**1. In a festival season sale of 30 days, the discount offered on the first day is 30%. Every day, it decreases by 1% till it reaches to 1% on last day of sale. Because of the discount, selling price of a geyser on 23rd day is Rs. 1360 more than that on 7th day. What is the marked price of geyser?**

A) Rs. 7500

B) Rs. 8000

C) Rs. 8500

D) Rs. 8400

E) Rs. 9000

**2. For every rupee that Ricky invests in simple interest, he also invests 2 rupees in compound interest. The rate of interest in case of simple interest is 12% per annum while for compound interest, it is 10% per annum. The total amount of money invested is Rs. 18000, and it is invested for 3 years. How much interest will be earned in total by Ricky?**

A) Rs. 6132

B) Rs. 6228

C) Rs. 6318

D) Rs. 6000

E) Rs. 6424

**3. The price of a perishable item decreases by 10% every day. Heath bought 20 kg of this item and sold it after 3 days. Still he made a profit of 20% over market price of this item on the day he sold them. If the price of 1 kg of this item was bought for Rs. 120, then at what price all 20 kg of it was sold?**

A) Rs. 2080

B) Rs. 2099.52

C) Rs. 2116.45

D) Rs. 2167.32

E) Rs. 2208.36

**4. P and Q are two points along the flow of a river. R is midpoint of line joining P and Q. Between P and R, speed of river is 2 km/hr. But just after R, speed of river becomes 4 km/hr till Q. The water is flowing from P to Q. The speed of boat in still water is 14 km/hr. If the boat takes 8 hours to cover distance P and Q, then find the distance between P and Q.**

A) 112.67 km

B) 120 km

C) 135.63 km

D) 145.21 km

E) 160 km

**5. Uranium and Thorium are mixed in the ratio 3 : 7 in order to make the mixture ready for nuclear fission. The cost of Uranium is Rs. 30000 per milligram. If the cost of a milligram of mixture is Rs. 25000, then what is the price of a milligram Thorium?**

A) Rs. 20675

B) Rs. 20000

C) Rs. 21672.33

D) Rs. 22857.14

E) Rs. 23168.67

**6. 80 workers are employed to do a work. The work will be finished in 30 days. However, after a few days, 20 workers left the work. After two more days, 20 more workers left. It took 37 days for remaining workers to finish the remaining work. For how many days all 80 workers worked together?**

A) 8

B) 10

C) 12

D) 15

E) 20

**7. In an examination, there are three different sections, A, B and C. Each of the section has 20 questions. A student needs to answer at least 18 questions in each section to clear the exam. In how many ways can he clear the exam?**

A) 3636361

B) 4848481

C) 72727271

D) 9393931

E) 8484841

**8. In a family of a husband, wife and son, son is 28 years younger than the father. 3 years back, the ratio of ages of wife and husband was 4 : 5. After 3 years, the ratio of ages of father and son will be 9 : 2. What will be the age of wife after 7 years?**

A) 25 years

B) 28 years

C) 31 years

D) 34 years

E) 40 years

**9. A metallic sphere with radius 10 cm is molten and recast to form a hollow cube. The side length of the cube formed is 20 cm. Find the volume of hollow space inside the cube (in cubic cm). (Take π = 3.14)**

A) 1906.66

B) 3813.33

C) 4705.12

D) 9410.24

E) 2352.61

**10. The transportation cost of one kg of flowers is 15% of the cost/kg of flowers itself. The cost of flowers decreases by 30%, while the total cost of transportation owing to hike in diesel prices, goes up by 45%. What is the percentage change in the total cost, if the total cost is the sum of cost of flowers and transportation cost (approx).**

A) 10.2%

B) 20.2%

C) 25.3%

D) 30.1%

E) 15.7%

**Solutions:**

**1. C)**Let marked price of geyser be Rs. M.

Discount decreases by 1%
every day.

The discount offered on
the first day is 30%. So, discount offered on 23

^{rd}day will be (30 – 22)%, i.e., 8%, and discount offered on 7^{th}day will be (30 – 6)%, i.e., 24%.
We know, Selling Price =
Marked price × (1 – (Discount Percentage)/100)

⇒ Selling price on 23

^{rd}day = M × (1 – 8/100) = 0.92M
And, Selling price on 7

^{th}day = M × (1 – 24/100) = 0.76M
⇒ 0.92M – 0.76M = 1360

⇒ M = 1360/0.16 = 8500

∴ Marked price of geyser is Rs. 8500.

**2. A)**For every rupee that Ricky invests in simple interest, he also invests 2 rupees in compound interest.

Suppose he invests Rs. T
in simple interest and Rs. 2T in compound interest.

Total amount invested is
Rs. 18000.

⇒ T + 2T = 18000

⇒ T = 18000/3 = 6000

⇒ Amount invested in simple interest is Rs. 6000 and in
compound interest is Rs. 12000.

The rate of interest in
case of simple interest is 12% per annum while for compound interest, it is 10%
per annum.

