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Q:

How many pairs (A, B) are possible in the number 479865AB if the number is divisible by 9 and it is given that the last digit of the number is odd?

A) 5 B) 6
C) 9 D) 11
 
Answer & Explanation Answer: A) 5

Explanation:
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5 1952
Q:

There are several peacocks and deers in a cage (with no other types of animals). There are 72 heads and 200 feet inside the cage. How many peacocks are there, and how many deers ?

A) 42 & 24 B) 38 & 26
C) 36 & 24 D) 44 & 28
 
Answer & Explanation Answer: D) 44 & 28

Explanation:

Let
x = peacocks
y = deers

2x + 4y = 200
x + y = 72
y = 72 - x
2x + 288 - 4x = 200
x = 44 peacocks
y = 28 deers

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5 3496
Q:

If in the following set of numbers, the first and the third digits are interchanged in each number, which number will be second from the end if arranged in ascending order after interchanging the digits ?

585  546  514  404  206  369

A) 404 B) 585
C) 415 D) 602
 
Answer & Explanation Answer: C) 415

Explanation:

Given numbers are 585 546 514 404 206 369
Then interchanging the 1st and 3rd digits of the given numbers, we get
585 645 415 404 602 963
Now, arranging them in ascending order
963 645 602 585 415 404
Then the last second number is 415.

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5 4925
Q:

In an arithmetic progression if 13 is the 3rd term, -­47 is the 13th term, then ­-17 is which term?

A) 9 B) 10
C) 7 D) 8
 
Answer & Explanation Answer: D) 8

Explanation:
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5 1142
Q:

If A = 1 – 10 + 3 – 12 + 5 – 14 + 7 + ... upto 60 terms, then what is the value of A?

A) – 360 B) – 310
C) – 240 D) – 270
 
Answer & Explanation Answer: D) – 270

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5 1298
Q:

What is the sum of first 40 terms of 1 + 3 + 4 + 5 + 7 + 7 + 10 + 9 + ....?

A) 1010 B) 1115
C) 1030 D) 1031
 
Answer & Explanation Answer: C) 1030

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5 1233
Q:

If N = 1 + 11 + 111 + 1111 + ... +111111111, then what is the sum of the digit's of N?

A) 45 B) 18
C) 36 D) 5
 
Answer & Explanation Answer: A) 45

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5 2270
Q:

Find the sum to 200 terms of the series 2 + 5 + 7 + 6 + 12 + 7 + ....

A) 30,400 B) 30,200
C) 34,600 D) 38,400
 
Answer & Explanation Answer: A) 30,400

Explanation:

we can treat every two consecutive terms as one.
So, we will have a total of 100 terms of the nature:
(2 + 5) + (7 + 6) + (12 + 7).... => 7, 13, 19,....

 

We know the sum of n terms  nn+12

 

Now, a= 7, d=6 and n=100
Hence the sum of the given series is
S= 100/2 x[2 x 7 + 99 x 6]
=> 50[608]
=> 30,400.

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5 4548