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Find out the wrong number: 3601 3602 1803 604 154 36 12
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3601 ÷ 1 + 1 = 3602 3602 ÷ 2 + 2 = 1803 1803 ÷ 3 + 3 = 604 604 ÷ 4 + 4 = 155 155 ÷ 5 + 5 = 36 36 ÷ 6 + 6 = 12 154 is written in place of 155.
Find out the wrong number: 4 12 42 196 1005 6066 42511
4 x 2 + (2)2 = 12 12 x 3 + (3)2 = 45 (42 is written in place of 45) 45 x 4 + (4)2 = 196 196 x 5 + (5)2 = 1005 1005 x 6 + (6)2 = 6066 6066 x 7 + (7)2 = 42511
Find out the wrong number: 2 8 12 20 30 42 56
2 + 4 = 6 (8 is written in place of 6) 6 + 6 = 12 12 + 8 = 20 20 + 10 = 30 30 + 12 = 42 42 + 14 = 56
Find out the wrong number: 32 16 24 65 210 945 5197.5
32 x 0.5 = 16 16 x 1.5 = 24 24 x 2.5 = 60 (65 is written in place of 60) 60 x 3.5 = 210 210 x 4.5 = 945 945 x 5.5 = 5197.5
Find out the wrong number: 7 13 25 49 194 385
7 + 6 = 13 13 + 12 = 25 25 + 24 = 49 49 + 48 = 97 97 + 96 = 193 (194 is written in place of 193) 193 + 192 = 385
What approximate value will come in place of the question mark (?) in the following questions ? 2371 ÷ 6 + (43 × 4.35) = ?
2371 ÷ 6 + (43 × 4.35) = ? ? ≈ 395 + 187 = 582
What approximate value will come in place of the question mark (?) in the following questions ? 3√(3380) + √1300 = ?
? =3√(3380) + √1300 ≈ 3√(3380) + √1296 ≈ 15 + 36 ≈ 51
What approximate value will come in place of the question mark (?) in the following questions ? (4.989)2 + (21.012)3 + √1090 = ?
? ≈ (5)2 + (21)3 + √1089 ≈ 25 + 9261 + 33 ≈ 9319
What approximate value will come in place of the question mark (?) in the following questions ? 7020 ÷ 2.99 × 13/29 = ?
? ≈ 7020/3 x 13/29 ≈ 1049 ≈ 1050.
What approximate value will come in place of the question mark (?) in the following questions ? 24.99% of 5001 – 65.01% of 2999 = ?
? ≈ (5000 x 25)/100 - (3000 x 65)/100 ≈ 1250 – 1950 ≈ – 700.
Study the following graph and table carefully and answer the questions given below :
Which of the following vehicles travelled at the same speed on both the days ?
B has the same speed on both days:
Speed on first day: 516/12 = 43 km/h
Speed on first day: 774/18 = 43 km/h
Directions for Below Question :
Que : What was the difference between the speed of vehicle A on day 1 and the speed of vehicle C on the same day ?
Speed of vehicle A on day 1 = 52 km/hr Speed of vehicle C on day 1 = 63 km/hr Difference = 63 – 52 = 11 km / hr
Que : What was the speed of vehicle C on day 2 in terms of meters per second ?
Speed of vehicle can day 2 = 45 km/hr => (45 x 5/18)m/sec = 12.5m / sec
Que : The distance travelled by vehicle F on day 2 was approximately what percent of the distance travelled by it on day 1 ?
Percentage = (Distance travelled by vehicle F on day 2)/(Distance travelled by vehicle F on day 1) x 100 = 636/703 x 100 ≈ 630/700 x 100 = 90%
Que : What is the respective ratio between the speeds of vehicle D and vehicle E on day 2 ?
Speed of vehicle D on day 2 = 51 Speed of vehicle E on day 2 = 39 Required ratio = 51/39 = 17/13 = 17:13.