We know, Simple interest
= (Principal × Rate × Time)/100

And, Compound Interest =
Principal × [(1 + R/100)

^{T}– 1]
⇒ Total interest earned in 3 years = (6000 × 12 × 3)/100 + 12000 × [(1+10/100)

^{3}– 1]
⇒ Total interest earned in 3 years = 2160 + 12000 × 0.331 = 2160 + 3972 = 6132

∴ Total interest earned is Rs. 6132.

**3. B)**It is given that 1 kg of item was bought for Rs. 120. Price decreases by 10% every day.

After 3 days, market price
of 1 kg of item = Rs. 120 × (1-10/100) × (1-10/100) × (1-10/100) = Rs. 87.48

If Heath made a profit of
20% over market price on the day he sold,

Selling Price = Cost
price × (1 + (Profit %)/100)

⇒ Selling price = 87.48 × (1 + 20/100) = 87.48 × 1.2 = 104.976

⇒ 1 kg of item was sold for Rs. 104.976.

∴ Price for which 20 kg of item was sold = Rs. 104.976 × 20 = Rs. 2099.52

**4. C)**Let the distance between P and Q be T km.

Since river flows from P
to Q, going from P to Q means going downstream.

Downstream speed between
P and R = (14 + 2) km/hr = 16 km/hr

Downstream speed between
R and Q = (14 + 4) km/hr = 18 km/hr

Since R is midpoint of P
and Q, distance from P to R will be T/2 km and between R and Q will be T/2 km.

Time taken to go from P
to Q = (T/2)/16 + (T/2)/18 = 8

⇒ T/32 + T/36 = 8

⇒ 17T/288 = 8

⇒ T = 2304/17 = 135.53

∴ Distance between P and Q is 135.63 km.

**5. D)**Let the price of a milligram of Thorium be Rs. T.

Let amounts of Uranium
and Thorium mixed be 3D milligram and 7D milligram, respectively.

Price of resultant
mixture = [Price of 3D milligram of Uranium + Price of 7D milligram of
Thorium]/(3D + 7D)

⇒ 25000 = [3D × 30000 + 7D × T]/10D

⇒ 250000 = 90000 + 7T

⇒ T = 160000/7 = 22857.14

∴ Price of a milligram of Thorium is Rs. 22857.14.

**6. B)**Let us suppose that 80 workers worked for N days.

⇒ Workers for the first N days = 80

Workers for next 2 days =
80 – 20 = 60

Workers from next 37 days
= 60 -20 = 40

80 workers can finish
work in 30 days.

Number of man days
required to finish work = 80 × 30 = 2400

⇒ 80 × N + 60 × 2 + 40 × 37 = 2400

⇒ 80N = 2400 – 1480 – 120 = 800

⇒ N = 10

∴ For 10 days, 80 workers worked.

**7. D)**As per given information,

A student needs to answer
at least 18 out of 20 questions in each section to clear the exam

Consider section A,
student can answer 18, 19 or 20 questions.

Number of ways in which
section A can be numbered =

^{20}C_{18 }+^{20}C_{19}+^{20}C_{20}
= 190 + 20 + 1 = 211

Similar is the case with
sections B and C.

⇒ Number of ways in which section B can be answered = 211

Number of ways in
which section C can be answered= 211

∴ Total number of ways in which this can be done = 211 × 211 × 211 = 9393931.

**8. D)**Let the age of wife be T years, and that of father be M years.

⇒ Age of son = (M – 28) years

After 3 years, the ratio
of ages of father and son will be 9:2.

⇒ (M+3)/(M – 28 + 3) = 9/2

⇒ 2M + 6 = 9M – 225

⇒ M = 231/7 = 33

3 years back, the ratio
of ages of wife and husband was 4:5.

⇒ (T – 3)/(M – 3) = 4/5

⇒ T – 3 = (30 × 4)/5 = 24

⇒ T = 27

Present age of wife is 27
years.

∴ After 7 years, age of wife will be 34 years.

**9. B)**Radius of sphere = 10 cm

Volume of sphere = (4/3)
π (cube of radius) = (4/3) × 3.14 × 10 × 10 × 10 = 4186.67 cubic cm

If this is used to form a
hollow cube of side length 20 cm,

Volume of cube formed =
Volume of sphere + Volume of hollow space inside cube

Cube of side length = 4186.67
+ Volume of hollow space inside cube

20 × 20 × 20 = 4186.67 +
Volume of hollow space inside cube

Volume of hollow space
inside cube = 8000 – 4186.67 = 3813.33 cubic cm

**10. B)**Let cost of flowers be Rs 100/kg,

Transportation cost = Rs
15 (i.e. 15% of 100)

∴ Total cost earlier = Rs (100 + 15) = Rs 115

Now, according to the
question:

New cost of flowers = 70%
of 100 = Rs 70/kg

New cost of
transportation = 145% of 15 = Rs 21.75

∴ Total new cost = Rs (70 + 21.75) = Rs 91.75

Percent change (decrease)
in total cost = (115 – 91.75)/115 x 100% = 20.2%

Hence, the required change
in the total cost price is 20.2%