**Q1. A number is multiplied by 5 and 25 is added to it. The resultant is divided by 5 and finally original number is subtracted from the same. The resultant will be **

(a) 2

(b) 3

(c) 4

(d) 5

**S1. Ans.(d)**

**Sol.** Let the number is x

According to question

(5x+25)/5-x = 5

**Q2. John has a camera that takes film that allows 24 exposures, whereas Nancy has a camera that takes film that allows 36 exposures. Both of them want to be able to take the same number of photographs and complete their rolls of film. How many rolls should each buy?**

(a) 12

(b) 72

(c) 3 and 2

(d) 6

**S2. Ans.(c)**

**Sol.** To take the same number of photograph no. of exposure will be LCM of (24, 36)

= 72

Number of roll buy by John = 72/24 = 3 rolls

Number of roll buy by Nancy = 72/36 = 2 rolls

Hence option (c) is correct.

**Q3. Three men start together to travel the same way around a circular track of 11 kms. Their speeds are 4, 5 1/2 and 8 kms per hour respectively. When will they meet at the starting point?**

(a) 22 hrs.

(b) 12 hrs.

(c) 11 hrs.

(d) 44 hrs.

**S3. Ans.(a)**

**Sol. **Time take by the first man to complete the circular track = 11/4 hour.

Time taken by second man = (11 × 2)/11 = 2 hr.

Time taken by the third man = 11/8 hour

Time taken by them to meet at starting point = LCM of (11/4, 2, 11/8)

= 22 hour

**Q4. In a class, 18 boys are there whose height is more than 160 cm. If they are three-fourth of the total number of boys and the total number of boys is two-third of the total number of students, then how many girls are there in the class?**

(a) 18

(b) 6

(c) 12

(d) 24

**S4. Ans.(c)**

**Sol.** Total no. of boys = 18×4/3

= 24

Total no. of student = (24 × 3)/2 = 36

Total no. of girls = 36 – 24

= 12

**Q5. A vessel, full of water, weighs 16.5 kg. When the vessel is 1/4 full, it weighs 5.25 kg. The weight of the empty vessel (in kg) is:**

(a) 1.125

(b) 4.5

(c) 1.5

(d) 3

**S5. Ans.(c)**

**Sol.** Let the weight of empty vessel = x kg

Weight of water when the vessel in full = y kg

According to question,

x + y = 16.5 kg … (1)

x + 1/4 y = 5.25 kg … (2)

3/4 y = 11.25

y = 11.25 ×4/3

y = 15 kg.

weight of empty vessel = 16.5 – 15.0

= 1.5 kg

**Q6. A fraction becomes 1/2 when denominator is increased by 4. The same fraction becomes 1/8 when the numerator is reduced by 5. The fraction is**

(a) 4/6

(b) 3/5

(c) 6/8

(d) 5/8

**S6. Ans.(c)**

**Sol.** Assume the fraction be x/y, then

x/(y + 4)=1/2 ⇒ 2x – y =4

(x – 5)/y=1/8 ⇒ 8x – y = 40

(2) – (1), we get 6x – 36 ⇒ x = 6

Put x = 6 in equation (1), y = 8

∴ fraction = x/y=6/8

**Q7. Fifteen years hence, a man will be just four times as old as he was 15 years ago. What is his present age?**

(a) 25 years

(b) 40 years

(c) 35 years

(d) 5 years

**S7. Ans.(a)**

**Sol.** Assume, present age of man = x years.

His age 15 years hence = x + 15

His age 15 years ago = x – 15

(x + 15) = 4 (x – 15) ⇒ x = 25

His present age = 25 years

**Q8. Deepak is 12 years older than Pratap and Pratap is 22 years younger than Ravi. Which of the following represents the difference between the age of Ravi and Deepak? (P stands for age of Pratap) **

(a) 34 years

(b) (34 + 2p) years

(c) Data inadequate

(d) 10 years

**S8. Ans.(d)**

**Sol.** Age of pratap = p year

Deepak age = p + 12 year

Ravi age = p + 22 year

Difference in Dipak and ravi age = p + 22 – (P + 12)

= 10 year.

**Q9. The diameter of a cylindrical jar is increased by 25%, what percent the height must be reduced so that volume remains the same**

(a) 54

(b) 25

(c) 10

(d) 36

**Q10. In a class of 52 students, 25% are rich and others are poor. There are 20 females in the class, of whom 55% are poor. How many rich males are there in the class?**

(a) 13

(b) 4

(c) 39

(d) 2

**S10. Ans.(b)**

**Sol**. Rich student = 52 ×25/100 = 13

Poor student = 52 ×75/100 = 39

Poor female = 20 ×55/100 = 11

⇒ rich female = 20 – 11 = 9 females

Rich boys = total rich students – rich female

= 13 – 9 = 4 boys

**Q11. A cloth merchant says that due to slump in the market, he sells the cloth at 10% loss, but he uses a false metre-scale and actually gains 15%. Find the actual length of the scale.**

(a) 78.25 cm

(b) 75cm

(c) 85cm

(d) 72.25cm

**Q12. If the selling price is doubled, the profit triples. The profit percent is**

(a) 66.66% gain

(b) 150%

(c) 100%

(d) 120%

**S12. Ans.(c)**

**Sol. **Let CP = x SP = y Profit = y – x

If SP = 2y Profit = 3(y – x)

⇒ 2y – x = 3 (y – x)

y = 2x

Profit = 100%

**Q13. A piece of cloth costs Rs. 35. If the length of the piece would have been 4m longer and each metre costs Rs. 1 less the cost would remain unchanged. How long is the piece?**

(a) 12 m

(b) 7 m

(c) 10 m

(d) 14 m

**S13. Ans.(c)**

**Sol.** Let each meter cost = x rupees

Length of piece of cloth = y meter

According to the question,

xy = 35 … (1)

(x + 4) (y – 1) = xy … (2)

By solving these two equation.

x = 10 m & y = 3.5 Rs

**Q14. A train has a length of 150 meters . it is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train.**

(a) 182 km/hr

(b) 180 km/hr

(c) 152 km/hr

(d) 169 km/hr

**S14. Ans.(a)**

**Sol.** Length of the train, l = 150m

Speed of the man, Vm= 2 km/hr

Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr

Relative Speed = Speed of train – Speed of man (As both are moving in the same direction)

=> 180 = Vt – 2

=> Vt = 180 + 2 = 182 km/hr

**Q15. In a 100 m race, A beats B by 10 m and C by 13 m. In a race of 180 m, B will beat C by:**

(a) 5.4 m

(b) 4.5 m

(c) 5 m

(d) 6 m

**S15. Ans.(d)**

**Sol.** A: B = 100: 90.

A: C = 100: 87.

B/C = B/A x A/C = 90/100 x 100/87 = 30/29.

When B runs 30 m, C runs 29 m.

When B runs 180 m, C runs (29/30 x 180) m = 174 m.

B beats C by (180 – 174) m = 6 m.