Answer : D

Solution :

Three boxes contains 6X, 4X and 3X kg of mixtures respectively.
1st box contains mixture in the ratio 5:3.
Type 1 in 1st mixture = 6X x 5/8 = 15X/4 kg.
And type 2 in 1st mixture = 6X – 15X/4 = 9X/4 kg

2nd box contains mixture in the ratio 4:3.
Type 1 in 2nd mixture = 4X x 4/7 = 16X/7 kg.
And type 2 in 2nd mixture = 4x – 16X/7 = 12X/7 kg

3rd box contains mixture in the ratio 3:1.
Type 1 in 3rd mixture = 3X x 3/4 = 9X/4 kg.
And type 2 in 3rd mixture = 3X – 9X/4 = 3X/4 kg

Totally, type 1 in final mixture = 15X/4 + 16X/7 + 9X/4 = 232X/28 = 58X/7
Totally, type 2 in final mixture = 9X/4 + 12X/7 + 3X/4 = 132X/28 = 33X/7
Required ratio = 58X/7 : 33X/7 = 58:33

Answer : D

Solution :

Let the effective upstream speed be x and the effective downstream speed be y
AK Rehman can row 180 km upstream and 120 km downstream in 28 hours.
Distance while travelling upstream / Effective upstream speed + Distance while travelling downstream / Effective downstream speed = Total time taken for upstream and downstream
Or 180/x + 120/y = 28 …(1)

Also it is given that he can row 90 km upstream and 90 km downstream in 16 hours
Similar to eq 1 we can form a new equation as below
90/x + 90/y = 16 ….(2)
Multiply by 2 on both sides of eq 2 we get,
180/x + 180/y = 32….. (3)
(1) – (3) = 180/x + 120/ y – 180/x – 180/y = 28 – 32 = -4
-60/y = -4
Effective downstream speed = y = 15
Effective upstream speed = x = 9

Speed of boat can be calculated using the below formula :
Speed of Boat = Effective upstream speed + Effective downstream speed / 2 = (15 +9)/2 = 12.
Speed of current can be calculated using the below formula :
Speed of Current = Effective upstream speed – Effective downstream speed / 2 = (15 – 9)/2 = 3.

Answer : A

Solution :

Note that, Area of a triangle = 1/2 x base x height.

Given ratio of base = p:q
Let their altitudes be h1 and h2.
Ratio of their area’s = [1/2 x p x h1] / [1/2 x q x h2]…(1)

Given ratio of area’s = r/s …(2)
From (1) & (2), we have,
[1/2 x p x h1] / [1/2 x q x h2] = r/s
q x r x h2 = p x s x h1
h1/h2 = qr/ps
Hence, required ratio is qr:ps.

Answer : C

Solution :

Given series is 11, 17, 23,…
This is an A.P. in which a = 11 and d = 6.

Let the A.P. has n terms.
Then the sum of n terms = 11 + 17 + 23 +…+ tn = 256.
We have to find the n.

The formula for the first n terms of an A.P. is given by Sn = (n/2)[2a + (n-1)d]
Here, Sn = 256 = (n / 2) [2×11 + (n-1)6].
(n / 2) [22 + (n-1) 6] = 256
22n + 6n2 – 6n = 512.
6n2 + 16n – 512 = 0.
3n2 + 8n – 256 = 0.

This is a quadratic equation of the form ax2 + bx + c = 0 and the roots of this equation is obtained by using the formula x = [-b(+/-)sqrt(b2 – 4ac)]/2a.
Here, a = 3, b = 8, c = -256 and x = n.
n = [-8 (+/-)sqrt (82 + 4 x 3 x 256) ] / 2×3.
= [-8 (+/-)sqrt (64 + 3072) ] / 2×3.
= [-8 (+/-) sqrt(3136) ] / 6.
= [-8 (+/-)56] / 6
= [-4 (+/-)28] / 3
i.e., n = (-4 + 28)/3 or (-4 – 28)/3
n is either 8 or -32/3
Since n is a non-negative whole number, n value will be 8.
So the number of terms of the series to sum 256 is 8

Answer : B

Solution :

Let the highest score be X.
Given difference of highest and lowest is 86.
Then, his lowest score = X – 86
And, the aggregate of highest and lowest = X + X – 86 = 2X – 86.

The average of 20 tests = 50
Then, his total score = 50 x 20.

And, the total score excluded = 50 x 20 – (2X – 86)….(1).

Given that, the average of 18 tests (excluded highest and lowest) = 46 marks.
Total of 18 tests = 46 x 18 …(2)

From (1) and (2),
We have, 50 x 20 – (2X – 86) = 46 x 18
1000 + 86 – 2X = 828
2X = 258
X = 129.
Hence, the highest score is 129.

