*Question 1 ]*

**Let A be a natural number consisting only 1. B is another natural number which is equal to quotient when A is divided by 13. C is yet another natural number equal to the quotient when B is divided by 7. Find B-C.**

A. 7236

B. 7362

C. 7326

D. None of these

__Explanation :__

According to question, A contain only 1

Let A = 111111 when which is divide by 13 we get quotient 8547 and remainder=0 {111111 is 1st no. which is divisible by 13}.

this quotient is equal to i.e B=8547

and C equal to quotient of B when it is divided by 7 , then we get C=1221

Now, B-C = 8547-1221=7326.

*Question 2 ]*

**A biker notices a certain number(2 digits number) on the milestone before starting the journey. After riding for an hour he notices a milestone with reversed digits of the previous number. Now after riding for another hour he notices that the number on a new milestone had same digits as the first one (in the same order) but with a "0" between the 2 digits. If the rider maintains a constant speed throughout, Calculate his speed.**

A. 45 kmph

B. 50 kmph

C. 35 kmph

D. None of these

__Explanation :__

Let first two digit number is xy = 10x + y

after one hour, the number is reversed yx = 10y + x

after another one hour, it x0y = 100x + y.

Now xy < yx < x0y

so, (10y + x) - (10x + y) = (100x + y) - (10y + x)

or, 9(y - x) = 9(11x - y)

or, 12x = 2y

The above equation will be true only if x = 1 and y = 6.

Hence, his speed is 45.

*Question 3]*

**PT Usha and Shelly John decide to run a marathon between Ramnagar and Jamnagar. Both start from Ramnagar at 1 pm. On the way are two towns Ramgarh and Rampur, separated by a distance of 15 km. PT Usha reaches Ramgarh in 90 minutes running at a constant speed of 40 kmph. She takes additional 30 minutes to reach Rampur. Between Rampur and Jamnagar she maintains an average speed of V kmph (where V is a whole number),Shelly John being a professional marathon runner, maintains a constant speed of 18 kmph. They both reach Jamnagar together after ‘n’ hours, ‘n’ being a whole number. What could be total time taken by PT Usha?**

A. 5 hours

B. 15 hours

C. 41 hours

D. All of the above

__Explanation :__

According to the question,

Ramnagar-------60km------Ramgrah-------15km-----Rampur----(n-2)vkm----Jamnagar

Total distance covered by shelly =18n

Total distance covered by usha =60+15+(n-2)v=75+nv-2v

18n =75+nv-2v

18n-nv= 75-2v

n(18-v) =75-2v

n =75-2v/18-v

As n being a natural number the only satisfying above equation is when v =5

Then,n = 65/13 = 5.

*Question 4]*

**If 1/a + 1/b + 1/c = 1 / (a + b + c); where a + b + c 1 0; abc 1 0, then what is the value of ( a + b ) ( b + c ) ( c + a )?**

A. Equal to 0

B. Greater than 0

C. Less than 0

D. Cannot be determined

__Explanation :__

Given, 1/a+1/b+1/c=1/(a+b+c)

or (bc+ac+ab)/(abc)=1/(a+b+c)

or (ab+bc+ac)(a+b+c)=abc

or abc+(a^2)c+(a^2)b+(b^2)c+abc+(b^2)a+(c^2)b+(c^2)a+abc=abc

or 2abc+(a^2)c+(a^2)b+(b^2)c+(b^2)a+(c^2)b+(c^2)a=0

also,

=> (a+ b )(b + c)(c + a)= 2abc + (a^2)c + (a^2)b + (b^2)c + (b^2)a + (c^2)b + (c^2)a

or(a + b)(b + c)(c + a) = 0.

Question 5]

**Find the number of ways you can fill a 3 x 3 grid (with 4 corners defined as a, b, c, d), if you have 3 white marbles and 6 black marbles.**

A. 75

B. 84

C. 80

D. None of these

__Explanation :__

The area of the grid is = 9.

