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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The Product Of Two Numbers Is 436 And The Sum Of Their Squares Is 186. The Difference Of The Numbers Is: |
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Answer» 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) = 8. 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) = 8. |
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| 2. |
A Two Digit Number Is Such That The Product Of The Digits Is 6. When 45 Is Added To The Number, Then The Digits Are Reversed. The Number Is: |
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Answer» Let the ten's and unit digit be X and 8/x respectively. Then, 10x + 6/x + 45 = 10 x 6/x + x => 10x2 + 6 + 45x = 60 + x2 => 9X2 + 45x 54 = 0 => x2 + 5X 6 = 0 => (x + 6)(x 1) = 0 => x = 1 So the number is 16 Let the ten's and unit digit be x and 8/x respectively. Then, 10x + 6/x + 45 = 10 x 6/x + x => 10x2 + 6 + 45x = 60 + x2 => 9x2 + 45x 54 = 0 => x2 + 5x 6 = 0 => (x + 6)(x 1) = 0 => x = 1 So the number is 16 |
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| 3. |
Two Times The Second Of Three Consecutive Odd Integers Is 6 More Than The Third. The Third Integer Is? |
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Answer» Let the three INTEGERS be X, x + 2 and x + 4. Then, 2(x+2) = (x + 4) + 6 => x = 6. Third INTEGER = x + 4 = 10. Let the three integers be x, x + 2 and x + 4. Then, 2(x+2) = (x + 4) + 6 => x = 6. Third integer = x + 4 = 10. |
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| 4. |
Find The Number Which Is Nearest To 457 And Is Exactly Divisible By 11. |
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Answer» On dividing 457 by 11, remainder is 6. Required number is EITHER 451 or 462. On dividing 457 by 11, remainder is 6. Required number is either 451 or 462. Nearest to 456 is 462. |
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| 5. |
A Number When Divided By The Sum Of 333 And 222 Gives Three Times Their Difference The Quotient And 62 As The Remainder. The Number Is? |
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Answer» REQUIRED NUMBER = (333+222)×3×111+62 = 184877 Required number = (333+222)×3×111+62 = 184877 |
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| 6. |
The Difference Between The Place Values Of Two Eights In The Numeral 97958481 Is? |
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Answer» REQUIRED DIFFERENCE = (8000 80) = 7920 Required difference = (8000 80) = 7920 |
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| 7. |
The Sum Of First 75 Natural Numbers Is? |
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Answer»
Here n=75. Formula is n(n+1)/2, Here n=75. So the answer is 2850 |
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| 8. |
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 ? |
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Answer» Let P = 5x + 2. Then 3P = 15X + 6 Thus, when 3P is divided by 5, the REMAINDER is 1. Let P = 5x + 2. Then 3P = 15x + 6 = 5(3x + 1 ) + 1 Thus, when 3P is divided by 5, the remainder is 1. |
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| 9. |
Which Natural Number Is Nearest To 6475, Which Is Completely Divisible By 55 ? |
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Answer» (6475/55) 647540=6435 (6475/55) Remainder =40 647540=6435 |
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| 10. |
Which Of The Following Numbers Will Completely Divide (36^11 1) ? |
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Answer» => (XN 1) will be divisible by (x + 1) only when n is even. => (36^11 1) = {(6^2)^11 1} = (6^22 1),which is divisible by (6 +1) i.e., 7. => (xn 1) will be divisible by (x + 1) only when n is even. => (36^11 1) = {(6^2)^11 1} = (6^22 1),which is divisible by (6 +1) i.e., 7. |
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| 11. |
If The Number 13 * 4 Is Divisible By 6, Then * = ? |
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Answer» 6 = 3 x 2. Clearly, 13 * 4 is DIVISIBLE by 2. REPLACE * by x. Then, (1 + 3 + x + 4) must be divisible by 3. So, x = 1. 6 = 3 x 2. Clearly, 13 * 4 is divisible by 2. Replace * by x. Then, (1 + 3 + x + 4) must be divisible by 3. So, x = 1. |
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| 12. |
