**1).The incomes of A and B are in the ratio of 2:3 and their expenditures are in the ratio of 1:3. If each of them saves RS.12000, find A’s income.**

a) RS.20000

b) RS.24000

c) RS.8000

d) RS.16000

e) RS.12000

**2).The length of a tree increases at the rate of 5% every year. If it is 20 feet high now, how much will it increase after three years?**

a) 23.1525 ft

b) 4.1525 ft

c) 3.1525 ft

d) 6.2515 ft

e) None of these

**3).If A exceeds B by 80% and B is less than C by 20%, find the ratio of A to C.**

a) 36:25

b) 25:29

c) 25:36

d) 29:36

e) 36.29

4).A man sells two articles for the same price-one on 25% gain and the other on 20% gain. If the first article costs RS.8000 then what will be the cost price of the second article?

4).A man sells two articles for the same price-one on 25% gain and the other on 20% gain. If the first article costs RS.8000 then what will be the cost price of the second article?

a) RS.7330.63

b) RS.8333.33

c) RS.6333.33

d) RS.9333.33

e) RS.9000.53

**5).The difference between a 40% discount on RS.1000 and successive discounts of 30% and 5% on the same amount is**

a) RS. 31

b) RS. 32

c) RS. 30

d) RS. 40

e) RS. 65

**6).A man takes a loan of RS.20000 partly from a bank at 8% pa and the remaining from another bank at 10% pa. He pays a total interest of RS.1900 per annum. The amount of loan taken from the first bank is**

a) RS.7000

b) RS.10000

c) RS.15000

d) RS.7500

e) RS.5000

**7).If the interest is calculated half-yearly, the compound interest on RS.18000 at10% per annum at the end of 1(1/2) years is**

a) RS.2700

b) RS.2037.5

c) RS.2043.7

d) RS.2837.5

e) RS.2187.5

**8).30 men working 16 hours per day make a dam in 20days. In how many days will 24 men working 20 hours per day make it?**

a) 5 days

b) 10 days

c) 20 days

d) 25 days

e) 40 days

**9).There are two leakages in the bottom of a cistern. Both together empty the cistern in 12 hours. If the first leakage alone empties it in 30 hours, then in how many hours will the second leakage alone empty it?**

a) 20 hours

b) 25 hours

c) 30 hours

d) 15 hours

e) 12 hours

10).The diameter of a wheel of a vehicle is 56 cm. It completes 1000 rotations in 1 minute. Find the speed (in m/s) of the vehicle.

10).The diameter of a wheel of a vehicle is 56 cm. It completes 1000 rotations in 1 minute. Find the speed (in m/s) of the vehicle.

a) 29.33 m/s

b) 59.33 m/s

c) 58.66 m/s

d) 70 m/s

e) 29.33 m/s

Answers:

Solutions

1). Let the incomes of A and B be 2x and 3x respectively. Each of them saves RS.12000.

Now, (2x-12000)/ (3x-12000) =1/3

Or, 6x-36000=3x-12000 or, 3x=24000

∴x=RS.8000

Hence income of A=2x=8000×2= RS.16000

Answer: d)

2). Height of the tree now=20 feet Height of the tree after 3 years at the rate of 5% increase per year=20×(105/100)×(105/100)×(105/100) =23.1525 feet

∴ Required increase= (23.1525-20) feet=3.1525 feet

Answer: c)

3). Let C be 100, then B=80

∴A= (80×180)/100=144

∴ Required ratio=144:100=36:25

Answer: a)

4). If two articles bought on different prices, are sold for the same price at % gain or % loss, then

(CP of First article / CP of Second article)

= (100±% gain or loss on the second article / 100±% gain or loss on the first article)

Or, 8000 / x = (100+20) / (100+25) or, 8000/x = 120/125

∴ x= (8000×125) / 120

∴ x=RS.8333.33

Answer: b)

5). CP=RS.1000

SP at 40% discount= (100×60)/100=RS.600

∴ Discount=1000-600=400

SP at two successive discounts of 30% and 5%

=1000×(70/100)×(95/100) =RS.665

Discount=1000-665=335

∴ Required difference=400-335=RS.65

Answer: e)

6). Let the loan taken from the first bank be RS. x.

Loan from the second bank= RS.(20000-x)

Interest of the first bank+ interest of the second bank=1900

Now,(x×8×1) / 100 + [(20000-x)×10×1] / 100=1900

Or, 8x+200000-10x =190000

or, 2x=10000

∴ x=RS.5000

Answer: e)

7). P=RS.18000

R=10% pa=(10/2)% half-yearly

∴r=5% half-yearly

Time will be double when interest is calculated half-yearly, ie 1(1/2) years × 2 = 3 years

∴Amount=18000(I+(r/100))t

=18000(1+[5/100)]3

=18000×(105/100)×(105/100)×(105/100)

=RS.20837.25

∴CI=20837.25-18000=RS.2837.5

Answer: d)

8). M1D1H1= M2D2H2

Now, 30×16×20=24×D×20

Or, D= (30×16×20) / (20×24)

∴ D =20 days

Answer: c)

9). Required time= (Product of both times / Difference of both times)

= (12×30) / (30-12)

= (12×30)/18 =20 hours

Answer: a)

10). Circumference of the wheel=2πr

=2× (22/7)× 28cm =176 cm

The distance covered by the wheel in 1 minute= (176×1000) cm =176000cm

Speed of the wheel=176000 / (60×100) m/s

≈29.33 m/s

Answer: c)