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

### Set A

1.If the total area under bajra was three hundred acres, then the total area (in hundred acres) under Rice and Barley together is.

(a) 18

(b) 12

(c) 15

(d) 13

2.The combination of three crops which contribute to more than 50% of the total area under the good crops is

(a) wheat, rice and maize

(b) wheat, rice and jowar

(c) wheat, rice and bajra

(d)rice, barley and maize.

3.The ratio of the land used for rice and barley is –

(a) 3 : 2

(b) 1 : 2

(c) 2 : 1

(d) 3 : 1

4.If 10% of the land reserved for rice be distributed to wheat and barley in the ratio 2 : 1, then the angle corresponding to wheat in the new pie-chart will be-

(a) 38.40

(b) 76.80

(c) 75.60

(d) 45.50

5.If the production of rice is 5 times that of Jowar and the production of Jowar is 2 times that of Bajra, then the ratio between the yield per acre of rice and Bajra is –

(a) 5 : 2

(b) 3 : 1

(c) 4 : 1

(d) 6 : 1

Directions (06-10): In the following table the number of students passed in different subjects is given.

6.What per cent of students passed in Science in 2001?

(a) 23 1⁄3 %

(b) 21 1⁄9 %

(c) 22 1⁄3 %

(d) 27 1⁄9 %

7.What is the average Commerce students passed in 2001-2005?

(a) 101

(b) 103

(c) 105

(d) 106

8.In which year percentage of students passed in Science is more than the total percentage of passed in Art and Commerce together?

(a) 2001 and 2003

(b) 2003

(c) 2001 and 2005

(d) 2003 and 2005

9.What is the ratio of total students and total passed students in 2003?

(a) 40 : 21

(b) 41 : 21

(c) 44 : 23

(d) 44 : 21

10.In which year minimum percentage of students failed?

(a) 2001

(b) 2002

(c) 2003

(d) 2004

1.(a) 180 area under Bajra = 300 acre

=> 720 + 360 area under Rice & Barley = 300/18×(720+360)

= 300/18×108

= 18 hundred Acres.

2.(a) : It is clear from the pie chart that wheat,

rice and maize contribute to more than 50% of the total area.

3.(c) : Ratio of the land used for rice & barley

= 720 : 360

= 720/360 : 360/3600

= 2 : 1

4.(b) : 10% of land reserved for rice =

10% of 720 = 10/100× 720 = 7.20

7.20 is distributed among wheat & barley in ratio = 2 : 1

2n + n = 7.20

3n = 7.20

n =2.40

distribution of land among wheat & barley will be 4.80 and 2.40 resp.

∴ The corresponding angle to wheat in new pie chart will be 72 + 4.80. = 76.80

5.(a) : Let the total production of Bajra be T tones and let Z acres of land

be put under Bajra production.

Then, the total production of Bajra =(T) tones.

Production of Jowar = 2T

And Production of Rice = 10T

Then, Area under Rice production = 4Z (Area Under Rice Production /Area Under Bajra Production

=> 72/18 = 4)

And therefore,Area under Rice = 4*Area under Bajra = (4Z)acres

Now, yield per acre for Rice = (10T/4Z) tones/acre = (5T/2Z) tones/acre

And yield per acre for Bajra = (T/Z) tones/acre

Require ratio = 5T/2Z/T/Z = 5 : 2

6.(a)

7.(d);

Average = (115+100+110+121+84)= 530/5= 106

8.(c)

9.(b);

Ratio = (820/420) = 41 : 21

10.(c);

48.78% failed.

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

### Set B

1.What is the least number of square tiles required to pave the floor of a room 9 m 99 cm long and 4 m 7 cm broad?

(a) 247

(b) 277

(c) 297

(d) 307

2.A person bought 5 tickets from a station P to a station Q and 10 tickets from the station P to a station R. He paid Rs. 350/-. If the sum of a ticket from P to Q and a ticket from P to R is Rs. 42/- then what is the fare from P to Q?

(a) Rs. 12/-

(a) Rs. 14/-

(a) Rs. 16/-

(a) Rs. 18/-

3.A trader marked a watch 40% above the cost price and then gave a discount of 10%. He made a net profit of Rs. 468/- after paying a tax of 10% on the gross profit. What is the cost price of the watch?

(a) Rs. 1,200/-

(a) Rs. 1,800/-

(a) Rs. 2, 000/-

(a) Rs. 2,340/-

4.The sum of two number is 80. If the larger number exceeds four times the smaller by 5, what is the smaller number?