If The Product 5465 X 6k4 Is Divisible By 15, Then The Value Of K Is |
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Answer» 5465 is divisible by 5. So 6K4 must be divisible by 3. So (6+K+4) must be divisible by 3. K = 2 5465 is divisible by 5. So 6K4 must be divisible by 3. So (6+K+4) must be divisible by 3. K = 2 |
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| 13. |
How Many Natural Numbers Are There Between 17 And 84 Which Are Exactly Divisible By 6? |
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Answer» REQUIRED NUMBERS are 18,24,30,.....84 This is an A.P a=18,d=6,l=84 84=a+(N1)d n=12 Required numbers are 18,24,30,.....84 This is an A.P a=18,d=6,l=84 84=a+(n1)d n=12 |
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| 14. |
(?) + 2763 + 1254 1967 =26988 |
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Answer» X = 28955 4017 = 24938. x = 28955 4017 = 24938. |
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| 15. |
How Many Of The Following Numbers Are Divisible By 132 ? 264, 396, 462, 792, 968, 2178, 5184, 6331 |
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Answer»
So, if the NUMBER DIVISIBLE by all the three number 4, 3 and 11, then the number is divisible by 132 also. 264,396,792 are divisible by 132. Required answer =3 132 = 4 x 3 x 11 So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also. 264,396,792 are divisible by 132. Required answer =3 |
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| 16. |
96 X 96 + 84 X 84 = ? |
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Answer» = 96 X 96 + 84 x 84 = (96)2 + (84)2 = (90 + 6)2 + (90 6)2 = 2 x [(90)2 + (6)2] =16272 = 96 x 96 + 84 x 84 = (96)2 + (84)2 = (90 + 6)2 + (90 6)2 = 2 x [(90)2 + (6)2] =16272 |
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| 17. |
If N Is A Natural Number, Then (7(n2) + 7n) Is Always Divisible By: |
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Answer» (7n2 + 7N) = 7n(N + 1), which is always divisible by 7 and 14 both, since n(n + 1) is always EVEN. (7n2 + 7n) = 7n(n + 1), which is always divisible by 7 and 14 both, since n(n + 1) is always even. |
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| 18. |
The Difference Of Two Numbers Is 1097. On Dividing The Larger Number By The Smaller, We Get 10 As Quotient And The 17 As Remainder. What Is The Smaller Number ? |
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Answer» Let the smaller number be x. Then LARGER number = (x + 1097) 9x = 1080 x = 120 Let the smaller number be x. Then larger number = (x + 1097) x + 1097 = 10x + 17 9x = 1080 x = 120 |
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| 19. |
3621 X 137 + 3621 X 63 = ? |
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Answer» 3621 X 137 + 3621 x 63 = 3621 x (137 + 63) = (3621 x 200) = 724200 3621 x 137 + 3621 x 63 = 3621 x (137 + 63) = (3621 x 200) = 724200 |
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| 20. |
1004*1004+996*996= |
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Answer» = (1004)2+(996)2=(1000+4)2+(10004)2 = (1000)2 + (4)2 + 2*1000*4 + (1000)2 + (4)2 2*100*4 = 2000000 +32 = 2000032 = (1004)2+(996)2=(1000+4)2+(10004)2 = (1000)2 + (4)2 + 2*1000*4 + (1000)2 + (4)2 2*100*4 = 2000000 +32 = 2000032 |
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| 21. |
If The Number 24*32 Is Completely Divisible By 6. What Is The Smallest Whole Number In The Place Of *? |
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Answer» The number is DIVISIBLE by 6 MEANS it must be divisible by 2 and 3. Since the number has 2 as its end DIGIT it is divisible by 2. Now, 2+4+x+3+2=11+x which must be divisible by 3. Thus x=1 The number is divisible by 6 means it must be divisible by 2 and 3. Since the number has 2 as its end digit it is divisible by 2. Now, 2+4+x+3+2=11+x which must be divisible by 3. Thus x=1 |
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| 22. |
What Is The Least Number That Must Be Subtracted 2458 So That It Becomes Completely Divisible By 13? |
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Answer» DIVIDE 2458 by 13 and we GET REMAINDER as 1. Then 131=12. Adding 12 to 2458 we get 2470 which is divisible by 13. Thus ANSWER is 1. Divide 2458 by 13 and we get remainder as 1. Then 131=12. Adding 12 to 2458 we get 2470 which is divisible by 13. Thus answer is 1. |
