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  1. Asked: 3 years agoIn:

    There are 5 cards. Five positive integers (may be different or equal) are written on these cards, one on each card. Abhiram finds the sum of the numbers on every pair of cards. He obtains only three different totals 57, 70, 83. Find the largest integer written on a card.

    0abcc

    As the three sum are obtained 57,70,83We have combination of three numbers same and two numbers differenta, a, a, b, ca + b = 57b + c = 70c + a = 832a = 70a = 35b = 22c = 48Largest integer is 48

    As the three sum are obtained 57,70,83

    We have combination of three numbers same and two numbers different

    a, a, a, b, c

    a + b = 57

    b + c = 70

    c + a = 83

    2a = 70

    a = 35

    b = 22

    c = 48

    Largest integer is 48

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  2. Asked: 3 years agoIn:

    The number of ways in which 26 identical chocolates be distributed between Amy, Bob, Cathy and Daniel so that each receives at least one chocolate and Amy receives more chocolates than Bob is _______.

    faad1

    Total number of natural solution of equation x1 + x2 + x3 + x4 = 26 is 25C3 = 2300. total number of natural solution of equation 2x1 + x3 + x4 = 26 = 1 + 3 + ……… + 23 = 144Total number of natural solution of equation x1 + x2 + x3 + x4 = 26 when x1 > x2 is 2300 - 144/2 = 1078

    Total number of natural solution of equation x1 + x2 + x3 + x4 = 26 is 25C3 = 2300. total number of natural solution of equation 2x1 + x3 + x4 = 26 = 1 + 3 + ……… + 23 = 144

    Total number of natural solution of equation x1 + x2 + x3 + x4 = 26 when x1 > x2 is 2300 – 144/2 = 1078

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  3. Asked: 3 years agoIn:

    a, b, c are three natural numbers such that a x b x c = 27846. If a/6 = b + 4 = c – 4, find a + b + c.

    564f3

    a ×b × c = 27846     ...... (1)a/b = b + 4 = c - 4a = 6b + 24b = bc = b + 8Put value of a, b, c in (1)(6b + 24) (b) (b + 8) = 278466(b + 4) (b) (b + 8) = 27846(b + 4) (b) (b + 8) = 4641 = 13 × 17 × 3 × 7 = 13 × 17 × 21∴ b = 13a = 6 × 13 + 24= 78 + 24= 102c = b + 8 = 13 + 8 = 21∴ a + b + c = 102 + 13Read more

    a ×b × c = 27846     …… (1)

    a/b = b + 4 = c – 4

    a = 6b + 24

    b = b

    c = b + 8

    Put value of a, b, c in (1)

    (6b + 24) (b) (b + 8) = 27846

    6(b + 4) (b) (b + 8) = 27846

    (b + 4) (b) (b + 8) = 4641 = 13 × 17 × 3 × 7 = 13 × 17 × 21

    ∴ b = 13

    a = 6 × 13 + 24

    = 78 + 24

    = 102

    c = b + 8 = 13 + 8 = 21

    ∴ a + b + c = 102 + 13 + 21 = 136

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  4. Asked: 3 years agoIn:

    ABC is a triangle in which the angles are in the ratio 3 : 4 : 5 . PQR is a triangle in which the angles are in the ratio 5 : 6 : 7. The difference between the least angle of ABC and the least angle of PQR is aº. Then a = _________

    deb97

    3x + 4x + 5x = 180°12x = 180x = 15Least angle of triangle ABC is = 3 × 15 = 45°5x + 6x + 7x = 180°18x = 180°x = 10Least angle of triangle PQR = 5 × 10 = 50Difference a° = 50 – 45a° = 5

    3x + 4x + 5x = 180°

    12x = 180

    x = 15

    Least angle of triangle ABC is = 3 × 15 = 45°

    5x + 6x + 7x = 180°

    18x = 180°

    x = 10

    Least angle of triangle PQR = 5 × 10 = 50

    Difference a° = 50 – 45

    a° = 5

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  5. Asked: 3 years agoIn:

    A cube has edge length x (an integer). three faces meeting at a corner are painted blue. The cube is then cut into smaller cubes of unit length. If exactly 343 of these cubes have no faces painted blue, then the value of x is ________ .

    7d05d

    (x – 1)3 = 343⇒ x – 1 = 7⇒ x = 8

    (x – 1)3 = 343

    ⇒ x – 1 = 7

    ⇒ x = 8

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  6. Asked: 3 years agoIn:

    The numbers 2, 3, 12, 14, 15, 20, 21 may be divided into two sets so that the product of the numbers in each set is the same. What is this product?

    (A) 420 

    (B) 1260 

    (C) 2520 

    (D) 6720

    e2e8c

    Correct option: (C) 2520Explanation:P2 = 2 × 3 × 12 × 14 × 15 × 20 × 21=> P = 2520

    Correct option: (C) 2520

    Explanation:

    P2 = 2 × 3 × 12 × 14 × 15 × 20 × 21

    => P = 2520

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  7. Asked: 3 years agoIn:

    The sum of the ages of a man and his wife is six times the sum of the ages of their children. Two years ago the sum of their ages was ten times the sum of the ages of their children. Six years hence the sum of their ages will be three times the sum of the ages of their children. How many children do they have?

