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Number System -Test 2

Number System

Online exam in Number System For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLAT

Number System -Test 2

Subject :- Quantitative Aptitude

Chapter :- Number System-Test 1

Questions :- 25

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Question 1 of 25

1. Question
1 points
The product of two consecutive even numbers is 9408. Which is the greater of the two numbers ?
- 96
- 98
- 94
- 92
- None of these
Correct

Here, let the first even number be x.

Then, second even number is x + 2.

And, x(x+2) = 9408
$o r x^{2} + 2 x = 9408$ $\Rightarrow x^{2} + 2 x - 9408 = 0 (b y s p l i t t i n g t h e m i d d l e t e r m) \Rightarrow x^{2} + 98 x - 96 x - 9408 = 0 \Rightarrow x (x - 96) + 98 (x - 9408) = 0 \Rightarrow (x + 98) (x + 96) = 0 \Rightarrow x = 96$
Hence, the greater number is 98.

Incorrect

Here, let the first even number be x.

Then, second even number is x + 2.

And, x(x+2) = 9408
$o r x^{2} + 2 x = 9408$ $\Rightarrow x^{2} + 2 x - 9408 = 0 (b y s p l i t t i n g t h e m i d d l e t e r m) \Rightarrow x^{2} + 98 x - 96 x - 9408 = 0 \Rightarrow x (x - 96) + 98 (x - 9408) = 0 \Rightarrow (x + 98) (x + 96) = 0 \Rightarrow x = 96$
Hence, the greater number is 98.
Question 2 of 25

2. Question
1 points
The product of two successive even numbers is 6888. Which is the greater of the two numbers ?
- 78
- 82
- 86
- 90
- None of these
Correct

Here, let the first even number be x.

Then, second even number is (x+2)

And x(x + 2) = 6888
$o r x^{2} + 2 x = 6888 \Rightarrow x^{2} + 2 x - 6888 = 0 \Rightarrow x^{2} + 84 x - 82 x - 6888 = 0$
(by splitting the middle term)
$\Rightarrow x (x - 82) + 84 (x - 82) = 0 \Rightarrow (x + 84) (x - 82) = 0 \Rightarrow x = 82$
and second number is 84.

Hence, Greater number is 84.

Incorrect

Here, let the first even number be x.

Then, second even number is (x+2)

And x(x + 2) = 6888
$o r x^{2} + 2 x = 6888 \Rightarrow x^{2} + 2 x - 6888 = 0 \Rightarrow x^{2} + 84 x - 82 x - 6888 = 0$
(by splitting the middle term)
$\Rightarrow x (x - 82) + 84 (x - 82) = 0 \Rightarrow (x + 84) (x - 82) = 0 \Rightarrow x = 82$
and second number is 84.

Hence, Greater number is 84.
Question 3 of 25

3. Question
1 points
The sum of two odd numbers is 38 and their product is 325. What is three times the larger number ?
- 42
- 39
- 75
- 72
- 78
Correct

Here, let the first odd number be x.

Then, second odd number is y.

Then x + y = 38

And x + y = 325
$(x + y)^{2} = (x + y)^{2} - 4 x y = (38)^{2} - 4 * 325 = 1444 - 1300 x - y = \sqrt{144}$
or x – y =12

By solving equations (i) and (ii), we get

2x = 50
$\Rightarrow x = 25 a n d y = 13$
Hence, Required number = 3*25 = 75

Incorrect

Here, let the first odd number be x.

Then, second odd number is y.

Then x + y = 38

And x + y = 325
$(x + y)^{2} = (x + y)^{2} - 4 x y = (38)^{2} - 4 * 325 = 1444 - 1300 x - y = \sqrt{144}$
or x – y =12

By solving equations (i) and (ii), we get

2x = 50
$\Rightarrow x = 25 a n d y = 13$
Hence, Required number = 3*25 = 75
Question 4 of 25

4. Question
1 points
The average of four consecutive odd numbers is 36. What is the smallest of these numbers ?
- 31
- 35
- 43
- 47
- None of these
Correct

Here, let the first odd number be x.

Second odd number is x + 2.