Answer : C

Solution :

Given that the combinations of toys are 5 different dresses, 4 different eye colors, 4 different hairstyles, and 3 different hair colors

Let’s start with the different dress.This gives us 5 different toys.

Now, each of those 5 toys could have 4 different eye colors, so we multiply 5 x 4 for 20 different toys so far.

Each of those toys can have 4 different hairstyles, so that’s 20 x 4 for 80 different toys.

And, each of those 80 toys can have 3 different hair colors, so now we multiply 80 x 3 for 240 different toys.

Hence, we would draw 240 different toys with given combinations.

Answer : C

Solution :

Let AB be the first number of the addition and CD be the second one.
By observing, we can conclude that questions are in the general form AB + CD = (AB – CD)/8
That is,
128 + 8 = 128 – 8 /8 = 120 /8 = 15
82 + 10 = 82-10 /8 = 72/8 = 9
316 + 12 = 316-12 / 8 = 304/8 = 38
624 + 8 = 624–8 /8 = 616/8 = 77.
Hence, the answer is 77.

Answer : C

Solution :

Total score of 11 students = (28 x 11) points = 308 points.

Let the least score of the team be Y and the leader’s score be C. Then,

I. C = Y + 11 => C – Y = 11 …. (1)

II. Total score of 9 students (except leader and least scorer) = [ (25 x 3) + (28 x 3) + (30 x 3)] points = 249 points.
C + Y = (308 – 249) = 59 …. (2)

From (1) and (2), we get C = 35.

Thus, I and II together gives the answer.

Answer : B

Solution :

a = fixed cost and

k = variable cost and n = number of boarders

total cost when 25 boarders c = 25*70 = 1750 i.e. 1750 = a + 25k

total cost when 50 boarders c = 50*60 = 3000 i.e. 3000 = a + 50k

solving above 2 eqns, 3000-1750 = 25k i.e. 1250 = 25k i.e. k = 50

therefore, substituting this value of k in either of above 2 eqns we get

a = 500 (a = 3000-50*50 = 500 or a = 1750 – 25*50 = 500)

so total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500

so cost per head = 5500/100 = 55

Answer : B

Solution :

Let the speed of the man in still water be X km/hr.

Given that, speed of the stream = 6 km/hr.
Therefore, speed in downstream = (X+6) km/hr (by using formula 2)
And, speed in upstream = (X-6) km/hr

Distance covered in downstream in 2 hours = (X+6)2 km

Distance covered in upstream in 3 hours = (X-6)3 km

Therefore, (X+6)2 = (X-6)3
2X+12 = 3X-18
X = 30km/hr.

Answer : D

Solution :

Let the C.P of the item be Rs.100
Marked price is 20% more than it’s cost price = Rs.120

Salesman made a profit of 12%.
Then its selling price = (CP + Profit) = Rs.112

Discount = marked price – selling price
Rs.120 – Rs.112 = Rs.8
Required discount % = 8/120 x 100 = 20/3 %
Hence the answer is 20/3%.

Answer : D

Solution :

Let the total number of chocolates be X.
Total number of kids in the school(when everyone is present) = 350.
If everyone is present, number of chocolates per kid = X/350 chocolates …(1)

But, 140 kids are absent.
Therefore, total number of kids present = 350 – 140 = 210
Therefore, number of chocolates per present kid = X/210 chocolates …(2)

210 kids were present and each kid gets 3 extra.

i.e (2)-(1)=3

Or X/210 – X/350=3

350X – 210X / (210 x 350)=3

140X / 210 x 350=3

X = 3 x 210 x 350 / 140 = 1575

Then the required answer is 1575.

Answer : D

Solution :

We know that, “A perfect cube when written in its composite form will have all the powers divisible by 3”

Now, we check with given numbers:

The prime factorization of 10648 = 2 x 2 x 2 x 11 x 11 x 11 = 23 x 113.
The powers of 2 and 11 are divisible by 3.
Therefore, 10648 is a perfect cube.

The prime factorization of 91125 = 5 x 5 x 5 x 32 x 32 x 32 = 53 x 36.
The powers of 3 and 5 are divisible by 3.
Therefore, 91125 is a perfect cube.

The prime factorization of 74088 = 8 x 27 x 343 = 23 x 33 x 73.
The powers of 2, 3 and 7 are divisible by 3.
And so, 74088 is a perfect cube.

The prime factorization of 19683 = 27 x 27 x 27 = 33 x 33 x 33 = 39
The power of 3 is divisible by 3.
So, 19683 is a perfect cube.

All the given numbers are perfect cubes.
Hence, the answer is option D.

