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Q:

According to the National Agricultural Aviation Society (NAAS), without the use of crop protection products to control insects, weeds, and diseases, crop yields per acre will drop by more than 50 percent. The first aerial appligreion of insecticide occurred in 1921, and it was a huge success. By contrast, in today's economy all aircraft that are classified as aerial appligreors do more than just apply insecticide; today, they also spread seed and apply fertilizer. 

 

From the information given above it CANNOT be validly concluded that

 

A) In today's economy, if an aerial appligreor is used, then it will be able to spread seed and to apply fertilizer B) according to the NAAS, if crop yields per acre never drop by more than 50 percent, then crop protection products have been used to control insects, weeds, and diseases
C) in today's economy, any aircraft that cannot be used to apply fertilizer cannot be classified as an aerial appligreor D) according to the NAAS, if crop yields per acre drop by more than 50 percent, then crop protection products have not been used to control insects, weeds, and diseases.
 
Answer & Explanation Answer: D) according to the NAAS, if crop yields per acre drop by more than 50 percent, then crop protection products have not been used to control insects, weeds, and diseases.

Explanation:
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Filed Under: Deriving Conclusions From Passages
Exam Prep: GRE

Q:

The total weight of a tin and the cookies it contains is 2 pounds. After ¾ of the cookies are eaten, the tin and the remaining cookies weigh 0.8 pounds. What is the weight of the empty tin in pounds?

A) 0.2 B) 0.3
C) 0.4 D) 0.5
 
Answer & Explanation Answer: C) 0.4

Explanation:

Let the weight of the empty tin = w

(2 - w) = weight of the cookies; and one quarter of the cookies weigh (2 - w)/4

One quarter of the cookies + tin = 0.8 = w + (2 - w)/4

Multiply through by 4; 3.2 = 4w + 2 - w

1.2 = 3w; w = 0.4

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Filed Under: Problems on Numbers
Exam Prep: GRE

Q:

Courier charges for packages to a certain destination are 65 cents for the first 250 grams and 10 cents for each additional 100 grams or part thereof. What could be the weight in grams of a package for which the charge is $1.55 ?

A) 1155 B) 1145
C) 1040 D) None
 
Answer & Explanation Answer: B) 1145

Explanation:

The weight will be 250g plus (1.55 - 0.65)/0.10 units of 100g

250 + 900 = 1150

This is the maximum weight that can be sent at that price. But, weights exceeding

250 + 800 will also get charged this amount (that is what the ‘part thereof’ implies).

Hence a package weighing 1145 will be charged $1145

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Filed Under: Numbers
Exam Prep: GRE

Q:

How many cubes of 10cm edge can be put in a cubical box of 1m edge

A) 10 B) 100
C) 1000 D) 10000
 
Answer & Explanation Answer: C) 1000

Explanation:

Number of cubes= (100x100x100) / (10x10x10)= 1000

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Filed Under: Volume and Surface Area
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

The cost of the paint is rs.36.50 per kg. if 1kg of paint covers 16sq.ft, how much will it cost to paint outside of a cube having 8 feet each side

A) Rs.962 B) Rs.672
C) Rs.546 D) Rs.876
 
Answer & Explanation Answer: D) Rs.876

Explanation:

surface area of a cube= 6 x (8 x 8) = 384 sq.ft  

quantity of paint required=(384/16)=24kg  

cost of painting= 36.5 x 24 = Rs.876

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Filed Under: Volume and Surface Area
Exam Prep: Bank Exams
Job Role: Bank PO

Q:

Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?

A) 3 B) 6
C) 4 D) 2
 
Answer & Explanation Answer: A) 3

Explanation:

Let the number of Rose plants be ‘a’.
Let number of marigold plants be ‘b’.
Let the number of Sunflower plants be ‘c’.
20a+5b+1c=1000; a+b+c=100

 

Solving the above two equations by eliminating c,
19a+4b=900

b = (900-19a)/4 

b = 225 - 19a/4----------(1)


b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)

Substituting (1) in (2),

 0 < 225 - 19a/4 < 99

225 <  -19a/4 < (99 -225)

=> 4 x 225 > 19a > 126 x 4

=> 900/19 > a > 505

 

a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.


Hence possible values of a are (28,32,36,40,44)


For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40


Three solutions are possible.

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Filed Under: Permutations and Combinations

Q:

There are five cards lying on the table in one row. Five numbers from among 1 to 100 have to be written on them, one number per card, such that the difference between the numbers on any two adjacent cards is not divisible by 4. The remainder when each of the 5 numbers is divided by 4 is written down on another card (the 6th card) in order. How many sequences can be written down on the 6th card ?

A) 4 x 3^4 B) 3^4
C) 4^3 D) 3 x 4^3
 
Answer & Explanation Answer: A) 4 x 3^4

Explanation:

The remainder on the first card can be 0,1,2 or 3 i.e 4 possibilities.
The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier).

For each value on the card the remainder can have 3 possible values.

The total number of possible sequences is: 4*3^4

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Filed Under: Permutations and Combinations

Q:

If Aug 15th,2012 falls on Thursday then June 11th,2013 falls on which day ?

A) Wednesday B) Saturday
C) Monday D) Tuesday
 
Answer & Explanation Answer: A) Wednesday

Explanation:

First,we count the number of odd days for the left over days in the given period.

Here,given period is 15.8.2012 to 11.6.2013

Aug Sept Oct Nov Dec Jan Feb Mar Apr May Jun

16   30   31  30   31   31   28  31   30   31   11(left days)

2 + 2 + 3 + 2 + 3 + 3 + 0 + 3 + 2 + 3 + 4 (odd days) = 6 odd days

So,given day Thursday + 6 = Wednesday is the required result.

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Filed Under: Calendar