(a) 5

(b) 15

(c) 20

(d) 25

5.If salary of X is 20% more than salary of y, then by how much percentage is salary of y less than X?

(a) 25

(b) 20

(c) 50/3

(d) 65/4

6.Two vessels are full of milk with water ratio 1 : 3 and 3 : 5 respectively. If both are mixed in the ratio 3 : 2, what is the ratio of milk and water in the new mixture?

(a) 4 : 15

(b)3 : 7

(c) 6 : 7

(d) 3 : 10

7.The ratio between the ages of A and B is 2 : 5. After 8 year, their ages will be in the ratio 1 : 2. What is the difference between their present ages?

(a) 20 year

(b) 22 year

(c) 24 year

(d) 25 year

8.A person borrowed Rs. 7, 500/- at 16% compound interest. How much does he have to pay at the end of 2 years to clear the loan?

(a) Rs. 9, 900/-

(a) Rs. 10, 092/-

(a) Rs. 10, 902/-

(a) Rs. 10, 920/-

9.A can do a piece of work in 24 days. If B is 60% more efficient than A. then how many days does B require to do the same work?

(a) 12

(b) 15

(c) 16

(d) 18

10. The dimensions of a rectangular box are in the ratio 2 : 3 : 4 and the difference between the cost of covering it with a sheet of some metal at Rs. 8/- and Rs. 10/- per sq m is Rs. 234/-. What are the length, breadth and height of the box respectively?

(a) 2 m, 3 m, 4 m

(b) 3 m, 4.5 m, 6 m

(c) 4 m, 6 m, 8 m

(d) 5 m, 7.5 m, 10 m

ANSWERS AND SOLUTION

1.(c); Side of tiles = HCF of 999 cm and 407 cm

= 37

so Required no. of tiles = (999 * 407)/ (37 * 37) = 297

2.(b); Let the fare from p to Q = Rs. X

And the fare form P to R = Rs. Y

According to question,

5x + 10y = 350 …….(i)

x + y = 42 ……………..(ii)

Solving (i) and (ii), we get

x = Rs. 14/- y = Rs. 28/-

3.(c) lest cost price =x

marked price = x * 140/100 = 14x/10

Selling price = 14x/10 * 90/100

= 126x/100

profit= sp-cp = 126x/100 – x= 26x/100

Net profit = 26x/100 * 90/100

so 468 = (26*90)/(100*100)x

x = 2000

4.(b) Let smaller number = x

so Larger number = 80 – x

Accroding to question,

80 – x = 4x + 5

5x = 75

=> X = 15

5.(c); Let salary of x = Rs.x

Salary of y = x*100/120

= Rs. 10x/12

Salary of X – salary of Y = x – 10x/12 = 2x/12 = x/6

so Required salary = x/6/x * 100 = 50/3%

6.(d)

By law of mixture

1/4 3/8

x

3 2

so (3/8 -x)/(x-1/4) = 3/2

3/4-2x=3x-3/4

5x=6/4 = 3/2 =>x = 3/10

7.(c) Let A’s age = 2x

so B’s age = 5x

According to question,

(2x + 8)/(5x + 8) = 1/2

4x + 16 = 5x + 8

X = 8

so Required differecne = (5x – 2x)

= 3x

= 3 * 8 = 24 years

8.(b) A = P (1 + r/100)^n

A = 7500 (1 + 16/100)^2

= 7500 * 29/25 * 29/25

= Rs. 10,092

9.(b) A’s one day work = 1/24

so B’s one day work = 1/24 * 160/100 = 1/15

so B’s can do a piece of work in 15 days.

10.(b) ; Let the dimensions by 2x, 3x and 4x

According to question

2(2x * 3x + 3x * 4x + 4x * 2x)(10 – 8) = 234

=> 4 * 26x^2 = 234

=> X^2 = 234/104 = 2.35

X = 1.5

Dimensions are 3m, 4.5 m and 6 m.

