Tuesday, May 28, 2013

Number System - Solved Problems

Q1 :In the following, which is the greatest number ?

(1). (4)²
(2). (2×2×2)²
(3). [(2 + 2)²]²
(4). (2 + 2 + 2)²


Answer: (3). [(2 + 2)²]² 

Explanation:
1. (4)²                =          4 x 4 =                              16
2. (2×2×2)² = (8)² = 8 X 8        =                     64
3. [(2 + 2)²]² = [(4)²]² = [16]²=16 x16 =  256 ( and hence this the answer)
4. (2 + 2 + 2)² = (6)² =                                        36


Q2. 407928 is read as

(1) Four lakh seventy nine thousand twenty eight
(2) Forty seven thousand nine hundred twenty eight
(3) Forty thousand nine hundred twenty eight
(4) Four lakh seven thousand nine hundred twenty eight


Answer: (4) Four lakh seven thousand nine hundred twenty eight 

Q3.If an operator θ is defined as

4 θ 3 = 4 + 5 + 6
5 θ 4 = 5 + 6 + 7 + 8
6 θ 4 = 6 + 7 + 8 + 9

what will n θ 8 be equal to ?
(1) n + 28
(2) 8n + 28
(3) 8n + 36
(4) n + 36


Answer:(2) 8n + 28

Explanation:From the examples given in the question, the first number is the starting number of the series and the second number represents the number of numbers in the series inclusive the starting number of the series.
Using the above rule, we can expand n θ 8 to n +(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)+(n+7)
= 8n + (1 + 2 + 3 +4 +5 +6 +7)
= 8n +28


Q4. How many 4-digit numbers are there in the Hindu-Arabic Numeration System ?
(1) 99
(2) 8999
(3) 9999
(4) 9000


Answer:(4) 9000
explanation:
Don't get scared by "Hindu-Arabic Numeration" argon. Now to get at the answer just subtracted the lowest 4 digit number from the highest 4 digit number ( and add + 1 since both the ranges are included ) to arrive at the answer.
ie. 9999 - 1000 + 1 = 8999 + 1 = 9000

Q5. The number 49532 rounded off to the nearest thousand is
(1) 49000
(2) 49500
(3) 41000
(4) 50000


Answer: (4) 50000 

Explanation: The thousand part in the given number is 49 thousand.Now we will look at the rest of the number i.e. 532 which is greater than 500 and hence can be rounded off to 1000.
Therefore, the answer is 49,000+1000 = 50000

Q6. When faced with word problems, Rajan usually asks ‘‘Should I add or subtract ?’’ ‘‘Should I multiply or divide ?’’. Such questions suggest

(1) Rajan seeks opportunities to disturb the class
(2) Rajan has problems in comprehending language
(3) Rajan lacks understanding of number operations
(4) Rajan cannot add and multiply


Answer:(2) Rajan has problems in comprehending language
Explanation: From his words it is clear that he is confused about using the mathematical operators.


Q7. Sum of place value of 6 in 63606 is
(1) 6606
(2) 6066
(3) 18
(4) 60606


Answer: (4) 60606

Explanation:
The number is 63606. Place value of first six in the number, from left to right, is 60000, place value of second six is 600 and last six is 6.
Adding place values of all the sixes we have,
60000 + 600 + 6 = 60606

Q8. The difference of 5671 and the number obtained on reversing its digits is
(1) 3916
(2) 7436
(3) 3906
(4) 4906


Answer:(3) 3906
Explanation: The number is 5671 and the number obtained by reversing the digits is 1765.
i.e. 5671 - 1765 = 3906 

Q9.Study the following pattern :
1 X 1 = 1
11 x 11 = 121
111 x 111 = 12321
.... .... What is 11111 X 11111? (1) 12345421
(2) 123453421
(3) 1234321
(4) 123454321


Answer:(4) 123454321
Explanation: From the pattern given in examples, we can observe that:
1. The numbers of digit in the answer = Sum of number of digits in each multiplier - 1.
2. The resulting number have digits starting from 1 to digit having value equal to the number of digits in either number and then decreases from there to 1.
i.e. 111 X 111 = 12321. The Digits in the answer start from 1 and goes to 3 ( Number of digits in 111) and decreases from there till digit 1 .
Applying the same rules to 11111 X 11111, we know the right answer is 123454321. 