Third odd number is x – 2

Fourth odd number is x + 4

Now,

x + x + 2 + x + 2 + x + 4 = 4 * 36

or 4x + 4 = 4 * 36
$\Rightarrow x + 1 = 36 \Rightarrow x = 35$
Hence, smallest odd number

= 35 – 2 = 33

Incorrect

Here, let the first odd number be x.

Second odd number is x + 2.

Third odd number is x – 2

Fourth odd number is x + 4

Now,

x + x + 2 + x + 2 + x + 4 = 4 * 36

or 4x + 4 = 4 * 36
$\Rightarrow x + 1 = 36 \Rightarrow x = 35$
Hence, smallest odd number

= 35 – 2 = 33
Question 5 of 25

5. Question
1 points
The sum of five consecutive odd numbers is 575. What is the sum of the next set of the consecutive odd numbers ?
- 615
- 635
- 595
- Cannot be determined
- None of these
Correct

Here, let the five consecutive odd numbers be x, x-2, x-4, x+2 and x+4

And,

x + x + 2 + x – 2 + x + 4 + x – 4 = 575
$\Rightarrow 5 x = 575 \Rightarrow x = 115$
So, from this, five odd numbers are 111, 113, 115, 117, 119.

By this, we cannot determine the next set of odd numbers.

Incorrect

Here, let the five consecutive odd numbers be x, x-2, x-4, x+2 and x+4

And,

x + x + 2 + x – 2 + x + 4 + x – 4 = 575
$\Rightarrow 5 x = 575 \Rightarrow x = 115$
So, from this, five odd numbers are 111, 113, 115, 117, 119.

By this, we cannot determine the next set of odd numbers.
Question 6 of 25

6. Question
1 points
If the numerator of a fraction is increased by 200% and the denominator is increased by 350%. The resultant fraction is 5/12. What was the original fraction ?
- 5/9
- 5/8
- 7/12
- 11/12
- None of these
Correct

Here, let the original fraction be
$\frac{x}{y}$ $T h e n n u m e r a t o r = x + x * \frac{200}{100} = 3 x a n d d e n o m i n a t o r = y + y * \frac{350}{100} o r \frac{450 y}{100} S o, f r a c t i o n = \frac{N u m e r a t o r}{D e n o m i n a t o r} = \frac{3 x * 100}{450 y} = \frac{2 x}{3 y} N o w, \frac{2 x}{3 y} = \frac{5}{12}, \frac{x}{y} = \frac{15}{24} = \frac{5}{8}$

Incorrect

Here, let the original fraction be
$\frac{x}{y}$ $T h e n n u m e r a t o r = x + x * \frac{200}{100} = 3 x a n d d e n o m i n a t o r = y + y * \frac{350}{100} o r \frac{450 y}{100} S o, f r a c t i o n = \frac{N u m e r a t o r}{D e n o m i n a t o r} = \frac{3 x * 100}{450 y} = \frac{2 x}{3 y} N o w, \frac{2 x}{3 y} = \frac{5}{12}, \frac{x}{y} = \frac{15}{24} = \frac{5}{8}$
Question 7 of 25

7. Question
1 points
If the numerator of a fraction is increased by 400% and the denominator is increased by 500%. The resultant fraction is 20/27. What was the original fraction ?
- 9/8
- 11/12
- 3/4
- Cannot be determined
- None of these
Correct

Here, let the original fraction be x/y.
$T h e n \frac{x * 500}{y * 600} = \frac{20}{27}$ $\Rightarrow \frac{x}{y} = \frac{120}{135} \Rightarrow \frac{x}{y} = \frac{8}{9}$

Incorrect

Here, let the original fraction be x/y.
$T h e n \frac{x * 500}{y * 600} = \frac{20}{27}$ $\Rightarrow \frac{x}{y} = \frac{120}{135} \Rightarrow \frac{x}{y} = \frac{8}{9}$
Question 8 of 25

8. Question
1 points
If the numerator of a fraction is increased by 500% and the denominator is increased by
$300 % . T h e r e s u l \tan t f r a c t i o n i s 2 \frac{4}{7 .} W h a t w a s t h e o r i g i n a l f r a c t i o n ?$
- 4/7
- 12/7
- 15/4
- 6/5
- None of these
Correct

