## Basic Arithmetic Aptitude questions and answers for interview preparation

Q1. A four Digit Number 8a43 Is Added To Another 4 Digit Number 3121 To Give A 5 Digit Number 11b64, Which Is Divisible By eleven, Then (a+b)=?

A+1=b

=> ba=1.

And 11b64 is divisible through eleven

=> (four+b+1)(6+1)=zero

=> b2=zero

=> b=2.

So, a=1

=>(a+b)= three.

Q2. If The Number 24*32 Is Completely Divisible By

The number is divisible by 6 me it need to be divisible by way of 2 and @Since the quantity has 2 as its quit digit it's far divisible by way of @Now, 2+four+x+three+2=eleven+x which have to be divisible by means of @Thus x=1

Q3. The Difference Of Two Numbers Is 109

Let the smaller range be x.

Then larger quantity = (x + 1097)

x + 1097 = 10x + 17

9x = 1080

x = 120

Q4. If The Sum Of 1st N Integers Is fifty five Then What Is N?

Sum=n(n+1)/2

sum=55

n^2+n=55*2

n^2+n110=zero

(n10)(n+eleven)=zero

n=10,eleven,forget poor

wer =10

Q5. A Number When Divided By The Sum Of 333 And 222 Gives Three Times Their Difference The Quotient And sixty two As The Remainder. The Number Is?

Required range = (333+222)×3×111+sixty two

= 184877

Q6. The Sum Of First 75 Natural Numbers Is?

Formula is n(n+1)/2,

Here n=seventy five.

So the wer is 2850

Q7. How Many Natural Numbers Between 23 And 137 Are Divisible By 7?

These numbers are 28, 35, forty two,…., 133.

This is in A.P. In which a= 28, d=(3528)= 7 and L=133.

Let the variety of there phrases be n. Then, Tn=133

a+(n1)d=133 by means of solving this we can get n=sixteen.

Q8. Which Of The Following Is Not A Prime Number?

133 is divisible by means of 7.

Rest of numbers is not divisible via any numbers except itself and 1.

Q9. The Difference Between The Place Values Of Two Eights In The Numeral 97958481 Is?

Required difference = (8000 80)

= 7920

Q10. Find The Number Which Is Nearest To 457 And Is Exactly Divisible By

On dividing 457 through eleven, the rest is 6.

Required wide variety is both 451 or 462.

Nearest to 456 is 462.

Q11. What Is The Least Number That Must Be Subtracted 2458 So That It Becomes Completely Divisible By thirteen?

Divide 2458 by means of thirteen and we get remainder as 1.

Then 131=12.

Adding 12 to 2458 we get 2470 that's divisible with the aid of 13.

Thus wer is 1.

Q12. The Product Of Two Numbers Is 436 And The Sum Of Their Squares Is 18

Let the numbers be x and y.

Then, xy = 186 and x2 + y2 = 436.

=> (x y)

2 = x2 + y2 2xy

= 436 (

2 x 186)

= 64

=> x y

= SQRT(64)

= eight.

Q13. Two Times The Second Of Three Consecutive Odd Integers Is 6 More Than The Third. The Third Integer Is?

Let the three integers be x, x + 2 and x + four.

Then, 2(x+2) = (x + 4) + 6

=> x = 6.

Third integer = x + 4 = 10.

Q14. If The Product 5465 X 6k4 Is Divisible By 15, Then The Value Of K Is

5465 is divisible via five.

So 6K4 ought to be divisible through 3.

So (6+K+four) need to be divisible through three.

K = 2

Q15. Which Natural Number Is Nearest To 6475, Which Is Completely Divisible By fifty five ?

(6475/55)

Remainder =forty

647540=6435

Q16. How Many Natural Numbers Are There Between 17 And 84 Which Are Exactly Divisible By 6?

Required numbers are 18,24,30,.....Eighty four

This is an A.P a=18,d=6,l=84

84=a+(n1)d

n=12

Q17. How Many 4 Digit Numbers Are Completely Divisible By 7?

4digit

numbers divisible by way of 7 are: 1001, 1008, 1015….. 9996.

This is an A.P. In which a=1001, d=7, l=9996.

Let the range of phrases be n.

Then Tn=999@.'. A+(n1)d=9996

=> 1001+(n1)7= 9996

=>(n1)7=8995

=>(n1)=8995/7= 1285

=> n=1286.

.'. Number of terms =1286.

Q18. It Is Being Given That (5^32+1) Is Completely Divisible By A Whole Number. Which Of The Following Numbers Is Completely Divisible By This Number?

Let five^32=x.

Then (five^32+1)=(x+1). Let (x+1) be absolutely divisible by means of the complete number Y.

Then (five^96+1)=[(5^32)^3+1]=>(x^three+1)=(x+1)(x^2x+1) that is absolutely divisible through Y.

Due to the fact that (x+1) is divisible via Y.

Q19. 96 X 96 + eighty four X eighty four = ?

= ninety six x 96 + eighty four x eighty four = (ninety six)2 + (eighty four)2

= (ninety + 6)2 + (90 6)2

= 2 x [(90)2 + (6)2]

=16272

Q20. If The Number 13 * 4 Is Divisible By 6, Then * = ?