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| 23. |
Which Of The Following Is Not A Prime Number? |
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Answer» 133 is DIVISIBLE by 7. Rest of NUMBERS is not divisible by any numbers EXCEPT itself and 1. 133 is divisible by 7. Rest of numbers is not divisible by any numbers except itself and 1. |
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| 24. |
The Sum Of Two Numbers Is 30. The Difference Between The Two Numbers Is 20. Find The Product Of Two Numbers? |
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Answer» => x+y=30 => x+y=30 |
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| 25. |
The Product Of Two Numbers Is 20. The Sum Of Squares Of The Two Numbers Is 81.find The Sum Of The Numbers.? |
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Answer» Let the NUMBERS be x,y. => x2+y2=81, => (x+y)2=81+40=121, => x+y=sqrt(121)=11 Let the numbers be x,y. => x2+y2=81, => 2(x+y)=40, => (x+y)2=81+40=121, => x+y=sqrt(121)=11 |
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| 26. |
The Sum Of Digits Of A Two Digit Number Is 13,the Difference Between The Digits Is 5. Find The Number.? |
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Answer» => x+y=13, xy=5 => x=9, y=4. Thus the NUMBER is 94 => x+y=13, xy=5 Adding these 2x =18 => x=9, y=4. Thus the number is 94 |
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| 27. |
Two Third Of Three Fourth Of A Number Is 24. Then One Third Of That Number Is? |
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Answer»
=> x=48,1/3x = 16 => (2/3)*(3/4)*x = 24 => x=48,1/3x = 16 |
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| 28. |
If (55^55+55) Is Divided By 56, Then The Remainder Is:? |
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Answer» (x^n+1) is divisible by (x+1), when n is ODD. .'. (55^55+1) is divisible by (55+1)=56. when (55^55+1)+54 is DIVIDED by 56, the remainder is 54. (x^n+1) is divisible by (x+1), when n is odd. .'. (55^55+1) is divisible by (55+1)=56. when (55^55+1)+54 is divided by 56, the remainder is 54. |
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| 29. |
Here The Sum Of The Series Is 4+8+12+16+….. =612. Find How Many Terms Are There In The Series? |
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Answer» This is an A.P. in which a=4, d=4 and Sn=612 Then, n/2[2a+(N1)d]=612 => n/2[2*4+(n1)*4]=612 => 4n/2(n+1)=612 => n(n+1)=306 => n^2+n306=0 => n^2+18n17n306=0 => n(n+18)17(n+18)=0 => (n+18)(n17)=0 => n=17. Number of terms=17. This is an A.P. in which a=4, d=4 and Sn=612 Then, n/2[2a+(n1)d]=612 => n/2[2*4+(n1)*4]=612 => 4n/2(n+1)=612 => n(n+1)=306 => n^2+n306=0 => n^2+18n17n306=0 => n(n+18)17(n+18)=0 => (n+18)(n17)=0 => n=17. Number of terms=17. |
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| 30. |
(11/n)+( 12/n)+(13/n)+…… Up To N Terms=? |
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Answer» Given SUM=(1+1+1+…. to n TERMS)(1/n+2/n+3/n+…. to n terms) = n(n(n+1)/2)/n = n(n+1)/2=1/2(n1). Given sum=(1+1+1+…. to n terms)(1/n+2/n+3/n+…. to n terms) = n(n(n+1)/2)/n = n(n+1)/2=1/2(n1). |
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| 31. |
How Many 4 Digit Numbers Are Completely Divisible By 7? |
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Answer» 4digit NUMBERS divisible by 7 are: 1001, 1008, 1015….. 9996. This is an A.P. in which a=1001, d=7, l=9996. Let the number of terms be N. Then Tn=9996. .'. a+(n1)d=9996 => 1001+(n1)7= 9996 =>(n1)7=8995 =>(n1)=8995/7= 1285 => n=1286. .'. number of terms =1286. 4digit numbers divisible by 7 are: 1001, 1008, 1015….. 9996. This is an A.P. in which a=1001, d=7, l=9996. Let the number of terms be n. Then Tn=9996. .'. a+(n1)d=9996 => 1001+(n1)7= 9996 =>(n1)7=8995 =>(n1)=8995/7= 1285 => n=1286. .'. number of terms =1286. |
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| 32. |
A 4 Digit Number 8a43 Is Added To Another 4 Digit Number 3121 To Give A 5 Digit Number 11b64, Which Is Divisible By 11, Then (a+b)=? |
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Answer» a+1=b => ba=1. => (4+b+1)(6+1)=0 => b2=0 => b=2. so, a=1 =>(a+b)= 3. a+1=b => ba=1. and 11b64 is divisible by 11 => (4+b+1)(6+1)=0 => b2=0 => b=2. so, a=1 =>(a+b)= 3. |
