### Find solutions for all the questions at the end of each set:

### Set A

1. A tradesman sold an article at a loss of 20%. If the selling price had been increased by Rs. 100, there would have been a gain of 5%. The cost price of the article was?

(A) Rs. 200

(B) Rs. 25

(C) Rs. 400

(D) Rs. 250

2. A man took a loan from a bank at the rate of 12% per annum at simple interest. After 3 years he had to pay Rs. 5,400 as interest only for the period. The principal amount borrowed by him was

(A) Rs. 2,000

(B) Rs. 10,000

(C) Rs. 20, 000

(D) Rs. 15,000

3. A sum of money at simple interest amounts to Rs. 1,012 in 2(1/2) years and to Rs. 1,067.20 in 4 years. The rate of interest per annum is

(A) 2.5%

(B) 3%

(C) 4%

(D) 5%

4. The value of √32 – √128 + √50 Correct to 3 places of decimals is

(A) 1.732

(B) 1.141

(C) 1.414

(D) 1.441

5. A man can row at 5 km/hr in still water. If the velocity of current is 1 km/hr and it takes him 1 hour to row to a place and come back, how far is the place?

(A) 2.5 km

(B) 3 km

(C) 2.4 km

(D) 3.6 km

6. A trader marked his goods at 20% above the cost price. He sold half the stock at the marked price, one quarter at a discount of 20% on the marked price and the rest at a discount of 40% on the marked price. His total gain is

(A) 2%

(B) 4.5%

(C) 13.5%

(D) 15%

7. A man had a certain amount with him. He spent 25% of that to buy an article and 10% of the remaining on clothes. Then he donated Rs. 531.25. If he is left with Rs. 8,000, the amount he spent on the clothes is

(A) Rs. 6562.50

(B) Rs. 2625

(C) Rs. 948

(D) Rs. 1968.75

8. A tap can fill a cistern in 40 minutes and a second tap can empty the filled cistern in 60 minutes. By mistake without closing the second tap, the first tap was opened. In how many minutes will the empty tank be filled?

(A) 72

(B) 84

(C) 108

(D) 120

9. A man goes from Mysore to Bangalore at a uniform speed of 40 km/hr and comes back to Mysore at a uniform speed of 60 km/hr. His average speed for the whole journey is :

(A) 48 km/hr

(B) 50 km/hr

(C) 54 km/hr

(D) 55 km/hr

10. A boat goes 12 km downstream and comes back to the starting point in 3 hours. If the speed of the current is 3 km/hr, then the speed (in km/hr) of the boat in still water is

(A) 12

(B) 9

(C) 8

(D) 6

Solutions

1. (c); Let x = initial C.P. of the circle.

C.P. S.P.

1st condition x 20% loss 0.8x

2nd condition x 5% profit 0.8x + 100

=1.05x

⇒ 1.05x = 0.8x + 100

⇒ x = 100/0.25 = 400

2. (d); Total S.I. = (3 × 12)% of the principal amount

= 36% of the principal amount

= Rs. 5400

So, the principle amount (i.e. 100%)

= Rs. (5400/36) ×100

= Rs. 15000

3. (c); S.I. in 4 years – S.I. in 2.5 years

= S.I. in 1.5 years

⇒ S.I. in 1.5 years = Rs. (1067.20 – 1020) = Rs. 55.2

⇒ S.I. Per year = Rs. (55.2/1.5) = Rs. 36.8

So, Principle money = Rs. (1067.2 – Rs.36.8 × 4) = Rs. 920

So, Rate of interest per annum = (36.8/920) × 100% = 4%

4. (c); √32 – √128 + √50

= √(2 × 16) – √(2 × 64) + √(2 × 25)

= √2 = 1.414

5. (c); Required distance = (Average speed × Total Time)/2

= {(2 × (5 + 1)(5 – 1))/((5 + 1) + (5 – 1)) ×1}/2] km

= 4.8/2 km = 2.4 km

6. (a); Total % discount on M.P.

= (½× 0 + ¼ × 20 + ¼ × 40)%

=15%

Now,

If C.P. = x

Then, M.P. = 20% above C.P.

= 1.2x

⇒ M.P. after discount = 8.5% of 1.2x

= 1.02x

So,

Total gain% = [(1.02x – x)/x] × 100 %

=2%

7. (c); Money spent on article = 25% of total amount

Money spent on cloths = 10% of remaining (75%) amount

= 7.5% of total amount

⇒ (25% + 7.5%) of amount + Rs. 531.25 = Total amount – Rs. 8000

⇒ Total (100%) amount – 32.5% of total amount = Rs. 8000 + 531.25

= Rs. 8531.25

⇒ 67.5% of the total amount = Rs. 8531.25

So,

Money spent on clothes

i.e. 7.5% of the total amount = (8531.25/67.5)×7.5= Rs. 948

8. (d); Required time = (60 × 40)/(60 – 40) minutes

= 120 minutes

9. (a); Average Speed = (2 × 40 × 60)/ (40 + 60) km/hr

= 48 km/hr

10. (b) total distance covered=24km

Total time taken= 3 hours

Avg. speed=24/3=8km/hr

so,Speed in still water=2*(x+3)(x-3)/(x+3)+(x-3)=9 km/hr

### Find solutions for all the questions at the end of each set:

### Set B

1. Each edge of a cube is increased by 40%. What is the percentage increase in its volume?

(a) 120%

(b) 40%

(c) 174.4%

(d) 146%

2. The volume of a height circular cone is 1232cm3 and its vertical height is 24 cm. Its curved surface area is –

(a) 154 cm2

(b) 550 cm2

(c) 604 cm2

(d) 704 cm2

3. A circle and a rectangle have the same perimeter. The sides of the rectangle are 26 cm and 18 cm. The area of the circle is [take π = 22/7]

