(37÷44) + (78÷34) x (480÷85) = ? a) 11.769 b) 12.756 c) 13.796 d) 12.796 e) none

(37÷44) + (78÷34) x (480÷85) = ?
a) 11.769
b) 12.756
c) 13.796
d) 12.796
e) none
by

1 Answer

Answer :

(37÷44) + (78÷34) x (480÷85) = ?
Sol: 0.841 + 2.29412 x 5.64706 =?
 0.841 + 12.955 =?
 ? = 13.796
Answer: c)
by

Related questions

(37÷44) + (78÷34) x (480÷85) = ? a) 11.769 b) 12.756 c) 13.796 d) 12.796 e) none

Description : (37÷44) + (78÷34) x (480÷85) = ? a) 11.769 b) 12.756 c) 13.796 d) 12.796 e) none

Last Answer : (37÷44) + (78÷34) x (480÷85) = ? Sol: 0.841 + 2.29412 x 5.64706 =? 0.841 + 12.955 =? ? = 13.796 Answer: c)

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

39.98 × 85.01 – 79.9 × 34.9 + 175.03 × 12.08 = ? a) 4800 b) 4300 c) 2700 d) 2100 e) 3400

Description : 39.98 × 85.01 – 79.9 × 34.9 + 175.03 × 12.08 = ? a) 4800 b) 4300 c) 2700 d) 2100 e) 3400

Last Answer : ? = 40 × 85 – 80 × 35 + 175 × 12 = 3400 – 2800 + 2100 = 2700 Answer: c)

(37) 0.75 x [(7056) 0.25] ÷ (64) – 2 = (8) ? (Take 3 digits after the decimal point) a) 95.357 b) 69.567 c) 96.257 d) 78.584 e) 88.598

Description : (37) 0.75 x [(7056) 0.25] ÷ (64) – 2 = (8) ? (Take 3 digits after the decimal point) a) 95.357 b) 69.567 c) 96.257 d) 78.584 e) 88.598

Last Answer : (37) 0.75 x [(7056) 0.25] ÷ (64) – 2 = (8) ? Sol: 27.75 x 1764 ÷ 64 – 2 = 8 x ? 27.75 x 27.5625 – 2 = 8 x ? 762.859/8=? ? = 95.357 Answer: a)

67,74,78,?,92,94 a) 86 b) 85 c) 83 d) 84 e) 88

Description : 67,74,78,?,92,94 a) 86 b) 85 c) 83 d) 84 e) 88

Last Answer : The next term is obtained by adding the unit’s digit to the number. Hence the missing number =78+8=86 Answer: a)

A batsman in his 17th innings makes a score of 85, and thereby increases his average by 3. What is his average after 17 innings? A.30 B.37 C.40 D.45 E.None of these

Description : A batsman in his 17th innings makes a score of 85, and thereby increases his average by 3. What is his average after 17 innings? A.30 B.37 C.40 D.45 E.None of these

Last Answer : Answer – B (37) Explanation – Let the average after 16th innings be a, then total score after 17th innings => 16a+85 = 17 (a+3) a = 85-51 = 34 Average after 17 innings = a + 3 = 34 + 3 = 37

A man sold a cycle at a loss of 15%. Had he sold it for Rs.241.87 more he would have gained 24%. For what value should he sell it in order to earn a profit of 30%? a) Rs. 782 b) Rs. 792 c) Rs. 796 d) Rs. 800 e) Rs. 806

Description : A man sold a cycle at a loss of 15%. Had he sold it for Rs.241.87 more he would have gained 24%. For what value should he sell it in order to earn a profit of 30%? a) Rs. 782 b) Rs. 792 c) Rs. 796 d) Rs. 800 e) Rs. 806

Last Answer : Let the cost price be Rs. X :. Selling price = 85X/100 If profit is 24% then selling price = 124X/100 Now, (124X/100) – (85X/100) = 241.8 Or. X= (24180/39) = Rs.620 For 30% gain SP = (130/100)×620 = Rs. 806 Answer is: e)

Deepa decided to donate 16% of her monthly salary to an NGO. On the day of donation she changed her mind and donated Rs. 6,567/- which was 75%of what she had decided earlier. How much is Deepa’s monthly salary? a) Rs. 54,725/- b) Rs. 58,756/- c) Rs. 56,700/- d) Rs. 55,696/- e) Rs. 52,696/-

