280 IBPS Mocks for Rs. 299 – Enroll Now<\/a><\/p><\/div>\n Take a free mock test for IBPS PO Download IBPS PO Previous Papers PDF<\/a><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n Question 1:\u00a0<\/b>What is the last digit of the sum $131^{23}+147^{45}+153^{52}+139^{32}$ ?<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a07<\/p>\n d)\u00a06<\/p>\n e)\u00a04<\/p>\n Instructions<\/b><\/p>\n There are 450 students in a college. Each student has to choose one or more elective out of management, history and physics. Further following information is also known – Question 2:\u00a0<\/b>How many students selected both management and physics as elective?<\/p>\n a)\u00a084<\/p>\n b)\u00a0107<\/p>\n c)\u00a075<\/p>\n d)\u00a089<\/p>\n e)\u00a098<\/p>\n Instructions<\/b><\/p>\n In each of the following questions a question is asked followed by three statements.You have to study the questions and all the given three statements and all the statements and decide whether any information provided in the statement are redundant and can be dispensed with while answering the questions ?<\/p>\n Question 3:\u00a0<\/b>what is the area of the given right angle triangle ? a)\u00a0I only<\/p>\n b)\u00a0II only<\/p>\n c)\u00a0II or III only<\/p>\n d)\u00a0II and III both<\/p>\n e)\u00a0Data inadequate<\/p>\n IBPS PO Quant Formulas & Shortcuts<\/a><\/p>\n IBPS PO Previous papers<\/a><\/p>\n Question 4:\u00a0<\/b>A train running at speed of 120 kmph crosses a signal in 15 seconds .What is the length of the train in meters?<\/p>\n a)\u00a0300<\/p>\n b)\u00a0200<\/p>\n c)\u00a0500<\/p>\n d)\u00a0Cannot determined<\/p>\n e)\u00a0None of these<\/p>\n Question 5:\u00a0<\/b>Rs 2520 is distributed among W,X,Y and Z and W:X=3:5, X:Y=7:8 ,Y:Z =4:3 then what is the share of X ?<\/p>\n a)\u00a0Rs.700<\/p>\n b)\u00a0Rs.420<\/p>\n c)\u00a0Rs.800<\/p>\n d)\u00a0Rs.600<\/p>\n e)\u00a0None of these<\/p>\n Instructions<\/b><\/p>\n Compare the values and select the correct option.<\/p>\n Question 6:\u00a0<\/b>X:$x^{2}+4x-221$=0 a)\u00a0X < Y<\/p>\n b)\u00a0X > Y<\/p>\n c)\u00a0X $\\geq$ Y<\/p>\n d)\u00a0X $\\leq$ Y<\/p>\n e)\u00a0X = Y or they cannot be compared<\/p>\n IBPS PO Free Mock Test<\/a><\/p>\n Question 7:\u00a0<\/b>A shopkeeper buys a dozen fruits for Rs. 3. At what price must he sell the fruits if he wants to realize a profit of 20 %?<\/p>\n a)\u00a09 fruits for 30 rupees<\/p>\n b)\u00a06 fruits for 15 rupees<\/p>\n c)\u00a010 fruits for 6 rupees.<\/p>\n d)\u00a090 fruits for 30 rupees<\/p>\n e)\u00a050 fruits for 15 rupees<\/p>\n Question 8:\u00a0<\/b>What is the probability of selecting a number from 1 to 100 which is divisible by either 2 or 3 ?<\/p>\n a)\u00a00.83<\/p>\n b)\u00a00.67<\/p>\n c)\u00a00.81<\/p>\n d)\u00a00.33<\/p>\n e)\u00a00.50<\/p>\n Instructions<\/b><\/p>\n What comes in the place of the ?<\/p>\n Question 9:\u00a0<\/b>6,35,143,323,?<\/p>\n a)\u00a0391<\/p>\n b)\u00a0567<\/p>\n c)\u00a0609<\/p>\n d)\u00a0437<\/p>\n e)\u00a0667<\/p>\n IBPS PO Previous Papers (Download Pdf)<\/a><\/p>\n Question 10:\u00a0<\/b>A work can be finished In 14 days by 36 workers. If the work were to be finished in 8 days, how many additional workers would be required ?<\/p>\n a)\u00a029<\/p>\n b)\u00a033<\/p>\n c)\u00a023<\/p>\n d)\u00a031<\/p>\n e)\u00a027<\/p>\n Question 11:\u00a0<\/b>A is 60% more efficient than B. In how many days will A and B together complete a piece of work if A alone can complete the work in 15 days?<\/p>\n a)\u00a0$9\\frac{7}{13}$<\/p>\n b)\u00a0$8\\frac{9}{13}$<\/p>\n c)\u00a0$8\\frac{4}{13}$<\/p>\n d)\u00a0$10\\frac{1}{13}$<\/p>\n e)\u00a0$9\\frac{3}{13}$<\/p>\n Question 12:\u00a0<\/b>A and B together can complete a particular task in 8 days. If B alone can complete the same task in 10 days, how many days will A take to complete the task if he works alone?