Here, let the original fraction be x/y.
$T h e n \frac{x * 600}{y * 400} = \frac{18}{7} \Rightarrow \frac{x}{y} = \frac{72}{42} \Rightarrow \frac{x}{y} = \frac{12}{7}$

Incorrect

Here, let the original fraction be x/y.
$T h e n \frac{x * 600}{y * 400} = \frac{18}{7} \Rightarrow \frac{x}{y} = \frac{72}{42} \Rightarrow \frac{x}{y} = \frac{12}{7}$
Question 9 of 25

9. Question
1 points
If the numerator of a faction is increased by 200% and the denominator of the fraction is increased by 150%, the resultant fraction is 7/10. What is the original fraction ?
- 3/4
- 7/12
- 7/11
- 9/11
- None of these
Correct

Let the original fraction be x/y.
$T h e n \frac{x * 300}{y * 250} = \frac{7}{10} \Rightarrow \frac{x}{y} = \frac{7 * 5}{10 * 6} \Rightarrow \frac{x}{y} = \frac{7}{12}$

Incorrect

Let the original fraction be x/y.
$T h e n \frac{x * 300}{y * 250} = \frac{7}{10} \Rightarrow \frac{x}{y} = \frac{7 * 5}{10 * 6} \Rightarrow \frac{x}{y} = \frac{7}{12}$
Question 10 of 25

10. Question
1 points
If the numerator of a faction is increased by 20% and its denominator by 25%, then the fraction so obtained is 3/5. What is the original fraction ?
- 3/5
- 3/8
- 5/8
- Cannot be determined
- None of these
Correct

Here, let the original fraction be x/y.
$T h e n \frac{x * 120}{y * 125} = \frac{3}{5} R e q u i r e d a n s w e r i s \Rightarrow \frac{x}{y} = \frac{3 * 25}{5 * 24} \Rightarrow \frac{x}{y} = \frac{5}{8}$

Incorrect

Here, let the original fraction be x/y.
$T h e n \frac{x * 120}{y * 125} = \frac{3}{5} R e q u i r e d a n s w e r i s \Rightarrow \frac{x}{y} = \frac{3 * 25}{5 * 24} \Rightarrow \frac{x}{y} = \frac{5}{8}$
Question 11 of 25

11. Question
1 points
The difference between two numbers is 18 and their HCF and LCM are 6 and 168, respectively. What is the sum of squares of two numbers ?
- 2280
- 2260
- 2420
- 2340
- 2380
Correct

Let one number be x.

Then, another number is x+18

Now, product of two numbers = HCF * LCM
$\Rightarrow x * (x + 18) = 6 * 168 \Rightarrow x^{2} + 18 x = 1008 \Rightarrow x^{2} + 18 x - 1008 = 0 \Rightarrow x^{2} + 42 x - 24 x - 1008 = 0 \Rightarrow x (x - 24) + 42 (x - 24) = 0 \Rightarrow x = 24 H e n c e, a n o t h e r n u m b e r i s 42 . R e q u i r e d s u m = (24)^{2} + (42)^{2} = 576 + 1764 = 2340$

Incorrect

Let one number be x.

Then, another number is x+18

Now, product of two numbers = HCF * LCM
$\Rightarrow x * (x + 18) = 6 * 168 \Rightarrow x^{2} + 18 x = 1008 \Rightarrow x^{2} + 18 x - 1008 = 0 \Rightarrow x^{2} + 42 x - 24 x - 1008 = 0 \Rightarrow x (x - 24) + 42 (x - 24) = 0 \Rightarrow x = 24 H e n c e, a n o t h e r n u m b e r i s 42 . R e q u i r e d s u m = (24)^{2} + (42)^{2} = 576 + 1764 = 2340$
Question 12 of 25

12. Question
1 points
LCM of two numbers is 120 and their HCF is 10. Which of the following can be the sum of those two numbers ?
- 140
- 80
- 60
- 70
- None of these
Correct

Here, LCM of two numbers = 120

HCF of these two numbers = 10

Let the two numbers be 10x and 10y

Then, LCm of 10x and 10y = 120

10xy = 120

xy = 12

Here, xy can be (3,4)

