**Q1. Find the number of lead balls of diameter 1 cm each, that can be made from a sphere of diameter 16 cm**

(a) 4096

(b) 512

(c) 1028

(d) 2048

**Q2. The area of the base of rectangular tank is 6500 cm2 and the volume of water contained in it is 2.6 cubic metres. The depth of the water in tank is: **

(a) 4 m

(b) 6 m

(c) 8 m

(d) 10 m

**Q3. A right cylindrical vessel is full with water. How many right cones having the same diameter and height as that of the right cylinder will be needed to store that water? (Take 𝜋 = 22/7) **

(a) 4

(b) 2

(c) 3

(d) 5

**Q4. The heights of a cone, cylinder and hemisphere are equal. If their radii are in the ratio 2 : 3 : 1, then the ratio of their volumes is **

(a) 2 : 9 : 2

(b) 4 : 9 : 1

(c) 4 : 27 : 2

(d) 2 : 3 : 1

**Q5. A tire has three punctures. The first puncture alone would have made the tire flat in 20 minutes, the second alone would have done it in 24 minutes, and the third alone would have done it in 30 minutes. If air leaks out at a constant rate, how long does it take for all the three punctures together to make it flat?**

(a) 10 minutes

(b) 12 minutes

(c) 8 minutes

(d) 11 minutes

**S5. Ans.(c)**

**Sol.** One minute’s work of all the three punctures will be

= (1/20+1/24+1/30)=1/8,

So, all the three punctures will make the tire flat in 8 minutes.

**Q6. A man, a woman or a boy can do a job in 20 days, 40 days and 80 days respectively. How many boys must assist 2 men and 8 women to do the work in 2 days?**

(a) 15 boys

(b) 18 boys

(c) 16 boys

(d) None of these

**S6. Ans.(c)**

**Sol.** Let T boys assist 2 men and 8 women to do the work in 2 days, but from given information we can conclude that 1M = 2W = 4B.

Therefore, 2Men + 8Women + T boys will be equal to (2+8/2+T/4) Men.

Now applying time and work rule we get the result as 1 × 20 = (2+4+T/4)× 2

⇒ T = 16.

**Q7. A can do 150% more work as B can do in the same time. B alone can do a piece of work in 35 hours. A, with help of B, can finish the same work in how many hours?**

(a) 12

(b) 8

(c) 40/3

(d) 10

**S7. Ans.(d)**

**Sol.** From given information, A is 1.5 times more efficient than B or efficiency of A is 2.5 times of B, hence A = 2.5 B or A and B together equals the efficiency of 3.5 B, hence the time taken by A and B together will be 35/3.5 = 10 hours

**Q8. A alone would take 9 days more to complete the job than both A and B would work together. If B worked alone, he took 4 days more to complete the job than A and B worked together. What time would they take, if both A and B worked together?**

(a) 7 days

(b) 5 days

(c) 4 days

(d) 6 days

**S8. Ans.(d)**

**Sol.** Let A and B working together take x days.

Therefore (A + B)’s 1 day work = 1/x

A’s 1 day work = 1/(x + 9)

B’s 1 day’s work = 1/(x + 4)

⇒ 1/(x + 9)+1/(x + 4)=1/x

⇒ x = 6 days

**Q9. A contractor undertook to do a piece of work in 9 days. He employed certain number of laborers but 6 of them were absent from the very first day and the rest could finish the work in 18 days. Find the number of men originally employed.**

(a) 15

(b) 6

(c) 12

(d) 9

**S9. Ans.(c)**

**Sol.** Let the number of men originally employed be x then we get, 9x = 18 (x – 6) or x = 12

**Q10. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero? **

(a) 0

(b) 1

(c) 10

(d) 19

**S10. Ans.(d)**

**Sol. **Average of 20 numbers = 0

∴ Sum of 20 numbers = (0 × 20) = 0.

It is quite possible that 19 of these numbers may be

positive and if their sum is a, then 20th number is (-a)

**Q11. If the mean of a, b, c is M and ab + bc + ca = 0, then the means of a^2+b^2+ c^2 is: **

(a) M^2

(b) 3M^2

(c) 6M^2

(d) 9M^2

**Q12. The average of 7 consecutive numbers is 20. The largest of these numbers is:**

(a) 20

(b) 22

(c) 23

(d) 24

**S12. Ans.(c)**

**Sol.** Let the numbers be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6.

Then (x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6))/7=20

Or 7x + 21 = 140 or 7x = 119 or x = 17.

∴ Largest number = x + 6 = 23.

**Q13. The average age of the boys in a class is 16 years and that of the girls is 15 years. The average age for the whole class is: **

(a) 15 years

(b) 15.5 years

(c) 16 years

(d) Cannot be computed with the given information

**S13. Ans.(d)**

**Sol.** Clearly, to find the average, we ought to know the

number of boys, girls or students in the class, neither of which has been given.

So, the data provided is inadequate.

**Q14. The average annual income (in Rs.) of certain agricultural workers is S and that of other workers is T. The number of agricultural workers is 11 times that of other workers. Then the average monthly income (in Rs.) of all the workers is: **

(a) (S+T)/2

(b) (S+11T)/2

(c) 1/11S+T

(d) (11S+T)/12

**S14. Ans.(d)**

**Sol**. Let the number of other workers be x.

Then, number of agricultural workers = 11x

Total number of workers = 12x

∴ Average monthly income =(S × 11x + T × x)/12x=(11S + T)/12

**Q15. In a class, 20% of the boys are same as 1/2 of the number of girls and there are 20 girls. Total number of students in the class is: **

(a) 70

(b) 120

(c) 220

(d) None of these

**S15. Ans.(a)**

**Sol. **20% of number of boys =20/2 = 10, hence number of boys = 50. Therefore total number of students will be equal to = 50 + 20 = 70