Q10 Which of the following is correct?

(1) predecessor of predecessor of 1000 is 999
(2) successor of predecessor of 1000 is 1001
(3) successor of predecessor of 1000 is 1002
(4) predecessor of sucessor of 1000 is 1000


Answer: (4) predecessor of sucessor of 1000 is 1000
Explanation:
1. predecessor of predecessor of 1000 is 999 => predecessor of (1000 - 1 )=> 1000- 1 - 1 = 998 ≠ 999.Rejected
2. successor of predecessor of 1000 => successor of (1000 - 1) => 1000 - 1 + 1 => 1000 ≠ 1001 Rejected!
3. successor of predecessor of 1000 => successor of (1000 - 1 ) => 1000 - 1 + 1 => 1000 ≠ 1002 Rejected!
4. predecessor of sucessor of 1000 => predecessor of (1000 + 1) => 1000 + 1 -1 => 1000 = 1000 Success!


Q11.A shop has 239 toys. Seventy more toys were brought in.Then 152 of them were sold.The number of toys left was
(1) 239 - 70 + 152
(2) 239 + 70 -152
(3) 239 -70 -152
(4) 239 +70 +152


Answer: 239 + 70 -152

Explanation: Let put a plus sign on the incoming toys and negative sign on all outgoing toys ( negative signs because outgoing decreases the inventory). Then, we have 239 + 70 = 152


Q12.In the product 3759 x 9573, the sum of tens' digit and units' digit is
(1) 9
(2) 16
(3) 0
(4) 7


Answer: (4) 7

Explanation: Just multiply the last two digits from each number and you will get the tens' digit and the ones digit of the resultant. Add the digit to get the answer.

Q13. 19thousands + 19 hundreds + 19 ones is equal to
(1) 20919
(2) 19919
(3) 191919
(4) 21090


Answer:(1) 20919

Explanation: We can take one thousand out 19 hundreds as 19 hundreds makes 1 thousand and 900 hundred.Add 1 to 19,000 to get 20,000. Then add remaining 900 and 19 to get the answer, 20919.easy !!

Q14. If 567567567 is divided by 567, the quotient is 

(1) 10101
(2) 1001001
(3) 3
(4) 111


Answer: 1001001 
Explanation: The first impulse is to tick 111. nope. Just do it in your head or on piece of paper. your will get 1 quotient for first 567 and then you have to add 0 and 0 to quotient then 1 to clear 567 and then again 0 and 0 and then 1. 

Q15.How many 1/8 are there in 1/2?

(1) 16
(2)8
(3)4
(4)2


Answer: (3) 4

Solution: Divide 1/2 by 1/8 to get the answer.
Therefore, No. of 1/8 in 1/2 = (1/2)/(1/8) = (1/2)x(8)  = 4


16. A pencil costs two and half rupees. Amit buys one and half dozen pencils and gives hundred rupee note to the shop keeper.The money he will get back is

(1)55
(2)45
(3)65
(4)30

Answer: (1)55

Solution:
Cost of 1 pencil = 2.5
Cost of one and half dozen ( 1 dozen = 12) or 18 pencil = 2.5 x 18 = 45
Since he gave 100 rupee note to the shop keeper, he will get ( 100 - 45 ) = 55 rupees from the him.



Q17. When 90707  is divided by 9, the remainder is 
(1)3
(2)5
(3)6
(4)7

Answer: (2) 5
Solution:

Divide the 90707 by 9 and you will get 5 as remainder and 10078 as quotient. 

Quicker Method:    Try Rule of divisibility for 9. According to this rule if the sum of the all the digits of the dividend is divisible by 9 then the number itself is divisible by 9.
Here, 
                      Sum of digits of 90707 = 9 + 7 + 7 = 23

 and if subtract 5 from this number we will get 23 - 5 = 18  therefor it should be the remainder of the division.