Description : Deepa decided to donate 16% of her monthly salary to an NGO. On the day of donation she changed her mind and donated Rs. 6,567/- which was 75%of what she had decided earlier. How much is Deepa’s monthly salary? a) Rs. 54,725/- b) Rs. 58,756/- c) Rs. 56,700/- d) Rs. 55,696/- e) Rs. 52,696/-

Last Answer : Let the Deepa Monthly Salary = x Deepa decided to donate 16% of her monthly salary. On the day of donation she changed. So, 75%of she had decided earlier of 16% 16 × 75/100 = 12% Therefore = 12%= Rs. 6567 100% = Rs .6567 × 100 / 12 x = Rs. 54725 Answer: a)

Five years ago, the age of Arun was 4 times the age of Sarmi after 10 years, Arun will be twice as old as Sarmi. Find the Present ages of Arun and sarmi? a) 30 years, 10.5 years b) 32 years, 11.5 years c) 34 years, 14.5 years d) 35 years, 12.5 years e) None of these

Description : Five years ago, the age of Arun was 4 times the age of Sarmi after 10 years, Arun will be twice as old as Sarmi. Find the Present ages of Arun and sarmi? a) 30 years, 10.5 years b) 32 years, 11.5 years c) 34 years, 14.5 years d) 35 years, 12.5 years e) None of these

Last Answer : Five years ago, x – 5/y – 5 = 4/1 x – 4y = – 15 —-(i) After 10 years, x + 10/y + 10 = 2/1 x – 2y = 10 —–(ii) From eqn (i) and (ii) => x = 35, y = 12.5 Answer: d)

19/14 x 12/72 x 18/29 x 767 = ? (Take 2 digits after the decimal point) a) 110.69 b) 118.78 c) 111.56 d) None e) 108.98

Description : 19/14 x 12/72 x 18/29 x 767 = ? (Take 2 digits after the decimal point) a) 110.69 b) 118.78 c) 111.56 d) None e) 108.98

Last Answer : 19/14 x 12/72 x 18/29 x 767 =? Sol: 1.35 x 0.16 x 0.62 x 767 = ? 102.71 = ? Answer: d)

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of theseSix bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of these

Description : Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? A.8 B.11 C.13 D.16 E.None of these

Last Answer : Answer – D (16) Explanation – L.C.M. of 2, 4, 6, 8, 10, 12 is 120.So, the bells will toll together after every 120 seconds, i.e, 2 minutes.In 30 minutes, they will toll together 30/2 + 1 = 16

(331/30)+( 661/60) +(704/11) -35.013 + 36.026 =? a) 69 b) 67 c) 83 d) 89 e) 85

Description : (331/30)+( 661/60) +(704/11) -35.013 + 36.026 =? a) 69 b) 67 c) 83 d) 89 e) 85

Last Answer : ?=331/30 + 661/60 + 704/11 – 35.013 + 36.026 =11 + 11 + 64 -1.013 =86-1 ≈ 85 Answer: e)

In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Description : In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Last Answer : A)11:77:7

12 17 32 57 ? 137 a) 98 b) 95 c) 37 d) 92 e) none of these

Description : 12 17 32 57 ? 137 a) 98 b) 95 c) 37 d) 92 e) none of these

Last Answer : The pattern of the number series is 12+5×1=17, 17+5×3=32, 32+5×5=57 57+5×7=92, 92+5×9=137 Answer: d)

The labelled price of a product is Rs 750. If it is sold at a 20% discount, the dealer earns a 25% profit. What is the cost price? 1. Rs 550 2. Rs 450 3. Rs 435 4. Rs 480

Description : The labelled price of a product is Rs 750. If it is sold at a 20% discount, the dealer earns a 25% profit. What is the cost price? 1. Rs 550 2. Rs 450 3. Rs 435 4. Rs 480

Last Answer : 1. Rs 550

If x:y =3:4,in the ratio of (7x-4y) : ( 3x + y) a) 2:11 b) 4:9 c) 5:13 d) 6:5 e) 13:5

Description : If x:y =3:4,in the ratio of (7x-4y) : ( 3x + y) a) 2:11 b) 4:9 c) 5:13 d) 6:5 e) 13:5

