The average of 100 numbers is 44. The average of these 100 n

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The average of 100 numbers is 44. The average of these 100 numbers and 4 other new number is 50. The average of the four new numbers will be-

A800

B200

C176

D24

Nithin K ? Nov 20 '2016 at 21:32

Answer is : 200

Explanation:

Sum of four new numbers  \(=50 \times 104 – 100 \times 44 \)

\(= 5200 – 4400 \)

\(= 800 \)

\(\therefore\text{ Average}= \frac{800}{4} =200\)

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Rs.395 are divided among A, B and C in such a manner that B

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Rs.395 are divided among A, B and C in such a manner that B gets 25 percent more than A and 20 percent more than C. The share of A will be-

ARs.195

BRs.180

CRs.98

DRs.120

Nithin K ? Nov 20 '2016 at 21:7

Answer is : Rs.120

\(\text{Let A gets Rs x} \)

\(\therefore \text{B gets} = \frac{125x}{100} = \text{Rs }\frac{5x}{4}\)

\(\text{C gets }= \frac{100}{120} \times \frac{5x}{4} = \frac{25}{24}x\)

\(\therefore x + \text{ }\frac{5}{4}x+\text{ }\frac{25}{24}x =395 \)

\(=>\frac{24x+30x+25x}{24} =395\)

\(=>\frac{79}{24}x=395\)

\(x = \frac{395 \times 24}{79}\)

\(x = 120\)

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In an innings of a cricket match, three players A, B and C s

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In an innings of a cricket match, three players A, B and C scored a total of 361 runs. If the ratio of the number of runs scored by A to that scored by B and also number of runs scored by B to that scored by be 3 : 2. The number of runs scoredby A was-

A171

B181

C185

D161

Nithin K ? Nov 20 '2016 at 20:48

Answer is : 171
Explanation:

Given,
(A:B):(B:C) = 3:2
A : B = (3 : 2) 
B : C = (3 : 2)

On Multiplying (A:B) with 3 and (B:C with 2)
A : B = 9 : 6
B : C = 6 : 4
Now,
A : B :C = 9 : 6 : 4
Total run = 361
9x + 6X + 4X = 361
19x = 361
x = 19
Number of Runs Scored by A = 9x =9*19 = 171 runs.

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The price of certain item is increased by 15%. If a consumer

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The price of certain item is increased by 15%. If a consumer wants to keep his expenditure on the item the same as before, how much percent must he reduce his consumption of that item?

A

B

C

D

Nithin K ? Nov 20 '2016 at 20:44

Answer is : 13 (1/23)

Explanation:
Formula
Percentage decrease = \(\frac{p\times m}{p+m},\) where m is the original percentage and p is price.

Let us Consider Initial price is 100
and m is 15% => 15
Therefore
Percentage decrease in sugar case
\(=\frac{100 \times 15}{100+15}\)
\(=\frac{1500}{115}\)
\(=\frac{300}{23}\)
=13 + \(\frac{1}{23} \)% decrease.

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The speed of a motor - boat is that of the current of water

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The speed of a motor - boat is that of the current of water as 36 : 5.The boat goes along with the current in 5 h 10 min. It will come back in -

A5 h 50 min

B6 h

C6 h 50 min

D12 h 10 min

Nithin K ? Nov 20 '2016 at 19:45

Answer is : 6 h 50 min

Explanation:

Let the speed of the motor boat = 36 kmph and 
speed of current stream= 5 kmph. 
The boat goes along with the current in 5 hours 10 minutes = \(\frac{31}{6} hour\)

First We Need To Calculate upstream and downstream for above condition

Speed downstream = (36x+5x) = 41x kmph
Since distance = speed × time, we have
distance = \(41x \times \frac{31}{6} \) km. ..................(i)

Speed upstream =(36x-5x) = 31x kmph
speed = 31x kmph .........................................(ii)

Hence, time taken to come back 
\(Time = \frac{distance}{speed}\)
\(= \frac{equation (i) }{ equation (ii)}\)
\(= \frac{41x \times \frac{31}{6}}{31x}\)
\(= \frac{41}{6} hours\)

= 6 hours 50 minutes.

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In an examination, 60% of the candidates passed in English a

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In an examination, 60% of the candidates passed in English and 70% of the can- didates passed in Mathematics, but 20% failed in both of these subjects. If 2500 candidates passed in both the subjects, the number of candidates that appeared at the examination was-

A3000

B3500

C4000

D5000

Nithin K ? Nov 20 '2016 at 19:36

Answer is : 5000

Explanation:

Students passed in English = 60%
Students failed in both subjects = 20%
therefore Students passed only Mathematics is 20%

Students passed in Mathematics= 70%
Students failed in both subjects = 20%
therefore Students passed only English is 10%

Total Appeared In Exam = all fail + only English pass + only Mathematics pass + both English and Mathematics pass.
Total Appeared In Exam (100%) = 20% + 10% + 20% + both English and Mathematics pass.
Total Appeared In Exam(100%) = 50% + both English and Mathematics pass.
Total Appeared In Exam (100%) = 50% + 2500
Total Appeared In Exam (100%) = 2500 + 2500
= 5000

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(45 + 46 + 47 + ....... + 113 + 114 + 115) is equal to

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(45 + 46 + 47 + ....... + 113 + 114 + 115) is equal to-

A5600

B5656

C5680

D4000

No answer description available for this question. Let us discuss.

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