    49d44

    Let present age of man = MLet present age of wife = WLet the no. of children = xLet the sum of the ages of children = CATQM + W = 6 C         ... (1)M - 2 + W - 2 = 10 (C - 2x )        ... (2)M + 6 + W + 6 = 3 (C + 6x)     ... (3)From (2) M + W - 4 = 10 C - 20 xBy using (1) 6 C - 4 = 10 C - 20 x- 4CRead more

    Let present age of man = M

    Let present age of wife = W

    Let the no. of children = x

    Let the sum of the ages of children = C

    ATQ

    M + W = 6 C         … (1)

    M – 2 + W – 2 = 10 (C – 2x )        … (2)

    M + 6 + W + 6 = 3 (C + 6x)     … (3)

    From (2) M + W – 4 = 10 C – 20 x

    By using (1) 6 C – 4 = 10 C – 20 x

    – 4C + 20x = 4

    -C + 5x = 1       …. (5)

    From (3) M + W + 12 = 3C + 18x

    By using (1) 6C + 12 = 3C + 18x

    3C – 18x = – 12

    C – 6x = – 4       …. (6)

    From (5) & (6)

    – C + 5x = 1

    C – 6x = – 4

    ∴ – x = – 3

    x = 3

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  8. Asked: 3 years agoIn:

    The difference between the biggest and the smallest three digit number each of which has different digits is

    (A) 864 

    (B) 875 

    (C) 885 

    (D) 895

    96ef8

    Correct option: (C) 885Explanation:Biggest three digit number with distinct digit = 987Smallest three digit number with distinct digit = 102Difference = 987 – 102 = 885.Option C is correctThe biggest three  digit number having three different digit is 987The smallest three  digit number having threeRead more

    Correct option: (C) 885

    Explanation:

    Biggest three digit number with distinct digit = 987

    Smallest three digit number with distinct digit = 102

    Difference = 987 – 102 = 885.

    Option C is correct

    The biggest three  digit number having three different digit is 987

    The smallest three  digit number having three different digit is 102

    So the difference = 987-102=885

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  9. Asked: 3 years agoIn:

    We have four sets S1, S2, S3, S4 each containing a number of parallel lines. The set Si contains i + 1 parallel lines i = 1,2,3,4. A line in Si is not parallel to lines in Sj when i ≠ j. In how many points do these lines intersect ?

    (A) 54 

    (B) 63 

    (C) 71

    (D) 95

    08b75

    Correct option: (C) 71Explanation:Number of point of intersection = 2C1 (3C1 + 4C1 + 5C1) + 3C1 (4C1 + 5C1) + 4C1 × 5C1 = 2(12) + 3(9) + 20= 24 + 27 + 20= 71

    Correct option: (C) 71

    Explanation:

    Number of point of intersection = 2C1 (3C1 + 4C1 + 5C1) + 3C1 (4C1 + 5C1) + 4C1 × 5C1 

    = 2(12) + 3(9) + 20

    = 24 + 27 + 20

    = 71

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  10. Asked: 3 years agoIn:

    Let a, b, c be real numbers, not all of them are equal. Prove that if a + b + c = 0, then a2 + ab + b2 = b2 + bc + c2 = c2 + ca + a2.

    Prove the converse, if a2 + ab + b2 = b2 + bc + c2 = c2 = ca + a2, then a + b + c = 0.

    f948d

    I. a2 + ab + b2 = (– b – c)2 + (– b – c) b + b2     {as a + b + c = 0, a = – b – c}= (b2 + c2 + 2bc) – b2 – bc + b2= b2 + c2 + bcSimilarly b2 + c2 + bc = c2 + ca + a2∴ a2 + ab + b2 = b2 + c2 + bc = c2 + a2 + ac.II. a2 + ab + b2 = b2 + bc + c2a2 + ab – bc – c2 = 0a2 – c2 + b (a – c) = 0(a – c) (a + cRead more

    I. a2 + ab + b2 = (– b – c)2 + (– b – c) b + b2     {as a + b + c = 0, a = – b – c}

    = (b2 + c2 + 2bc) – b2 – bc + b2

    = b2 + c2 + bc

    Similarly b2 + c2 + bc = c2 + ca + a2

    ∴ a2 + ab + b2 = b2 + c2 + bc = c2 + a2 + ac.

    II. a2 + ab + b2 = b2 + bc + c2

    a2 + ab – bc – c2 = 0

    a2 – c2 + b (a – c) = 0

    (a – c) (a + c + b) = 0

    a – c = 0 or a + b + c = 0

    as a – c = 0 is not possible as a, b, c are not equal.

    ∴ a + b + c = 0.

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