Answer : B

Solution :

Length of the two trains are 400 meters and 300 meters and their respective speeds are 80 km/hr and 90 km/hr.
Since the trains are moving in the same direction,the time required to cross each other = (a + b)/(u – v) sec

Here, a = 300 m, b = 400 m and u = 90 km/hr, v = 80 km/hr.
a + b = 300 + 400 = 700 m.
And, u – v = (90 – 80)km/hr = 10 km/hr.

Converting the unit of speed into m/Sec:
10 km/hr = 10 x 5/18 m/sec = 50/18 m/s.
Now, the required time = 700 x 18/50 sec = 252 sec.

Answer : A

Solution :

Given that, 5% of 68.25 + 45% of 95% of 1755 = 35% of ? % of 6350.
This can be written as, 5/100 x 68.25 + 45/100 x 95/100 x 1755 = 35/100 x ? /100 x 6350.

Converting the decimal 68.25 into fraction, we get
5/100 x 6825/100 + 45/100 x 95/100 x 1755 = 35/100 x ? /100 x 6350.
5x 6825/10000 + 45x 95/10000 x 1755 = 35 x ? /10000 x 6350

Cancelling out the denominator (10000) on both sides, we get
5 x 6825+ 45 x 95 x 1755 = 35 x ? x 6350

Dividing each term by 25 on both sides, we get
5 x 273 + 9 x 19 x 1755 = 35 x ? x 254
273 + 9 x 19 x 351 = 7 x ? x 254
60924 = 1778 x ?
? = 60924/1778 = 33.91
Hence the answer is 33.91.

Answer : B

Solution :

Let h and r be the height and radius of the cone
Then 3h and r/3 are the height and radius of new cone respectively.
We know that, the volume of a right circular cone of height h and radius r is 1/3 (pi)(r2)h unit3.

Now, the volume of the cone = 1/3 (pi) (r2)h.
Volume of new cone = 1/3 (pi) (r / 3)2 (3h) = 1/9 (pi) (r2) h
= 1/3 (1/3 (pi) (r2)h) = 1/3 x volume of of the cone.

Hence the answer is option B.

Answer : D

Solution :

8 typists type 900 files in 10 hours.

Then, 660 files typed in 12 hours by X typists.

Typists          Files             Time(hrs)

18               900              10

X                660              12

Here, the number of files has been decreased, so the number of typists also decreases – Directly proportional

Time has been increased but the number of typist is decreased which is inversely proportional.

X = 660 / 900 x 10 / 12 x 18 (Since time is inversely proportional to typists 12 / 10 is changed reciprocally as 10 / 12)

X = 11.

The number of typist required is 11.

Answer : A

Solution :

The numbers in the given series are obtained as follows:

1st number: 15

2nd number: 1st x 2 – 2 = 15 x 2 – 2 = 30 – 2 = 28
3rd number: 2nd x 3 – 2 = 28 x 3 – 2 = 84 – 2 = 82
4th number: 3rd x 2 – 2 = 82 x 2 – 2 = 164 – 2 = 162
5th number: 4th x 3 – 2 = 162 x 3 – 2 = 486 – 2 = 484
6th number: 5th x 2 – 2 = 484 x 2 – 2 = 968 – 2 = 966 (note that, there is 1006 instead of 966).

Hence the wrong number is 1006.

Answer : B

Solution :

Let their ages be a, b, c and d respectively.
Given average = a + b + c + d / 4 = 22.5 …(1)

Age of A is thrice the age of B then a = 3b
B’s age is 1/4 of C, then b = c/4
C’s age is 7 times D, then c = 7d
Sub. c value in b, we have b = 7d/4.
Sub. b value in a, we have a = 21d/4.

Now, substitute a, b and d values in (1), we have
(21d/4) + (7d/4) + 7d + d = 22.5 x 4
(21d + 7d + 32d) = 22.5 x 16
60d = 22.5 x 16
d = 22.5 x 16 / 60 = 6.

Sub. d value, we get
c = 7d = 42,
b = 7d/4 = 10.5
and a = 3b = 31.5.
Therefore, the eldest age is 42.

It is difficult to reconcile the ideas of different schools of thought on the question of education. Some people maintain that pupils at school should concentrate narrow range of subjects which will benefit them directly in their subsequent careers.

Others contend that they should study a wide range of subjects so that they have not only the specialized knowledge necessary for their chosen careers but also sound general knowledge about the world they will have to work and live in.

Supporters of the first theory state that the greatest contributions to civilization are made by those who are most expert in their trade profession. Those on the other side say that, unless they have a broad general education, the experts will be too narrow in their outlook to have sympathy with their follows or a proper sense of responsibility towards humanity as a whole. 

main() 

int x=11,y=6,z; 
z=x==5 || !=4; 
printf(“z=%d”,z); 
}

Answer : 1

main()
{
int i=5;
if(i>5);
{
i=100;
printf(” %d “,i);
}
}

Answer : 100

Answer : B