Bank PO Exam Questions with Solution - Top Method

Bank PO Exam Questions with Answer

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Q. 1 The Length of Rectangle is three fifth of the side of a square. The radius of a circle is equal to side of the square. The circumference of the circle is 132 cm. What is the area of the rectangle if the breadth of the rectangle is 8cm ?

Solution

2∏r = 132

2 * 22/7 * r = 132

r  = (132 *7 )/( 2 * 22)  = 21 cm

side of square is = 21 cm

length of rectangle  = (3 * 21)/ 5 cm

so area of rectangle is = (3 * 21)/ 5 * 8 = 100.8 Sq. cm Ans.

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Q. 2 Five Fifth of a number is equal to twenty five percent of second number.  Second Number is equal to one fourth of third number. The value of third number is 2960. What is 30 % of the First number?

Solution

Second Number = ¼ * 2960 = 740

First number  * 5/9  = (740* 25)/100

So first number = 740/4 * 9/5  = 333

So 30% of 333  = (333* 30)/ 100  = 99.9 Ans.

Q. 3 The respective Ratio B/W present age of Manoj and Wasim is 3  :11. Wasim is 12 years younger than Rehana. Rehana’s age after 7 years will be 85 years. What is the present age of Manoj”s father who is 25 years older than manoj?

Solution

Rehana’s present age is  = 85 -7 = 78 years

Wasim’s present age is  = 78 – 12 = 66 years

Manoj’s present age is  = 3/11 * 66 = 18 years

Manoj’s father’s present age is   = 25 + 18  = 43 years ans.

Q. 4 Dinesh’s monthly icome is four times suresh’s monthly income. Suresh’s monthly income is 20 % more than joyti’s monthly income is Rs 22, 000. What is the dinesh’s monthly income?

Solution

Suresh’s monthly income  = ( 22000* 120)/100  = Rs 26400

So dinesh monthly income 

= ( 4 * 26400)  = Rs ( 105600) ans.

Q. 5 Smallest side of a right angled triangle is 8 cm less than the side of a squre of perimeter 56 cm. Second largest side of the right angled triangles is 4 cm less than the length of a rectangle of area 96 square cm and breadth 8 cm. What is the largest side of the right angled triangle?

Solution

Side of square = perimeter/4  = 56/4 = 14 cm

So smallest side of right angled  = 14 -8 = 6cm.

Length of rectangle  = area/breadth

                                   = 96/8  = 12 cm.

So second side of the triangle  = 12 -4 = 8cm

Then hypotenuse of the right angled triangle  = √ 6² + 8²

                                                                                     = √ 36 + 64

                                                                                  = √100  = 10cm ans.

Q. 6 The Ratio between the adjacent angles of a parallelogram is 7 : 8 respectively. Also the ratio between the angles of quadrilateral is 5 : 6: 7 : 12. What is the sum of the smaller angle of parallelogram and second largest angle of quadrilateral?

Solution 

Let the adjacent angles be 7x degree and 8 x degree

Then 7 x + 8 x = 180

=>   15x = 80

=>    x = 12

So that smaller angle  = 7 * 12  = 84 degree

Again 5y + 6y + 7y + 12y = 360 degree

30y = 360 degree

Y = 360/30  = 12 degree

Then second largest angle of the quadrilateral  = 7 * 12 = 84 degree

Required sum  = 84 + 84  = 168 degree Ans.

Q. 7  Raju runs 1250 metre on Monday and Friday. Other days he runs 1500 meters except for Sunday. How many kilometers will he run in 3 weeks ( first day starting from Monday)?

Solution

Total distance covered is = 3 ( 2 * 1250 + 4 *1500) metre

                                              =  3 ( 2500 + 6000) metre

                                               25500 metre   = 25.5 km Ans.

Q. 8 The sum of consecutive odd numbers of set – A is 621. What is the sum of different set of six consecutive even numbers whose lowest number is 15 more than the lowest number of set – A ?

Solution

Fifth number of set A  = 621/9  = 69

Smallest number of Set A  = 61

Then smallest number of set B  = 61 + 15  = 76

Required sum  = 76 + 78 + 80 + 84 + 86  = 486 ans.

Q. 9 In a school there are 250 students out of whom 12 percent are girls. Each girls monthly fee is Rs 450 and each boy’s monthly fee is 24% more than a girl. What is the total monthly fee of girls and boys together?

Solution

Monthly fee of each boy is  =( 450 * 124 )/100  = Rs 558

Number of girls  = (250 *12)/100  = 30

Number of boys  = Rs 220

Total monthly fee  = Rs ( 220 * 558 + 30 * 450)

                          = ( 122760 + 13500)  = 136260 Rs ans

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Q. 10 The average speed of a train is 1^3/7 times the average speed of a car. The car covers a distance of 588 km in 6 hours. How much distance will the train cover in 13 hours?

Solution

Average speed of car  = distance/time  = 588/6  = 98kmph

Average speed of train is  = (98 * 10 )/7  = 140 kmph

Distance covered by train in 13 hours  = speed * time

                                                                      = 140 * 13  = 1820 Km.

