The sum of 2 numbers is 216, and their HCF is 27. How many such pairs of numbers exist? (A pair a unique set of 2 numbers, so (a, b) and (b, a) will be considered as 1 pair). a) 0 b) 1 c) 2 d) 3 e) 4

The sum of 2 numbers is 216, and their HCF is 27. How many such pairs of numbers exist? (A pair a unique set of 2 numbers, so (a, b) and (b, a) will be considered as 1 pair).
a) 0
b) 1
c) 2
d) 3
e) 4
by

1 Answer

Answer :

Let the numbers be a,b
A = 27m, b = 27n where m, n are co-prime
27(m + n) = 216, or m+n = 8
M = 1, n = 7 and m = 3, n = 5 are the only 2 co – prime pairs possible.
So we have 2 pairs: (27 , 189) and ( 81, 135)
Answer: c)
by

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Description : The LCM of two numbers is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, then the other number is? a) 100 b) 125 c) 250 d) 225 e) 180

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Last Answer : Let the age of Ram be R, brother be B R=3B And (R + 9 ) =1.5(B + 9) 3B + 9 =1.5B + 13.5 1.5B =4.5 or B = 3 years R= 9 years. Father current age = 4R(Since father is 3 times older than Ram, his ... R + 3 R, and not 3R) Father age when Ram is 1.5 times the age of his brother = 4R + 9 = 45. Answer : c

Ram is 3 times as old as his brother. Ram’s father is 3 times older than Ram. In 9 years , Ram will be 1.5 times the age of his brother. What will be the age of Ram’s father when Ram is 1.5 times the age of his brother? a) 27 b) 36 c) 45 d) 54 e) 63

Description : Ram is 3 times as old as his brother. Ram’s father is 3 times older than Ram. In 9 years , Ram will be 1.5 times the age of his brother. What will be the age of Ram’s father when Ram is 1.5 times the age of his brother? a) 27 b) 36 c) 45 d) 54 e) 63

Last Answer : Let the age of Ram be R, brother be B R=3B And (R + 9 ) =1.5(B + 9) 3B + 9 =1.5B + 13.5 1.5B =4.5 or B = 3 years R= 9 years. Father current age = 4R(Since father is 3 times older than Ram, his ... R + 3 R, and not 3R) Father age when Ram is 1.5 times the age of his brother = 4R + 9 = 45. Answer : c

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(3333÷30) + (665 ÷ 25) + (3771 ÷ 27) = ? a) 357.637 b) 458.558 c) 229.357 d) 277.367 e) 498.627

Description : (3333÷30) + (665 ÷ 25) + (3771 ÷ 27) = ? a) 357.637 b) 458.558 c) 229.357 d) 277.367 e) 498.627

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18.4 x 17.5 + 27.6 x 13.5 = ? ÷ 4 a) 2778.4 b) 2786.4 c) 2866.4 d) 2986.3 e) 3256.8

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Last Answer : 18.4 x 17.5 + 27.6 x 13.5 = ? ÷ 4 Sol: 322 + 372.6 = ? ÷4 694.6 x 4 = ? ? = 2778.4 Answer: a)

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Last Answer : According to question, l×b = ¶r2 l ×22 = 22/7 ×14 × 14 l = 22/7 ×14 × 14/22 cms = 28 cms. Answer: b)

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Last Answer : Cannot be determined Answer: d)

Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 seconds. What is the length of the fast train? A.27 7/9 m B.28 m C.29 m D.30 2/7 m E.None of these

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Suresh and Ramesh are twins. In a form by mistake, Suresh reverses the last 2 digits of his year of birth, and this makes him 9 years older than Ramesh. With this mistake, the sum of their ages in 2015 becomes 43. How old will Ramesh be 2021? a) 17 b) 19 c) 21 d) 23 e) 25

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The sum of present ages of a father and his son is 36 years. When the son reaches father’s present age, the sum of their ages will be 80 years. What is the present age of the son? a) 4 years b) 7 years c) 9 years d) 12 years e) None of these

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The sum of ages of Mohan and his father is 35 years. When Mohan’s age will be equal to present age of his father, then sum of their ages will be 85 years. Find present age of the father. a) 20 years b) 25 years c) 30 years d) 32 years e) None of these

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Last Answer : Sum of present ages of A, B and C is 72 years. sum of their ages (4 years ago) = 72 – (3 × 4) = 60 years 4 years ago, ratio of ages of A, B and C was 1 : 2 :3. A’s age, 4 years ago = (1/6) ×60 = 10 years A’s present age = 10 + 4 = 14 years Answer: d)

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Description : Sum of present ages of A and B is 60 years. 5 years hence, their ages will be in the ratio 3:4. Find A’s present age. a) 25 years b) 30 years c) 35 years d) 40 years e) None of these

Last Answer : Sum of their present ages = 60 years sum of their ages, 5 years hence = 60 + (5 × 2) = 70 years. after 5 years, ratio of ages of A and B will be 3 : 4. A’s age after 5 years = (3/7) × 70 = 30 years A’s present age = 30 – 5 = 25 years Answer: a)

One year ago the ratio between Sam and Ash’s age was 4:3. One year hence the ratio of their ages will be 5:4. What is the sum of their present ages in years? A.14 B.16 C.18 D.12 E.None of these

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Last Answer : Answer - B (16) Explanation - Let one year ago Sam's age be 4X years And, Ash's age be 3X years Present age of Sam = (4X + 1) years Present age of Ash= (3X + 1) years One year hence Sam's age = (4X + 2) ... = 2 Sum of their present ages = 4X + 1 + 3X + 1 = 7X + 2 = 7 x 2 + 2 = 16 years.