Last Answer : An easy way to solve this question is use x = 3, y=4 7x – 4y = 21 – 16 = 5 3x + y = 9+4 = 13 Required ratio = 5:13 Answer: c)

f the mass is of 10 Kg, find the natural frequency in Hz of the free longitudinal vibrations. The displacement is 0.01mm. a) 44.14 b) 49.85 c) 43.43 d) 46.34

Description : f the mass is of 10 Kg, find the natural frequency in Hz of the free longitudinal vibrations. The displacement is 0.01mm. a) 44.14 b) 49.85 c) 43.43 d) 46.34

Last Answer : b) 49.85

(?)2 + (9.99)3 = (45.02)2 – (4.01)3 a) 34 b) 19 c) 26 d) 31 e) 12

Description : (?)2 + (9.99)3 = (45.02)2 – (4.01)3 a) 34 b) 19 c) 26 d) 31 e) 12

Last Answer : ?2 + 1000 = 2025 – 64 ?2 = 961  ? = 31 Answer: d)

What is the total number of illiterate people in city B & C together? 1) 57.34 lakh 2) 58.12 lakh 3) 59.64 lakh 4) 60.62 lakh 5) 61.32 lakh

Description : What is the total number of illiterate people in city B & C together? 1) 57.34 lakh 2) 58.12 lakh 3) 59.64 lakh 4) 60.62 lakh 5) 61.32 lakh

Last Answer : 4) 60.62 lakh

A man who makes a profit of 25 percent by selling sugar at Rs. 4.25/kg., lowers his price so as to gain only 5 paise per Kg. In what ratio must his sales be increased so that his total profit may be the same as before? a) 1 : 5 b) 1 : 11 c) 1 : 13 d) 1 : 17 e) 1 : 21

Description : A man who makes a profit of 25 percent by selling sugar at Rs. 4.25/kg., lowers his price so as to gain only 5 paise per Kg. In what ratio must his sales be increased so that his total profit may be the same as before? a) 1 : 5 b) 1 : 11 c) 1 : 13 d) 1 : 17 e) 1 : 21

Last Answer : 125% C.P. of sugar per kg = Rs. 4.25 C.P of the sugar per kg = [4.25×100] / 125 = Rs. 3.40. thus, the profit on 1 kg 85 paise by reducing price to 5 paise, to make same profit he must sell, 85/5 = 17kg so required ratio = 1 : 17. Answer: d)

24 14 26 ? 28 16 30 a) 13 b) 17 c) 15 d) 11 e) 20

Description : 24 14 26 ? 28 16 30 a) 13 b) 17 c) 15 d) 11 e) 20

Last Answer : The series consists of two series 1 and 2: Series 1 : 24 26 28 30 +2 +2 +2 Series 2 : 14 15 16 +1 +1 Answer: c)

(32.3)2 ÷ 4 + √361 = ?2 + 50 a) 15 b) 13 c) 11 d) 17 e) None of these

Description : (32.3)2 ÷ 4 + √361 = ?2 + 50 a) 15 b) 13 c) 11 d) 17 e) None of these

Last Answer : 32*32 = 1024 1024/4 + 19 = 256+19= 275 Answer: a)

If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is? a) 13 b) 15 c) 19 d) 21 e) None of these

Description : If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is? a) 13 b) 15 c) 19 d) 21 e) None of these

Last Answer : If the distance be x km, then x/40-x/50=6/60 x/4-x/5=1 x=20 km. Required time = (20/40) hour – 11 minutes = (1/2×60-11) minutes = 19 minutes Answer: c)

34.8% x 950 – 33.2% of 850 = ? a) 58.4 b) 48.4 c) 56.4 d) 68.9 e) 76.3

Description : 34.8% x 950 – 33.2% of 850 = ? a) 58.4 b) 48.4 c) 56.4 d) 68.9 e) 76.3

Last Answer : 34.8% x 950 – 33.2% of 850 = ? Sol: 0.348 x 950 – 0.332 x 850=? 330.6 – 282.2=? ?= 48.4 Answer: b)

What least number would be subtracted from 427398 so that the remaining number is divisible by 15? 1. 13 2. 3 3. 16 4. 11 5. 14

Description : What least number would be subtracted from 427398 so that the remaining number is divisible by 15? 1. 13 2. 3 3. 16 4. 11 5. 14