Now, No. of ways we can arrange 3 white balls in 9 places = 9c3.

and 6 black balls in (9-3) = 6 places = 6c6 = 1.

So total = 9c3 * 1 = 9c3 = 9 * 8 * 7/3 * 2 *1 = 84.

Question 6]

**A man can hit the target once in four shots. If he fires four shots in succession, what is the probability that he will hit the target?**

A. 1

B. 1/256

C. 81/256

D. 175/256

__Explanation :__

In four shots he can hit once,twice,thrice,All hit.

the probability of hitting the target isp(1hit out of 4)+P(2hit out of 4)+p(3hit out of 4)+p(All hit)

it is total probability-probability of not hitting yhr target

=>1-(3/4*3/4*3/4*3/4)

=>175/256.

*Question 7 ]*

**If N = 82^3 - 62^3 - 203 then N is divisible by:**

A. 31 and 41

B. 13 and 67

C. 17 and 7

D. none

__Explanation :__

Value of N is = 82^3 - 62^3 -20^3 = 3*82*62*20 [a+b+c=0 => a^3+b^3+c^3=3abc]

= 3*(2*41)*(2*31)*20 divisible by both 41 & 31.

*Question 8]*

**In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of locality. In how many ways can he plan such that no two houses are next to each other?**

A. 56

B. 73

C. 80

D. 100

__Explanation :__

Assume that the thief start from 1st house then 3rd house and then 5th house to any of remaining 5th houses.

1st step:the thief can stole from 1st house ,3rd and from 5 house; there are 6+5+4+3+2+1=21 ways.

2nd step:the thief can stole from 2nd house,4th house and from 6th house ;there are 5+4+3+2+1=15 ways.

3rd step:the thief can stole from 3rd ,5th and from 7th; there are 4+3+2+1=10 ways.

4th step: the thief can stole from 4th , 6th and from 8th ;there are 3+2+1=6 ways.

5th step:the thief can stole from 5th ,7th and from 9th ;there are 2+1=3 ways.

6th step:the thief can stole from 6th , 8th and from 10th ;there is 1 way.

So, total=(1+3+6+10+15+21)=56.

*Question 9]*

**A hexagon of side a cm is folded along its edges to obtain another hexagon What is the % decrease in the area aof orignal hexagon ?**

A. 70%

B. 75%

C. 80%

D. 60%

__Explanation :__

As we know, area of hexagon is = (3√3)(a^2)/2

Now, lets take a=2

then area = 6√3

after fold it will side become half ; a=1

new area = 1.5√3

percentage= 6√3/ 1.5√3* 100=25 %

So decrease in area =100 - 25 = 75%.

*Question 10]*

**If v,w,x,y, and z are non negative integers, each less than 11, then how many distinct combinations of (v,w,x,y,z) satisfy v(11^4) + w(11^3) +x(11^2) + y(11) + z =151001 ?**

A. 0

B. 1

C. 2

D. 3

__Explanation :__

Taking 11 common from LHS separating z and breaking 151001 as a factor of 11 with some remainder in RHS

=> 11(113v + 112w + 11x + y) + z = 11x13727 + 4 [hence; z = 4]

Equating z =4 and cancelling 11 from each side we get the following simplified equation and again repeating the above procedure;

=> 11(112v + 11w +x) + y =11x1247 + 10 [y = 10]

=> 11(11v + w ) + x = 11x113 + 4 [x = 4]

=> 11v + w = 11x10 + 3 [v=10 , w=3]

So, we get the unique solution for the above equation.