(a) 125 cm2

(b) 230 cm2

(c) 550 cm2

(d) 616 cm2

4. From a point in the interior of an equilateral triangle, perpendiculars draw to the three sides are 16 cm, 25 cm & 28 cm respectively. Find the area of the triangle.

(a) 1587√3 cm2

(b) 1587 cm2

(c) 2116√3 cm2

(d) 2116 cm2

5. If a solid cone of volume 27 π cm3 is kept inside a hollow cylinder whose radius and height are that of the cone, then the volume of water needed to fill the empty space is –

(a) 3 π cm3

(b) 18 π cm3

(c) 54 π cm3

(d) 81 π cm3

6. A right angled isosceles triangle is inscribed in a semi-circle of radius 7 cm. The area enclosed by the semi-circle but exterior to the triangle is –

(a) 14 cm2

(b) 28 cm2

(c) 44 cm2

(d) 68 cm2

7. The lengths of three medians of a triangle are 9cm, 12 cm and 15 cm. The area (in sq. cm) of the triangle is –

(a) 24

(b) 72

(c) 48

(d) 144

8. The height of a right prism with a square base is 15 cm. If the area of the total surfaces of the prism is 608 sq. cm., its volume is –

(a) 910 cm3

(b) 920 cm3

(c) 960 cm3

(d) 980 cm3

9. The area of a circle is increased by 22 cm2 when its radius is increased by 1 cm. The original radius of the circle is –

(a) 3 cm

(b) 5 cm

(c) 7 cm

(d) 9 cm

10. 2 cm of rain has fallen on a square km of land. Assuming that 50% of the raindrops could have been collected and contained in a pool having a 100 m x 10 m base, by what level would the water level in the pool have increased?

(a) 1 km

(b) 10 m

(c) 10 cm

(d) 1 m

Solutions

1. (c); Volume of cube = a3 (where a = length of a edge)

When each edge is increased by 40%

⇒ Length of the new edge = 1.4a

⇒ Volume of new cube = (1.4a)3 = 2.744a3

⇒ Required % increase = [(2.744a3 – a3)/a3] × 100%

= 174.4%

2. (b); Volume of cone = 1232 cm3

⇒ 1/3 πr2h = 1232 cm3

R=7 cm

l = √(h2 + r2)

= 25 cm

So, Curved surface area of cone = πrl

= 22/7 × 7 × 25 cm2 = 550 cm2

3. (d); Perimeter of the rectangle = 2 × (26 + 18) cm= 88 cm

=2πr ⇒ r = (88 × 7)/(22 × 2) = 14 cm

⇒ Area of the circle i.e. πr2

= 22/7 × 14 × 14 = 616 cm2

4. (a); Area of ∆ABC = Area of (∆AOB + ∆AOC + ∆BOC)

⇒ (√3/4) a2 = ½ × a (16 + 25 + 28)

⇒ a = 46√3 cm

So, Area of ∆ABC = (√3/4) × (46√3)2

= 1587√3 cm3

5. (c); Volume of water needed to fill the empty space

= Volume of cylinder – Volume of cone

= πr2h – 1/3 πr2h

= 2 × (1/3 πr2h)

= 2 × 27π cm3

= 54π cm3

6. (b); In right angled isosceles ABC, AB = BC

Also,

AD = CD = BD = 7cm = circumradius (where BD⊥AC)

Now, Required area

= Area of semicircle – Area of ∆ABC

= ((22 × 7 × 7)/(7 × 2) – ½ × 14 × 7) cm2

= 28 cm2

7. (b); In the given ∆ABC,

AD, BE and CF are medians and they cut one another at G.

⇒ AG/GD = BG/GE = CG/GF = 2/1

Here,

AD = 9 cm, BE = 12 cm and CF = 15 cm

∵ AG + GD = AD = 9 cm

⇒ AG = 6 cm and GD = 3 cm

Also,

BG + GE = BE = 12 cm

⇒ BG = 8 cm and GE = 4 cm

Also,

CG + GF = CF = 15 cm

⇒ CG = 10 cm and GF = 5 cm

⇒ Area of ∆ AGB = ½ × 6 × 8 = 24 cm2

So, Area of ∆ABC = 3 × ∆AGB

= 2 × 24 cm2

= 72 cm2

8. (c); Total surface area of prism = C.S.A. + 2 × Area of base

⇒ 608 = Perimeter of base × height + 2 × Area of base

⇒ 608 = 4x × 15 + 2 × 2 (where x = side of square)

