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1. If a number is chosen at random from the set {1, 2, 3…, 100}, then the probability that the chosen number is a perfect cube is

a. 1/25 b. 1/2 c. 4/13 d. 1/10

2. What is the probability of getting at least one six in a single throw of three unbiased dice?

a. 1/6 b. 125/216 c. 1/36 d. 81/216 e. 91/216

3. In a simultaneous throw of two dice, what is the probability of getting a doublet?

a. 1/6 b. 1/4 c. 2/3 d. 3/7

4. A bag contains 4 red balls, 5 green balls and 6 white balls. A ball is drawn at

random from the box. What is the probability that the ball drawn is either red or green?

a. 2/5 b. 3/5 c. 1/5 d. 7/15

5. When 4 dice are thrown, what is the probability that the same number appears on each of them?

a. 1/36 b. 1/18 c. 1/216 d. 1/5

6. The probability that it is Friday and that a student is absent is 0.03. Since there are 5 school days in a week, the probability that it is Friday is 0.2. What is the probability that a student is absent given that today is Friday?

a. 10% b. 15% c. 12% d. 13%

7. Two dice are rolled. The probability of getting a sum of at least 9 is

a. 13/36 b. 5/18 c. 35/36 d. 11/36

8. If four cards are drawn at random from a well shuffled pack of cards, what is the probability that each card is an ace?

a. 6/52C4 b. 4/52C4 c. 1/52C4 d. 3/52C4

9. A person tosses an unbiased coin. When head turns up, he gets Rs.8 and tail turns up he loses Rs.4. If 3 coins are tossed, what is probability that the gets a profit of Rs.12?

a. 3/8 b. 5/8 c. 3/4 d. 1/8

10. A number n is chosen from {2, 4, 6 … 48}. The probability that ‘n’ satisfies the equation (2x – 6) (3x + 12) (x – 6) (x – 10) = 0 is

a. 1/24 b. 1/12 c. 1/8 d. 1/6

Directions for questions: 11 to 13: These questions are based on the following data.

A box contains 12 mangoes out of which 4 are spoilt. If four mangoes are chosen at random, find the probability that

11. All the four mangoes are spoiled.

a. 1/495 b. 494/495 c. 1/395 d. 394/395

12. Not all the mangoes are spoiled.

a. 1/495 b. 394/395 c. 494/495 d. 1/395

13. Exactly three are not spoiled.

a. 116/495 b. 224/495 c. 129/495 d. 187/495

14. A number is selected at random from first thirty natural numbers. What is the chance that it is a multiple of either 3 or 13?

a.17/30 b. 2/5 c. 11/30 d. 4/15

Directions for questions: 15 to 17: These questions are based on the following data.

If the numbers 1 to 100 are written on 100 pieces of paper, (one on each) and one piece is picked at random, then

15. What is the probability that the number drawn is neither prime nor composite?

a. 1/50 b. 1/25 c. 1/100 d. 1

16. Find the probability that the number drawn is a multiple of 6 and 8.

a. 3/50 b. 2/25 c. 1/50 d. 1/25

17. Find the probability that the number drawn is a factor of 50.

a. 1/25 b. 1/50 c. 3/25 d. 3/50

18. Out of 7 fruits in a basket, 2 are rotten. If two fruits are drawn at random from the basket, the probability of both being rotten is

a. 1/21 b. 10/21 c. 20/21 d. 11/21

19. The probability that a number selected at random from first 50 natural numbers is a composite number is

a. 21/25 b. 17/25 c. 4/25 d. 8/25

20. If six persons sit around a table, the probability that some specified three of them are always together is

a. 1/20 b. 3/10 c. 1/5 d. 4/5

Answer & Explanations

1. Exp: We have 1, 8, 27 and 64 as perfect cubes from 1 to 100. Thus, the probability of picking a perfect cube is 4/100 = 1/25

2. Exp: Find the number of cases in which none of the digits show a ‘6’.

i.e. all three dice show a number other than ‘6’, 5 * 5 *5 = 125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a ‘6’ = Total possible outcomes when three dice are thrown – Number of outcomes in which none of them show ‘6’.

= 216 – 125 = 91.

The required probability = 91/216.

3. Exp: In a simultaneous throw of two dice, n(S) = (6 x 6)= 36.

Let E = event of getting a doublet = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}.

Therefore, P(E) = n (E)/ n(S) = 6/36 = 1/6.

4. Exp: Total number of balls = (4 + 5 + 6) = 15.

Therefore, n(S) = 15.

Let E1 = event of drawing a red ball.

and E2 = event of drawing a green ball.

Then, E1 n E2 = f.

P (E1 n E2) = P(E1) + P(E2) = (4/15 + 5/15) = 9/15 = 3/5.

