Q: A dealer sold two of his cattle for Rs. 500 each. On one of them he lost 10% on the other, he gained 10%. His gain or loss percent in the entire transaction was:
Solution: Loss%= (common gain or loss % / 10)2 = (10/10)² % = 1%. The correct answer is: B) 1% loss
Q: If the arithmetic mean of seventy-five numbers is calculated, it is 35. if each number is increased by 5, then mean of new numbers is :
Solution: Let the numbers are x1,x2,.......x75 By Hypothesis ,x1+x2+....x7575=35 ......(i) Mean of new numbers = x1+5+x2+5+.......x75+575 x1+x2+......x75+5*7575 x1+x2+......x7575+37575=35+5=40 By (i) The correct answer is: B) 40
Q: In a simultaneous throw of pair of dice. Find the probability of getting the total more than 7. Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12. }
Solution: Here n(S) = (6 x 6) = 36 Let E = event of getting a total more than 7 = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)
The correct answer is: B) 5/12Q: The sum of the number of boys and girls in a school is 150. if the number of boys is x, then the number of girls becomes x% of the total number of students. The number of boys is :
Solution: We have : x + (x% of 150 )= 150 => x+(x/100)*150] = 150 => 52x = 150 => x = (150*2)/5 = 60 The correct answer is: A) 60
Q: In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of:
Solution: To obtain Rs. 10, investment = Rs. 96. To obtain Rs. 650, investment = Rs.96*65010 = Rs. 6240. The correct answer is: B) 6240
Q: A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:
Solution: The correct answer is: B) 1 dm
Q: Find the missing number in the following Series ? 18, 97, 26, 91, ? , 83
Solution: The given series follows the pattern that, 18, 97, 26, 91, ?, 83 => 18 + 8 = 26 => 97 - 8 = 91 => 26 + 8 = 34 => 91 - 8 = 83 Hence, the missing number is 34. The correct answer is: D) 34
Q: In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys ?
Solution: The toys are different; The boxes are identical If none of the boxes is to remain empty, then we can pack the toys in one of the following ways a. 2, 2, 1 b. 3, 1, 1 Case a. Number of ways of achieving the first option 2 - 2 - 1 Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be selected in 3C2 ways and the last toy can be selected in 1C1 way. However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2 Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways 5C2*3C2= 15 ways Case b. Number of ways of achieving the second option 3 - 1 - 1 Three toys out of the 5 can be selected in 5C3 ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way. Therefore, total number of ways of getting the 3 - 1
Q: Which of the following can be used to illustrate that not all prime numbers are odd?
Solution: The only even numbers in the list are 2 and 4, but 4 is not a prime. So 2 can be used to illustrate the statement that all primes are not odd. The correct answer is: B) 2
Q: If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle.
Solution: let length = x and breadth = y then 2(x+y) = 46 => x+y = 23 x²+y² = 17² = 289 now (x+y)² = 23² => x²+y²+2xy= 529 => 289+ 2xy = 529 => xy = 120 area = xy = 120 sq.cm The correct answer is: B) 120
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