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| 33. |
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? |
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Answer» LET 5^32=x. Then (5^32+1)=(x+1). Let (x+1) be COMPLETELY divisible by the WHOLE NUMBER Y. then (5^96+1)=[(5^32)^3+1]=>(x^3+1)=(x+1)(x^2x+1) which is completely divisible by Y. since (x+1) is divisible by Y. Let 5^32=x. Then (5^32+1)=(x+1). Let (x+1) be completely divisible by the whole number Y. then (5^96+1)=[(5^32)^3+1]=>(x^3+1)=(x+1)(x^2x+1) which is completely divisible by Y. since (x+1) is divisible by Y. |
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| 34. |
If The Sum Of 1st N Integers Is 55 Then What Is N? |
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Answer» sum=n(n+1)/2 sum=55 n^2+n=55*2 n^2+n110=0 (N10)(n+11)=0 n=10,11,neglect NEGATIVE ans answer =10 sum=n(n+1)/2 sum=55 n^2+n=55*2 n^2+n110=0 (n10)(n+11)=0 n=10,11,neglect negative ans answer =10 |
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| 35. |
The Difference Of The Cubes Of Two Consecutive Even Integers Is Divisible By Which Of The Following Integers? |
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Answer» LET take 2 CONSECUTIVE EVEN numbers 2 and 4. => (4*4*4)(2*2*2)=648=56 which is divisible by 4. let take 2 consecutive even numbers 2 and 4. => (4*4*4)(2*2*2)=648=56 which is divisible by 4. |
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| 36. |
What Is The Smallest Number Should Be Added To 5377 So That The Sum Is Completely Divisible By 7? |
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Answer» DIVIDE 5377 with 7 we GET REMAINDER as 1. so, add 6 to the GIVEN number so that it will divisible by 7. Divide 5377 with 7 we get remainder as 1. so, add 6 to the given number so that it will divisible by 7. |
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| 37. |
597**6 Is Divisible By Both 3 And 11. The Nonzero Digits In The Hundred’s And Ten’s Places Are Respectively? |
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Answer» Let the given number be 597xy6. Then (5+9+7+x+y+6)=(27+x+y) must be divisible by 3 And, (6+x+9)(y+7+5)=(xy+3) must be EITHER 0 or divisible by 11. xy+3=0 => y=x+3 27+x+y) =>(27+x+x+3) =>(30+2x) => x = 3 and y = 6. Let the given number be 597xy6. Then (5+9+7+x+y+6)=(27+x+y) must be divisible by 3 And, (6+x+9)(y+7+5)=(xy+3) must be either 0 or divisible by 11. xy+3=0 => y=x+3 27+x+y) =>(27+x+x+3) =>(30+2x) => x = 3 and y = 6. |
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| 38. |
How Many Natural Numbers Between 23 And 137 Are Divisible By 7? |
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Answer» These numbers are 28, 35, 42,…., 133. This is in A.P. in which a= 28, d=(3528)= 7 and L=133. Let the number of there TERMS be N. then, Tn=133 a+(n1)d=133 by solving this we will get n=16. These numbers are 28, 35, 42,…., 133. This is in A.P. in which a= 28, d=(3528)= 7 and L=133. Let the number of there terms be n. then, Tn=133 a+(n1)d=133 by solving this we will get n=16. |
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| 39. |
Find The Remainder When 3^27 Is Divided By 5? |
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Answer» 3^27= ((3^4)^6) * (3^3) = (81^6) * 27 then unit DIGIT of (81^6) is 1 so on MULTIPLYING with 27, unit digit in the result will be 7. now, 7 when divided by 5 gives 2 as REMAINDER. 3^27= ((3^4)^6) * (3^3) = (81^6) * 27 then unit digit of (81^6) is 1 so on multiplying with 27, unit digit in the result will be 7. now, 7 when divided by 5 gives 2 as remainder. |
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| 40. |
On Dividing A Certain Number By 234, We Get 43 As Remainder. If The Same Number Is Divided By 13, What Will Be The Remainder? |
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Answer» suppose that on DIVIDING the given number by 234, we get quotient=x and REMAINDER= 43 then, number= 234*x+43----->(1). => (13*18X)+(13*3)+4 => 13*(18x+3)+4. So, the number when divided by 13 GIVES remainder=4. suppose that on dividing the given number by 234, we get quotient=x and remainder= 43 then, number= 234*x+43----->(1). => (13*18x)+(13*3)+4 => 13*(18x+3)+4. So, the number when divided by 13 gives remainder=4. |
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