So, numbers are 10*3 and 10*4

Sum of these two numbers = 30+40 = 70

Incorrect

Here, LCM of two numbers = 120

HCF of these two numbers = 10

Let the two numbers be 10x and 10y

Then, LCm of 10x and 10y = 120

10xy = 120

xy = 12

Here, xy can be (3,4)

So, numbers are 10*3 and 10*4

Sum of these two numbers = 30+40 = 70
Question 13 of 25

13. Question
1 points
Find the maximum number of girls, among whom 2923 bags and 3239 eyeliners can be distributed in such a way that each girl gets the same number of bags and eyeliners.
- 80
- 79
- 78
- 81
- None of the above
Correct

Here, we have to find maximum number. Hence, here we have to find out the HCF of 2923 and 3239.

So, find the HCF of these two numbers by factorization method.

i.e., 2923 = 1, 37, 79

3239 = 1, 41, 79

Therefore, HCF = 79

Hence, maximum number of girls = 79

Incorrect

Here, we have to find maximum number. Hence, here we have to find out the HCF of 2923 and 3239.

So, find the HCF of these two numbers by factorization method.

i.e., 2923 = 1, 37, 79

3239 = 1, 41, 79

Therefore, HCF = 79

Hence, maximum number of girls = 79
Question 14 of 25

14. Question
1 points
The sum of two numbers is 45 and their product is 500. Their HCF is
- 5
- 9
- 10
- 15
- None of these
Correct

Here, the sum of two numbers = 45

or x+y = 45

And x*y = 500

So, going through the options 5 is the only common factor which satisfies both the equations.

Incorrect

Here, the sum of two numbers = 45

or x+y = 45

And x*y = 500

So, going through the options 5 is the only common factor which satisfies both the equations.
Question 15 of 25

15. Question
1 points
LCM of two numbers is 28 times their HCF. The sum of HCF and LCM is 1740. If one of these numbers is 240, the sum of digits of the other number is
- 4
- 5
- 6
- 7
- None of these
Correct

Here, let LCM of the two numbers = x

Then, HCF of the two numbers = x/28
$N o w, x + \frac{x}{28} = 1740 \Rightarrow \frac{29 x}{28} = 1740 \Rightarrow x = \frac{1740 * 28}{29} = 1680$
So, LCM of these two numbers = 1680

Again, 240 * y = 1680*60

(First number * Second number)

= (LCM * HCF)

or y = 7 * 60y = 420

Hence, other number is 420.

or sum of digits = 4 + 2 + 0 = 6

Incorrect

Here, let LCM of the two numbers = x

Then, HCF of the two numbers = x/28
$N o w, x + \frac{x}{28} = 1740 \Rightarrow \frac{29 x}{28} = 1740 \Rightarrow x = \frac{1740 * 28}{29} = 1680$
So, LCM of these two numbers = 1680

Again, 240 * y = 1680*60

(First number * Second number)

= (LCM * HCF)

or y = 7 * 60y = 420

Hence, other number is 420.

or sum of digits = 4 + 2 + 0 = 6
Question 16 of 25

16. Question
1 points
HCF of two numbers each of 4 digits is 103 and their LCM is 19261. Sum of the numbers is
- 2884
- 2488
- 4288
- 4882
- None of these
Correct

Here, HCF of two 4 digits numbers = 103

Therefore, Let the numbers be 103a and 103b.

And LCM of 103a and 103b = 19261

or 103(a*b) = 19261
$\Rightarrow a * b = 187$
or a= 11 or 17 and b = 11 or 17

Therefore, Numbers = 103*11 = 1133

= 103*17 = 1751

Therefore, Sum of these two numbers = 2884

Incorrect

Here, HCF of two 4 digits numbers = 103

Therefore, Let the numbers be 103a and 103b.

