Quantitative Aptitude Practice Question and Answer
8Q: A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B ? 1704 05b5cc6e7e4d2b4197774ed6f
5b5cc6e7e4d2b4197774ed6f- 110 yrstrue
- 211 yrsfalse
- 312 yrsfalse
- 413 yrsfalse
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Answer : 1. "10 yrs"
Explanation :
Answer: A) 10 yrs Explanation: Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years. (2x + 2) + 2x + x = 27 5x = 25 => x = 5 Hence, B's age = 2x = 10 years.
Q: Raju sold an article by giving a discount of 8% for Rs. 17,940 and earn profit of 19.6 %. If he did not give discount then how much profit percentage he gets ? 1701 05b5cc6cde4d2b4197774e1a9
5b5cc6cde4d2b4197774e1a9- 127%false
- 230%true
- 332%false
- 425%false
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Answer : 2. "30%"
Explanation :
Answer: B) 30% Explanation: Let the M.R.P of an article be 100 % Selling price of article =Rs. 17940 If he give a discount of 8 % then S.P = 92 % 92 % = 17490 100 % = 19500 Now C.P of the article is 17490 x 100/119.6 = 15000 If he did not give the discount, then Required profit percentage =
Q: In how many different ways can the letters of the word 'HAPPYHOLI' be arranged? 1699 05b5cc6a7e4d2b4197774ce0e
5b5cc6a7e4d2b4197774ce0e- 189,972false
- 290,720true
- 372,000false
- 481,000false
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Answer : 2. "90,720"
Explanation :
Answer: B) 90,720 Explanation: The given word HAPPYHOLI has 9 letters These 9 letters can e arranged in 9! ways. But here in the given word letters H & P are repeated twice each Therefore, Number of ways these 9 letters can be arranged is 9!2! x 2! = 9 x 8 x 7 x 6 x 5 x 4 x 32 = 90,720 ways.
Q: At present, the ratio between the ages of Arun and Harish is 4:3. After 6 years, Arun's age will be 26 years. What is the age of Harish at present ? 1699 05b5cc6e1e4d2b4197774ebfc
5b5cc6e1e4d2b4197774ebfc- 115 yearstrue
- 216 yearsfalse
- 318 yearsfalse
- 421 yearsfalse
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Answer : 1. "15 years"
Explanation :
Answer: A) 15 years Explanation: Let the present ages of Arun and Harish be 4x and 3x years respectively. Then, 4x + 6 = 26 => x = 5 Harish's age = 3x = 15 years.
Q: P can do a piece of work in 5 less days than Q. If both of them can do the same work in 1119days, in how many days will Q alone do the same work ? 1699 05b5cc6c2e4d2b4197774dbcd
5b5cc6c2e4d2b4197774dbcd- 124 daysfalse
- 225 daystrue
- 320 daysfalse
- 419 daysfalse
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Answer : 2. "25 days"
Explanation :
Answer: B) 25 days Explanation: Let Q complete that work in 'L' days => 1L + 1L-5 = 9100 => 9L2-245L+500=0 L = 25 days.
Q: Prasanna invested certain amount in three different schemes X, Y and Z with the rate of interest 10% p.a, 12% p.a and 15% p.a respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme Z was 150% of the amount invested in Scheme X and 240% of the amount invested in Scheme Y, what was the amount invested in Scheme Y by Prasanna ? 1699 05b5cc6c5e4d2b4197774ddb9
5b5cc6c5e4d2b4197774ddb9- 1Rs.6000false
- 2Rs.4500false
- 3Rs.7500false
- 4Rs.5000true
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Answer : 4. "Rs.5000"
Explanation :
Answer: D) Rs.5000 Explanation: Let a, b and c be the amounts invested in schemes X, Y and Z respectively. Then, As we know: Simple interest (S.I.) = PTR/100 (a × 10 × 1/100) + (b × 12 × 1/100) + (c × 15 × 1/100) = 3200 = 10a + 12b + 15c = 320000 .........(1) Now, c = 240% of b = 12b/5 .........(2) And, c = 150% of a = 3a/2 => a = 2/3 c = (2 × 12)b/(3 × 5) = 8b/5 .......(3) From (1), (2) and (3), we have 16b + 12b + 36b = 320000 => 64b = 320000 => b = 5000 ∴ Sum invested in Scheme Y = Rs.5000.
Q: The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is 1697 05b5cc7b4e4d2b419777508a1
5b5cc7b4e4d2b419777508a1- 17! x 7!false
- 27! x 6!true
- 36! x 6!false
- 47! x 5!false
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Answer : 2. "7! x 6!"
Explanation :
Answer: B) 7! x 6! Explanation: The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in 7P6ways. 7 teachers can sit in (7-1)! ways. Required no.of ways is = 7P6.6! = 7!.6!
Q: What should come in place of x in the following equation? x128 = 162x 1697 05b5cc6bee4d2b4197774d9ab
5b5cc6bee4d2b4197774d9ab- 113false
- 212true
- 317false
- 416false
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Answer : 2. "12"
Explanation :
Answer: B) 12 Explanation: x128 = 162x Then x2 = 162x128 = Sqrt of 82 x 62 x 32 = 8 x 6 x 3 x2 = 144. x = 12.