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   Exercise 2
   Exercise 1
1.
A number when divided by 119 leaves 19 as remainder. If the same number is divided by 17, the remainder obtained is
 
A.
10

B.

7
C.
3
D.
2
   
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Answer: D
Explanation :
Let the number be x.
X=119y+19
(where y is a constant)
= 17 (7y) + 17 + 2
= 17(7y + 1) + 2
Thus, if the number is divided by 17, it leaves 2 as remainder.






 
2.
In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 46, find the dividend
 
A.
5388

B.

5343
C.
5336
D.
5391
   
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Answer: C
Explanation :
Divisor= 5 x 46 = 230
Also, 10 x Quotient = 230
:. Quotient= 23 We know that,
Dividend = (Divisor x Quotient) + Remainder Dividend= (230 x 23) + 46
= 5290 + 46 = 5336






 
3.
The missing digit which makes 345*02 divisible by 11, is
 
A.
8

B.

7
C.
6
D.
2
   
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Answer: D
Explanation :
If the number 345 * 02 is divisible by  11, then [(3 + 5 + 0)- (4 + * +2)] =multiple of 11 or zero
(18 - 6 - *I)= multiple of 11 or  zero (12 - * )  =multiple of 11 or zero
Thus, the possible value of * is only 2.






 
4.
If 3 4 a 6 8 is to be divisible by 9, then the value of missing digit a should be
 
A.
6

B.

5
C.
4
D.
3
   
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Answer: A
Explanation :
If 34a68 is to be divisible by 9, then the sum of all digits must be divisible by 9.
3 + 4 + a + 6 + 8 = multiple of 9
21 + a= multiple of 9 Thus, the possible value of a is 6.






 
5.
On Teacher's Day, sweets were to be equally distributed amongst 540 children. But on that particular day 135 children remained absent, hence each child got 2 sweets extra. How many sweets was each child originally supposed to get
 
A.
4

B.

8
C.
10
D.
6
   
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Answer: D
Explanation :
Let sweets are X.
540X = 405 (X + 2)
540X = 405 X + 810
X(540 - 405) = 810
X  x 135 =810
X =6






 
6.
The divisor is 25 times the quotie remainder. If the quotient is 16, the nt  and  5  times  the dividend is amongst  540 children. But on that particular  day 135 children remained absent, hence each child got 2 sweets extra.  How  many  sweets  was  each  child  originally supposed to get?
 
A.
2

B.

3
C.
1
D.
6
   
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Answer: D
Explanation :
Let sweets are x.
540x = 405 (x + 2)
540x = 405 x + 810
x(540 - 405) = 810
X  x 135 =810
X =6






 
7.
The first 100 multiples of 10, i.e., 10, 20, 30, ...... 1000 are multiplied together. How many zeroes will be there at the end of the product?
 
A.
122

B.

120
C.
200
D.
130
   
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Answer: A
Explanation :
We are not able to find any quicker method for this question. We suggest that you understand the working of its detailed solution.
Total zeroes at the end are produced  due to three factors:
1) Due to zeroes at the end of numbers.
2) Due to 5 at the ten-place of numbers followed by zero at the unit place.
3) Due to 5 at the hundred-place of numbers followed by zeroes at the end.
Due to factor (1):
Number of numbers with single zero= 90 Number of numbers with double zeroes= 9 Number of numbers with triple zeroes=  1 Thus, the total no. of zeroes due to this factor
= 90 X   1 + 9 X  2 + 1X  3 = 111
Due to factor (2):
When 5 is multiplied by any even number, a zero is produced. There are 10 numbers which have 5 at the ten-place followed by zero (like 50,  150,250, ...., 950).
Thus, 10 zeroes are produced due to factor (2).
 
 
Due to factor (3): Following the same reasoning, we may say that 500 x 200 will produce an extra zero.
Thus, there are total Ill + 10 + 1 = 122 zeroes.
 






 
8.
In a question ; on ;division ;with zero remainder, a candidate took 12 as divisor instead of 21. The quotient obtained by him was 35. Find the correct quotient.
 
A.
10

B.

12
C.
20
D.
15
   
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Answer: C
Explanation :
Number= 35 x 12 = 420
Required correct Quoteient = 420/21 = 20






 
9.
Consider a 21-digit number created by writing side by side the natural numbers as follows:
N = 123456789101112 ........
What will be the remainder when above number N is divided by 11?
 
A.
3

B.

1
C.
2
D.
4
   
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Answer: A
Explanation :
N = 1234 1415
Sum of digits at odd places
=1+3+5+7+9+0+1+2+3+4+5=40
Sum of digits at even places
=2+4+6+8+1+1+1+1+1+1=26
For the number to be divisible by 11, difference of sums of digits at odd and even places should be multiple of 11. Here the difference is 40 - 26 = 14, which should be 11. As the last digit of N is at odd place, so if we reduce the digit at 21st place by 3 the number will be divisible by 11. In other words, the number N is 3 more than the multiple of 11. That is, if we divide the number by 11 we get a remainder  3.
 






 
10.
When a certain number is divided by 223, the remainder is 79.
When that number is divided by 179, the quotient is 315; then what will be the remainder?
 
A.
113

B.

123
C.
145
D.
152
   
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Answer: A
Explanation :
When 179 x 315 = 56385 is divided by 223, we have the remainder
189. Therefore the required remainder will be 79 + (223 -189)
=79+34=113.
Detail Method: Let the number be N. Then,
N = 223 Q + 79. Where Q is the quotient when N is divided by 223 Again N = 179 x 3 15 + R
= 56385 + R
= 223 X  252 + 189 + R
or, 223 Q + 79 = 223 x 252 + 189 + R
:. R = 223 (Q- 252) + 79- 189
Since R can't be more than 179, Q- 252 = 1 ie R = 223 X  1 + 79 - 189 = 113
 

 







 

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