Last Answer : Answer- 2 ( 3) Explanation:- On dividing 427398 by 15 we get the remainder 3, so 3 should be subtracted 

Two pipes A and B can fill a tank in 5 and 6hrs.Pipe C can empty it in 12hrs.If all the 3 pipes are opened together,then the tank will be filled in 1)1(13/17)hrs 2)2(8/11)hours. 3)3(9/17)hrs 4)4(1/2)hrs

Description : Two pipes A and B can fill a tank in 5 and 6hrs.Pipe C can empty it in 12hrs.If all the 3 pipes are opened together,then the tank will be filled in 1)1(13/17)hrs 2)2(8/11)hours. 3)3(9/17)hrs 4)4(1/2)hrs

Last Answer : 3)3(9/17)hrs Exp:Net part filled in 1hr=>(1/5+1/6-1/12)=17/60 Tank filled in=60/17=3(9/17).

5287 – 176.22 – 78.584 = ? a) 5023.196 b) 5032.196 c) 5302.196 d) 5203.196 e) None of these

Description : 5287 – 176.22 – 78.584 = ? a) 5023.196 b) 5032.196 c) 5302.196 d) 5203.196 e) None of these

Last Answer : 5287 – 176.22 – 78.584 = 5032.196 Answer: b)

In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Description : In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Last Answer : 2 boys, 1 gi(Prl = (15c2×10c1) / 25c3 = 1050/2300 1 boy, 2 girls = (15c1×10c2) / 25c3 = 675/2300 Difference = (1050 - 675)/2300 = 375/2300 = 15/92 Answer: c)

1 15 16 31 47 78 125 ? a) 172 b) 203 c) 139 d) 167 e) None of these

Description : 1 15 16 31 47 78 125 ? a) 172 b) 203 c) 139 d) 167 e) None of these

Last Answer : 15 + 1 = 16 15 + 16 = 31 16 + 31 = 47 31 + 47 = 78 47 + 78 = 125 78 + 125 = 203 Answer: b)

55 60 67 78 91 108 ? a) 125 b) 121 c) 127 d) 89 e) None of these

Description : 55 60 67 78 91 108 ? a) 125 b) 121 c) 127 d) 89 e) None of these

Last Answer : 55 + 5 = 60 60 + 7 = 67 67 + 11 = 78 78 + 13 = 91 91 + 17 = 108+108 + 19 = 127 Answer: c)

3, 9, 17, 27, 39, 53, ? a) 64 b) 69 c) 74 d) 78 e) 81

Description : 3, 9, 17, 27, 39, 53, ? a) 64 b) 69 c) 74 d) 78 e) 81

Last Answer : Series is 1× 2 + 1, 2 × 3 + 3, 3 × 4 + 5, 4 × 5 + 7, 5 × 6 + 9 Answer: b)

The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C,D and E becomes 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg

Description : The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C ... 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg

Last Answer : A+B+C = 3×84=252 A+B+C+D= (4×80)=320 D = (320-252)=68 and E = (68+3)=71 Now, B+C+D+E = (4×79)=316 B+C+D=(316-71)=245 kg So, A = (320-245)=75 kg Answer: c)

.12 X 13+105% of 993 +879/18 +15 (1)1150 (2)1170 (3)1185 (4)1200 (5) 1215 .12 X 13+105% of 993 +879/18 +15 (1)1150 (2)1170 (3)1185 (4)1200 (5) 1215

Description : .12 X 13+105% of 993 +879/18 +15 (1)1150 (2)1170 (3)1185 (4)1200 (5) 1215

Last Answer : (4)1200

3 7.5 15 37.5 75 167.5 375 a) 167.5 b) 75 c) 37.5 d) 15 e) 7.5

Description : 3 7.5 15 37.5 75 167.5 375 a) 167.5 b) 75 c) 37.5 d) 15 e) 7.5

Last Answer : The series is ×2.5, ×2 alternately Answer: a)

5 10 17 27 37 50 65 a) 10 b) 17 c) 27 d) 37 e) 50

Description : 5 10 17 27 37 50 65 a) 10 b) 17 c) 27 d) 37 e) 50

Last Answer : The series is +5, +7, +9, +11, .... Answer: c)

