APTITUDE TEST (APTITUDE Problems On Numbers)

1. What could be the maximum value of Q in the following equation? 5PQ+3R7+2Q8=1114
  • 9
  • 5
  • 1
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)
2. Evaluate 313*313+287*287
  • 147581
  • 180338
  • 146532
  • 174585
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
3. Find the total number of prime factors in 411 *7 5 *112 ?
  • 18
  • 58
  • 29
  • 41
Solution: 411 7 5 112= (2*2) 11 *7 5 *112 = 222 *7 5 *112 Total number of prime factors=22+5+2=29
4. What least value must be assigned to * so that th number 197*5462 is divisible by 9?
  • 2
  • 7
  • 4
  • 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.
5. Find the smallest number of 6 digits which is exactly divisible by 111?
  • 10011
  • 10048
  • 10000
  • 10068
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.
6. A number when divided by 342 gives a remainder 47.When the same number is divided by 19 what would be the remainder?
  • 2
  • 9
  • 7
  • 1
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.
7. Find the remainder when 231 is divided by 5?
  • 6
  • 3
  • 7
  • 1
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.
8. Find the sum of all odd numbers up to 100?
  • 4500
  • 7800
  • 2500
  • 4700
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.
9. 19.2+22+23+24+25……….+28=?
  • 157
  • 510
  • 467
  • 728
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.
10. 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
  • 75
  • 47
  • 22
  • 14
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.

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Posted on April 16, 2011, in APTITUDE Test. Bookmark the permalink. Leave a comment.

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