Quantitative Aptitude

Let a, b, and c denote pencil, pen and eraser respectively.

Given that the cost of one pencil, two pens and four erasers is Rs.22

This can be written as,

a + 2b + 4c = 22 ..........(1)

Similarly, the cost of 5 pencils, 4 pens and 2 erasers is Rs. 32

This implies ....

5a + 4B + 2c = 32 ......... (2)

Adding (1) and (2), we get

6a + 6b + 6c = 54

We need to find the value of 3a + 3b + 3c.

Dividing by 2, we get

3a + 3b + 3c = 27.

Hence, the cost of 3 pencils, 3 pens, and 3 erasers is Rs.27

Divisibility rule of 4 is last two digits are divisible by 4

so in 5 digit no last two digit should always be 12,24,32,44,52;

(any one of them may be)

so for last two digits 5 ways are possible ;

since repetion are allowed then for 3rs place form right there are also 5 ways so in 3rd place either of 1,2,3,4,5 is possible;

simillarly at 4 rth place either of 1,2,3,4,5 is possible so again 5 ways are possible

simillarly at 5th place also 5 ways are possible

so total no are: 5*5*5*5=625

ya the answer is 5sec..

converting 72kmph in m/sec we get 20m/sec

so 25-20=5sec/m

solution--- t1= time for forword journey. t2 =time 4 return journey. t=t1 + t2 =8hrs; x= destination; t1 + t2 = x / 40 + x / 30 = 8; x( 7/120 ) = 8; x= 8*120/ 7 = 137 milesie answer is(b)

let x be a no

now 1/3 of a no is x/3

and again

x/3=x/6+3

x/3-x/6=3.

x=18

Jalia is 38 years old and qurban is 19 years old.

38-4=34 jalia younger

19+3=22 qurban older the difference btw them is 34-22=12.

Let each mechanism solution is 100 litre.

In "A" mechanism, wastage is (100-30)/100 = 70 %

In "B" mechanism, wastage is (100-40)/100 = 60 %

In "C" mechanism, wastage is (100-10)/100 = 90 %

If all are combined,total solution = 300 litres

Total wastage = 220 litres

Total used fuel = 80 litres

So, wastage in 100 litres = 80/3 litres

% wastage = 26.6666.... %

= 27 % (approx)

Well ship A moves at 2blks per hr and ship B moves at 4 blks per hr..assuming they move towards each we should have the answer between h and i. Lets take one hr time, ship A will be at I and B at H, it means they have alredy met so just moving them both back we will have their meeting between H&I.

b) 36 min

d=s*t

s=40 km/hr=40/60km/min

t=45min

d=(4/6)*45=30km

now increased speed= 50 km/hr=50/60 km/min

d=30 km

t=d/s

= 36 min

answer is 48(c)

answer is 31143

answer is 1234

Ans is 3212 but how explain

Given g(0) = g(1) = 1.

To find: g(6)

g(n) = g(n-1) + g(n-2)

g(2) = g(1) + g(0) = 1 + 1 = 2

g(3) = g(2) + g(1) = 2 + 1 = 3

g(4) = g(3) + g(2) = 3 + 2 = 5

g(5) = g(4) + g(3) = 5 + 3 = 8

g(6) = g(5) + g(4) = 8 + 5 = 13.

7

There is a mathematical formula for calculating standard deviation, but you don't need it for this question. StDev is a measurement of how much variation there is among different numbers in a series. Since choices b and c contain the same number 6 times, there is no variation. choice d also doesn't show much variation. Number 1 does vary from one point to the next, so it has variance, and therefore would have a higher StDev.

ans: choice (a)

night temp=n,day temp=d

n=1/3*d/2+d/2==4d/6

n+d=100(given)

hence, 4d/6+d=100

d==60

n==40 deg

Threshold speed = x mph

let us assume that the 1st person was travelling at (x+10)mph

so it's obvious that the 2nd person travels at 2*(x+10)mph = 35mph; which means x = 7.5 which is the threshold or the speed limit.

So according to me the ans is 7.5mph.

To find A-C, We will find 2's compliment of C and them add it with A,

That will give us (A-C)

2's compliment of C=1's compliment of C+1

=10101010+1=10101011

A-C=10101011+00001111

=10111010

Now (A-C) U B is .OR. logic operation on (A-C) and B

10111010.OR . 00110011

The answer is = 10111011,

Whose decimal equivalent is 187.

From 1 to 40 there are 37 possible combinations of 4 consecutive numbers.

Total number of possible combinations= 40C4.

Probability of outcome to be consecutive = 37/(40C4)= 1/2470

Probability of outcome not to be consecutive = 1-(1/2470)= 2469/2470

(buc)=100110110

(i) we know that (a-b)=a intersection b'

therefor a-(buc)=a intersection (buc)'

(buc)'=011001001

a intersection (buc)'=00000001=a-(buc)

(ii)similarly au(b-c)=100110011