1824 ÷ 32 + 5040 ÷ ? – 5600 ÷ 35 = 37 a) 36 b) 48 c) 60 d) 72 e) 84

Description : 1824 ÷ 32 + 5040 ÷ ? – 5600 ÷ 35 = 37 a) 36 b) 48 c) 60 d) 72 e) 84

Last Answer : 57 + 5040 ÷ ? – 160 = 37 5040 ÷ ? = 37 – 57 + 160 = 140 = 5040 ÷ 140 = 36 Answer: a)

40 39 32 37 16 ? -8 a) 35 b) 36 c) 33 d) 32 e) 40

Description : 40 39 32 37 16 ? -8 a) 35 b) 36 c) 33 d) 32 e) 40

Last Answer : The series is + (1^2 - 2), – (2^2 + 3), + (3^2 - 4), –(4^2 + 5), ..... Answer: a)

A is 2 years older than B who is twice as old as C. If sum of ages of A, B and C is 37 years, find the age of A. a) 7 years b) 9 years c) 14 years d) 16 years e) None of these

Description : A is 2 years older than B who is twice as old as C. If sum of ages of A, B and C is 37 years, find the age of A. a) 7 years b) 9 years c) 14 years d) 16 years e) None of these

Last Answer : Let the age of C be x years. Then age of B = 2x and age of A = 2x + 2 sum of ages of A, B and C = (2x + 2) + (2x) + x = 37 (5x + 2) = 37  x = 7 Age of A = 2x + 2 = 16 years Answer: d)

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: A.124 B.100 C.111 D.175 E.None of these

Description : The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: A.124 B.100 C.111 D.175 E.None of these

Last Answer : Answer – C (111) Explanation – Let the numbers be 37a and 37b. Then, 37a x 37b = 4107 ab = 3. Now, co-primes with product 3 are (1, 3). So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is: a) 10 b) 14 c) 12 d) 13 e) 15

Description : Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is: a) 10 b) 14 c) 12 d) 13 e) 15

Last Answer : (A+B+C) - (A+B) (A+B+C) =1/6 and (A+B+C) in 2hrs=2/6 and remaining part 1-2/6=2/3. So (A+B) in 7 hrs is 2/ (3*7) =2/21. 1/6-2/21=1/14 so answer is 14. Answer: b)

A man drives from his house to the station. If he drives at the rate of 10 kms per hour, he reaches the station at 6 pm. if he drives at 15 kms per hour, he would reach the station at 4 pm. At what speed, in kms per hour, should he drives so as to reach the station at 5 pm? a) 12 b) 12.33 c) 5√5 d) 12.5 e) 13

Description : A man drives from his house to the station. If he drives at the rate of 10 kms per hour, he reaches the station at 6 pm. if he drives at 15 kms per hour, he would reach the station at 4 pm. At what speed, in kms ... he drives so as to reach the station at 5 pm? a) 12 b) 12.33 c) 5√5 d) 12.5 e) 13

Last Answer : Let x' be the distance between house and station. Time taken to travel a distance of x at the speed of 10km/hr, T1 = x/10 hours Time taken to travel a distance of x at the speed of 15km/hr, T2 = x/15 hours we ... the station at 5 pm, speed = 60/5 = 12 km/hr [12 pm - 5pm, time = 5 hours] Answer: a)

8/15 of 3/17 of 12/13 of 15470 = ? a) 1242 b) 1344 c) 1464 d) 1504 e) 1624

Description : 8/15 of 3/17 of 12/13 of 15470 = ? a) 1242 b) 1344 c) 1464 d) 1504 e) 1624

Last Answer : b) 1344

When a ball bounces, it rises to 2/3 of the height from which it fell. If the ball is dropped from a height of 36m, how high will it rise at the third bounce? a) 10 2/3 m b) 13 2/3 m c) 12 1/3 m d) 12 2/3m e) None of these

Description : When a ball bounces, it rises to 2/3 of the height from which it fell. If the ball is dropped from a height of 36m, how high will it rise at the third bounce? a) 10 2/3 m b) 13 2/3 m c) 12 1/3 m d) 12 2/3m e) None of these

Last Answer : Height at third bounce =36 × (2/3)3 = 10 2/3 metre Answer: a)