5. Exp: Sample space (Denominator): When 4 dice are thrown simultaneously, then the total number of possible outcomes is 64 = 1296.

Event (Numerator): The chances that all the dice show same number (1,1,1,1),

(2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)} is 6.

Probability = Event/ Sample space = 6/64 = 1/63 = 1/216.

6. Exp: P(Absent|Friday) = P(Friday and abscent) = 0.03/0.2 = 15%

P(Friday)

7. Exp: Sum of 9 = {(3,6) (6,3)} = 2ways.

Sum of 10 = {(5,5) (6,4) (4,6) (5,5)} = 4 ways.

Sum of 11 = {(6,5) (5,6)} = 2 ways.

Sum of 12 = {(6,6) (6,6)} = 2 ways.

Therefore, Favourable cases = 10

Total cases = 6 x 6 = 36.

Therefore, Probability = 10/36 = 5/18.

8. Exp: Four cards can be drawn from a pack in 52C4 ways.

Let E be the event of each card being an ace.

This can be done in 4C4 i.e, 1 way.

So P(E) = 1/52C4

9. Exp: When a person tosses two heads and one tail, he will get Rs.12. When three coins are tossed, total outcomes = 23 = 8. Favourable out comes i.e, two heads and one tail is = {HHT, HTH, THH} = 3ways. Therefore, required probability = 3/8.

10. Exp: Given, set is {2, 4, 6 … 48}

n(s) = 24

The roots of given equations are 3,4,4,10. The number of chosen from the set are 4, 10, which are the roots of given equation.

n(E) = 2

Therefore, required probability = 2/24 = 1/12.

11. Exp: Out of 12,8 are good and 4 are spoiled.

Required probability = 4C4/12C4 = 1/495.

12. Exp: Required probability = 1 – 1/495 = 494/495.

13. Exp: Required probability = 8C3 . 4C1 = 56 x 4 = 224/495.

12C4 495

14. Exp: The probability that the number is a multiple of 3 is 10/30. (Since 3*10 = 30).

Similarly the probability that the number is a multiple of 13 is 2/30. {Since 13*2 =

26).

Neither 3 nor 13 has common multiple from 1 to 30. Hence these events are mutually exclusive events. Therefore chance that the selected number is a multiple of 3 or 13 is (10+2)/30 = 2/5.

15. Exp: There are 25 primes, 74 composite numbers from 1 to 100. The number

which is neither prime nor composite is 1.

Therefore, required probability = 1/100.

16. Exp: From 1 to 100 there are 4 numbers which are multiples of 6 and 8. (i.e,

multiples of 24)

Therefore, required probability = 4 /100 = 1/25

17. Exp: The factors of 50 are 1, 2, 5, 10, 25, 50

Therefore, required probability = 6/100 = 3/50.

18. Exp: The number of exhaustive events = 7C2 = 21.

Let E be event of the 2 fruits being rotten. The number of favourable cases are

2C2 = 1 way. \ Required probability = 1/21.

19. Exp: The number of exhaustive events = 50C1 = 50. We have 15 primes from 1 to 50.

Number of favourable cases are 34.

Therefore, Required probability = 34/50 = 17/25.

20. Exp: There are six persons and three of them are grouped together. Since it is a circle, this can be done in 3! 3! ways. Therefore, P = 3!3!/5 = 3/10

VProgression-Exercise Questions updated on Apr 2018

23631

1). Find the 15th term of an arithmetic progression whose first term is 2 and the common difference is 3.

a.45

b.38

c.44

d.40

2). Find the number of terms in an arithmetic progression with the first term 2 and the last term being 62, given that common difference is 2.

a.31

b.40

c.22

d.27

3). The first term of an arithmetic progression is 3 and the 10 th term is 21. Find 15 th and 22 nd terms.

a.21,35

b.31,45

c.30,46

d.29,40

4). The 5 th term and 21 st term of a series in A.P are 10 and 42 respectively. Find the 31 st term.

a.50

b.55

c.65

d.62

5). Five times the fifth term of an A.P is equal to six times the sixth term of the A.P, What is the value of the eleventh term?

a.1

b.5

c.0

d.2

6). Find the sum of all 2 digit numbers divisible by 3.

a.2000

b.1665

c.1300

d.1448

7). The sum of n terms of a series in A.P is 6n2 + 6n. What is the 4 th term of the series?

a.38

b.49

c.60

d.48

8). How many terms are there in 2,4,8,16,………..1024?

a.10

b.6

c.9

d.8

9). The sum of the first five terms of a G.P is 363. If the common ratio is 1/3 find the first term.

a.323

b.243

c.232

d.332

10). Find the sum of the following series,

1,4/5,16/25,64/125 ……………..

a.10

b.6

c.9

d.5

11). There are n arithmetic means between 3 and 45 such that the sum of these arithmetic means is 552. find the value of n.

