7
Q:

# The difference between a positive proper fraction and its reciprocal is 9 / 20. Then the fraction is :

 A) 3/5 B) 3/10 C) 4/5 D) 5/4

Explanation:

Let the required fraction be x. Then, (1 / x )- x = 9/20
1 - x^(2) / x = 9 / 20  =>  20 - 20 * x^(2) = 9 * x.
20 * x^(2) + 9 *x - 20 = 0.
=> (4 * x + 5) (5 * x - 4) = 0.
=> x = 4 / 5.

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

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.

1 8
Q:

How many zeros are there from 1 to 10000 ?

 A) 2893 B) 4528 C) 6587 D) 4875

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

2 28
Q:

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

 A) 4 B) 1 C) 5 D) 3

Explanation:

(3x + 2) (2x - 5) = $\inline \fn_jvn \small ax^{2}+kx+n$ ...............(1)

But (3x + 2)(2x - 5) = $\inline \fn_jvn \small 6x^{2}-11x-10$........(2)

so by comparing (1) & (2),

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

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

2 14
Q:

What is the smallest number by which 2880 must be divided in order to make it into a perfect square ?

 A) 2 B) 3 C) 4 D) 5

Explanation:

By trial and error method, we get
2880/3 = 960 is not a perfect square
2880/4 = 720 is not a perfect square
2880/5 = 576 which is perfect square of 24
Hence, 5 is the least number by which 2880 must be divided in order to make it into a perfect square.

3 30
Q:

K men agree to donate a gift of Rs. L to trust. If three men drop out how much more will each have to contribute towards the purchase of the gift ?

 A) L / (k-3) B) k / (L-3) C) 2K / 3L-K D) 3L / K(K-3)

The amount more to pay in contribution = $\inline \fn_jvn \small \frac{L}{K-3}-\frac{L}{K}$
$\inline \fn_jvn \small \frac{3L}{K(K-3)}$