of Votes Were Counted, The Total Number Was Found To

Since 20% of the votes had been invalid, eighty% of the votes were legitimate = 80% of 7500 = 6000 votes had been legitimate.

One candidate were given 55% of the whole legitimate votes, then the second one candidate must have forty five% of the votes = zero.45 * 6000 = 2700 votes.

Q28. A Whole Number N Which When Divided By four Gives three As Remainder. What Will Be The Remainder When 2n Is Divided By four?

According to the query, n = 4q + three.

Therefore, 2n = 8q + 6 or 2n = 4(2q + 1) + 2.

Thus, we get when 2n is split by way of 4, the remainder is two.

Q29. Find The Largest Number, Smaller Than The Smallest Four-digit Number, Which When Divided By four,5,6and 7 Leaves A Remainder 2 In Each Case?

Take LCM of 4,five,6,@It is 420

BUt the no ought to leave the rest 2 in each case, so the no is of the shape: 420k + 2.

The smallest 4-digit no is 10@So keeping okay=0,1,2,three.

We get that the largest no smaller than the smallest 4 -digit no is 842.