a.11

b.15

c.17

d.23

12). Find the last term of a G.P whose first term is 9 and common ratio is ( 1/3 ) if the sum of the terms of the G.P is ( 40/3 )

a.1/3

b.2/5

c.¼

d.2/3

13). Find the common ratio of three numbers in G.P whose product is 216 and the sum of the products taken in pairs is 114.

a.2 or ½

b.2/3 or 3/2

c.¾ or 4/3

d.4 or ¼

14). In an A.P consisting of 27 terms, the sum of the first three terms is 21 and that of the three middle terms is 93. Find the first term and the common difference.

a.6,3

b.6,23

c.7,3

d.5,2

15). Find the first term and the common ratio of a G.P whose fourth term is 250 and seventh term is 31,250

a.2/5,25

b.4,5/2

c.1,16

d.2,5

16). The sum of the first eight terms of a geometric series is 10,001 times the sum of its four terms . Find the common ratio of these series

a.1,000

b.10

c.10

d.100

17). What is the sum of the first 15 terms of an A.P whose 11 th and 7 th terms are 5.25 and 3.25 respectively

a.56.25

b.60

c.52.5

d.None of these

18). If sum of three numbers in A.P is 33 and sum of their squares id 491, then what are the three numbers.

a.5,11,17

b.7,11,15

c.9,11,13

d.3,11,19

19). Four angles of a quadrilateral are in G.P. Whose common ratio is an intiger. Two of the angles are acute while the other two are obtuse. The measure of the smallest angle of the quadrilateral is

a.12

b.24

c.36

d.48

20). The sum of the terms of an infinite G.P is 7 and the sum of the cubes of the terms is 1,225. Find the first term of the series.

a.35/3

b.35/2

c.15/2

d.9/4

Answer & Explanations

1. Exp. n th term of A.P = a +(n-1) *d

= 2+(15-1)*3 , = 2 + 42 = 44.

2. Exp. The n th term = a +(n-1) *d

62 = 2 +(n- 1) *2, 62 – 2 = (n-1) *2, n = 60/2 +1 = 31.

3. Exp. a = 3, T 10 = a +9d =21, 3 +9d = 21, d =18/9 = 2

T 15 = a + 14d =3 +14*2 = 31, T22 = a + 21d = 3+ 21*2 = 45

4. Exp. a + 4d = 10 ………….. (1)

a + 20d = 42 ………….. (2)

Eqn (2) – Eqn (1) gives 16d = 32 , d = 2

Substituting d =2 in either (1) or (2), a = 2.

31 st term = a + 30d = 2 + 30*2 = 62

5. Exp. 5( a + 4d ) = 6 ( a + 5d ), 5a + 20d = 6a + 30d, a + 10d = 0,

i.e 11 th = 0

6. Exp. All 2 digit numbers divisible by 3 are 12,15,18,21,……….. 99

This is an A.P with a = 12, and d = 3, Let it contain n terms

Then, 12 + (n – 1 ) *3 = 99, or n = ( 99-12) /3 + 1 = 30

Required sum = 30/2*( 12 + 99 ) = 15*111 = 1665

7. Exp. 4 th term = sum of 4 terms – sum of 3 terms

= ( 6* 42 + 6*4 ) -( 6* 32 +6*3 ), = ( 96+24 ) – ( 54+18 )

= 120 – 72 = 48

8. Exp. Let the number of terms be n. Then,

2 * 2 n – 1 = 1024 , 2 n-1 = 512 = 2 9

n – 1 = 9, n = 10

9. Exp. Sum of the first n terms of a G.P = a (1 -rn ) ,

1 – r

a [ 1 – (1/3) 5 ]

1 – 1/3 = 363

a = 363 [ 1-1/3 ] = ( 363* 2/3) / ( 1- 1/242 ) = 242/ ( 242/243 )

1- (1/3) 5

a = 243

10. Exp. It is a G.P with infinite terms. The common ratio is (4/5) and this is less than 1. The sum to infinity of a series with first term ‘a ‘ and common ratio ‘r’ =[ a/( 1-r) ], The sum to infinity = 1/ (1 – 4/5 ) = 5

11. Exp. Arithmetic mean = 552/ n = 1/2( 3 +45 ) , = 24

n = 552/24 =23

12. Exp. Sum of the G.P . = ( First term – r * last term)/1 – r

40/3 = 9 – 1/3 ( last term )

2/3

Last term = ( -40/3*2/3+9 ) * 3 = – 80/3 + 27 = 1/3

13. Exp. Let the terms be a/r, a , ar

a/r *a * ar = a 3 =216, a =6

(a/r*a )+ (a*ar) + (ar*a/r ) =114, a 2/r + a 2 r + a 2 =114

a 2 ( 1/r + r +1 ) = 114, 36[( 1+ r 2 +r )/r] =114

6 [( 1 +r 2+ r )/r ]= 19, 6 r 2 -13 r + 6 = 0,

On solving, r = 2/3 or 3/2

14. Exp. As the sum of the first three terms will be thrice the second term, 3 ( a + d ) =21, also the sum of 13 th,14th and 15 th = 3 ( a+ 13d) = 93

