SNAP Number Systems Questions

$$k^4 + \frac{1}{k^4} +2 = 49 => (k^2+\frac{1}{k^2})^2 = 7^2 => k^2 +\frac{1}{k^2} = 7$$.

Similarly, $$k^2 + \frac{1}{k^2} + 2 = 9 => k+\frac{1}{k} = 3.$$

$$(k+1/k)^3 = k^3 + 1/k^3 + 3(k+1/k)=> k^3 +1/k^3 = 27 - 3*3 = 18.$$

In Roman numerals,

The values of M = 1000, D = 500, C = 100, L=50, X = 10, V = 5, 1 = 1.

The value of MMXVIII = 1000+1000+10+5+1+1+1 = 2018

2018 is the correct answer.

The least number that is divisible by 14,16,18 will be the LCM of three which is 1008 but it is not a perfect square .

$$1008=2^4\cdot3^2\cdot7^{ }$$ To make it a perfect square you need to multiply it by 7.

1008*7=7056

All of them will clap together after LCM(20,30,40,50) minutes = 600 minutes=10hrs

Therefore they will clap together again at 8pm.

Going in the reverse order 

When a number is divided by 6, the remainder is 2 => the number is of the form 6k+2

When 6k+2 is divided by 5,the remainder is 3 => the number is of the form 5(6k+2)+3 = 30k+13

When 30k+13 is divided by 3, the remainder is 1.

The remaining is 3(10k+4), so when 10k+4 is divided by 4, the remainder is 0 or 2

Among the given options 1, 2 satisfies this condition.

The number of zeros in n! = highest power of 5 in n! 

Highest power of 5 in 1024! = [$$\frac{1024}{5}$$] + [$$\frac{1024}{25}$$] +[$$\frac{1024}{125}$$] + [$$\frac{1024}{625}$$] where [ ]is the greatest integer function.

Highest power of 5 in 1024! = 204+ 40+ 8+ 1 = 253

For calculating the units digit ,you only take last digit in each number and start multiplying them and when ever a 2 digit number is generated,you take only the units digit of that as well and go on.

Therefore 13*27*63*51*98*46 = 3*7*3*1*8*6 = 21*3*1*8*6 = 1*3*1*8*6 = 24*6 = 4*6 = 24 = 4

The total number of multiples of 4 is 16. (04, 08, 12, ...., 64)

The numbers, which are multiples of 4, have their unit digit 0/2/4/6/8

The multiples of 4, with their unit digit as 8, are (08,28,48), and the multiples of 4, with their unit digit as 6, are (16, 36, 56).

The multiples of 4, with their unit digit as 4, are (04, 24, 44, 64), and the multiples of 4, with their unit digit as 2, are (12, 32, 52), the multiples of 4, with their unit digit as 0, are (20, 40, 60).

It is given that the number present in the units places and tens places is swapped, post which they are written in ascending order, then which of the following numbers will be at 10th place from the last => (16+1-10) = 7th from the beginning.

In ascending order (after swapping the digits), the least value will be those numbers that had '0' as their unit place. After that, the numbers had '2' as their unit place, and so on.

7th from the beginning = (3 numbers that had 0 as their unit place) + (3 numbers that had 2 as their unit place) + (1st number that had '4' as their unit place)

Hence, the first number that had 4 as their unit place now becomes (04 => 40). Hence, 40 is the correct answer.

If the question would have been "can be" , then all the options would have been correct but since it has been given as 'is' , only option d satisfies.
Option A: The polynomial can start with any power of P. The order of powers of P does not make any difference. 
Option B:It is possible only when P = a or b or c ...... or z
Option C:The polynomial can start with any power of P. The order of powers of P does not make any difference.
Option D:This is true in all the cases.

$$16a^2 - 12a$$ = $$\left(4a\right)^2-\left(2\right)\left(4a\right)\left(\ \frac{\ 3}{2}\right)$$

by adding $$\left(\ \frac{\ 3}{2}\right)^2\ $$ we can write it as $$\left(\ 4a-\frac{\ 3}{2}\right)^2\ $$

Hence $$\left(\frac{\ 3}{2}\right)^2\ =\ \ \frac{\ 9}{4}$$ is required answer

Let smaller number be $$x$$

The bigger number will be $$6x+15$$

Their difference = $$5x+15 = 1365$$

$$x = 270$$ is required answer

While finding the last 2 digits, we can ignore the hundredth digit of each number and the last two digit will be the same as last two digit in case of $$22\times\ 23\times\ 25\times\ 27\times\ 29$$.

25*22= 550, which is an important result. 

The rest numbers, 23, 27 and 29 when multiplied by 550, will result in the same last 2 digits as odd number multiplied by 5 will always give 5 as the last digit and we already have 0 at the end of it. So, odd number* 50= X50 always.

So, the last 2 digits will always be 50.

Let the original fraction be $$\ \frac{\ a}{b}$$.

Given that Numerator, a is increased by 200% or it becomes 3 times. So, new numerator= 3a.

The denominator increases by 200%, and it also becomes 3 times. So, new denominator= 3b.