Q18. The sum of the place value and the face value of the number 3 in 12345 is
(1)  0
(2) 295
(3) 297 
(4) 305

Answer: (3) 297
Solution:
Place value of 3 in the number 12345 is 3 X 100 = 300
And the face value of 3 is 3 itself.
Then the difference  = 300  - 3 =297



 Q19. Look at the following pattern:
(9 - 1) / 8 = 1
(98 - 2) / 8 = 12
 (987 - 3) / 8 = 123
(9876 - 4) / 8 = 1234
 ..
(987654 - 6) / 8 =

(1) 12345
(2)123456
(3)123465
(4)  123467

Answer:(2)123456

 Solution:

Looking at the pattern, it is evident that the answer will have :
(a) 6 digits  because there are six digit in the number beginning with 98.. and digit 6 is subtract from it
(b) starting from 1 and ending on 6

Therefore the number is 123456 
 

Q 20 The numbers of integers less than -3 and greater than -8 are :
(1)  2
(2) 3
(3) 4
(4)  6

Answer: (3) 4
Solution:
Numbers of integers greater than -8 and less than -3 are -7,-6,-5,-4. And hence 4 in numbers.
   

Q21. The value of 0.001 + 1.01 + 0.11 is 
   (1)  1.111                                      (2)  1.101
   (3)  1.013                                     (4)   1.121

Answer: (4)   1.121
Solution:
Aligning the decimal point of all the number given in the problem , we have

                          0.001
                          1.010
                          0.110
                          1.121       

 Q22.  In 1999, the population of a country was 30.3 million. The number which is the same as 30.3 million is 

   (1)  303000000                                      (2) 30300000
   (3) 3030000                                           (4)  3030000000

Answer:            (2) 30300000
Solution:
1 Million = 1000000
Therefore, 30.3 million = 30.3 x 1000000 = 30300000
 
_________________________________________________________________
 Q23. If a^3 = 1 + 7 , 3^3 = 1 + 7+ b and 4^3 =1 + 7 + c, where a, b and c are different positive integers, then the value of a + b + c is 

   (1) 58      (2)  68
   (3) 77     (4)  79

Answer:
Solution:
taking first equation,   a^3 = 1 + 7 = 8
                                              a^3 = 2^3
                                      => a = 2
Taking second equation,   3^3 = 1 + 7+ b
                                                           27 = 8 + b
                                                            19 = b
Taking third equation,  4^3 =1 + 7 + c
                                                  64   = 8 + c
                                                      56 = c

Therefore value of a,b and c came out to be 2,19 and56 respectively.

And a + b +c = 56 + 2 + 19 =  77                   
________________________________________________________________
Q 24: We call a number perfect if it is the sum of all its positive divisors , except itself. For example 28  is a perfect number because 28 = 1 + 2 + 4 + 7 +14. Which of the following is a perfect number?

   (1) 13      (2) 10
   (3) 9        (4) 6

 
Answer: (4) 6
Solution:
(1) factors of 13other than itself are 1. Not this number.
(2) factors of 10 other than itself are 1, 2, 5 and their sum = 1 + 2 + 5 = 8. Not this number.
(3) factors of 9 other than itself are 1, 3. Not this number.
(4) factors of 6 other than itself are 1,2,3 and their sum = 1 + 2 + 3 = 6.This is number.
________________________________________________________________
 Q25.  The product of two whole numbers is 24. The smallest possible sum of these numbers is 

   (1) 8           (2) 9
   (3) 10        (4)12

Answer:    (3) 10  
Solution:
First let find all pairs of number that can give 24 upon multiplication. They are ( 8, 3) ,(12,2),(6,4) and upon addition they give out 11, 14 and 10 respectively. 10 is the smallest.

_______________________________________________________________

Q26. The value of 
                                       ( 3^502 - 3^500 + 16 ) /( 3^500 + 2)
is 

   (1) 2         (2) 4
   (3) 8         (4) 16

Answer: (3) 8 
Solution:
Dividing both numerator and denominator by 3^500, we get

                                     {  ( 3^2 - 1) + 16/3^500} / (1 + 2/3^500)

                                 Both 16^/3^500 and 2/3^500 are negligible values and therefore can be ignored.