On solving the two eqns, d = 2, 3 (a+2 ) = 21, a = 5

15. Exp. a*r 3 = 250, a *r 6 =31,250, r 3 =31,250/250 = 125

r = 5, a* 125 = 250, a = 2

16. Exp. a (r 8 – 1)/(r – 1 ) = 10001 a (r 4 –1)/(r – 1)

r 4 +1 = 10001, r =10

17. Exp. a +10d = 5.25, a+6d = 3.25, 4d = 2, d = ½

a +5 = 5.25, a = 0.25 = ¼, s 15 = 15/2 ( 2 * ¼ + 14 * ½ )

= 15/2 (1/2 +14/2 ) = 15/2 *15/2 =225/ 4 = 56.25

18. Exp. a + (a + d ) + ( a + 2d ) = 3( a + d ) = 33,

a + d = 11, or second term = 11 , first term = 11– d ,

Then ( 11- d ) 2 + 11 2 + ( 11 + d ) 2 = 491

2 d 2 = 491 – ( 3 * 121 ) = 491 – 363 = 128

d 2 = 64, d = 8, a = 3,

19. Let the angles be a, ar, ar 2, ar 3.

Sum of the angles = a ( r 4- 1 ) /r -1 = a ( r 2 + 1 ) ( r + 1 ) = 360

a< 90 , and ar< 90, Therefore, a ( 1 + r ) < 180, or ( r 2 + 1 ) > 2

Therefore, r is not equal to 1. Trying for r = 2 we get a = 24

Therefore, The angles are 24, 48, 96 and 192.

20. Exp. S a = a/ ( 1 – r ) = 7 …………. (1)

Sum to infinity of the cubes = a 3 / 1 –r 3 = 1,225

From (1) a 3 / ( 1 – r ) 3 = 7 3 = 343

Therefore, ( 1 – r ) 3 / 1 – r 3 = 1225/343,

(1+ r 2 -2r)/ ( 1 + r 2 + r ) = 25/7

7 + 7 r 2 –14r = 25 + 25 r + 25 r 2

18 r 2 +39r + 18 = 0, on solving r = -3/2 or -2/3

for an infinite G.P | r | < 1, r = – 2/3

Therefore, a / [ 1 – ( – 2/3 ) ] = 7, a = 7 * 5/3 = 35/3

Profit and Loss-Exercise Questions updated on Apr 2018

155253

1. The profit obtained by selling an article for Rs. 56 is the same as the loss obtained by selling it for Rs. 42. What is the cost price of the article?

a. Rs. 40

b. Rs. 50

c. Rs. 49

d. None of these

2. The C.P of 21 articles is equal to S.P of 18 articles. Find the gain or loss percent.

a. 10%

b. 18 1/3%

c. 16 2/3

d. 20%

3. An article is sold at a certain price. By selling it at 2/3 of that price one loses 10%. Find the gain percent at original price.

a. 15%

b. 35%

c. 25%

d. 50%

4. A man bought a horse and a carriage for Rs. 3000. He sold the horse at a gain of 20% and the carriage at a loss of 10%, thereby gaining 2% on the whole. Find the cost of the horse.

a. 2200

b. 1800

c. 1200

d. 1000

5. The price of a jewel, passing through three hands, rises on the whole by 65%. If the first and second sellers earned 20% and 25% profit respectively, find the percentage profit earned by the third seller.

a. 10%

b. 22%

c. 18%

d. 12%

6. At what percentage above the C.P must an article be marked so as to gain 33% after allowing a customer a discount of 5%?

a. 38%

b. 40%

c. 43%

d. 48%

7. A grocer purchased 80 kg of rice at Rs. 13.50 per kg and mixed it with 120 kg rice at Rs. 16 per kg. At what rate per kg should he sell the mixture to gain 16%?

a. Rs. 19

b. Rs. 20.5

c. Rs. 17.4

d. Rs. 21.6

8. On an article ,the manufacturer gains 10%, the wholesale dealer 15%, and the retailer 25%, If its retail price is 1265, what is the cost of its production?

a. 1000

b. 800

c. 1100

d. 900

9. A dealer professing to sell his goods at cost price, uses 900gm weight for 1 kg. His gain percent is

a. 13%

b 12 1/3%

c. 11 1/9%

d.10%

10. A trader has 50 kg of rice, a part of which he sells at 14% profit and rest at 6% loss. On the whole his loss is 4% . What is the quantity sold at 14% profit and that at 6% loss?