6 = three x 2.

Clearly, thirteen * four is divisible by means of 2.

Replace * by using x.

Then, (1 + three + x + 4) need to be divisible through three.

So, x = 1.

Q21. (?) + 2763 + 1254 1967 =26988

x = 28955 4017

= 24938.

Q22. Two Third Of Three Fourth Of A Number Is 2

=> (2/three)*(3/four)*x = 24

=> x=48,1/3x = sixteen

Q23. 1004*1004+996*996=

= (1004)2+(996)2=(1000+4)2+(10004)2

= (1000)2 + (four)2 + 2*a thousand*four + (a thousand)2 + (4)2 2*100*four

= 2000000 +32 = 2000032

Q24. If N Is A Natural Number, Then (7(n2) + 7n) Is Always Divisible By:

(7n2 + 7n) = 7n(n + 1), which is always divisible by 7 and 14 each, due to the fact n(n + 1) is usually even.

Q25. What Is The Smallest Number Should Be Added To 5377 So That The Sum Is Completely Divisible By 7?

Divide 5377 with 7 we get remainder as @so, add 6 to the given number on the way to divisible by using 7.

Q26. Which Of The Following Numbers Will Completely Divide (36^eleven 1) ?

=> (xn 1) will be divisible by (x + 1) handiest when n is even.

=> (36^eleven 1)

= (6^2)^11 1

= (6^22 1),that's divisible by using (6 +1)

i.E., 7.

Q27. The Product Of Two Numbers Is 2

Let the numbers be x,y.

=> x2+y2=eighty one,

=> 2(x+y)=forty,

=> (x+y)2=eighty one+40=121,

=> x+y=sqrt(121)=eleven

Q28. 597**6 Is Divisible By Both 3 And

Let the given variety be 597xy6.

Then (five+nine+7+x+y+6)=(27+x+y) have to be divisible via 3

And, (6+x+9)(y+7+five)=(xy+three) have to be both 0 or divisible through @xy+three=0

=> y=x+3 27+x+y)

=>(27+x+x+3)

=>(30+2x)

=> x = 3 and y = 6.

Q29. How Many Of The Following Numbers Are Divisible By 132 ? 264, 396, 462, 792, 968, 2178, 5184, 6331

132 = 4 x 3 x eleven

So, if the wide variety divisible by way of all of the three quantity 4, 3 and 11, then the variety is divisible via 132 also.

264,396,792 are divisible through 132.

Required wer =3

Q30. If (fifty five^fifty five+55) Is Divided By fifty six, Then The Remainder Is:?

(x^n+1) is divisible by way of (x+1), when n is ordinary.

.'. (55^fifty five+1) is divisible through (55+1)=5@whilst (fifty five^55+1)+54 is split by way of fifty six, the remainder is fifty four.

Q31. P Is A Whole Number Which When Divided By 5 Gives 2 As Remainder. What Will Be The Remainder When 3p Is Divided By 5 ?

Let P = 5x + 2.

Then 3P = 15x + 6

= five(3x + 1 ) + 1

Thus, whilst 3P is divided via five, the remainder is 1.

Q32. The Difference Of The Cubes Of Two Consecutive Even Integers Is Divisible By Which Of The Following Integers?

Allow take 2 consecutive even numbers 2 and 4.

=> (four*4*four)(2*2*2)=648=56 which is divisible with the aid of four.

Q33. A Two Digit Number Is Such That The Product Of The Digits Is

Let the 10's and unit digit be x and 8/x respectively.

Then, 10x + 6/x + forty five = 10 x 6/x + x

=> 10x2 + 6 + 45x = 60 + x2

=> 9x2 + 45x fifty four

= 0

=> x2 + 5x 6

= zero

=> (x + 6)(x 1)

= zero

=> x = 1

So the variety is sixteen

Q34. The Sum Of Digits Of A Two Digit Number Is 13,the Difference Between The Digits Is

=> x+y=13, xy=5

Adding those 2x =18

=> x=9, y=4.

Thus the number is 94

Q35. Find The Remainder When 3^27 Is Divided By five?

Three^27= ((3^4)^6) * (3^three) = (81^6) * 27 then unit digit of (eighty one^6) is 1 so on multiplying with 27, unit digit inside the end result might be @now, 7 while divided via 5 gives 2 as remainder.

Q36. 3621 X 137 + 3621 X 63 = ?

3621 x 137 + 3621 x 63 = 3621 x (137 + 63)

= (3621 x 2 hundred)

= 724200

Q37. On Dividing A Certain Number By 234, We Get forty three As Remainder. If The Same Number Is Divided By 13, What Will Be The Remainder?

Suppose that on dividing the given quantity by 234,

we get quotient=x and the rest= 43

then, quantity= 234*x+forty three----->(1).

=> (thirteen*18x)+(thirteen*3)+four

=> 13*(18x+three)+4.

So, the quantity whilst divided by using 13 gives remainder=four.

Q38. The Sum Of Two Numbers Is three

=> x+y=30

=> xy=20

=> (x+y)2(xy)2 = 4xy

=> 4xy=302202=500

=> xy=500/4=125