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
