Aptitude Test

∴ M1*D1/W1 = M21*D2/W2
M*12 / 1 = 2*M*D / 1/2
M*12 = 4*M*D
D= 12/4 = 3

1000+N>1000N
clearly N=1

Converting into decimal

13/22=>.59~ ,15/24=>.62~ , 17/26=>.65~ and 19/28=> .67~

there .59~ < .62~ < .65~ < .67~ ,So correct answer is 13/22

Speed= 54 km/hr =54 *(1000/3600)m/sec = 15 m/sec.
Distance = Time * Speed
Length of the train = (Speed * Time) = 15 * 30= 450 m.

Let the two candidates are A and B
the ratio is 13:19
So,
A's votes = 13x
B's votes = 19x

As is given
19X - 13X = 312
6X = 312
X = 52
Hence A's votes= 13X = 13*52 =676
and B's votes= 19X = 19*52 = 988

the second candidate is secured 676 votes

If the speed downstream is x kmph and the speed upstream is y kmph, then:
Rate of stream = 1/2 (x - y) kmph.

= 1/2(21 - 15) km/h
= 3 km/h

The diagonals meet in the middle at a right angle.
All sides of rhombus are equal i.e. 10cm
(Side)²=(half of one diagonal)²+(half of second diagonal)²
by using this formula
(10)²=(8)²+(half of second diagonal)²
(half of second diagonal)² = 36
half of second diagonal=6
second diagonal=12cm

Let C = x,
Then B = (3/2)* x and A = (3/2)*(3/2)* x = (9/4) * x
A : B : C = (9/4) * x : (3/2)* x : x = 9 : 6 : 4
So, C's capital = Rs. (1900×4/19)
= Rs. 400

If X can do a piece of work in n days, then X's 1 day's work =1/n
If A and B, working together,and can complete a piece of work in X days. If A, working alone, can
complete the work in Y days, then B working alone, will complete the work in XY/(Y-X) days.

Then ,B alone can complete = (12 * 8) / (12 - 8) = 96/4 = 24 days

Suppose x years later the father will be twice as old as his son.
Then, x + 41 = 2 (x + 16)
⇒ x = 41 - 32 = 9 years

Hence the father be twice as old as his son in 9 years

Let marked price was X. then
X - 10% of X = 45000
0.9X = 45000
X = 45000/0.9
X = 50000

The marked price price was Rs. 50000

Sample space is, S : {1,2,3,4,5,6} =6

Favorable outcomes,even numbers E : { 2,4,6 } = 3

P(getting an event number)=n(E)/n(S)=3/6=1/2

Let us assume Ramu & Somu ages, 1 year ago, be 6x and 7x respectively.
Given that, four years hence, this ratio would become 7:8
⇒(6x+5):(7x+5)=7:8
⇒8(6x+5)=7(7x+5)
⇒48x+40=49x+35
⇒x=5
Somu's present age = 7x+1= 7×5+1=36 years

Remember : In such a case , there is always a loss .
Loss%=[common loss or gain%/10]²
=(16/10)²=2.56%

The ratio is
R/T = 5/9 = 5*6/9*6
T/A = 6/7 = 6*9/7*9

So, rotio of their runs
R:T:A = 5*6:9*6:7*9
= 5*2:9*2:7*3
= 10:18:21

As is given
21X - 10X = 187
11X = 187
X = 17

Therefore
Raju's runs= 10X = 10*17 = 170
Tendulkar's runs= 18X = 18*17 = 306
Azar's runs= 21X = 21*17 = 357

The required number, which is exactly divisible by each one of the given numbers, should be divisible by LCM of (6, 9, 12, 17) i.e. 612
The Largest number of four digits is 9999.by dividing 9999 to 612 get 207 as remainder.
So, the greatest number of 4 digits divisible by any numbers from 6, 9, 12 or 17 = (9999-207) = 9792.

But we should get the remainder as 2,so we need to add 2 to 9792.So the number PQRS is 9794.

Thus, required value (P + R)/(Q + S) = (9+9)/(7+4)
=>18/11

Ratio of speeds of A:B =100 :90
Ratio of speeds of A :C=100 :72
Ratio of speeds of B:C =B/A *A/C
=90/100*100/72 =90/72
When B runs 90 C runs 72
when B runs 100 C runs =72*100/90 = 80 m
So, B beats C by (100-80)= 20m

Let the sum is 100 then
S.I. = 100* 4* 2 / 100 = 8
C.I = 100 *(1 + 4/100)² - 100 = 100*( 26/25) ² - 100 = 204/25
Difference between C.I and S.I = 204/25 - 8 = 4/25 = .16

So, 0.16 : 10 : : 100 : P
∴ P = (10* 100) / 0.16 = 6250

As is given, area of square = a² = 121 cm²
So, side of square, a = 11 cm

length of the wire = Perimeter of square of side a = 4a = 4*11 = 44cm

Let the radius of circle be r cm. Then,
Perimeter of circle = Length of wire
=> 2πr = 44 cm
=> r = 44/2π = 7 (π = 22/7)

Now, Area of circle = πr² = (22/7) *7*7 = 154 cm²

Sum=1+2+3+⋯+n= n(n+1)/2
Hence, 1+2+3+⋯+12= 12(12+1)/2 = 6*13 = 78