Q. ( 11 – 15) What will come in place of x in the following questions?

Q. 11 (21)²  - 3717 ÷ 59 = x * 8

Solution

(21)²  - 3717 ÷ 59 = x * 8

441 – 3717/59  = 8x

441 – 63 = 8x

X =  378/8

  = 47.25 Ans.

Q. 12 2 ¹/8  - 1 ¹/17 = x + 1 ¹/ 32 – 1⁹/64

Solution

X = 2 – 1 – 1 + 1 + ( 1/8 + 1/16 – 1/32 + 9/64)

X = 1 + 11/64 = 1 ^11/64 Ans.

Q. 13 ( 0.64)⁴ ÷ (0. 512)³ * (0.8)⁴ = (0.8)^{x + 3}

Solution

((0.8)²)⁴  ÷ ( 0.8³)³  * (0.8)⁴   =  (0.8)^x+3

(0.8)^{8 -9 +4}   = (0.8)^x+3

X + 3  = 3

X = 0 Ans.

Q. 14 34.5 % of 1800 + 12.4 % of 1500 = (x)³ + 78

Solution

(1800 * 34.5)/ 100  +  (1500 * 12.4)/100   =   (x)³ + 78

621 + 186  = (x)³ + 78

(x)³  =   807 – 78  = 729

X = 9 Ans.

Q. 15  √(15)² * 12 ÷ (9) – 125 + 21 = x

Solution

√ (225 * 12)/9  - 125 + 21

√ 300 – 125 + 21

√196 = 14 Ans.

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SBI PO Math Sample Paper 2017

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Q. 1 On republic day sweets were to be equally distributed among 450 children. But on that particular day 150 children remained absent. Thus each got 3 sweets extra, how many sweets did child get?

Solution

Number of sweets for 150 absent children

= 300 * 3 = 900

Number of sweets each student gets = 900/150  + 3

                                                                   = 9 Ans.

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Q. 2 In how many different ways can the letters of the word “trust” be arranged?

Solution

The word trust consists of five letters in which T comes twice

Number of arrangement = └5/ L2

= ( 5 * 4 * 3 * 2 * 1)/(2 *1)

= 60 Ans.

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Q. 3 Amit, Sucheta and neeti strat running around a circular track and complete one round in 18, 24 and 32 seconds respectively. In how many seconds will the three meet again at the starting point if they all started running at the same time?

Solution

Required time   = LCM of 18, 24 and 32

                            = 288 Seconds Ans.

Q. 4 A bsket contains three blue and four red balls. If three balls are drawn at random from the basket, what is the probability that all three are either blue or red?

Solution

Number of possible outcomes

N(s)  = 7c₃   = (7 *6 * 5)/(1 *2 *3)

= 35

Favorable number of cases

N (e) = 3c₃ + 4c₃

=          1 + 4 = 5

Required probability

N (e)/n (S)  = 5/35

= 1/7 ans.

Q. 5 A man buys a single apple for Rs 25. If he were to buy  a dozen apples, he would have to pay a total amount of Rs 250. What would be the approximate percent discount he would get on buying a dozen apples?

Solution

Required percentage = (50 *100)/ 300 = 17ans

Q. 6 6 men can complete a piece of work  in 12 days. 8 women can complete the same piece of work in 18 days wheres 18 childeren can complete the piece of work in 10 days. 4 men , 12 women and 20 chilkdren work together for 2 days. If only men were to complete the remaing work in 1 day how many men would be required totally.

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Solution

72 men = 144 women = 180 childeren

2 men = 4 women = 5 children

4 men +12 women + 20 children

= 4 men + 6 Men + 8 Men = 18 men

When 6 men can complete a work in 12 days

18 men will do the same in 4 days

Then remaining work = ½

M1 D1/W1 = M2 D2/W2

= ½ * 6 * 12 = 1 * m2

M2 = 36 number of men

Q. 7 Percent Profit Earned six company’s over the years is given table

Company/Year	P	Q	R	S	T	U
2004	11	12	3	7	10	6
2005	9	10	5	8	12	6
2006	4	5	7	13	12	5
2007	7	6	8	14	14	7
2008	12	8	9	15	13	5
2009	14	12	11	15	14	8

(Q 8 -13) Study the above table carefully to answer the questions that given below

Q. 8 If the profit Earned by company R in the year 2008 Rs 18.9 lakhs, what was the income in that year?

Solution

If the expenditure be Rs X lacs then

18.9/x * 100  = 9

=> x = (18.9*100)/9

= 210 lakh

So income = Rs ( 210 + 18.9) lakh

Rs 228.9 lacs ans.

Q. 9 What is the percentage rise in percent profit of company T in year 2009 from the year 2004?

Solution

Percentage increase is

= 4/10 * 100

= 40% ans.

Q. 10 If the profit earned by company P in the year 2007 was Rs 2.1 lakhs, what was its expenditure in that year?

Solution

Required expenditure of the company

= (2.1 * 100)/7

= Rs 30 lacs Ans.