So, the new fraction= $$\ \frac{\ 3a}{3b}$$, which is same as $$\ \frac{\ a}{b}$$, the old fraction. 

The value of the fraction is given to be $$\ 2\frac{\ 4}{5}=\ \ \frac{\ 14}{5}$$, which is also the old fraction.

Since, none of the first three options match. Option D is the correct choice.

We always follow BODMAS 
Now first taking numerator 
which $$\frac{1}{2}divided\ by\ \frac{1}{2\ }\ of\ \frac{1}{2}$$
so we get $$\frac{\frac{1}{2}}{\frac{1}{2}}\times\ \frac{1}{2}=\frac{1}{2}$$
Now taking denominator we get $$\frac{1}{2}+\frac{1}{2}\times\ \frac{1}{2}=\frac{3}{4}$$
Taking ratio 
we get : $$\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$$

For divisibility by 6, number should be divide by 3 and even.

Sum of number = 4 + 4 + 5 + 6 = 19

For divisible by 3, 2 number should be added so,

Number = 4456 + 2 = 4458 

Let the total number of chocolates be C

let chocolates received by each child = x

After distributing the sweets equally among 25 children, the number of chocolates left = 8
25x+8=C                - Eq 1

Had there been 28 children, the number of chocolates that would be required = 22

25x+8-22 should be divisible by 28            - Eq 2

25x-14 should be divisible by 28

28x-(3x+14) should be divisible by 28

(3x+14) should be divisible by 28.

==> x=14

The total number of chocolates = 25x+8 = 25*14+8=358

C is the correct answer.

Minimum distance will be LCM of 40 cm, 42 cm and 45 cm

40 = 2³ x 5

42 = 2 x 3 x 7

45 = 3² x 5

LCM = 2³ x 3² x 5 x 7 = 2520 cm

=25 m 20cm

A is the correct answer.

The least number which when divided by 6, 7, 8, 9 and 10 leaves a remainder of 4, 5, 6 7 and 8 respectively is LCM(6, 7, 8, 9, 10) -2

=2520-2= 2518

The number will be in the form of k * LCM(6, 7, 8, 9, 10) -2

When k = 3, the value of x =7558

When k = 4, the value of x =10078

So the value of x = 7558

Sum of the digits of x = 7+5+5+8= 25

A is the correct answer.

Given, 

The average of the four prime numbers = 35.

a + b + c + d = 35 * 4 = 140.

Since a and d are equidistant from 36.

a + d = 72 --- Eq (1)

b + c = 68 --- Eq (2)

a + b = 60 --- Eq (3) and c + d = 80 --- Eq (4)

Using the equation (3) let us look for the prime values of a and b and the corresponding values of c and d using Eq 2 and 1.

Also given that a < b < c < d.

 (a, b, c, d) = 29, 31, 37, 43

d - a = 43 - 29 = 14

B is the correct answer.

36 = 9*4 Since 9 and 4 are co-prime to each other, we can say that for a number to be divisible by 36 it must be divisible by both 9 and 4.

For G236G0, to be divisible by 4, last two digits should be divisible by 4.

Hence G0 should be a multiple of 4.

Possible values of G are 2, 4, 6, 8, 0

For the number to be divisible by 9, the sum of the digits should be a multiple of 9

G+2+3+6+G+0 = 11+2G should be multiple of 9

If G = 0, 11+2G is not a multiple of 9

If G = 2, 11+2G is not a multiple of 9

If G = 4, 11+2G is not a multiple of 9

If G = 6, 11+2G is not a multiple of 9

If G = 8, 11+2G is a multiple of 9.

Hence 8 is the correct answer.

On adding the units digits of the numbers, we get sum as 10.

$$\therefore\ $$ The summation will be divisible by 10.

2 is the only even prime number. Rest are odd. If both X and Y are greater than 2, they must be odd and sum of 2 odd nos. can never be odd. So option '2' is correct.

one-tenth of 400 = 40

one-quarter of 40 = 10

one half of 10 = 5

Option B

We have R 
When multiplied by 8 we get 8R 
Now taking square we get $$64R^2$$
Dividing by 4 we get $$16R^2$$
Taking square root we get 4R

There are 2 prime nos. between 60 and 70 which are 61 and 67. Their sum is 128.

In the given series, the difference between the terms is in A.P. with a common difference of 2.

30-20 = 10

42-30 = 12

So, next term will be 42+14 = 56

1st term - 1^2:1^3

2nd term - 2^3:2^2

3rd term - 3^2:3^3 ...and so on.

The missing term is 6^3:6^2

Alternate:

Let us assume n is the nth term of the series-

When n is odd, the term is $$n^2:n^3$$

When n is even, the term is $$n^3:n^2$$

For n = 1,

1:1

For n = 2,

8:4

For n = 3,

9:27

For n = 4

64:16

Similarly, for n = 6

216:36

We have :
$$\sqrt{\ \frac{11025}{100}}\times\ \sqrt{\ \frac{1}{100}}\ \ \div\sqrt{\ 0.0025}-\sqrt{\ 420.25}$$
Using BODMAS, we get
= $$10.5\times\ \frac{0.1}{0.05}-\sqrt{\ \frac{42025}{100}}$$
= 21-20.5
= 0.5

We have the series as : 2,3,13,37,86,167,288
Now taking difference between 2 consecutive terms :
3-2 =1 
13-3 =10
37-13=24
86-37 = 49
167-86 = 81
288-167 = 121
Now if you observe the differences are squares of consecutive odd numbers except 13-3 and 37-13 and so we can say in place of 13 , it should have been 12 
Hence 13 is the wrong number .