a. 5 and 45 kg

b. 10 and 40 kg

c. 15 and 35 kg

d. 20 and 30 kg

11. The cost price of two types of tea are Rs. 180 per kg and Rs. 200 per kg respectively. On mixing them in the ratio 5:3, the mixture is sold at Rs. 210 per kg . In the whole transaction, the gain percent is

a. 10%

b. 11%

c. 12%

d. 13%

12. A trader marks his product 40% above its cost. He sells the product on credit and allows 10% trade discount. In order to ensure prompt payment, he further gives 10% discount on the reduced price. If he makes a profit of Rs. 67 from the transaction, then the cost price of the product is

a. Rs. 300

b. Rs. 400

c. Rs. 325

d. Rs. 500

13. A retailer sold two articles at a profit percentage of 10% each. The cost price of one article is three – fourth that of the other. Find the ratio of the selling price of the dearer article to that of the cheaper one

a 4:3

b.3:4

c.41:31

d.51:41

14. If the S.P of Rs. 24 results in a 20% discount on the list price, What S.P would result in a 30% discount on the list price?

a. Rs. 27

b. Rs..21

c. Rs.20

d. Rs. 9

15. Anil bought a T.V with 20% discount on the labeled price . Had he bought it with 25% discount, he would have saved Rs. 500. At what price did he buy the T.V?

a. Rs. 16000

b. Rs. 12000

c. Rs. 10000

d. Rs. 5000

16. A single discount equivalent to a series of 30%, 20%, and 10% is

a.50%

b. 49.6%

c. 49.4%

d. 51%

17. Ramya sells an article at three- fourth of its list price and makes a loss of 10%. Find the profit percentage if she sells at the list price.

a.20%

b.25%

c.15%

d. None of these

18. The ratio of the selling prices of three articles is 5:6:9 and the ratio of their cost prices is 4:5:8 respectively. What is the ratio of their respective percentages of profit, if the profit on the first and the last articles is the same?

a. 4:5:6

b. 10:8:5

c. 5;6;9

d. Cannot be determined

19. With the money I have , I can buy 50 pens or 150 pencils. I kept 10% aside for taxi fare. With the remaining , I purchased 54 pencils and P pens. What is the value of P?

a. 32

b. 30

c. 27

d. None of these

20. The selling price of 13 apples is the same as the cost price of 26 mangoes . The selling price of 16 mangoes is the same as the cost price of 12 apples. If the profit on selling mangoes is 20%, What is the profit on selling apples?

a. 20%

b.25%

c.40%

d. Cannot be determined

Answer & Explanations

1. Exp. S.P 1- C.P = C.P – S.P 2

56 – C.P = C.P – 42

2 C.P = 56 + 42;

C.P = 98/2 = 49

2. Exp. Let C.P of each article be Re. 1.

Then C.P of 18 articles = Rs. 18,

S.P of 18 articles = Rs. 21.

Gain % = ( 3/18 * 100 ) % = 16 2/3

3. Exp. Let the original S.P be Rs. X. Then new S.P = Rs. 2/3 X, Loss =10%

So C.P = Rs. [100/90*2/3 X ] = 20 X/27.

Now C.P = Rs. 20X/27, S.P =Rs. X, Gain = Rs. [ X -20X/27 ] =Rs.7X/27.

Gain % = [ 7X/27 *27/20X *100 ]% =35%

4. Exp.Let the C.P of the horse be Rs. X, Then, C.P of the carriage = Rs.(3000- x).

20% of x – 10% of ( 3000-x ) =2% of 3000 = 60,

x/5 – ( 3000-x )/10 = 60, 3x – 3000 = 600, 3x =3600, x = 1200.

Hence, C.P of the horse = Rs. 1200

5. Exp. Let the original price of the jewel be Rs. P and let the profit earned by the third seller

be x%.

Then, ( 100 +x )% of 125% of 120% of P = 165% of p

[ ( 100+x )/100*125/100*120/100* P ] = [ 165/100* P ]

( 100 + X ) = 165*100*100 = 110, X = 10%.

