Quantitative Aptitude – Modern Maths – Progressions – If a1 = 1/(2*5), a2 = 1/(5*8)

January 1st, 2020 by

Quantitative Aptitude - Modern Maths - Progressions - If a1 = 1/(2*5), a2 = 1/(5*8) Quantitative Aptitude - Modern Maths - Progressions Question If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is A) 25/151 B) 1/2 C) 1/4 D) 111/55 Answer Option (A) Solution From CAT 2017 - Quantitative Aptitude - Modern Maths - Progressions, we can see that, The 100th term will be 1/299*302 The series is: 1/(2*5) + 1/(5*8) + 1/(8*11) + ……………+ 1/(299*302) It can also be written as 3[1/2

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Quantitative Aptitude – Modern Maths – Progressions – An infinite GP a1, a2, a3,….

January 1st, 2020 by

Quantitative Aptitude - Modern Maths - Progressions - An infinite GP a1, a2, a3,.... Quantitative Aptitude - Modern Maths - Progressions Question An infinite geometric progression a1, a2, a3,... has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +….... = 32, then a5 is A) 1/32 B) 2/32 C) 3/32 D) 4/32 Answer Option (C) Solution From CAT 2017 - Quantitative Aptitude - Modern Maths - Progressions, we can see that, For any n ≥ 1, an = 3 (a(n+1) + a(n+2) + ……..) So, a1 =

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Quantitative Aptitude – Modern Maths – P&C – How many four digit numbers

January 1st, 2020 by

Quantitative Aptitude - Modern Maths - P&C - How many four digit numbers Quantitative Aptitude - Modern Maths - Permutation and Combination Question How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position? Answer 50 Solution From CAT 2017 - Quantitative Aptitude - Modern Maths - Permutation and Combination, we can see that, For a number to be divisible by 6, it should have both 2 and 3 as factors.

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Quantitative Aptitude – Modern Maths – P&C – In how many ways can 8 identical

January 1st, 2020 by

Quantitative Aptitude - Modern Maths - P&C - In how many ways can 8 identical Quantitative Aptitude - Modern Maths - Permutation and Combination Question In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens? Answer 6 Solution From CAT 2017 - Quantitative Aptitude - Modern Maths - Permutation and Combination, we can see that, Number of pens that Amal gets = a+1 Number of pens that Bimal gets = b+2 Number of pe

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Quantitative Aptitude – Modern Maths – Progressions – Let a1, a2, a3, a4, a5

January 1st, 2020 by

Quantitative Aptitude - Modern Maths - Progressions - Let a1, a2, a3, a4, a5 Quantitative Aptitude - Modern Maths - Progressions Question Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3. If the sum of the numbers in the new sequence is 450, then a5 is Answer 51 Solution From CAT 2017 - Quantitative Aptitude - Modern Maths - Progressions, we can see that, 5 consecutive odd numbers are a1 , a2 , a3 , a4 , a5. 5 consecutive even number

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Quantitative Aptitude – Modern Maths – Progressions – Let a1, a2,……..a3n be an arithmetic progression

December 30th, 2019 by

Quantitative Aptitude - Modern Maths - Progressions - Let a1, a2,……..a3n be an arithmetic progression Quantitative Aptitude - Modern Maths - Progressions Question Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830? A) 8 B) 9 C) 10 D) 11 Answer Option (B) Solution From CAT 2017 - Quantitative Aptitude - Modern Maths - Progressions, we can see that, a = 3 a + d = 7 => d=4 Apply

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Quantitative Aptitude – Modern Maths – Permutation and Combination – In how many ways can 7 identical erasers

December 30th, 2019 by

Quantitative Aptitude - Modern Maths - Permutation and Combination - In how many ways can 7 identical erasers Quantitative Aptitude - Modern Maths - Permutation and Combination Question In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers? A) 16 B) 20 C) 14 D) 15 Answer Option (A) Solution From CAT 2017 - Quantitative Aptitude - Modern Maths - Permutation and Combination, we can see that, a + b + c + d = 7 Since, each

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Quantitative Aptitude – Modern Maths – Progressions – If the square of the 7th term

December 30th, 2019 by

Quantitative Aptitude - Modern Maths - Progressions - If the square of the 7th term Quantitative Aptitude - Modern Maths - Progressions Question If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is A) 2 : 3 B) 3 : 2 C) 3 : 4 D) 4 : 3 Answer Option (A) Solution From CAT 2017 - Quantitative Aptitude - Modern Maths - Progressions, we can see that, (a+6d)^2 = (a+2d)(a+16d) a^2 + 12

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Quantitative Aptitude – Modern Maths – Progressions – Let a(base1), a(base2)

December 23rd, 2019 by

Quantitative Aptitude – Modern Maths - Progressions – Let a(base1), a(base2) Slot -2 – Quantitative Aptitude – Modern Maths - Progressions – Let a(base1), a(base2), ...a(base52)  Let a(base1), a(base2), ... , a(base52) be positive integers such that a(base1) < a(base2) < ... < a(base52). Suppose, their arithmetic mean is one less than the arithmetic mean of a(base2), a(base3), ..., a(base52). If a(base52) = 100, then the largest possible value of a(base1) is? a) 45 b) 20 c) 48 d) 23 Answer: d) 23

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Quantitative Aptitude – Modern Maths – Sequence and Series – Let t1, t2,… be real numbers

December 23rd, 2019 by

Quantitative Aptitude – Modern Maths - Sequence and Series – Let t1, t2,… be real numbers Slot -2 – Quantitative Aptitude – Modern Maths - Sequence and Series – Let t1, t2,… be real numbers Let t1, t2,… be real numbers such that t1+t2+…+tn = 2n2+9n+13, for every positive integer n ≥ 2. If tk=103, then k equals? Answer: 24 Solution: Given, t(base1)+t(base2)+…+t(base n) = 2n^2+9n+13 So t(base1) + t(base2) = 2×2^2+9×2+13=39 t(base1) + t(base2) + t(base3) = 2×3^2+9×3+13=58 means t(base 3) = 58 –

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