*Question 11 ]*

**If 1/a + 1/b + 1/c = 1/(a+b+c) where a+b+c !=(not equals) 0, abc != 0 , then what is the value of (a+b) (b+c) (c+a)?**

A. equal to 0

B. greater than 0

C. less than 0

D. can't determine

__Explanation :__

Given, 1/a + 1/b + 1/c =1/ (a+b+c)

=> 1/a + 1/b =1/ (a+b+c) - 1/c

=> (b+a)/ab = (c - (a+b+c)) /c(a+b+c)

Cross multiplying (Since, a,b,c !=0 and a+b+c != 0)

=> c(a+b+c)(b+a) = - ab(a+b)

=> (a+b)(ca +cb +cc + ab) = 0

=> (a+b)(ca + cc + cb + ab) = 0

=> (a+b)(b+c)(c+a) = 0

*Question 12]*

**How many no can be formed using digits (1,2,3,4,5,6,7,8,9) such that they are in increasing order (eg:0 12345, 345, 6789, 123456789)?**

A. 36

B. 30

C. 25

D. 32

__Explanation :__

According to the question,

2 digit no.- 8 (12,23,34,45,56,67,78,89)

3 digit no.- 7(123,234,345,456,567,678,789),

4 digit no.- 6, 5 digit no -5, 6 digit no -4, 7 digit no-3, 8 digit no-2, 9 digit no -1(123456789)

So, total =1+2+3+.....+8=36.

*Question 13]*

**How many 3 digit numbers can be formed with 2 consecutive same numbers ?**

A. 160

B. 171

C. 180

D. 256

__Explanation :__

if we fix the 2nd and 3rd position which can be done in 10 ways and 1st position can take value from 1-9 therefore total ways comes out to b 9*10=90

And, if you fix the 1st and 2nd position which can be done in 9 ways and 3rd position can take 10 values i.e from 0-9, that makes it 9*10 = 90.

Total=90+90=180

Now minus the common part like 111,222 to 999 i.e total of 9 numbers.

Therefore, the required three digit number finally 180-9 = 171.

*Question 14 ]*

**If y=x^4+4, where x=any 5 digit number. Find probability that y is divisible by 5??**

A. 4/5

B. 2/5

C. 3/5

D. 1/5

__Explanation :__

N can be any number so it will be 0.1,2,.......9.

x = -----(5 digit no)

y = (x^4+4)/5;

So the value of x can be as a last digit 1,2,3,4,6,7,8,9

Total possible value is (0 to 9)=10

So probability is =8/10=4/5

Testing divisibility :

(0)^4+4=4(not divisible)

(1)^4+4=5

(2)^4+4=20;

(3)^4+4=85;

(4)^4+4=260;

(5)^4+4=___9(not divisible)

(6)^4+4=__0;

(7)^4+4=__5;

(8)^4+4=__0;

(9)^4+4=__5;

*Question 15]*

**Two pyramids are thrown with faces 1,2,4,6. Find the probability to get a sum of even number on an their faces?**

A. 5/8

B. 3/4

C. 1/8

D. 3/5

__Explanation :__

__1 pyramid can show 4 different kind of arrangements__

1,2,4=7

1,2,6=9

1,4,6=11

2,4,6=12

So, out of 4 outcomes, 3 are odd and 1 is even.

In question given two pyramids are thrown.

So, to make some of their faces as even,

Condition 1: some of both individual faces of each pyramid is odd

=> 3/4 * 3/4 = 9/16

Condition 2: some of both individual faces of pyramid is even

=> 1/4 + 1/4 = 1/16

Total such conditions = 9/16 + 1/16

= 10/16 = 5/8.

Question 16]

**How many 6 digit no. can be formed using digits 0 to 5,without repetition such that number is divisible by digit at its unit place?**

A. 420

B. 426

C. 432

D. none of above

__Explanation :__

when unit place is 1 then numbers which are divisible by 1 = 96 (4*4*3*2*1*1 ways)

when unit place is 2 then numbers which are divisible by 2 = 96 (4*4*3*2*1*1 ways)

when unit place is 3 then numbers which are divisible by 3 = 96 (4*4*3*2*1*1 ways)

when unit place is 3 then numbers which are divisible by 4 = 42 (3*3*2*1*1*1 ways when last two digits are 24 and 4*3*2*1*1*1 ways when last two digits are 04)

when unit place is 3 then numbers which are divisible by 5 = 96 (4*4*3*2*1*1 ways)

Hence, the total number of way = 426.