125*120

6. Exp. Let C.P = Rs. 100, Then S.P = Rs.133.

Let the marked price be x

Then , 95% of x = 133, 95 x/ 100 =133, x = 133*100/95 =140

Marked price = 40% above C.P

7. Exp. C.P OF 200 kg of mix = Rs.[ 80*13.50+120*16 ] =Rs.3000

S.P =116% of Rs.3000 = Rs. 116/100*3000 = 3480

Rate of S .P of the mixture = Rs. [ 3480/200] per kg

= Rs. 17.40 per kg

8. Exp. 110/100*115/100*125/100*C.P =1265, 11/10*23/20*5/4 C.P =1265

C.P = 800

9. Exp. Gain % = Error *100 %

( True value) – Error

= 1000gm -900gm * 100 % = 100/900*100% = 100/9

1000 -100

= 11 1/9 %

10. Exp. Alligation Method

Ratio of quantities sold at 14% profit and 6% loss = 1: 9

Quantity sold at 14% profit = 50/1+9 *1 = 5 kg

Quantity sold at 6% loss = 50/1+9 *9 = 45kg

11. Exp. Let 5kg of first kind of tea be mixed with 3 kg of second kind

C.P of 8 kg of tea = Rs. ( 180*5 + 200*3 ) = Rs. 1500

S.P of 8 kg of tea = Rs. (210 * 8) = Rs. 1680

Gain = Rs. ( 1680 – 1500 ) = Rs. 180

Gain% = ( 180/1500*100 )% = 12%

12. Exp. M.P = C.P *1.4

Profit = S.P – C.P = C.P ( 1.4 ) ( 0.9 ) ( 0.9 ) – C.P = 67

C.P ( 1.134 – 1 ) = 67, C.P = 500

13. Exp. Let C.P of one of the article be X, Then C.P of the other = ¾X,

S.P 1 = 11X/10, S.P2 =3/4*11/10X,

S.P1/S.P2 = 11X/10 *40/33X = 4/3

S.P1: S.P2 = 4 : 3

14. Exp. Let the list price be Rs. X,

80/100*x = 24, x = 24*100/80 = 30

Required S.P = 70% of Rs. 30 = 70*30/100 =21

15. Exp. Let the labelled price be Rs. X,

S.P = 80/100*X =4X/5

New S.P = 75/100*X = 3X/4

4X/5 –3X/4 500, X = 10000

16. Exp. Let the marked price be Rs. 100

Then S.P = 90% of 80% of 70% of 100

= ( 90/100*80/100*70/100*100 ) =50.4

Single discount = ( 100- 50.4 )% = 49.6%

17. Exp. Let the list price be x,

S.P =3/4X, S.P = (100 – loss%) /100*C.P = 0.9 C.P

3/4x =0.9C.P, C.P =3X/3.6

If S.P = X, Profit % = (x- 3x/3.6 ) / ( 3x/3.6 ) *100= 60/3 = 20%

18. Exp. Given that the selling prices of three articles,

S.P1 =5X, S.P2 =6X, S.P3 =9X,

And their cost prices are C.P1 =4Y, C.P 2 =5Y, C.P3 =8Y

Given that , S.P1 –C.P1 =S.P2 –C.P2, 5X-4Y =9X-8Y, X =Y,

Their profit percentages are, p1 =(5-4)/4*100 = 25% ,

p2 = (6-5)/5*100 = 20%, p3 = (9-8)/8*100 = 12 1/2 %

Ratio of the percentages is 25:20:12 1/2 = 10:8:5

19. Exp. Since cost of, 50 pens = 150 pencils, With the cost of 3 pencils I can buy 1 pen. After

putting aside 10% for taxi I was left with 90% of the money , with which I can buy 135

pencils (90% of 150) or 45(90% of 50) pens, I bought 54 pencils and P pens, or I could have

bought ( 54 +3P ) pencils ,

54 +3P =135, 3P = 135-54 =81, P =27

20. Exp. Given that S.P of 13 apples = C.P of 26 mangoes

S.P of an apple = 2*C.P of the mango

S.P of 16 mangoes = C.P of 12 apples

C.P of the apple = 4/3 *S.P of the mango

Mango Apple

C.P x 4/3*y

S.P y 2x

Given that y = 1.2x

C.P of apple = 4/3*1.2x =1.6x

Profit on each apple = (S.P – C.P)/C.P*100,

= (2x –1.6x)/1.6*100 = 0.4/1.6 *100 =25%

Profit and Loss-Exercise Questions updated on Apr 2018

155253

1. The profit obtained by selling an article for Rs. 56 is the same as the loss obtained by selling it for Rs. 42. What is the cost price of the article?

a. Rs. 40

b. Rs. 50

c. Rs. 49

d. None of these

2. The C.P of 21 articles is equal to S.P of 18 articles. Find the gain or loss percent.

a. 10%

b. 18 1/3%

c. 16 2/3

d. 20%

3. An article is sold at a certain price. By selling it at 2/3 of that price one loses 10%. Find the gain percent at original price.

a. 15%

b. 35%

c. 25%

d. 50%

4. A man bought a horse and a carriage for Rs. 3000. He sold the horse at a gain of 20% and the carriage at a loss of 10%, thereby gaining 2% on the whole. Find the cost of the horse.

a. 2200

b. 1800

c. 1200

d. 1000

5. The price of a jewel, passing through three hands, rises on the whole by 65%. If the first and second sellers earned 20% and 25% profit respectively, find the percentage profit earned by the third seller.

