[Download PDF] 2013 RRB Allahabad Paper with Solutions

Instructions

For the following questions answer them individually

2013 RRB Allahabad - Question 51

The average height of 35 students in a class is 4’2”. Three students of average height moved to new section while 6 students of total height 33’4” joined the class The average height of the student in the class is now

Video Solution

Solution

After the students left, there will be 32 students with the same average height. 4 ft and 2 inches

Total height = 32 x 4.17 = 133.44

Total height of 6 students is 33 ft and 4 inches. = 33.33

New total height = 133.44 + 33.33 = 166.77

New average height = 166.77/ 38 = 4 feet 4 inches

2013 RRB Allahabad - Question 52

If 16 sheep or 12 horses eat the grass of a field in 20 days then in how many days will 5 sheep and 4 horses eat it ?

Video Solution

Solution

16 sheep can feed in 20 days

So, 5 sheep can feed in 20 x 16/5 = 64 days

12 horses can feed in 20 days

So, 4 horses can feed in 20 x 3 = 60 days

Together they can feed in

$$ \frac {1}{\frac{1}{60} + \frac{1}{64}}$$ = 30 days approximately

So, answer is (a)

2013 RRB Allahabad - Question 53

The smallest positive integer n with 24 divisors considering 1 and n as divisors is

Video Solution

Solution

For any given number, that can be represented as $$ A^{x} \times B ^ {y} $$, etc

The number of factors is denoted by (x+1) x ( y+1), etc

360 = $$ 2 ^ {3} \times 3 ^ {2} \times 5^ {1}$$

So the number of factors = (3 +1) x (2+ 1) (1+ 1) = 4x3x2 = 24

For 240, it is $$ 2 ^ {4} \times 3 ^ {1} \times 5^ {1}$$

Number of factors = 5 x 2 x 2 = 20 only

2013 RRB Allahabad - Question 54

The diagonal of a rectangular field is 50 m and one of the sides is 48 m If the cost of cutting the grass of the field is Rs 24 per square metre then the total cost of cutting all grass of the rectangular field is

Video Solution

Solution

The sides of the rectangle are 48 and 14 (by Pythagoras theorem)

Total cost of grass cutting is 48 x 14 x 24 = 16,128

2013 RRB Allahabad - Question 55

The ratio of the ages of X and Y three years ago was 4:5 and that after three years will be 5:6. Then the sum of the ages of X and Y is

Video Solution

Solution

Let the ages of X and Y 3 years ago be 4x and 5x. Ages 6 years from then are 4x+6 and 5x+6

But 4x+6:5x+6 = 5:6

Solving we get the answer as x = 6

Sum of today's ages = 4x + 5x + 6 = 9x + 6 = 60 years

2013 RRB Allahabad - Question 56

What is the formula of Sodium Zincate ?

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2013 RRB Allahabad - Question 57

Bronze is an alloy It’s constituents are

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2013 RRB Allahabad - Question 58

Which compound (given below) is not a peroxide ?

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2013 RRB Allahabad - Question 59

In Bayer’s process the Bauxite ore (for purification) is digested in

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2013 RRB Allahabad - Question 60

The magnitude of a vector is never

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CAT Content

2013 RRB Allahabad

2013 RRB Allahabad

2013 RRB Allahabad - Question 51

2013 RRB Allahabad - Question 52

2013 RRB Allahabad - Question 53

2013 RRB Allahabad - Question 54

2013 RRB Allahabad - Question 55

2013 RRB Allahabad - Question 56

2013 RRB Allahabad - Question 57

2013 RRB Allahabad - Question 58

2013 RRB Allahabad - Question 59

2013 RRB Allahabad - Question 60

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