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Q:

The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero ?

A) 14th element B) 9th element
C) 12th element D) 7th element
 
Answer & Explanation Answer: C) 12th element

Explanation:

If we consider the third term to be ‘x”
The 15th term will be (x + 12d)
6th term will be (x + 3d)
11th term will be (x + 8d) and
13th term will be (x + 10d).
Thus, as per the given condition, 2x + 12d = 3x + 21d.Or x + 9d = 0.
x + 9d will be the 12th term.

Thus, 12th term of the A.P will be zero.

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Filed Under: Chain Rule
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Q:

A man has only 20-paise and 25-paise coins in a bag. If he has 50 coins in all totaling to Rs.10.25, then the number of 20-paise coins is 

A) 42 B) 45
C) 38 D) 36
 
Answer & Explanation Answer: B) 45

Explanation:

Let number of 20 ps coins = x and

number of 25 ps coins = y

Given total coins in the bag = 50

x + y = 50.......(1)

But the total money in the bag = Rs. 10.25

0.20x + 0.25y = 10.25

20x + 25y = 1025.........(2)

Now multiplying (1) by 25 we get

25x+25y=1250.............(3)

By solving (2) and (3)

20x + 25y = 1025;

=> x = 45;

Then, the no. of 20 ps coins are 45.

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Filed Under: Simplification
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Q:

Equal quantities of three mixtures of milk and water are mixed in the ratio 1:2, 2:3 and 3:4. The ratio of water and milk in the mixture is ?

A) 193 : 122 B) 97 : 102
C) 115 : 201 D) 147 : 185
 
Answer & Explanation Answer: A) 193 : 122

Explanation:

Given the three mixtures ratio as (1:2),(2:3),(3:4)

(1+2),(2+3),(3+4)

Total content = 3,5,7

 

Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105

105/3 = 35 , 105/5 = 21 , 105/7 = 15

 

Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)

 i.e (35,70), (42,63), (45,60)

 

So overall mixture ratio of milk and water is

35+42+45 : 70+63+60

122:193

 

But in the question asked the ratio of water to milk = 193 : 122

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Filed Under: Alligation or Mixture
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Q:

Two boys are playing on a ground. Both the boys are less than 10 years old. Age of the younger boy is equal to the cube root of the product of the age of the two boys. If we place the digit representing the age of the younger boy to the left of the digit representing the age of the elder boy, we get the age of the father of the younger boy. Similarly, we place the digit representing the age of the elder boy to the left of the digit representing the age of the younger boy and divide the figure by 2, we get the age of the mother of the younger boy. The mother of the younger boy is younger than his father by 3 years. Then, what are the ages of elder and younger boys ?

A) E = 15 & Y = 3 B) E = 14 & Y = 12
C) E = 40 & Y = 22 D) E = 4 & Y = 2
 
Answer & Explanation Answer: D) E = 4 & Y = 2

Explanation:

Let the the age of the elder boy = E

 

Let the the age of the younger boy = Y

 

Given that Y = cube root of EY

 


=> Y3 = EY => E = Y2 .....(1)
By the condition of number replacement the age of the father is YE

 

The Mother's age = EY/2

 

But she is 3 years less than father => EY/2 + 3 = YE
2YE = EY + 6 ......(2)

 

Then now from the given options we can identify which satisfies the all the conditions.

 

Here Y =2 and E = 4 satisfies all the conditions.

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Filed Under: Problems on Ages
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Q:

Three strategies P, Q and R have been initiated for cost cutting in a company producing respectively 20%, 30% and 10% savings. Assuming that they operate independently, what is the net saving achieved ?

A) 49.6% B) 50.4%
C) 33.67% D) 66.66%
 
Answer & Explanation Answer: A) 49.6%

Explanation:

As these three strategies P, Q and R are independent so these will cut cost one after the other.
If initial cost is Rs 100, then
20% cost is cut after initializing strategy P, then cost will remain 80% = 80
further 30% cost is cut after strategy Q, then cost will remain 70% of 80 = 56
further 10% cost is cut after strategy R, then cost will remain 90% of 56 = 50.4

Thus final cost remains 50.4 % of the original cost. Hence net saving is 100 - 50.4 = 49.6 %.

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Filed Under: Percentage
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Q:

A computer assisted method for the recording and analyzing of existing or hypothetical systems is

A) Data transmission B) Data capture
C) Data flow D) Data processing
 
Answer & Explanation Answer: C) Data flow

Explanation:
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Q:

Two parallel chords on the same side of the centre of a circle are 5 cm apart. If the chords are 20 and 28 cm long, what is the radius of the circle? 

A) 14.69 cm B) 15.69 cm
C) 18.65 cm D) 16.42 cm
 
Answer & Explanation Answer: B) 15.69 cm

Explanation:

circle11487663359.jpg image

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm

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Filed Under: Simplification
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Q:

What are intents, shared preference in android ?

Answer

An Android Intent is an abstract description of an operation to be performed. It can be used with startActivity to launch an Activity, broadcastIntent to send it to any interested BroadcastReceiver components, and startService(Intent) or bindService(Intent, ServiceConnection, int) to communicate with a background Service.The intent itself, an Intent object, is a passive data structure holding an abstract description of an operation to be performed.
Android provides many ways of storing data of an application. One of this way is called Shared Preferences. Shared Preferences allow you to save and retrieve data in the form of key,value pair.


In order to use shared preferences, you have to call a method getSharedPreferences() that returns a SharedPreference instance pointing to the file that contains the values of preferences.

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