a. 10%

b. 22%

c. 18%

d. 12%

6. At what percentage above the C.P must an article be marked so as to gain 33% after allowing a customer a discount of 5%?

a. 38%

b. 40%

c. 43%

d. 48%

7. A grocer purchased 80 kg of rice at Rs. 13.50 per kg and mixed it with 120 kg rice at Rs. 16 per kg. At what rate per kg should he sell the mixture to gain 16%?

a. Rs. 19

b. Rs. 20.5

c. Rs. 17.4

d. Rs. 21.6

8. On an article ,the manufacturer gains 10%, the wholesale dealer 15%, and the retailer 25%, If its retail price is 1265, what is the cost of its production?

a. 1000

b. 800

c. 1100

d. 900

9. A dealer professing to sell his goods at cost price, uses 900gm weight for 1 kg. His gain percent is

a. 13%

b 12 1/3%

c. 11 1/9%

d.10%

10. A trader has 50 kg of rice, a part of which he sells at 14% profit and rest at 6% loss. On the whole his loss is 4% . What is the quantity sold at 14% profit and that at 6% loss?

a. 5 and 45 kg

b. 10 and 40 kg

c. 15 and 35 kg

d. 20 and 30 kg

11. The cost price of two types of tea are Rs. 180 per kg and Rs. 200 per kg respectively. On mixing them in the ratio 5:3, the mixture is sold at Rs. 210 per kg . In the whole transaction, the gain percent is

a. 10%

b. 11%

c. 12%

d. 13%

12. A trader marks his product 40% above its cost. He sells the product on credit and allows 10% trade discount. In order to ensure prompt payment, he further gives 10% discount on the reduced price. If he makes a profit of Rs. 67 from the transaction, then the cost price of the product is

a. Rs. 300

b. Rs. 400

c. Rs. 325

d. Rs. 500

13. A retailer sold two articles at a profit percentage of 10% each. The cost price of one article is three – fourth that of the other. Find the ratio of the selling price of the dearer article to that of the cheaper one

a 4:3

b.3:4

c.41:31

d.51:41

14. If the S.P of Rs. 24 results in a 20% discount on the list price, What S.P would result in a 30% discount on the list price?

a. Rs. 27

b. Rs..21

c. Rs.20

d. Rs. 9

15. Anil bought a T.V with 20% discount on the labeled price . Had he bought it with 25% discount, he would have saved Rs. 500. At what price did he buy the T.V?

a. Rs. 16000

b. Rs. 12000

c. Rs. 10000

d. Rs. 5000

16. A single discount equivalent to a series of 30%, 20%, and 10% is

a.50%

b. 49.6%

c. 49.4%

d. 51%

17. Ramya sells an article at three- fourth of its list price and makes a loss of 10%. Find the profit percentage if she sells at the list price.

a.20%

b.25%

c.15%

d. None of these

18. The ratio of the selling prices of three articles is 5:6:9 and the ratio of their cost prices is 4:5:8 respectively. What is the ratio of their respective percentages of profit, if the profit on the first and the last articles is the same?

a. 4:5:6

b. 10:8:5

c. 5;6;9

d. Cannot be determined

19. With the money I have , I can buy 50 pens or 150 pencils. I kept 10% aside for taxi fare. With the remaining , I purchased 54 pencils and P pens. What is the value of P?

a. 32

b. 30

c. 27

d. None of these

20. The selling price of 13 apples is the same as the cost price of 26 mangoes . The selling price of 16 mangoes is the same as the cost price of 12 apples. If the profit on selling mangoes is 20%, What is the profit on selling apples?

a. 20%

b.25%

c.40%

d. Cannot be determined

Answer & Explanations

1. Exp. S.P 1- C.P = C.P – S.P 2

56 – C.P = C.P – 42

2 C.P = 56 + 42;

C.P = 98/2 = 49

2. Exp. Let C.P of each article be Re. 1.

Then C.P of 18 articles = Rs. 18,

S.P of 18 articles = Rs. 21.

Gain % = ( 3/18 * 100 ) % = 16 2/3

3. Exp. Let the original S.P be Rs. X. Then new S.P = Rs. 2/3 X, Loss =10%

So C.P = Rs. [100/90*2/3 X ] = 20 X/27.

Now C.P = Rs. 20X/27, S.P =Rs. X, Gain = Rs. [ X -20X/27 ] =Rs.7X/27.

Gain % = [ 7X/27 *27/20X *100 ]% =35%

4. Exp.Let the C.P of the horse be Rs. X, Then, C.P of the carriage = Rs.(3000- x).

20% of x – 10% of ( 3000-x ) =2% of 3000 = 60,

x/5 – ( 3000-x )/10 = 60, 3x – 3000 = 600, 3x =3600, x = 1200.