Q. 11 What was the average percent profit of company S over all the years together?

Solution

Average percent profit

= (7 + 8 + 13 + 14 + 15 + 15)/6

= 12% Ans.

Q. 12 What is the difference between the percent profit earned by company Q in the year 2005 and the average profit earned by the remaining companies together in that year?

Solution

Required difference   = 10 – ( {9+5 + 8 + 12 +6)/5})

10 – 8 = 2 % ans. 

Time Speed & Distance Aptitude Questions with Solution - Best Method

Time Speed & Distance Problems with Solution

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Q. 1 In covering a certain distance, the speeds of A & B are in the ratio 3 : 4,  A takes 30 minutes more than B to reach the destination. The Time Taken by A to reach the destination is?

Ratio of speeds  = 3: 4

Then ratio of time taken  = 4 : 3

Suppose a takes 4x hrs and B takes 3x hrs to reach destitaion, then

4x  - 3x  = 30/60  = ½ or  x= ½

Time taken by A  = 4 *1/2  = 2hrs ans.

Q. 2 A car covers a distance of 715 Km at a constant speed, if the speed of the car would have been 10 km/hr more, then it would have taken 2 hours less to cover the same distance. What is the original speed of the car is?

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Answer

Let the original speed be x km/hr then 715/x – 715/(x +10)  = 2

2x(x +10)  = 7150

X² +10x- 3575 = 0

X = 55  km/hr Ans.

Q. 3 Excluding stoppage, the speed of a bus is 54 kmph and including stoppages it is 45 Kmph, for how many minutes does the bus stop per hour ?

Answer

Due to stoppage it covers 9 km less

Then, taken to cover 9 km  = 9/54 * 60

Min  = 10 minute ans

Q. 4 A train can travel 50% faster than a car, Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way however, the train lost about 12.5 minutes while stopping the station. The speed of the car is?

Answer

Let the speed of car x kmph then speed of the train

= 150/100x  = 3/2 kmph

75/x – 75/3/2x  = 125/(10*60)

= 75/x – 50/x = 5/24

X  = ( 25*25)/5 = 120kmps Ans.

Q. 5 A car travel from P to Q at a constant speed. If its speed were increased by 10 Km/hr, it would have taken one hour less to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. what is the distance between the two cities?

Let distance = x km and usual rate y kmph,

Then x/y – x/(y+10) = 1 or

Y ( y+10)  = 10x

And x/y – x/(y+2o)  = 7/4

Or y (y + 20)  = 80/7 x

On dividing 1 by II we get y = 60

Substituting  y = 60 we get X  = 420 km Ans

Q. 6 A man covered a certain distance at some speed. Had he moved 3 Kmph fater would have taken 40 minutes less. If he had moved 2kmph slower, he would taken 40 minute more, the diatnce in Km is?

Answer

Let distance  = Km snd usual rate  = Y

X/y – X/ (Y+3)  = 40/60

Or  2Y ( Y+3)  = 9X

And X/(Y-2) – X/Y  = 40/60

Or Y ( Y-2)  -  3X

On dividing 1 by II we get  X  = 40 Km ans

Q. 7 If a trains runs at 40 kmph, it reaches its destination late by 11 minutes but if it runs at 50 kmph, it is late by 5 minutes only. The correct time for the train to its journey is?

Answer

Req. distance  = (T1 + T2)/60 * (S1S2)/(S2 – S1)

(11+5)/60 * (40*50)/(50 – 40)

Distance  = (16/60) * (40 * 50)/10  = 160/3km

Actual Time  = (160/3 * 40) – 11/60 =  4/3 – 11/60

ð  (80-11)/60  = 69/60 hr   = 69 min Ans

Q .8 Two men start together to walk to a certain destination, one at 3 kmph and another at 3.75 kmph. The latter arrives half an hour before the former. The distance is?

Answer

Let the distance be X km, then

X/3  - X/3.75  = ½

=>  2.5X – 2X = 3.75

X = 3.75/0.50  =  15/2  = 7.5 Km Ans.

Q. 9 A train when moves at an average speed of 40kmph, reaches its destination. When its average speed becomes 35 kmph, then it reaches its destination 15 late, find the distance of journey?

Answer

Difference between timing  = 15 min  = ¼ hr

Let the length of journey is X km

Then X/35  - X/40   = ¼

= 8X  - 7 X  = 70

X = 70 Km ans

Q. 10 Walking 6/7^th of his actual speed, a man is 12 minutes too late,  The actual time taken by him to cover that distance is?

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Answer

New speed  = 6/7 of usual speed

New time = 6/7 of usual time

(7/6 of usual time) – ( usual time)

= 1/5 hr  = 1/6 of usual time = 1/5 hr

Usual time = 6/5 hr  = 1 hr  12 min

Q. 11 A man can reach a certain place in 30 hrs. if he reduces his speed 1/15^th in 10 km less in that time, find his speed?

Answer

Let the speed is km/hr then

30X – 30 * 14/15X = 10

= > 2X = 10

X  = 5 km/hr  Ans

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