And LCM of 103a and 103b = 19261

or 103(a*b) = 19261
$\Rightarrow a * b = 187$
or a= 11 or 17 and b = 11 or 17

Therefore, Numbers = 103*11 = 1133

= 103*17 = 1751

Therefore, Sum of these two numbers = 2884
Question 17 of 25

17. Question
1 points
When a number is divided by 2,3,4,5 or 6 remainder in each case is 1. But the number is exactly divisible by 7. If the number lies between 250 and 350, the sum of digits of the number will be
- 4
- 5
- 7
- 10
- None of these
Correct

Here, first of all we have to find the LCM of 2,3,4,5 and 6 = 60

Now, 6*5+1 = 7k

or 7k = 301

k = 43

So, sum of the digits of 301 = 3+0+1 = 4

Incorrect

Here, first of all we have to find the LCM of 2,3,4,5 and 6 = 60

Now, 6*5+1 = 7k

or 7k = 301

k = 43

So, sum of the digits of 301 = 3+0+1 = 4
Question 18 of 25

18. Question
1 points
Amit, Sucheta and Neeti start running around a circular track and complete one round in 18s, 24s and 32s, respectively.

In how many seconds, will the three meet again at the starting point as they all have started running at the same time ?
- 196 s
- 288 s
- 324 s
- Cannot be determined
- None of the above
Correct

Here, to find out the time after which they meet again at the starting point

= LCM of 18, 24 and 32

= 288 s

Therefore, After 288 s they will meet with one another.

Incorrect

Here, to find out the time after which they meet again at the starting point

= LCM of 18, 24 and 32

= 288 s

Therefore, After 288 s they will meet with one another.
Question 19 of 25

19. Question
1 points
$T h e L C M o f \frac{1}{3}, \frac{5}{6}, \frac{2}{9}, \frac{4}{27} i s$
- 1/54
- 10/27
- 20/3
- 27/3
- None of these
Correct

Here, to find the LCM of fractions.

We find the LCM of numerators and HCF of denominators.
$S o, \frac{L C M o f 1, 5, 2 a n d 4}{H C F o f 3, 6, 9 a n d 27} = \frac{20}{3}$

Incorrect

Here, to find the LCM of fractions.

We find the LCM of numerators and HCF of denominators.
$S o, \frac{L C M o f 1, 5, 2 a n d 4}{H C F o f 3, 6, 9 a n d 27} = \frac{20}{3}$
Question 20 of 25

20. Question
1 points
Punit, Vinit and Ajit begin to jog around a stadium. They complete their revolutions in 45s, 54s and 36s respectively. After how many seconds, will they be together at the starting point ?
- 360 s
- 600 s
- 540 s
- 450 s
- None of these
Correct

Here, to find out the time after which they will meet one another at the starting point, we have to find LCM of 45s, 36s and 54s.

So, LCM of 45, 36 and 54 = 540s

Incorrect

Here, to find out the time after which they will meet one another at the starting point, we have to find LCM of 45s, 36s and 54s.

So, LCM of 45, 36 and 54 = 540s
Question 21 of 25

21. Question
1 points
The sum of 8 consecutive odd numbers is 656. Also, average of four consecutive even numbers is 87. What is the sum of the smallest odd number and second largest even number ?
- 165
- 175
- 163
- Cannot be determined
- None of the above
Correct

Here, let the first odd number be x.

Then, 8 consecutive odd numbers are as

x-8, x-6, x-4, x-2, x.

x+2, x+4, x+6

Now, x-8 + x-6 + x-4 + x-2 + x + x+2 + x+4 + x+6 = 656
$\Rightarrow 8 x - 8 = 656 \Rightarrow 8 x = 656 + 8 \Rightarrow 8 x = 664 \Rightarrow x = 83$
Now, let four consecutive even numbers be y-2, y, y+2 and y+4.

Now, 4y+4 = 87*4
$\Rightarrow 4 y + 4 = 348 \Rightarrow 4 y = 344 \Rightarrow y = 86$
Now, smallest even number = 86-8

= 78

Second largest even number = 88

Therefore Sum of both the numbers = 88 + 78

= 166

Incorrect

Here, let the first odd number be x.

Then, 8 consecutive odd numbers are as

x-8, x-6, x-4, x-2, x.

x+2, x+4, x+6

Now, x-8 + x-6 + x-4 + x-2 + x + x+2 + x+4 + x+6 = 656
$\Rightarrow 8 x - 8 = 656 \Rightarrow 8 x = 656 + 8 \Rightarrow 8 x = 664 \Rightarrow x = 83$
Now, let four consecutive even numbers be y-2, y, y+2 and y+4.

