1. Three-fifth of a number is equal to 70% of another number. What is the ratio between the first number and second number ?

a) 4 : 7

b) 3 : 4

c) 7 : 8

d) 7 : 6

Explanation:

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{two}}\,{\text{numbers}}\,{\text{be}}\,X\,{\text{and}}\,Y \cr & X \times {\frac{3}{5}} = 70\% \,of\,Y \cr & \frac{{3X}}{5} = \frac{{70Y}}{{100}} \cr & \frac{X}{Y} = \frac{{ {70 \times 5} }}{{ {100 \times 3} }} = \frac{7}{6} \cr & X:Y = 7:6 \cr} $$

2. A virus known to cause a deadly disease, is also a rapid multiplier. It is known that it can double itself every 30 minutes. If a container is completely full of this virus after 50 hours, when was the container half empty?

a) 35 hours

b) 49.5 hours

c) 46 hours

d) 20 hours

Explanation: In every 30 minutes, Virus doubles it self. It means in 49.5 hours, it would be half empty and it will be full in 50 hours.

3. The LCM of two number is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, the other number is:

a) 215

b) 220

c) 225

d) 235

Explanation: LCM = 45 HCF

LCM + HCF = 1150

45 × HCF + HCF = 1150

46 × HCF = 1150

HCF = 25

LCM = 45 × 25 = 1125

Now, use the formula,

1

^{st}number × second Number = LCM × HCF

125 × second number = 1125 × 25

Second Number = 225

4. The smallest number of four digits which on division by 4, 6, 10 and 15 leaves a remainder 2 in each case is:

a) 1020

b) 1022

c) 1024

d) 1040

Explanation: First of all,we find the LCM of 4, 6, 10 and 15

LCM of 4, 6, 10, 10 = 60

The smallest four digit no. is 1000. We divide it by 60

$$\frac{{1000}}{{60}}$$ It leaves remainder 40

Now, the smallest four digit no which is divisible by 4, 6, 10, 15 is,

1000 + (60 - 40) = 1020

Required number (as it gives remainder 2 always) would be = 1020 + 2 = 1022

5. One-third of a tower is painted black, $$\frac{5}{{11}}$$th of the remaining part is painted red and the rest is painted white. If the white part measures 60 ft.,the total height of the tower is-

a) 3100 ft.

b) 154 ft.

c) 165 ft.

d) 110 ft.

Explanation: Let the tower height will be x ft.

Black part will be = $$\frac{1}{3}x$$

Remaining part = $$x - \frac{1}{3}x = \frac{2}{3}x$$

Red part will be = $$\frac{2}{3}x \times \frac{5}{{11}} = \frac{{10}}{{33}}x$$

White part will be = $$\frac{2}{3}x - \frac{{10}}{{33}}x = \frac{{12x}}{{33}}$$

$$\frac{{12x}}{{33}}$$ = 60

x = 165 ft.

6. The greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?

a) 15

b) 30

c) 40

d) 60

Explanation: Required Number can be given by:

HCF of (3026 - 11) and (5053 - 13)

HCF of 3015 and 5040 = 15

To find HCF, we break the given numbers in their prime Factors

3015 = 3 × 3 × 5 × 67

5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7

We take common multiples in these two given numbers to get required HCF

And Common multiples are: 3 × 5

Required HCF = 15

7. The difference between three times and seven times of a number comes to 36. What is the number?

a) 6

b) 8

c) 9

d) 10

Explanation: Let number be X

7X - 3X = 36

4X = 36

X = 9

Required Number is 9

8. Amitabh picks a random integer between 1 and 999, doubles it and gives the result to Sashi. Each time Sashi gets a number from Amitabh, he adds 50 to the number, and gives the result back to Amitabh, who doubles the number again. The first person, whose result is more than 1000, loses the game. Let 'X' be the smallest initial number that results in a win for Amitabh. The sum of the digit of 'X' is:

a) 3

b) 5

c) 7

d) 9

Explanation: Assume x is the required number.

Step 1 Amitabh 2x Sashi 2x + 50

Step 2 Amitabh 4x + 100. Sashi 4x + 150

Step 3 Amitabh 8x + 300. Sashi 8x + 350

Step 4 Amitabh 16x + 700 Sashi 16x + 750

16x + 750 > 1000

And, x can be 16

7 is the sum

9. Prof. Suman takes a number of quizzes for a course. All the quizzes are out of 100. A student can get an A grade in the course if the average of her scores is more than or equal to 90. Grade B is awarded to a student if the average of her scores is between 87 and 89 (both included). If the average is below 87, the student gets a C grade. Ramesh is preparing for his last quiz and he realizes that he must score a minimum of 97 to get an A grade. After the quiz, he realizes that he will score 70, and he will just manage a B. how many quizzes did Prof. Suman take?

a) 6

b) 7

c) 8

d) 9

Explanation: The difference between just getting an A grade and just getting a B grade would be equal to (number of quizzes × 3).

The difference between the required score to manage an A grade and the score achieved to just manage a B grade as given in the problem’s information is equal to 97 – 70 = 27

number of quizzes × 3 = 27

Number of quizzes is 9

10. A student got twice as many sums wrong as he got right. If he attempted 48 sums in all, how many did he solve correctly?

a) 12

b) 16

c) 24

d) 18

Explanation: Let he has solved correctly X no. of sums. Therefore incorrect no. of sums = 2X

X + 2X = 48

3X = 48

X = 16 sums he has done correctly.