What's left is    =  
( 3^2 - 1) / 1 = 9  - 1 = 8                                         _________________________________________________________________
Q27. If 800880 = 8 x 10^x + 8x10^y + 8x10^z where x, y and z are whole numbers, then the value of x + y + z is

   (1) 11         (2)8
   (3) 6         (4) 15

Answer:
Solution:
800880 can be broken down into
                                                         800880 =  800000 + 800 + 80 
                                                                           = 8x10^5 + 8x10^2 + 8x10^1
Comparing it with the equation given in the question , we have 
                                                            x =5, y =2 and z=1
and x + y + z = 5 + 2 + 1 = 8                          
 _________________________________________________________________                                                                 
Q28. The number n is doubled and then y is added to it.The result is then divided by 2 and the original number n is subtracted from it. The final result is 

   (1) y                 (2) y/2
   (3) n + y         (4) (n + y )/2

Answer:   (2) y/2
Solution:
Going step by step
1.Taking a number                =>  n
2. Double the number          => 2n
3. Add y                                      => 2n + y
4. Result is divided by 2     => n + y/2
5. Subtract n                            => y/2

__________________________________________________________________
Q 29: If 
                                 1957
                             -        a9
                                 18b8

 then sum of a and b is 


   (1) 15                  (2)14

   (3)13                  (4)12

Answer:  (2)14
Solution:

Starting from the right hand side of the subtraction. Since 7 is less than 9, 1 has to be borrowed from 5. Again 9 is reduced to 8 in the answer, therefore 1 has to borrowed from 9.This means 5 becomes 14 after borrowing from 9 and giving 1 to 7. 

And since we got b after subtracting a from 14 , this means a + b =14
 __________________________________________________________________

Q30. Which of the following is not a perfect square:

   (1)548543213             (2)548543251

   (3)548543215            (4)548543241

Answer:  (4)548543241
Solution:
  None of the perfect square will have 3 at the one's digit. This rules out option (1).

Again only a perfect square of a number that has digit 5 at ones place will have 5 at the unit digit but never alone. It will have 2 in the tenth place. This rules out option (3).

Try the division method for any one of the option (2) and (4) and you will get that only option (4) is a perfect square.

_________________________________________________________________
Q31. When an integer K is divided by 3, the remainder is 1, and when K + 1 is divided by 5, the remainder is 0. Of the following, a possible value of K is
(a) 62 (b) 63
(c) 64 (d) 65


Answer:
Solution:
Using Divisibility Rules, We know that any number that is divided by 5 has to have either 0 or 5 in the unit place. Therefore , (K+1) has 5 in the unit place and K should have either 4 or 9 in the unit place. While checking for options only option (c) has a number 4 in the unit place and hence the CORRECT answer. 

Q32. The least prime number is 
(a)  1           (b)  3
(c)  2           (d)   0

Answer: (c) 2
Solution:
2 is the lowest prime number and also 2 is the only even number.

Q33. If 999x + 888y =1332 and 888x +999y =555, then x² - y² is equal to 
(a) 7           (b) 8
(c) 9           (d) 5

Answer:(a) 7
Solution:
999x + 888y =1332 ----------(i)
888x + 999y =  555 ----------(ii)
Subtracting (ii) from (i)
=> 111x - 111y =    777
=>  (x - y) = 7   ----------(iii)
And adding (i) and (ii)
=>  1887x + 1887y = 1887
=>          x + y = 1  --------(iv)

Now, (x² - y²) = (x-y)(x+y) = (7)(1) = 7


Q34. If {1/2(a-b)}²  + ab = p(a + b)², then the value of p is (assume a≠b) 

(a) 1/2          (b) 1/4
(c)  1/8         (d) 1

Answer:
Solution:
Taking LHS
             
                      {1/2(a-b)}² + ab = 1/4(a-b)² + ab
                                                    = 1/4(a² + b²- 2ab) + ab
                                                    = 1/4((a² + b²)    -ab/2 + ab
                                                    = 1/4(a² + b²) + ab/2
                                                    =1/4(a² + b² + 2ab)
                                                    = 1/4(a+b)² 
Thus the  equation in the question becomes   1/4(a+b)² = p(a + b)²
Therefore. p = 1/4
   


Useful links:


Problems on Percentages

Solved problems on Unitary method




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