**Q3. The ratio between two numbers is 3 : 4. If each number is increased by 6, the ratio becomes 4 : 5. The difference between the numbers is:**

(a) 1

(b) 3

(c) 6

(d) 8

**S3. Ans. (c)**

**Sol.** Let the numbers be 3x and 4x. Then,

(3x+6)/(4x+6)=4/5

⇒ 5(3x + 6) = 4(4x + 6)

⇒ x = (30 – 24) = 6

Difference between the numbers = (4x – 3x) = x = 6

**Q5. Ram had to do a multiplication. Instead of taking 35 as one of the multipliers, he took 53. As a result, the product went up by 540. What is the new product? **

(a) 1050

(b) 1590

(c) 1440

(d) None of these

**S5. Ans.(b)**

**Sol. **Suppose one of the multiplier is x, then

53x – 35x = 540 ⇒ (18x = 540)

∴ x = 30

∴ New multiplication = 30 × 53 = 1590

**Q6. Find the number nearest to 2559 which is exactly divisible by 35. **

(a) 2535

(b) 2555

(c) 2540

(d) 2550

**Q7. A man riding his bicycle covers 150 metres in 25 seconds. What is his speed in km per hour? **

(a) 20 km/hr

(b) 21.6 km/hr

(c) 23 km/hr

(d) 25 km/hr

**S7. Ans.(b)**

**Sol.** Speed =150/25 m/sec =(150/25×18/5)km/hr

=108/5 km/hr = 21.6 km/hr

**Q8. A takes 2 hours more than B to walk d km. If A doubles his speed then he can make it in 1 hour less than B. how much time does B require for walking d km? **

(a) d/2 hrs

(b) 3 hrs

(c) 4 hrs

(d) 2d/3 hrs

**S8. Ans.(c)**

**Sol.** Suppose B takes x hours to walk d km

Then, A takes (x + 2) hours to walk d km

With Double of the speed,

A will take 1/2 (x+2) hrs

⇒ x-1/2 (x+2)=1

⇒ 2x – (x + 2) = 2

⇒ x = 4

Hence B takes 4 hours to walk d km.

**Q9. A train crosses a pole in 15 seconds, while it crosses a 100 m long platform in 25 seconds. The length of the train is: **

(a) 125 m

(b) 135 m

(c) 150 m

(d) 175 m

**S9. Ans.(c)**

**Sol.** Let the length of the train be x metres

Then x/15=((100 + x))/25 ⇒ 25x = 1500 + 15x

⇒ 10x = 1500 ⇒ x = 150

Hence, the length of the train is 150 m

**Q10. A began a business with Rs. 4500 and was joined afterwards by B with Rs. 5400. If the profits at the end of the year were divided in the ratio 2 : 1, B joined the business after: **

(a) 4

(b) 5

(c) 6

(d) 7

**S10. Ans.(d)**

**Sol. **Suppose B remained in the business for x months.

Then, A : B =(4500×12)/(5400×x)=2/1

⇒ 2x = 10 ⇒ x = 5

Thus, B remained in the business for 5 months. So, B joined the business after 7 months.

**Q11. Shruti took a loan at simple interest at 6% in the first year with an increase of 0.5% in each subsequent year. She paid Rs. 3375 as interest after 4 years. How much loan did she take? **

(a) Rs. 12500

(b) Rs. 15800

(c) Rs. 33250

(d) Cannot be determined

**S11. Ans.(a)**

**Sol. **Let the loan taken be Rs. x

Then, x×6/100×1+x×6.5/100×1+x×7/100×1+x×7.5/100×1=3375

⇒ (6+6.5+7+7.5)×x/100=3375

⇒ x=(3375×100/27)=12500

**Q12. At what rate of simple interest a certain sum will be double in 15 years?**

(a) 5 (1/2) p.a.

(b) 6% p.a.

(c) 6 (2/3)% p.a.

(d) 7 (1/2)% p.a.

S12. Ans. (c)

Sol. Let the sum be Rs. x

Then, S.I. = Rs. x

Rate =((100 × S.I.)/(Sum × Time))=((100 × x)/(x × 15))% p.a.

=20/3%=6 (2/3)% p.a.

**Q13. The difference between compound interest and simple interest on a sum for 2 years at 8% p.a. is Rs. 768. The sum is **

(a) Rs. 100000

(b) Rs. 110000

(c) Rs. 120000

(d) Rs. 170000

**Q14. The perimeter of the triangular base of a right prism is 15 cm and radius of the **

**incircle of the triangular base is 3 cm. If the volume of the prism be 270 cm3, then the height of the prism is **

(a) 6 cm

(b) 7.5 cm

(c) 10 cm

(d) 12 cm

**Q15. The base of a right pyramid is a square of side 40 cm long. If the volume of the pyramid is 8000 cm3, then its height is: **

(a) 5 cm

(b) 10 cm

(c) 15 cm

(d) 20 cm

**S15. Ans.(c)**

**Sol.** Area of the base = 40 × 40 = 1600 sq. cm

Volume of pyramid =1/3 × area of base × height

⇒ 8000=1/3×1600×h

⇒ h=(8000 × 3)/1600=15 cm