Hence, C.P of the horse = Rs. 1200

5. Exp. Let the original price of the jewel be Rs. P and let the profit earned by the third seller

be x%.

Then, ( 100 +x )% of 125% of 120% of P = 165% of p

[ ( 100+x )/100*125/100*120/100* P ] = [ 165/100* P ]

( 100 + X ) = 165*100*100 = 110, X = 10%.

125*120

6. Exp. Let C.P = Rs. 100, Then S.P = Rs.133.

Let the marked price be x

Then , 95% of x = 133, 95 x/ 100 =133, x = 133*100/95 =140

Marked price = 40% above C.P

7. Exp. C.P OF 200 kg of mix = Rs.[ 80*13.50+120*16 ] =Rs.3000

S.P =116% of Rs.3000 = Rs. 116/100*3000 = 3480

Rate of S .P of the mixture = Rs. [ 3480/200] per kg

= Rs. 17.40 per kg

8. Exp. 110/100*115/100*125/100*C.P =1265, 11/10*23/20*5/4 C.P =1265

C.P = 800

9. Exp. Gain % = Error *100 %

( True value) – Error

= 1000gm -900gm * 100 % = 100/900*100% = 100/9

1000 -100

= 11 1/9 %

10. Exp. Alligation Method

Ratio of quantities sold at 14% profit and 6% loss = 1: 9

Quantity sold at 14% profit = 50/1+9 *1 = 5 kg

Quantity sold at 6% loss = 50/1+9 *9 = 45kg

11. Exp. Let 5kg of first kind of tea be mixed with 3 kg of second kind

C.P of 8 kg of tea = Rs. ( 180*5 + 200*3 ) = Rs. 1500

S.P of 8 kg of tea = Rs. (210 * 8) = Rs. 1680

Gain = Rs. ( 1680 – 1500 ) = Rs. 180

Gain% = ( 180/1500*100 )% = 12%

12. Exp. M.P = C.P *1.4

Profit = S.P – C.P = C.P ( 1.4 ) ( 0.9 ) ( 0.9 ) – C.P = 67

C.P ( 1.134 – 1 ) = 67, C.P = 500

13. Exp. Let C.P of one of the article be X, Then C.P of the other = ¾X,

S.P 1 = 11X/10, S.P2 =3/4*11/10X,

S.P1/S.P2 = 11X/10 *40/33X = 4/3

S.P1: S.P2 = 4 : 3

14. Exp. Let the list price be Rs. X,

80/100*x = 24, x = 24*100/80 = 30

Required S.P = 70% of Rs. 30 = 70*30/100 =21

15. Exp. Let the labelled price be Rs. X,

S.P = 80/100*X =4X/5

New S.P = 75/100*X = 3X/4

4X/5 –3X/4 500, X = 10000

16. Exp. Let the marked price be Rs. 100

Then S.P = 90% of 80% of 70% of 100

= ( 90/100*80/100*70/100*100 ) =50.4

Single discount = ( 100- 50.4 )% = 49.6%

17. Exp. Let the list price be x,

S.P =3/4X, S.P = (100 – loss%) /100*C.P = 0.9 C.P

3/4x =0.9C.P, C.P =3X/3.6

If S.P = X, Profit % = (x- 3x/3.6 ) / ( 3x/3.6 ) *100= 60/3 = 20%

18. Exp. Given that the selling prices of three articles,

S.P1 =5X, S.P2 =6X, S.P3 =9X,

And their cost prices are C.P1 =4Y, C.P 2 =5Y, C.P3 =8Y

Given that , S.P1 –C.P1 =S.P2 –C.P2, 5X-4Y =9X-8Y, X =Y,

Their profit percentages are, p1 =(5-4)/4*100 = 25% ,

p2 = (6-5)/5*100 = 20%, p3 = (9-8)/8*100 = 12 1/2 %

Ratio of the percentages is 25:20:12 1/2 = 10:8:5

19. Exp. Since cost of, 50 pens = 150 pencils, With the cost of 3 pencils I can buy 1 pen. After

putting aside 10% for taxi I was left with 90% of the money , with which I can buy 135

pencils (90% of 150) or 45(90% of 50) pens, I bought 54 pencils and P pens, or I could have

bought ( 54 +3P ) pencils ,

54 +3P =135, 3P = 135-54 =81, P =27

20. Exp. Given that S.P of 13 apples = C.P of 26 mangoes

S.P of an apple = 2*C.P of the mango

S.P of 16 mangoes = C.P of 12 apples

C.P of the apple = 4/3 *S.P of the mango

Mango Apple

C.P x 4/3*y

S.P y 2x

Given that y = 1.2x

C.P of apple = 4/3*1.2x =1.6x

Profit on each apple = (S.P – C.P)/C.P*100,

= (2x –1.6x)/1.6*100 = 0.4/1.6 *100 =25%