*Question 17]*

**The contents of two vessels containing milk and water in the ratio of 1:2 and 2:3 are mixed in the ratio of 2:1. Find out the ratio after mixing the two mixture?**

A. 12:17

B. 13:30

C. 16:29

D. None of the above

__Explanation :__

Let's assume that volume of both the vessels are same and its equal to 15 Litre.

First Vessel has:

1/3 Milk and 2/3 Water = 5 Liter Milk and 10 Liter Water

Second Vessel has:

2/5 Milk and 3/5 Water = 6 Liter Milk and 9 Liter Water

MIXED IN THE RATIO OF 2:1.

As per above statement 2 times of first Vessel and 1 times of second Vessel are mixed.

So the final 2:1 mixture will have

2*(5 L Milk + 10 L Water) + 1*(6 L Milk + 9 L Water)

=> (10L Milk + 20 L Water) + (6 L Milk + 9 L Water)

=> 16L Milk + 29L Water

Hence, the required Ratio = 16:29.

*Question 18]*

**Find the sum of first 1230 terms of series 1 2 1 2 2 1 2 2 2 1 2 2 2 2 1 2 2 2 2 2 1 .......... ?**

A. 2541

B. 2401

C. 2411

D. None of these

__Explanation :__

We have to find the no. of 1's and no. of 2's in 1230 terms...

Lets take 1st two terms ... 1, 2

no. of 1's is 1 and no. of 2's is 1 => (1+1)

1st 5 terms ... 1,2,1,2,2

=> (1+1)+(1+2)

1st 9 terms ... 1,2,1,2,2,1,2,2,2

=> (1+1)+(1+2)+(1+3)

=> (1+1+1)+(1+2+3)

=> 3 + 3(3+1)/2

[Its of the form n + n(n+1)/2].

So in 1st n terms, no. of 1's is "n", no. of 2's is n(n+1)/2

So sum of 1st n terms ll be n*1 + n(n+1)

So, for the 1230 terms,

=> n + n(n+1)/2

=> n = 48

Thus, 48 + 48*49/2 = 1224 and next 6 no. are 1+2+2+2+2+2 = 11

So total sum will be = 48*1 + 48*49 + 11 = 2411.

*Question 19]*

**A mixture of 125 gallons of wine and water contains 20% of water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture?**

A. 10 gallons

B. 8.5gallons

C. 8gallons

D. 8.33gallons

__Explanation :__

Amount of water in mixture= 125*(20/100)=25 gallons

Amount of wine in mixture= 125-25=100 gallons

Amount of water in final mixture = 25% of the new mixture

so , amount of wine in final mixture = 75% of the new mixture

let , new mixture= x

so amount of wine in final mixture= (75/100 )*x..................(1)

also wine amount is constant because only extra water is added

so , wine amount = 100

From equation (1)

=> ( 75/100)*x =100

=> x=133.33

Amount of water = 133.33-125=8.33 gallons.

*Question 20]*

**A square was given. Inside the square there are white tiles and black tiles. Black tiles was among the diagonal of the square and dimensions of both white as well as black tiles is 1cmx1cm. If there are 81black tiles in the square. Then find the no. of white tiles in it.**

A. 1689

B. 1580

C. 1681

D. 1600

__Explanation :__

Total no of white tiles will be 1600.

Total no of tiles will be 1681(including black and white)

In a 3*3 square ,,no of black tiles will be 5.. Similarly in 5*5 square,, no of black tiles will be 4extra around every corner hence 9.

So we get two series..

1). 5,9,13......81

2). 3,5,7.....

From first series total no of term will be 20. Find the 20th term in second series.. It will be 41.. So 41*41 will be the total no of tiles i.e 1681 out of which 81 are black hence 1600 are white.

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