**Q1. Two sums of money are proportional to 8 : 13. If the first sum be Rs. 48, the addition of two sums are- **

(a) Rs. 128

(b) Rs. 120

(c) Rs. 124

**(d) Rs. 126**

**Sol. Let the addition of two sums be x.**

**∴ First sum = 48**

**∴ 8/(8 + 13)×x=48**

**⇒8/21 x=48**

**⇒ x = Rs. 126**

**Q2. The fourth proportional to**

**p^2-pq+q^2,p^3+q^3,p-q is-**

**(a) (p + q)**

(b) (p – q)

(c) (p^2-q^2 )

(d) (q^2-p^2 )

**Sol. Let the fourth ratio be x**

**∴(p^(2 )- pq + q^2)/(p^3 + q^3 )=(p – q)/x**

**⇒x=(p – q)(p^3 + q^3 )/(p^2 – pq + q^2 )=(p-q)(p+q)**

**Q3. The correct ascending order of magnitude, if the ratios 1 : 2, 2 : 5 and 3 : 4 is-**

**(a) 2 : 5 < 1 : 2 < 3 : 4**

(b) 1 : 2 < 2 : 5 < 3 : 4

(c) 3 : 4 < 1 : 2 < 2 : 5

(d) 1 : 2 < 3 : 4 < 2 : 5

**Sol. Given, ratios are 1/2,2/5 and 3/4**

**∴ LCM (2, 5, 4) = 20**

**∴ 1/2=1/2×10/10=10/20**

**2/5=2/5×4/4=8/20**

**And 3/4=3/4×5/5=15/20**

**Clearly, 8/20<10/20<15/20**

**⇒2/5<1/2<3/4**

**Q4. x is directly proportional to Y^2 and inversely proportional to Z^3. If X = 20, when Y = 10 and Z = 5, what is the value of Y when X = 10 and Z = 10 ?**

**(a) 20**

(b) 18

(c) 21

(d) 22

**Sol. Since X∝Y^2 and X∝1/Z^3**

**∴X∝Y^2/Z^3 ⇒X=(KY^2)/Z^3**

**We have X = 20, when Y = 10 and Z = 5**

**∴ 20=(k (10)^2)/5^3**

**⇒(20 × 125)/100=k**

**⇒ k = 25**

**New, when X = 10 and Z = 10, then**

**10=(25 (Y^2 ))/(10)^3**

**⇒Y^2=〖10〗^4/25=400**

**⇒ Y = 20**

**Q5. A container contained 80 L milk. From this container, 8L of milk was taken out and replaced by water. This process was further repeated two times. How much milk is now contained by the container?**

(a) 60 L

(b) 58.6 L

(c) 58 L

**(d) 58.32 L**

**Sol. Here, n = 3, x = 80, y = 8**

**∴ The amount of milk left**

**=80(1-8/80)^3=80(9/10)^3=58.32L**

**Q6. Average age of a group of five students is 14 yr. If a new student also admitted the group, new average age becomes 13 yr. Find the age of new student-**

(a) 13 yr

(b) 10 yr

(c) 7 yr

**(d) 8 yr**

**Sol. ∴ Average age of five students = 14**

**∴ Sum of ages of five students = 14 × 5 = 70**

**Again, average age of six students = 13**

**∴ Sum of ages of six students = 13 × 6 = 78**

**Thus, age of new students = 78 – 70 = 8 yr**

**Q7. In a cricket team, the average age of 11 players and the coach is 18 yr. If the age of the coach is not considered, the average decreases by 1 yr. Find out the age of coach-**

(a) 27 yr

(b) 28 yr

**(c) 29 yr**

(d) 31 yr

**Sol.Total age of 11 players + 1 coach = 12 × 18 = 216 yr**

**Total age of 11 players = 11 × 17 = 187 yr**

**∴ Age of coach = 216 – 187 = 29 yr**

**Q8. A camera is sold for Rs. 4600. If it was sold at a profit of 15%, what was its CP?**

(a) Rs. 4500

(b) Rs. 3500

**(c) Rs. 4000**

(d) Rs. 4200

**Sol.CP=(100/(100 + Profit %))×SP**

**=100/115×4600**

**= Rs. 4000**

**Q9 The fractional equivalent of 57.12% (approx.) is-**

(a) 359/625

**(b) 357/625**

(c) 347/625

(d) 349/625

**Sol.57.12%**

**=57.12/(100 × 100)**

**=(5712 ÷ 16)/(10000 ÷ 16)**

**(multiplying with the numerator and denominator’s HCF)**

**=357/625**

**Q10. 4/16-1/8=3/8**

**6/7-2/9=4/2**

**The above represents the work of a student. If this error pattern continues, the student’s answer to 5/11-2/7 will be-**

**(a) 3/4**

(b) 3/7

(c) 2/18

(d) 7/18

**Sol:5/11-2/7=3/4**