Now, 4y+4 = 87*4
$\Rightarrow 4 y + 4 = 348 \Rightarrow 4 y = 344 \Rightarrow y = 86$
Now, smallest even number = 86-8

= 78

Second largest even number = 88

Therefore Sum of both the numbers = 88 + 78

= 166
Question 22 of 25

22. Question
1 points
The sum of three consecutive odd numbers and three consecutive even numbers together is 231. Also, the smallest odd number is 11 less than the smallest even number. What is the sum of the largest odd number and the largest even number ?
- 82
- 83
- 74
- Cannot be determined
- None of these
Correct

Here, let the three consecutive odd numbers be x-2 and x+2 and three even numbers be y-2, y and y+2.

Now,

x-2 + x + x + 2 + y – 2 + y + y + 2

= 231
$\Rightarrow 3 x + 3 y = 231 \Rightarrow x + y = 77$
Again, y-2 – (x-2) = 11
$\Rightarrow y - 2 - x + 2 = 11 \Rightarrow y - x = 11 x + y = 77 y - x = 11$
2y = 88

y = 44 and x = 33

Therefore, Largest odd number = 35

And largest even number = 46

Therefore, Sum of these two numbers = 81

Incorrect

Here, let the three consecutive odd numbers be x-2 and x+2 and three even numbers be y-2, y and y+2.

Now,

x-2 + x + x + 2 + y – 2 + y + y + 2

= 231
$\Rightarrow 3 x + 3 y = 231 \Rightarrow x + y = 77$
Again, y-2 – (x-2) = 11
$\Rightarrow y - 2 - x + 2 = 11 \Rightarrow y - x = 11 x + y = 77 y - x = 11$
2y = 88

y = 44 and x = 33

Therefore, Largest odd number = 35

And largest even number = 46

Therefore, Sum of these two numbers = 81
Question 23 of 25

23. Question
1 points
The sum of five consecutive numbers is 270. What is the sum of the second and the fifth number ?
- 108
- 107
- 110
- Cannot be determined
- None of these
Correct

Here, let the five consecutive numbers be

x-2, x-1, x, x+1, x+2

Again,

x-2 + x-1 + x + x + 1 + x + 2 = 270
$\Rightarrow 5 x = 270 \Rightarrow x = 54 S e c o n d n u m b e r = 54 - 1 = 53$
and fifth number = 54+2

= 56

Therefore, Required sum = 56 + 53 = 109

Incorrect

Here, let the five consecutive numbers be

x-2, x-1, x, x+1, x+2

Again,

x-2 + x-1 + x + x + 1 + x + 2 = 270
$\Rightarrow 5 x = 270 \Rightarrow x = 54 S e c o n d n u m b e r = 54 - 1 = 53$
and fifth number = 54+2

= 56

Therefore, Required sum = 56 + 53 = 109
Question 24 of 25

24. Question
1 points
If the numerator of a fraction is increased by 1/4 and the denominator is decreased by 1/3, the new fraction obtained by 33/64. What was the original fraction ?
- 9/11
- 5/7
- 3/7
- 7/9
- None of these
Correct

Here, let the original fraction be x/y.
$T h e n, \frac{x + \frac{x}{4}}{y - \frac{y}{3}} = \frac{33}{64} \Rightarrow \frac{\frac{4 x + x}{4}}{\frac{2 y}{3}} = \frac{33}{64} \Rightarrow \frac{x}{y} = \frac{33 * 8}{320 * 3} \Rightarrow \frac{x}{y} = \frac{11}{40}$

Incorrect

Here, let the original fraction be x/y.
$T h e n, \frac{x + \frac{x}{4}}{y - \frac{y}{3}} = \frac{33}{64} \Rightarrow \frac{\frac{4 x + x}{4}}{\frac{2 y}{3}} = \frac{33}{64} \Rightarrow \frac{x}{y} = \frac{33 * 8}{320 * 3} \Rightarrow \frac{x}{y} = \frac{11}{40}$
Question 25 of 25

25. Question
1 points
What is true about prime numbers ?
- Prime numbers are not counting numbers
- Prime numbers are greater than 1
- Prime numbers are not natural numbers
- Prime numbers are less than 2
- None of the above
Correct

Incorrect

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Number System

Number System -Test 2

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