Bank Exams Practice Question and Answer
8Q: A class consists of both boys and girls along with a teacher. After a class, the teacher drinks 9 litres of water, a boy drinks 7 litres of water and a girl drinks 4 litres of water. If after a class 42 litres of water was consumed, find the number of girls in the class ? 1045 05b5cc745e4d2b4197774fa2c
5b5cc745e4d2b4197774fa2c- 18false
- 26false
- 35false
- 43true
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Answer : 4. "3"
Explanation :
Answer: D) 3 Explanation: Given teacher drinks 9 ltrLet number of boys be 'A'.Let number of girls be 'B'.Boy drinks 7 ltr and girl drinks 4 ltrAfter class total water consumed = 42 ltrThen,9 + 7A + 4B = 42=> 7A + 4B = 33By trial and error method,The only integers which satisfy the equation is A = 3 and B = 3Therefore, number of girls in the class = 3.
Q: A batsman scores 26 runs and increases his average from 14 to 15. Find the runs to be made if he wants top increasing the average to 19 in the same match ? 2725 05b5cc745e4d2b4197774fa22
5b5cc745e4d2b4197774fa22- 174true
- 279false
- 372false
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Answer : 1. "74"
Explanation :
Answer: A) 74 Explanation: Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12To raise the average by one (from 14 to 15), he scored 12 more than the existing average.Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.
Q: Out of sixty students, there are 14 who are taking Economics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class ? 1549 05b5cc745e4d2b4197774fa1d
5b5cc745e4d2b4197774fa1d- 18/15false
- 27/15true
- 31/15false
- 44/15false
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Answer : 2. "7/15"
Explanation :
Answer: B) 7/15 Explanation: Given total students in the class = 60Students who are taking Economics = 24 andStudents who are taking Calculus = 32Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4Students who are taking calculus only = 32 - 4 = 28probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.
Q: When I was married 10 years ago my wife is the 6th member of the family. Today my father died and a baby born to me.The average age of my family during my marriage is same as today. What is the age of Father when he died ? 4332 05b5cc745e4d2b4197774fa18
5b5cc745e4d2b4197774fa18- 150 yrsfalse
- 260 yrstrue
- 370 yrsfalse
- 465 yrsfalse
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Answer : 2. "60 yrs"
Explanation :
Answer: B) 60 yrs Explanation: Let the Father be x years when he died Average Age 10 years ago be A Total Age 10 years ago = 6*A Total Age after 10 years(Just before father's Death) = 6A + 6*10 = 6A + 60 Father Died and Baby was born => the Total number of people in the family is Same (6) Baby born today so age of baby = 0 (6A +60 - x)/6 = 6A/6 => A + 10 -(x/6) = A => x/6 = 10 => x = 60 Therefore we can conclude that the father was 60 years old when he died.
Q: A man covers a distance of 1200 km in 70 days resting 9 hours a day, if he rests 10 hours a day and walks with speed 1½ times of the previous in how many days will he cover 840 km ? 2242 05b5cc745e4d2b4197774fa0e
5b5cc745e4d2b4197774fa0e- 139 daysfalse
- 237 daysfalse
- 335 daystrue
- 433 daysfalse
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Answer : 3. "35 days"
Explanation :
Answer: C) 35 days Explanation: Distance d = 1200kmlet S be the speed he walks 15 hours a day(i.e 24 - 9)so totally he walks for 70 x 15 = 1050hrs. S = 1200/1050 => 120/105 = 24/21 => 8/7kmph given 1 1/2 of previous speedso 3/2 * 8/7= 24/14 = 12/7 New speed = 12/7kmph Now he rests 10 hrs a day that means he walks 14 hrs a day. time = 840 x 7 /12 => 490 hrs=> 490/14 = 35 days So he will take 35 days to cover 840 km.
Q: A simple interest earned on certain amount is triple the money when invested for 16 years.what is the interest rate offered ? 925 05b5cc745e4d2b4197774fa13
5b5cc745e4d2b4197774fa13- 113.33 %false
- 214.25 %false
- 316.98 %false
- 418.75 %true
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Answer : 4. "18.75 %"
Explanation :
Answer: D) 18.75 % Explanation: Given,S.I = 3 Principal Amount=> 3A = A x 16 x R/100By solving, we get => R = 18.75%
Q: Sony is one-fifth the age of her mother 15 years ago and Sony’s brother is three-fifth the age of his mother 10 years ago. If the sum of Sony and her brother’s ages is 31 then how old is Sony's mother ? 2253 05b5cc745e4d2b4197774fa04
5b5cc745e4d2b4197774fa04- 140false
- 250true
- 360false
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Answer : 2. "50"
Explanation :
Answer: B) 50 Explanation: Sony and her brother's age sum is 31Let Sony's mother's age be 'x'.Now the conditions given for both of their ages related to their mom is1/5(x-15) + 3/5(x-10) = 31x-15 + 3x-30 = 1554x-45 = 1554x = 200x = 50So, Mother's age = 50. Cross verification : 15 years ago age = 35 => Sony age = 710 years ago age = 40 => bro age = 24 Sum = 24 + 7 = 31.
Q: If x = 2 + 223 + 213 then find the value of x3-6x2+6x ? 1070 05b5cc745e4d2b4197774fa09
5b5cc745e4d2b4197774fa09- 11false
- 22true
- 33false
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Answer : 2. "2"
Explanation :
Answer: B) 2 Explanation: Given that x = 2+ 223 + 213 So x - 2 = 223 + 213 Now cubing on both sides, we get => x3-8-6x2+12x = 4 + 2 + 6(x-2) Therefore, x3-6x2+6x = -6 + 8 = 2