After interchanging the positions, A becomes 11th from the left. So B must have been 11th from the left and 9th from the right before the interchange. Hence the total number of people = 11+9-1=19

Let's assume $$x$$ to be product of $$a, a+1, a+2,$$ and $$a+3$$

Hence, $$x=a(a+1)(a+2)(a+3)$$

$$n=a(a+1)(a+2)(a+3)+1$$

$$n=a(a+3)(a+1)(a+2)+1$$

$$n=(a^2+3a)(a^2+3a+2)+1$$

Let's assume$$a^2+3=p$$

$$n=p(p+2)+1$$

$$n=p^2+2p+1$$

$$n=(p+1)^2$$

Hence, we can say that n is a perfect square. 

Hence, 3 is true and 2 is automatically false as perfect squares can't be prime. 

Product of any 4 consecutive integers is always even, as one of the numbers among the 4 consecutive integers will be even and any number multiplied by an even number is even. x is even, hence x+1 is odd. thus n is odd. 

Hence both 1 and 3 are true.

By looking into the option, we can see that we only need to check the divisibility of the given number by 3 and 11.

Divisibility test by 3 : The given number, 311311311311311311311 has seven 3s and fourteen 1s.

So, the sum of the digits in the number= (7*3)+(14*1)= 21+14=35.

Since, 35 is not divisible by 3, the given number is not divisible by 3.

Divisibility test by 11: The sum of digits at odd places= (1+3+1+1+3+1+1+3+1+1+3)= 19

The sum of digits at even places= Sum of all the digits- Sum of digits at odd places= 35-19= 16.

Since, the difference between the the sum of digits at odd places and the sum of digits at even places, i.e. 19-16 is not a multiple of 11.

So, the given number is also not divisible by 11.

Hence, option D

Here T(n) represent the $$n^{th}$$ term of series

T(1) - T(2) = 1

T(2) - T(3) = 3

T(3) -T(4) = 5

T(4)- T(5) = 9.. is incorrect since we notice that difference between consecutive terms are consecutive odd numbers.

T(5) = 34 is incorrect 

Given P =6Q

Required answer is

$$\ \frac{\ P-Q}{P}\times\ 100$$%

= $$\ \frac{\ 5Q}{6Q}\times\ 100$$% (since P=6Q)

= $$83\frac{\ 1}{3}$$ %

Family of  numbers when  divided by a,b,c and leaves remainders x,y,z respectively such that (a-x)=(b-y)=(c-z)=v are of the form N = k *LCM ( a,b,c) - v

Therefore  N = k*LCM(32,40,72) - 22 => N=1440k-22

Smallest such number is obtained when k=1 which is 1418

Option A : $$1\ +\ 2^{3\ }+\ 3^{3\ }\ -1\ -4^3=-29$$

Option B: $$1\ +\ 2^{3\ }+\ 3^{3\ }\ +1\ -4^3=-27$$

Option C: $$1\ -\ 2^{3\ }-3^{3\ }\ +1\ +4^3=31$$

Option D: $$1\ -\ 2^{3\ }-3^{3\ }\ -1\ +4^3=29$$

Total taking = number  of participants * cost per participant

 38950 =  $$2\times\ 5^2\times\ 19\times\ 41$$

50, 82, and 95 are the only factor of 38950 which lies between 45 to 100.

If, we check for 50,

Cost per participant = 38950/50 = 779

If, we check for 82,

Cost per participant = 38950/82 = 475

Hence 95 is number of participants.

Cost per participant = 38950/95 = 410

Which is option A.

T(n) be $$n^{th}$$ term

We see that T(n)= T(n-1) * T(n-2)

1, 6, 6, 36, 216, .....

T(1) = 1

T(2) = 6

T(3) = 6

T(4) = 36

T(5) = 216

T(6) = T(5)*T(4)

Hence next term is 36*216 = 7776

The sequence is of prime numbers. 2003 is the next prime number

Let the smaller number be denoted by $$x$$

The bigger number can be $$6x+15$$

Bigger number - smaller number = ($$6x+15$$) - ($$x$$)

$$5x+15$$ = 1365

$$5x$$ = 1350

$$x$$ = 270

Since the bus has numbers from 1 to 500 and is a perfect square it has to be one of the numbers among

1 , 4 , 9 , ..... 484($$22^2$$)

When the plate is put upside down then also a number is obtained which is a perfect square.

It is possible when the digit on the plate contains only  digits 0,1,6,8 and 9

Among the options, 196 ($$13^2$$)is a perfect square which upon putting upside down gives 961($$31^2$$) which is also a perfect square