Solve the equations: (I) x2 – 6x = 7 (II) 2y2 + 13y + 15 = 0
(I)x2 – 6x = 7 or, x2 – 6x – 7 = 0 or, (x – 7) (x + 1) = 0 or, x = 7, – 1 (II)2y2 + 13y + 15 = 0 or, 2y2 + 3y + 10y + 15 = 0 or, (2y + 3) (y + 5) = 0 or, y = –3/2, –5 Hence, x > y
Solve the equations: (I)3x2 – 7x + 2 = 0 (II)2y2 – 11y + 15 = 0
(I)3x2 – 7x + 2 = 0 or, 3x2 – 6x – x + 2 = 0 or, (x – 2) (3x – 1) = 0 or, x = 2, 1/3 (II)2y2 – 11y + 15 = 0 or, 2y2 – 6y – 5y+ 15 = 0 or, (2y – 5) (y – 3) = 0 or, y = 5/2, 3 Hence, y > x.
Solve the equations: (I)10x2 – 7x + 1 = 0 (II)35y2 – 12y + 1 = 0
(I)10x2 – 7x + 1 = 0 or, 10x2 – 5x – 2x + 1 = 0 or, (2x – 1) (5x – 1) = 0 or, x = 1/2, 1/5 (II)35y2 – 12y + 1= 0 or, 35y2 – 7y – 5y + 1 = 0 or, (5y – 1) (7y – 1) = 0 or, y = 1/5, 1/ 7 Hence, x >= y
Solve the equations: (I)4x2 = 25 (II)2y2 – 13y + 21 = 0
(I)4x2 = 25 or, x2 = 25/4, or x = ± 5/2 (II)2y2 – 13y + 21 = 0 or, 2y2 – 6y – 7y + 21 = 0 or, (y – 3) (2y –7) = 0 or, y = 3, 7/2 Hence, y > x
Solve the equations: (I)3x2 + 7x = 6 (II)6(2y2 + 1) = 17y
(I)3x2 + 7x – 6 = 0 or, 3x2 + 9 x – 2x – 6 = 0 or, (x + 3) (3x – 2) = 0 or, x = – 3, 2/3 (II)6(2y2 + 1) = 17y or, 12y2 + 6 – 17y = 0 or, 12y2 – 9y – 8y + 6 = 0 or, (4y – 3) (3y – 2) = 0 or, y = 3/4,, 2/3 Hence, y >= x
In the following table, the number of employees working in five companies and the corresponding ratio of male and female employees have been given. You are required to study the table carefully and answer the questions.
Que : What is the respective ratio between the number of females in company P and number of females in company L?
Females in company P = 5/12 x 1200 = 500 Females in company = 400 x 8/20 = 160 Required ratio = = 500 :160 = 25 : 8.
Que : The number of female employees working in company O is what percent of total employees working in that company?
Required percentage = 12/25 x 100 = 48
Que :What is the average number of employees in all companies together?
Required average = 4000/5 = 800
Que : The number of female employees in company M is
Females in company M = 5/20 x 600 = 150.
Que : The total number of male employees working in companies N and P together is
Males in company N and company P = 4/5 x 800 + 1200 x 7/12 = 640 + 700 = 1340.
When the numerator and the denominator of a fraction are increased by 1 and 2 respectively, the fraction becomes 2/3 and when the numerator and the denominator of the same fraction are increased by 2 and 3 respectively, the fraction becomes 5/7. What is the original fraction?
According to the question: (x+1)/(y+2) = 2/3 => 3x - 2y = 1 -----(i) and, (y+2)/(y+3) = 5/7 => 7x - 5y = 1 -----(ii) or, 3x - 2y = 7x - 5y => 3y = 4x => x/y = 3/4
The average marks in English subject of a class of 24 students is 56. If the marks of three students were misread as 44, 45 and 61 of the actual marks 48, 59 and 67 respectively, then what would be the correct average?
Correct Average = [(24 x 56) + (48 + 59 + 67) - ( 44 + 45 + 61)]/24 = (1344 + 174 - 150)/24 = 1368/24 = 57.