A manufacture undertakes to supply 2000 pieces of a particular component at Rs.25 per piece. According to his estimates, even if 5% fail to pass the quality tests, then he will make a profit of 25%. However as it turned out, 50% of the components were rejected. What is the loss to the manufacture? [TCS recruitment exam question] A. Rs 12,000 B. Rs 13,000 C. Rs 14,000 D. Rs 15,000 E. None f these

Description : A manufacture undertakes to supply 2000 pieces of a particular component at Rs.25 per piece. According to his estimates, even if 5% fail to pass the quality tests, then he will make a profit of 25%. However as it turned out, ... A. Rs 12,000 B. Rs 13,000 C. Rs 14,000 D. Rs 15,000 E. None f these

Last Answer : Answer - B. Rs 13,000 Explanation: 5% of 2000=100 so 2000-100=1900 so,if he sells 1900 he will get 25% profit cost per piece rs 25 so 25x 1900 don't solve here if 125% . ... /125=38000=CP if 50% rejected,only 1000 pieces sold so 1000 25=25000=SP loss=cp-sp= 38000-25000=13000

(54.78)^2= ? 1) 3000 2) 3300 3) 3500 4) 3700 5) 3900

Description : (54.78)^2= ? 1) 3000 2) 3300 3) 3500 4) 3700 5) 3900

Last Answer : 1) 3000

22.222 + 2220.2 + 202.20 + 0.0202 – 2.0022 = ? a) 2244.24 b) 2442.64 c) 2424.44 d) 2444.64 e) 2224.84

Description : 22.222 + 2220.2 + 202.20 + 0.0202 – 2.0022 = ? a) 2244.24 b) 2442.64 c) 2424.44 d) 2444.64 e) 2224.84

Last Answer : b) 2442.64

60 kg of an alloy A is mixed with 100 kg of alloy B. If alloy A has lead and tin in the ratio of 3 : 2 and alloy B has tin and copper in the ratio of I : 4 the amount of tin in the new alloy is a) 36 kg b) 44 kg c) 80 kg d) 53 kg e) 45 kg

Description : 60 kg of an alloy A is mixed with 100 kg of alloy B. If alloy A has lead and tin in the ratio of 3 : 2 and alloy B has tin and copper in the ratio of I : 4 the amount of tin in the new alloy is a) 36 kg b) 44 kg c) 80 kg d) 53 kg e) 45 kg

Last Answer : b) 44 kg

Saina spent Rs. 44,668/- on her air tickets, Rs.56,732 on buying gifts to the family members and the remaining 22% of the total amount she had as cash with her. What was the total amount ? a) Rs. 1,25,800/- b) Rs. 1,30,000/- c) Rs. 1,10,500/- d) Rs. 1,68,300/- e) Rs. 1,50,000/-

Description : Saina spent Rs. 44,668/- on her air tickets, Rs.56,732 on buying gifts to the family members and the remaining 22% of the total amount she had as cash with her. What was the total amount ? a) Rs. 1,25,800/- b) Rs. 1,30,000/- c) Rs. 1,10,500/- d) Rs. 1,68,300/- e) Rs. 1,50,000/- 

Last Answer : Saving = 22% Spent = a00 - 22) = 78% Spending money = 44668 + 56732 = 101400 78%- 101400 100% = 101400 100/78 = 130000 Answer: b)

9654 ÷ 21 +7638 ÷ 44 = ? a) 633 b) 600 c) 643 d) 621 e) 598

Description : 9654 ÷ 21 +7638 ÷ 44 = ? a) 633 b) 600 c) 643 d) 621 e) 598

Last Answer : 9654 ÷ 21 + 7338 ÷ 44 = ? = 459.714 + 173.59 = 633.304 = 633 (approx) Answer: a)

The average age of 30 boys of a class is equal to 14 years. When the age of the class teacher is included the average becomes 15 years. Find the age of the class teacher? A.50 B.44 C.45 D.42 E.None of these

Description : The average age of 30 boys of a class is equal to 14 years. When the age of the class teacher is included the average becomes 15 years. Find the age of the class teacher? A.50 B.44 C.45 D.42 E.None of these

Last Answer : Answer – C (45) Explanation – Total ages of 30 boys = 14 x 30 = 420 years Total age when class teacher is included = 15 x 31 = 465 years Age of class teacher = 465 – 420 = 45 years