⇒ x3 + 30x – 304 = 0

⇒ x = 8

⇒ Volume of prism = Area of base × height

= 8 × 8 × 15

= 960 cm3

9. (a); π(r+1)2 – πr2 = 22

⇒ 2πr + π = 22

⇒ 22/7 *(2r + 1) = 22

⇒ r = 3 cm

10. (b); Volume of water due to 2 cm rain on a square km land = 1 km × 1 km × 2 km

= 1000m + 1000m × (2/100) m

= 20000 m3

⇒ 50% of volume of rain drops

= 10000 m3

Now, Required level by which the water level in the pool will be increased = (10000 m3)/(100m × 10m) = 10m

### Find solutions for all the questions at the end of each set:

### Set C

1. Two men Ashwin and Brandon started a job in which Ashwin was thrice as good as Brandon and therefore took 60 days less than Brandon to finish the job. How many days will they take to finish the job, if they start working together?

(1) 15 days

(2) 20 days

(3) 22(1/2) days

(4) 25 days

2. A rectangular Buddha garden is 100 m × 80 m. There is a path along the garden and just outside it. Width of the path is 10m. The area of the path is

(1) 1900 sq m

(2) 2400 sq m

(3) 3660 sq m

(4) 4000 sq m

3. A dealer offered a LG Washing machine for sale for Rs. 27,500 but even if he had charged 10% less, he would have made a profit of 10%. The actual cost of the Washing machine is

(1) Rs. 22,000

(2) Rs. 24,250

(3) Rs. 22,500

(4) Rs. 22,275

4. In a Idea Telecom company an employer reduces the number of employees in the ratio 8 : 5 and increases their wages in the ratio 7 : 9. As a result, the overall wages bill is

(1) Increased in the ratio 56 : 69

(2) Decreased in the ratio 56 :45

(3) Increased in the ratio 13 : 17

(4) Decreased in the ratio 17 : 13

5. The average age of a jury of 5 is 40. If a member aged 35 resigns and a man aged 25 becomes a member, then the average age of the new jury is

(1) 30

(2) 38

(3) 40

(4) 42

6. With average speed of 40 km/hour, a train reaches its’ destination in time. If it goes with an average speed of 35 km/ hour, it is late by 15 minutes. The total journey is

(1) 30 km

(2) 40 km

(3) 70 km

(4) 80 km

7. Virat Kohli makes a profit of 20% on the sale by selling 20 articles fort Rs. 1. The number of articles he bought by Rs. 1 is

(1) 20

(2) 24

(3) 25

(4) 30

8. The number of seats in a PVR auditorium is increased by 25%. The price of a ticket is also increased by 12%. Then the increase in revenue collection will be

(1) 40%

(2) 35%

(3) 45%

(4) 48%

9. A Water ship is moving at a speed of 30 km/hr. To know the depth of the ocean beneath it, it sends a radio wave which travels at a speed 200 m/s. The ship receives the signal after it has moved 500 m. The depth of the ocean is (approx..)

(1)6 km

(2) 12 km

(3) 4 km

(4) 8 km

10. Vijay Mallya takes a loan of Rs. 10,000 partly from a bank at 8% p.a. and remaining from another bank at 10% p.a. He pays a total interest of Rs. 950 per annum. Amount of loan taken from the first bank (in Rs.) is

(1) 2500

(2) 5200

(3) 2050

(4) 5020

Solutions:

1. (c)If time taken by Ashwin be x days.

Then time taken by Brandon = 3x days

∴ 3x – x = 60

2x = 60

x = 30

Time taken by Brandon = 90 days

∴ (Ashwin + Brandon)’s 1 day’s work

(1/30+1/90=2/45)

∴ The work will be completed

In (45/2) i.e. 22 (1/2)days

2. (d)

Area of the shaded region

= (100 + 2 ×10)(80 + 2 × 10) – 100 × 80

= 120 × 100 – 8000

= 4000 sq. metre

3. (c)If the C.P. of Washing machine by Rs. x, then

(11x/10)=275*90

X=22500

4. (b) Required ratio = 8 × 7 : 5 × 9

= 56 : 45

5. (b) Required average=(40*5-35+25)/5=190/5 = 38 days

6. (c) If the total length of journey be x km, then

(x/35-x/40)=15/60

x= 70km

7. (b) C.P. of 20 articles

=100/120=5/6 Rs.

∴ Number of articles bought for Re. 1.=(6/5)*20=24

8. (a) Required increase=(25+12+(25*12/100))%

= 40%

9. (a)Speed of water ship = 30 kmph

= 30*5/18 m/sec. = 25/3 m/sec.

Time taken in covering 500 metre

=500*3/25= 60 seconds

Speed of radio waves

= 200/1000 km/sec. = 1/5 km/sec.

∴ (12 – x)2 = x2 + 1/4

x =575/(4*24)=6 km

10. (a) If the amount of loan taken from the first bank be Rs. x, then

(x*8*1/100)+((10000-x)*10)/100= 950

x = Rs. 2500