40
Q:

(51 + 52 + 53 + .........+100) is equal to 

Answer:
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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1 2
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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3 2
Q:

When a number is divided by 138 the remainder is 26. What will be the remainder if the same number is divided by 23 ?

A) 4 B) 6
C) 3 D) 1
 
Answer & Explanation Answer: C) 3

Explanation:

Number = quotient x divisor + remainder;
so, here
number = 138 k + 26
=> (23 x 6k) + (23+3)
=> 23(6k+1)+3
so, remainder is 3.

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3 8
Q:

How many zeros are there from 1 to 10000 ?

A) 2893 B) 4528
C) 6587 D) 4875
 
Answer & Explanation Answer: A) 2893

Explanation:

For solving this problem first we would break the whole range in 5 sections


1) From 1 to 9
Total number of zero in this range = 0


2) From 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)


3) From 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180


4) From 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893

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3 30
Q:

(3x + 2) (2x - 5) = ax² + kx + n. What is the value of  a - n + k ?

A) 4 B) 1
C) 5 D) 3
 
Answer & Explanation Answer: C) 5

Explanation:

(3x + 2) (2x - 5) =  ...............(1)

But (3x + 2)(2x - 5) = ........(2)

so by comparing (1) & (2),

we get a= 6, k= -11 , n= -10

(a - n + k) = 6 + 10 - 11 = 5.

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3 14