Quantitative Aptitude рдкреНрд░рд╢реНрди рдФрд░ рдЙрддреНрддрд░ рдХрд╛ рдЕрднреНрдпрд╛рд╕ рдХрд░реЗрдВ
рдкреНрд░: In how many different ways can the letters of the word 'ABYSMAL' be arranged ? 1645 05b5cc6c9e4d2b4197774df88
5b5cc6c9e4d2b4197774df88- 15040false
- 23650false
- 34150false
- 42520true
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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- SingleChoice
рдЙрддреНрддрд░ : 4. "2520"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: D) 2520 Explanation: Total number of letters in the word ABYSMAL are 7 ┬а Number of ways these 7 letters can be arranged are 7! ways ┬а But the letter is repeated and this can be arranged in 2! ways ┬а Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.
рдкреНрд░: The difference between the time taken by two trains to travel a distance of 350 km is 2 hours 20 minutes. If the difference between their speeds is 5 km/hr, what is the speed of faster train ? 2388 05b5cc6c9e4d2b4197774df6a
5b5cc6c9e4d2b4197774df6a- 136 kmphfalse
- 230 kmphtrue
- 334 kmphfalse
- 440 kmphfalse
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 2. "30 kmph"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: B) 30 kmph Explanation: Let the speed of the faster train be 'S' kmph ┬а Then speed of the slower train will be '(S-5)' kmph ┬а Time taken by faster train = 350/S hrs ┬а Time taken by slower train = 350/(S-5) hrs ┬а350s-5┬а-┬а350s┬а=┬а2┬аhrs┬а20┬аmin┬а=┬а213┬а=┬а73 ┬а => S = 30 km/hr.
рдкреНрд░: The price of 2 oranges, 3 bananas and 4 apples is Rs. 15. The price of 3 oranges, 2 bananas and 1 apple is Rs. 10. What will be price of 4 oranges, 4 bananas and 4 apples ┬а? 1405 05b5cc6c9e4d2b4197774df5a
5b5cc6c9e4d2b4197774df5a- 1Rs. 10false
- 2Rs. 15false
- 3Rs. 20true
- 4Rs. 25false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 3. "Rs. 20"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: C) Rs. 20 Explanation: Let the Oranges be 'O', Bananas be 'B' and Apples be 'A' From the given data, ┬а 2O + 3B + 4A = 15 ...... (I) 3O + 2B + A = 10 ..... (II) On adding (I) and (II), 5O + 5B + 5A = 25 O + B + A = 5 4O + 4B + 4A = 20.
рдкреНрд░: In the following number series only one number is wrong. Find out the wrong number ? 6, 15, 35, 63, 143, 221, 323 1864 05b5cc6c8e4d2b4197774df41
5b5cc6c8e4d2b4197774df41- 1143false
- 263true
- 3323false
- 435false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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- SingleChoice
рдЙрддреНрддрд░ : 2. "63"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: B) 63 Explanation: Here the series follows the pattern that, product of succesive prime numbers i.e, ┬а2 ├Ч 3 = 6 ┬а3 ├Ч 5 = 15 ┬а5 ├Ч 7 = 35 ┬а7 ├Ч 11 = 77 ┬а11 ├Ч 13 = 143 ┬а13 ├Ч 17 = 221 ┬а17 ├Ч 19 = 323 So here in place of 77 the number 63 is written. So the wrong number is 63.
рдкреНрд░: What is the difference between the number of uneducated females in city P and that of uneducated males in city S ? 1397 05b5cc6c8e4d2b4197774df37
5b5cc6c8e4d2b4197774df37- 19 lakhstrue
- 26 lakhsfalse
- 33 lakhsfalse
- 42 lakhsfalse
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 1. "9 lakhs"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: A) 9 lakhs Explanation: Educated females in city P = 25 lakhs Uneducated females in city P = 30 - 25 = 5 lakhs Educated males city S = 21 lakhs Educated males in city S = 35 - 21= 14 lakhs Required difference = 14 - 5 = 9 lakhs
рдкреНрд░: One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn find the probability the card drawn is red. 1379 05b5cc6c8e4d2b4197774df1e
5b5cc6c8e4d2b4197774df1e- 12/3false
- 21/2true
- 31/52false
- 413/51false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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- SingleChoice
рдЙрддреНрддрд░ : 2. "1/2"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: B) 1/2 Explanation: We know that: When one card is drawn from a pack of 52 cards The numbers of possible outcomes n(s) = 52 We know that there are 26 red cards in the pack of 52 cardsтЗТ The numbers of favorable outcomes n(E) = 26 Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S) тИ┤ required probability = 26/52 = 1/2.
рдкреНрд░: A fair coin is tossed 11 times. What is the probability that only the first two tosses will yield heads┬а? 1734 05b5cc6c8e4d2b4197774df23
5b5cc6c8e4d2b4197774df23- 1(1/2)^11true
- 2(9)(1/2)false
- 3(11C2)(1/2)^9false
- 4(1/2)false
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 1. "(1/2)^11"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: A) (1/2)^11 Explanation: Probability of occurrence of an event,┬а P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S) тЗТ Probability of getting head in one coin = ┬╜,┬а тЗТ Probability of not getting head in one coin = 1- ┬╜ = ┬╜,┬а Hence,┬а All the 11 tosses are independent of each other. тИ┤ Required probability of getting only 2 times heads =122├Ч129=1211
рдкреНрд░: The breadth of a rectangular field is 75% of its length. If perimeter of the field is 1050 m, what will be th area of field ? 1557 05b5cc6c8e4d2b4197774df0a
5b5cc6c8e4d2b4197774df0a- 165700 sq.mfalse
- 254500 sq.mfalse
- 378700 sq.mfalse
- 467500 sq.mtrue
- рдЙрддреНрддрд░ рджреЗрдЦреЗрдВрдЙрддреНрддрд░ рдЫрд┐рдкрд╛рдПрдВ
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рдЙрддреНрддрд░ : 4. "67500 sq.m"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :
Answer: D) 67500 sq.m Explanation: Let the length be 'l' and breadth be 'b'. b = l ├Ч 3/4__________(a) 2(l+b) = 1050 l+b = 525___________(b) From equations (a) and (b), l = 300m, b = 225 m Area = l ├Ч b = 300 ├Ч 225 = 67500 sq.m.