The digits of a two-digit number are in the ratio of 2 : 3 and the number obtained by interchanging the digits is bigger than the original number by 27.What is the original number? 1. 63 2. 48 3. 96 4. 69 5. 66

Description : The digits of a two-digit number are in the ratio of 2 : 3 and the number obtained by interchanging the digits is bigger than the original number by 27.What is the original number? 1. 63 2. 48 3. 96 4. 69 5. 66

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If the S.P of Rs.40 results in a 20% discount on list price. What S.P would result in a 30% discount on list price? A. Rs 18 B. Rs 20 C. Rs 35 D. Rs 27

Description : If the S.P of Rs.40 results in a 20% discount on list price. What S.P would result in a 30% discount on list price? A. Rs 18 B. Rs 20 C. Rs 35 D. Rs 27

Last Answer : Answer – C. Rs 35 Explanation – Let the list price be Rs. x 80/100x = 40 => x = 40 x 100/80 = 50 ∴ Required S.P. = 70% of Rs. 50=Rs.35

A dealer buys an article marked at Rs. 50,000 at 25% and 10% successive discount. He spends Rs. 5000 on its repair and sells it for Rs. 50,000. What is his loss or gain percentage? 1. 33.63% loss 2. 26.43% gain 3. 1.25% gain 4. 29.03% 5. 27.13% loss

Description : A dealer buys an article marked at Rs. 50,000 at 25% and 10% successive discount. He spends Rs. 5000 on its repair and sells it for Rs. 50,000. What is his loss or gain percentage? 1. 33.63% loss 2. 26.43% gain 3. 1.25% gain 4. 29.03% 5. 27.13% loss

Last Answer : CP of the article  

In 165 litres of mixtures of milk and water, water is only 28%. The milkman sold 40 litres of this mixture and then he added 30 litres of pure milk and 13 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture? 1) 29.35% 2) 28. 57% 3) 24. 57% 4) 27. 75% 5) 26. 57%

Description : In 165 litres of mixtures of milk and water, water is only 28%. The milkman sold 40 litres of this mixture and then he added 30 litres of pure milk and 13 litres of pure water in the remaining mixture. What is the percentage ... mixture? 1) 29.35% 2) 28. 57% 3) 24. 57% 4) 27. 75% 5) 26. 57%

Last Answer : 2) 28. 57%

One number is 3 more than another and the sum of their square is 269.Find the number. 1.10 2.12 3.15 4.20

Description : One number is 3 more than another and the sum of their square is 269.Find the number. 1.10 2.12 3.15 4.20

Last Answer : Answer-1 (10) Explanation:- x is one number x+3 the other x^2 +x^2 +6x+9 ; (that is (x+3)(x+3) =269 2x^2+6x -260=0 x^2+3x-130=0 , dividing by 2. (x+13)(x-10)=0 one number is 10, the other is 13 squared 10=100 square 13=169 sum is 269

In 10 years time, Ram will be 2 times older than his son. 5 years ago he was 9 times as old as his son. What will be the sum of the ages of Ram and his son 3 years from now? a) 46 b) 66 c) 86 d) 76 e) None of these

Description : In 10 years time, Ram will be 2 times older than his son. 5 years ago he was 9 times as old as his son. What will be the sum of the ages of Ram and his son 3 years from now? a) 46 b) 66 c) 86 d) 76 e) None of these

Last Answer : Let current age of Ram and his son be R,S (R + 10) = 3(S + 10) [2 times older = 3 times the age] R = 3S + 20 (R - 5) = 9(S - 5) 3S + 20 – 5= 9S – 45 3S + 15 = 9S – 45 or S = 10., R = 50 sum of ages 3 years from now = 10 + 3 + 50 + 3 = 66. Answer: b)

Two equal sums are lent at the same time at 6% and 5 % simple interest respectively. The former is recovered 24 months earlier than the latter. And the amount in each case is Rs. 2400. Find the sum of money lent initially. a) 1500 b) 2000 c) 2400 d) 3000 e) 3600

Description : Two equal sums are lent at the same time at 6% and 5 % simple interest respectively. The former is recovered 24 months earlier than the latter. And the amount in each case is Rs. 2400. Find the sum of money lent initially. a) 1500 b) 2000 c) 2400 d) 3000 e) 3600

Last Answer : Let the sum be S, time be ‘n’ years for 6%. n+2 years for 5% 2400 = S + S ×n × 6/100 2400 = S + S × (n+2) × 5/100 So, 6n = 5(n+2) or n = 10 years. 2400 = S + S × 10 × 6/100 = 1.6S S = 2400/1.6 = 1500 Answer: a)

The ratio between two number is 3:4 ,if each number be increased by 9, the ratio becomes 18:23 find the sum of the number a) 135 b) 105 c) 155 d) 165 e) None of these

Description : The ratio between two number is 3:4 ,if each number be increased by 9, the ratio becomes 18:23 find the sum of the number a) 135 b) 105 c) 155 d) 165 e) None of these

Last Answer : Let the two numbers be 3x and 4x. When they are increased by 9 they become 3x + 9 and 4x + 9. It is given that the ratio is 18:23 Thus, 3x + 9/4x + 9 = 18:23 23(3x + 9) = 18(4x + 9) 69x + 207 = 72x ... = -45 X = 15 Thus two numbers are 3x15 = 45 and 4 x 15 = 60 And the sum is 60+45 = 105 Answer: b)