BACKNUMBER SYSTEMSSOLVED PROBLEMS1.Simplify a.8888+888+88+8 b.11992-7823-456 Solution: a.8888 888 88 8 9872 b.11992-7823-456=11992-(7823+456) =11992-8279=3713 2.What could be the maximum value of Q in the following equation? 5PQ+3R7+2Q8=1114 Solution: 5 P Q 3 R 7 2 Q 8 11 1 4 2+P+Q+R=11 Maximum value of Q =11-2=9 (P=0,R=0) 3.Simplify: a.5793405*9999 b.839478*625 Solution: a. 5793405*9999=5793405*(10000-1) 57934050000-5793405=57928256595 b. 839478*625=839478*54=8394780000/16=524673750. 4.Evaluate 313*313+287*287 Solution: a²+b²=1/2((a+b)²+(a-b)²) 1/2(313+287)² +(313-287)²=1/2(600 ² +26 ² ) ½(360000+676)=180338 5.Which of the following is a prime number? a.241 b.337 c.391 Solution: a.241 16>√241.Hence take the value of Z=16. Prime numbers less than 16 are 2,3,5,7,11 and 13. 241 is not divisible by any of these. Hence we can conclude that 241 is a prime number. b. 337 19>√337.Hence take the value of Z=19. Prime numbers less than 16 are 2,3,5,7,11,13 and 17. 337 is not divisible by any of these. Hence we can conclude that 337 is a prime number. c. 391 20>√391.Hence take the value of Z=20. Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19. 391 is divisible by 17. Hence we can conclude that 391 is not a prime number. 6.Find the unit's digit n the product 2467 153 * 34172? Solution: Unit's digit in the given product=Unit's digit in 7 153 * 172 Now 7 4 gives unit digit 1 7 152 gives unit digit 1 7 153 gives 1*7=7.Also 172 gives 1 Hence unit's digit in the product =7*1=7. 7.Find the total number of prime factors in 411 *7 5 *112 ? Solution: 411 7 5 112= (2*2) 11 *7 5 *112 = 222 *7 5 *112 Total number of prime factors=22+5+2=29 8.Which of the following numbers s divisible by 3? a.541326 b.5967013 Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3. b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3. 9.What least value must be assigned to * so that th number 197*5462 is divisible by 9? Solution: Let the missing digit be x Sum of digits = (1+9+7+x+5+4+6+2)=34+x For 34+x to be divisible by 9 , x must be replaced by 2 The digit in place of x must be 2. 10.What least number must be added to 3000 to obtain a number exactly divisible by 19? Solution:On dividing 3000 by 19 we get 17 as remainder Therefore number to be added = 19-17=2. 11.Find the smallest number of 6 digits which is exactly divisible by 111? Solution:Smallest number of 6 digits is 100000 On dividing 10000 by 111 we get 100 as remainder Number to be added =111-100=11. Hence,required number =10011. 12.On dividing 15968 by a certain number the quotient is 89 and the remainder is 37.Find the divisor? Solution:Divisor = (Dividend-Remainder)/Quotient =(15968-37) / 89 =179. 13.A number when divided by 342 gives a remainder 47.When the same number is divided by 19 what would be the remainder? Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9. The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder. 14.A number being successively divided by 3,5,8 leaves remainders 1,4,7 respectively. Find the respective remainders if the order of divisors are reversed? Solution:Let the number be x. 3 x 5 y - 1 8 z - 4 1 - 7 z=8*1+7=15 y=5z+4 = 5*15+4 = 79 x=3y+1 = 3*79+1=238 Now 8 238 5 29 - 6 3 5 - 4 1 - 2 Respective remainders are 6,4,2. 15.Find the remainder when 231 is divided by 5? Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as 4*4*4 gives unit digit 4 unit digit of 231 is 8. Now 8 when divided by 5 gives 3 as remainder. 231 when divided by 5 gives 3 as remainder. 16.How many numbers between 11 and 90 are divisible by 7? Solution:The required numbers are 14,21,28,...........,84 This is an A.P with a=14,d=7. Let it contain n terms then T =84=a+(n-1)d =14+(n-1)7 =7+7n 7n=77 =>n=11. 17.Find the sum of all odd numbers up to 100? Solution:The given numbers are 1,3,5.........99. This is an A.P with a=1,d=2. Let it contain n terms 1+(n-1)2=99 =>n=50 Then required sum =n/2(first term +last term) =50/2(1+99)=2500. 18.How many terms are there in 2,4,6,8..........,1024? Solution:Clearly 2,4,6........1024 form a G.P with a=2,r=2 Let the number of terms be n then 2*2 n-1=1024 2n-1 =512=29 n-1=9 n=10. 19.2+22+23+24+25..........+28=? Solution:Given series is a G.P with a=2,r=2 and n=8. Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2. =2*255=510. 20.A positive number which when added to 1000 gives a sum , which is greater than when it is multiplied by 1000.The positive integer is? a.1 b.3 c.5 d.7 Solution:1000+N>1000N clearly N=1. 21.The sum of all possible two digit numbers formed from three different one digit natural numbers when divided by the sum of the original three numbers is equal to? a.18 b.22 c.36 d. none Solution:Let the one digit numbers x,y,z Sum of all possible two digit numbers= =(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y) = 22(x+y+z) Therefore sum of all possible two digit numbers when divided by sum of one digit numbers gives 22. 22.The sum of three prime numbers is 100.If one of them exceeds another by 36 then one of the numbers is? a.7 b.29 c.41 d67. Solution:x+(x+36)+y=100 2x+y=64 Therefore y must be even prime which is 2 2x+2=64=>x=31. Third prime number =x+36=31+36=67. 23.A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder .The number is? a.1220 b.1250 c.22030 d.220030. Solution:Number=(555+445)*(555-445)*2+30 =(555+445)*2*110+30 =220000+30=220030. 24.The difference between two numbers s 1365.When the larger number is divided by the smaller one the quotient is 6 and the remainder is 15. The smaller number is? a.240 b.270 c.295 d.360 Solution:Let the smaller number be x, then larger number =1365+x Therefore 1365+x=6x+15 5x=1350 => x=270 Required number is 270. 25.In doing a division of a question with zero remainder,a candidate took 12 as divisor instead of 21.The quotient obtained by him was 35. The correct quotient is? a.0 b.12 c.13 d.20 Solution:Dividend=12*35=420. Now dividend =420 and divisor =21. Therefore correct quotient =420/21=20. BACK |