Q1. Price of an article increases by 20%. As a result turn over increases by 12%. Find the decrease in quantity sold
(a) 5%
(b) 6.67%
(c) 8%
(d) 16%
Sol.
Let the quantity sold initially=1oo
Quantity sold after increase in price = ((100 + 12)100)/(100+20)=280/3
Decrease quantity sold =100-280/3=6.67%
Q2. Two pipes A and B can separately fill a tank in 2 h and 3 h respectively. If both the pipes are opened simultaneously in the empty tank, then the tank will be filled in
(a) 1 h 15 min
(b) 1 h 20 min
(c) 1 h 12 min
(d) 2 h 30 min
Sol.
pipe A fills a tank in 1 h = 1/2
pipe B fills a tank in 1 h = 1/3
Both pipe can fill a tank in 1 h = 1/2 + 1/3 = 5/6
So,both pipe can fill an empty tank = 6/5 h = 1 h 12 min
Q3. A person borrowed Rs.600 @ 3% per annum S.I and Rs.800 @ 4½ % per annum on the agreement that the whole sum will be returned only when the total interest becomes Rs.216. The number of years, after which the borrowed sum is to be returned, is
(a) 2 years
(b) 3years
(c) 4 years
(d) 5 years
Sol.
Let the time be x years. Then (600 *3* x)/100 + (800*9*x)/(2*100) = 216
18x+ 36x = 246 ; x = 216/54 = 4 years
Required time = 4 years
Q4. A sum of Rs.13000 is divided into three parts such that the simple interests accrued on them for two, three and four years respectively may be equal. Find the amount deposited for 4 years.
(a)5000
(b) 6000
(c)4000
(d)3000
Sol.
Let the amounts be x, y, z in ascending order of value. As the interest rate and interest accrued are same for 4 years 3 years and 2 years i.e. 4x = 3y = 2z = k.
L.C.M. of 4,3,2 = 12 So x:y:z: = 3000 : 4000 :6000
The amount deposited for 4 years = 3000
Q5. Anil had to do a multiplication. Instead of taking 35 as one of the multiplies, he took 53. As a result the product went up by 540. What is the new product?
(a) 1050
(b) 540
(c) 1440
(d) 1590
Sol.
Let the number be x.
Increase in product = 53x – 35x = 18x = 18x = 540 ⇒ x = 30
Hence, new product = 53 × 30 = 1590. Hence, option (d) is the answer.
Q6. What is the least number of soldiers in a regiment, such that they stand in rows of 18, 15 and 25 and also from a solid square?
(a) 1800
(b) 225
(c) 900
(d) None of these
Sol.
Number of soldiers should be a multiple of 18, 15 and 25 ⇒ Number should be a multiple of LCM (18, 15, 25)
Now lowest multiple of 450 that is a perfect square = 900. Hence option (c) is the answer.
Q7. The average weight of 5 apples is 100 g and the average weight of 4 oranges is 80 g. If 2 apples and 2 oranges are picked from them that average weight is found to be 85 g. What is the average weight of the remaining apples and oranges?
(a) 88 g
(b) 90 g
(c) 96 g
(d) 95 g
Sol.
Weight of 5 apples = 500 g
Weight of 4 apples = 320 g
Weight of 2 apples and 2 oranges picked = 340 g
So, weight of 3 apples and 2 oranges remaining = (500 + 320 – 340)g = 480 g
Average weight = 480/5 = 96 g
Q8. In an exam the average was found to be p marks. After deducting typographical error the average marks of 94 students reduced from 84 to 64. Due to his the average came down by 18.8 marks? What was the number of students who took the exam?
(a) 100
(b) 90
(c) 110
(d) 120
Sol.
Number of students = (84 – 64) × 94/18.8 = 100
Q9. The students in three sections of IIML are in the ratio 2 : 3 : 5. If 20 students are increased in each section, the ratio changes to 4 : 5 : 7. The total number of students in the three sections before the increase are-
(a) 75
(b) 90
(c) 100
(d) 150
Sol.
Go through the options. Since the sum of ratios = 10, so total number of students should be multiple of 10. Now check the options one by one.
Q10. A bag contains Rs 216 in the form of one rupee, 50 paise and 25 paise coins in the ratio of 2 : 3 : 4. The number of 50 paise coins is:
(a) 96
(b) 144
(c) 114
(d) 141
Sol.
Let number of one-rupee coins = 2 K
Number of 50-paise coins = 3 K
Number of 25-paise coins = 4 K
2K × Rs. 1 + 3 K × Rs. 0.5 + 4K × Rs. 0.25 = Rs. 216
⇒ 2 K + 1.5K + K = 216
K=48
The number of 50 paise coins=3K=144
