Q: Entry fee in an exhibition was Rs. 1. Later, this was reduced by 25% which increased the sale by 20%. The percentage increase in the number of visitors is :
Solution: Let the total original sale be Rs. 100. Then, original number of visitors = 100. New number of visitors = 120/0.75 = 160. Increase % = 60 %. The correct answer is: C) 60 %
Q: A box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains exactly one defective bulb.
Solution: Total number of elementary events = 10C5 Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is 3C1*7C4 So,required probability =3C1*7C4/10C5 = 5/12. The correct answer is: A) 5/12
Q: In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C will beat B by:
Solution: A : B = 200 : 169. A : C = 200 : 182. C/B = (C/A*A/B) = (182/200*200/169) = 182:169 When C covers 182 m, B covers 169 m. When C covers 350 m, B covers (169/180*350)m = 325 m Therefore, C beats B by (350 - 325) m = 25 m. The correct answer is: B) 25 m
Q: What is the next number in the given Number Series? 5, 24, 94, 279, ?
Solution: The given number series follows a pattern that 5 x 5 – 1 = 24 24 x 4 – 2 = 94 94 x 3 – 3 = 279 279 x 2 – 4 = 554 The correct answer is: B) 554
Q: In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?
Solution: The word EDUCATION is a 9 letter word, with none of the letters repeating.The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonantsAs the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways.Hence, the total number of ways = 4! × 5! The correct answer is: C) 4! x 5!
Q: A bag contains 50 P, 25 P and 10 P coins in the ratio 5: 9: 4, amounting to Rs. 206. Find the number of coins of each type respectively.
Solution: let ratio be x. Hence no. of coins be 5x ,9x , 4x respectively Now given total amount = Rs.206 => (.50)(5x) + (.25)(9x) + (.10)(4x) = 206 we get x = 40 => No. of 50p coins = 200 => No. of 25p coins = 360 => No. of 10p coins = 160 The correct answer is: C) 200, 360,160
Q: Find the odd man out. 6, 9, 15, 21, 24, 28, 30
Solution: Each of the numbers except 28, is a multiple of 3. The correct answer is: A) 28
Q: If the mean of a, b, c is M and ab + bc + ca = 0, then the mean of a2+b2+c23 is :
Solution: We have : ( a + b + c) / 3 = M or (a + b + c) = 3M. Now. a+b+c2=3M2=9M2 . <=>a2+b2+c2+2ab+bc+ca=9M2 a2+b2+c2=9M2 ,(ab+bc+ca=0) Required mean =a2+b2+c23=9M23=3M2 . The correct answer is: B) 3 M^2
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