A sum of money is divided among A, B, C and D in the ratio of 3 : 5 : 9 : 13 respectively. If the share of C is Rs. 2412 more than the share of A, then what is the total amount of money of B and D together?
Let the original sum be Rs. x. Sum of the Ratios = 3 + 5 + 9 + 13 = 30 .'. c's share = Rs. 9x/30 = Rs 3x/10 According to the question, => 3x/10 - x/10 = 2412 => 2x/10 = 2412 => x = 2412 × 5 = Rs. 12060 .'. Amount received by B and D together => ( 5 + 13)/30 x 12060 = Rs 7236.
There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls?
Total possible outcomes = 17C5 = (17 x 16 x 15 x 14 x 13)/(1 x 2 x 3 x 4 x 5) = 6188 Total favourable outcomes = 8C2 × 4C1 × 5C2 = (8 x 7)/(1 x 2) x 4 x (5 x 4)/(1 x 2) = 28 x 4 x 10 = 1120. Hence, the required probability = 1120/6188 = 280/1547.
The marked price of a machine is Rs. 18000. By selling it at a discount of 20%, the loss is 4%. What is the cost price of the machine?
Given marked price of machine = Rs. 18000 .'. Discount = 20/100 × 18000 = Rs. 3600 .'. SP = 18000 – 3600 = Rs. 14400. If loss of 4%, then CP = (100 x SP)/(100 - r) = ( 100 x 14400)/(100 - 4) = (100 x 14400)/96 = Rs 15000.
In a family, a couple has a son and daughter. The age of the father is three times that of his daughter and the age of the son is half of his mother. The wife is nine years younger to her husband and the brother is seven years older than his sister. What is the age of the mother?
Let the mother’s age be y years. .'. The age of father = (y + 9) years The age of son = y/2 years The age of daughter = (y/2 - 7) years Now according to the given condition, => (y + 9) = 3(y/2 - 7) => y + 9 = (3y - 42)/2 => 2y + 18 = 3y - 42 => y = 60 years.
A car travels a distance of 560 km in 9.5 hours partly at a speed of 40 kmh–1 and partly at 160 kmh–1. What is the distance it travel at the speed of 160 km/h?
Let the distance it travelled at the speed of 160 km/h be ‘x’ km. .'. x/160 + (560 - x)/40 = 9.5 => [x + 4(560 - x)]/160 = 9.5 => x + 2240 - 4x = 1520 => 3x = 720 => x = 240 km.
Find the rate of the stream, if a boat covers 120 km downstream and 40 km upstream in 4 hours.
Distance covered in downstream = 120 km Time taken in downstream = 4 hours. Rate of downstream = distance / time = a = 120km / 4 hours = 30 km/hr. Distance covered in upstream = 40 km Since it takes same time to cover this 40km, Time taken in upstream = 4 hours. Rate of upstream = distance / time = b = 40km / 4 hours = 10 km/hr. Speed of stream = (a - b)/2 = (1/2)(30 - 10) km/hr = 10 km/hr.
The average of the 9 consecutive positive integers is 63. The product of the largest and smallest integers is
Let positive consecutive integers are x, x + 1, x + 2 ..., x + 8. Average = [x + (x+1) + ......+ (x + 8)]/9 = 63 => (9x + 36)/9 = 63 => x + 4 = 63 => x = 59 Largest number, x + 8 = 59 + 8 = 67 Product of largest and smallest = 59 × 67 = 3953.
In two vessels A and B, there is mixture of milk and water. The ratio of milk and water in these vessels is 5 : 2 and 8 : 5 respectively. In what ratio these mixtures be mixed together so that the ratio of milk and water in the new mixture becomes 9 : 4 ?
Let C.P. of milk per litre be Rs. 1 Milk in 1 litre of A = 5/7 litre Milk in 1 litre of B = 8/13 litre Milk in 1 litre of mixture =9/13 litre C.P. of 1 litre of A C.P. of 2 litre of B 5/7 8/13 \ / Average Price 9/13 / \ 1/13 2/91 .'. Required ratio = 1/13 : 2/91 = 7 : 2
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