3
Q:

# $\inline \fn_cm 18800\div 470\div 20 = ?$

 A) 1 B) 2 C) 3 D) 4

Explanation:

Given Expression =$\inline \fn_cm \frac{18800}{470}\div 20$ =$\inline \fn_cm 40\div 20$=2

Q:

Simplify the following equation ?

2003 x 2004 - 2001 x 2002 = ?

 A) 8010 B) 8020 C) 8030 D) 8040

Explanation:

(2000 + 3)(2000 + 4) - (2000 + 1)(2000 + 2) = ?
Since (2000 x 2000) - (2000 x 2000) is equal to zero. ?
= (8000 + 6000 + 12) - (4000 + 2000 + 2)
=> 14012 - 6002 = 8010.

1 14
Q:

A man has only 20-paise and 25-paise coins in a bag. If he has 50 coins in all totaling to Rs.10.25, then the number of 20-paise coins is

 A) 42 B) 45 C) 38 D) 36

Explanation:

Let number of 20 ps coins = x and

number of 25 ps coins = y

Given total coins in the bag = 50

x + y = 50.......(1)

But the total money in the bag = Rs. 10.25

0.20x + 0.25y = 10.25

20x + 25y = 1025.........(2)

Now multiplying (1) by 25 we get

25x+25y=1250.............(3)

By solving (2) and (3)

20x + 25y = 1025;

=> x = 45;

Then, the no. of 20 ps coins are 45.

2 168
Q:

Two parallel chords on the same side of the centre of a circle are 5 cm apart. If the chords are 20 and 28 cm long, what is the radius of the circle?

 A) 14.69 cm B) 15.69 cm C) 18.65 cm D) 16.42 cm

Explanation:

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm

3 117
Q:

Hemavathi gets 3 marks for each right sum and loses 2 marks for each wrong sum. He attempts 35 sums and obtains 60 marks. The number of sums attempted correctly is ?

 A) 23 B) 24 C) 25 D) 26

Explanation:

Let, Hema attempted 'k' sum correctly, then

k x 3 -2 x(35-k) = 60
5k = 130
k = 26

so 26 correct sums.

3 146
Q:

The average temperature of Monday to Wednesday was 37 C and of Tuesday to Thursday was 34 C. If the temperature on Thursday was 4/5 th of that of Monday, the temperature on Thursday was ?

 A) 35 c B) 36 c C) 34 c D) 32 c

Explanation:

Monday + Tuesday + Wednesday = 37 x 3 -----------(1)
Tuesday + Wednesday + Thursday = 34 x 3---------(2)
Thursday = 4/5 of Monday ------------------(3)
subtract eqn (1) from (2) we get,
Thursday - Monday = -9
=> Monday - Thursday = 9.......(4)
From (3) & (4), we get

So,Thursday = 36 C