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

### Set C

1. A single discount, equivalent to the successive discounts of 30%, 20% and 10%, is

(a) 60%

(b) 49.60%

(c) 47.40%

(d) 40%

2. A work could be completed by certain workers in 22 days. However, due to 3 workers being absent, it was completed in 24 days. The original number of workers was

(a) 33

(b) 18

(c) 36

(d) 25

3. A and B together can do a piece of work in 18 days, B and C in 24 days, and A and C in 36 days. The number of days, all of them working together take to complete the work, is

(a) 18

(b) 16

(c) 17

(d) 15

4. A ladder 20 m long is placed in street so as to reach a window 16 m high. On turning the ladder on the other side of the street, it reaches a point 12 m high. The width of the street is

(a) 28.5 m

(b) 28 m

(c) 27 m

(d) 32 m

5. A person buys certain number of marbles at 20 per rupee are equal number of 30 per rupee. He mixes them and sells them at 25 per rupee. His gain or loss in the transaction is

(a) 2% loss

(b) 4% loss

(c) 2% gain

(d) 4% gain

6. A horse and a cow were sold for Rs. 12,000 each. This gave a loss of 20% on horse and a gain of 20% on cow. The entire transaction resulted in

(a) no loss no gain

(b) loss of Rs. 1,000

(c) gain of Rs. 10,000

(d) loss of Rs. 100

7. When an article is sold for Rs. 116, the profit percent is thrice as much as when it is sold for Rs. 92. The cost price of the article is

(a) Rs. 68

(b) Rs. 72

(c) Rs. 78

(d) Rs. 80

8. If the total cost of 73 articles having equal cost is Rs. 5,110 and the total selling price of 89 such articles if Rs. 5,607, then in the transaction, there will be

(a) a loss of 15%

(b) a gain of 10%

(c) a loss of 10%

(d) a gain of 15%

9. By selling an articles for Rs. 665, there is a loss of 5%. In order to make a profit of 12%, the selling price of the article must be

(a) Rs. 812

(b) Rs. 800

(c) Rs. 790

(d) Rs. 784

10. Nitin’s salary was reduced by 10% and then the reduced salary was increased by 10%. His new salary in comparison with his original salary is

(a) the same

(b) 1% more

(c) 1% less

(d) 5% less

Solutions

1. (b); Required single discount = (1 – 0.7 × 0.8 × 0.9) × 100%

= 49.60%

2. (c); Let x = original number of workers

According to question

22x = 24 (x – 3)

or, 22x = 24x – 72

⇒ x = 36

3. (b); The number of days taken by A, B and C to complete the work while working together

= ((L.C.M. of 18, 24 & 36) × 2/(L.C.M./18 + L.C.M./24 + L.C.M./36) ) days

= (72 × 2/(72/18 + 72/24 + 72/36) ) days

= 16 days

4. (b); In the given figure, width of the street = (x + y)

=√(202 –162) +√(202 – 122 )

= 28 m

5. (b); cost price of each marble @ 20 per rupees = Rs. 1/20

Cost price of each marble @ 30 per rupee = Rs. 1/30

⇒ Average cost price of each marble = Rs. (1/20 + 1/30)/2

[∴ no. of marbles are equal]

= Rs. 1/24

According to the question

Selling price of each marble @ 25 per rupee = Rs. 1/25

[∵ 1/25 < 1/24 ∵ S.P. < C.P. ⇒ loss ]

So, % loss = ((S.P. – C.P.) ×100)/ C.P.

= ((1/25 – 1/24) × 100%)/(1/24)

= – 4%

⇒ 4% loss

6. (b); Loss of 20% on one and gain of 20% on other.

⇒ there will be a loss (always)

and loss% = (20)2/100% = 4%loss (on total C.P.)

⇒ C.P. (100%) –4%_ S.P. (96%)

= 12,000 × 2 = 24000

So, loss in transaction (i.e. 4%)

= 24000 × 4/96 = 1000

⇒ loss of Rs. 1000

7. (d); 116 – 92 = 24 Let x = profit when S.P. = 92

i.e. 3x – x = 24 So, 3x = profit when S.P. = 116

⇒ x = 12

⇒ When S.P. = 92, Profit = Rs. 12

So, C.P. = 92 – 12 = Rs. 80

8. (c); C.P. of each article = Rs. 5110/73 = Rs. 70

S.P. of each article = Rs. 5607/89 = Rs. 63

Here, S.P. < C.P. ⇒ loss

and % loss = (S.P. – C.P.) × 100% /C.P.

= (63 – 70) × 100%/70

= – 10%

= 10% loss

9. (d); C.P. (100%) –5% S.P. (95%) = Rs. 665

S.P. at 5% loss i.e. 95% of C.P. = Rs. 665

So, SP at 12% profit i.e. 112% of C.P.

= Rs. 665 × 112/95= Rs. 784

10. (c); Less by ((10)2/100)% = i.e. less by 1%