<\/p>\n a)\u00a028<\/p>\n b)\u00a036<\/p>\n c)\u00a040<\/p>\n d)\u00a032<\/p>\n e)\u00a0None of these<\/p>\n Question 13:\u00a0<\/b>For equation 2a + 5b = 108 find the number of pairs of positive integers a and b that satisfy this equation?<\/p>\n a)\u00a015<\/p>\n b)\u00a08<\/p>\n c)\u00a010<\/p>\n d)\u00a012<\/p>\n e)\u00a09<\/p>\n Question 14:\u00a0<\/b>P starts business with Rs. 55000 and after 4 months, Q joins with P as his partner. After a year, the profit is divided in the ratio 5:4. What is Q’s capital contribution?<\/p>\n a)\u00a0Rs. 76000<\/p>\n b)\u00a0Rs. 80000<\/p>\n c)\u00a0Rs. 75000<\/p>\n d)\u00a0Rs. 66000<\/p>\n e)\u00a0Rs. 60000<\/p>\n Instructions<\/b><\/p>\n Question 15:\u00a0<\/b>Compound interest on a certain sum of money for 3 years at a rate of $16\\dfrac{2}{3}$% is Rs 12700. What is the simple interest on the same sum of money,same time period and same rate of interest ?<\/p>\n a)\u00a0Rs.11200<\/p>\n b)\u00a0Rs.10200<\/p>\n c)\u00a0Rs.10400<\/p>\n d)\u00a0Rs.10800<\/p>\n e)\u00a0Rs.10000<\/p>\n Question 16:\u00a0<\/b>A shopkeeper wants to buy a rod of certain length such that he can measure 140 cm, 420 cm and 280 cm precisely. What is the maximum possible length of the rod that the shopkeeper can buy?<\/p>\n a)\u00a070 cm<\/p>\n b)\u00a0140 cm<\/p>\n c)\u00a035 cm<\/p>\n d)\u00a07 cm<\/p>\n e)\u00a020 cm<\/p>\n Question 17:\u00a0<\/b>If two sides of a triangle are 6 cm and 4 cm then how many integral values are possible for the third side ?<\/p>\n a)\u00a06<\/p>\n b)\u00a07<\/p>\n c)\u00a08<\/p>\n d)\u00a09<\/p>\n e)\u00a010<\/p>\n Instructions<\/b><\/p>\n In each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and Question 18:\u00a0<\/b>What is the area of right angle triangle ? a)\u00a0If the data in statement I alone are sufficient to answer the question, while the data in statement<\/p>\n b)\u00a0If the data in statement II alone are sufficient to answer the question, while the data in statement<\/p>\n c)\u00a0If the data either in statement I alone or in statement II alone are sufficient to answer the question<\/p>\n d)\u00a0If the data given in both statements I and II together are not sufficient to answer the question and<\/p>\n e)\u00a0If the data in both statements I and II together are necessary to answer the question.<\/p>\n Instructions<\/b><\/p>\n Study the following table and answer the questions that follow. Some values are missing. You need to calculate them as per given information.<\/p>\n Question 19:\u00a0<\/b>If the average marks of B and D in Maths is 102.5, then find the average marks of all the students in Maths?<\/p>\n a)\u00a093.5<\/p>\n b)\u00a087.65<\/p>\n c)\u00a097<\/p>\n d)\u00a0102<\/p>\n e)\u00a0Can\u2019t be determined<\/p>\n Instructions<\/b><\/p>\n Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.<\/p>\n Question 20:\u00a0<\/b>Quantity I: Find the units digit of $9562^{2469}$. a)\u00a0Quantity I > Quantity II<\/p>\n b)\u00a0Quantity I < Quantity II<\/p>\n c)\u00a0Quantity I $\\geq$ Quantity II<\/p>\n d)\u00a0Quantity I $\\leq$ Quantity II<\/p>\n e)\u00a0Quantity I = Quantity II or The relationship between Quantity I and Quantity II cannot be established.<\/p>\n IBPS PO Important Questions PDF<\/a><\/p>\n Free Banking Study Material (15,000 Solved Questions<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n In the given terms $131^{23}$ has 1 in the units digit as any power which has 1 in units digit will return 1 in units digit. 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Let \u2018x\u2019 be the number of students who took all the three electives and \u2018y\u2019 be the number of students who took only management as an elective. Considering students who took management as an elective we have,<\/p>\n y + x + 84 + 75 = 238<\/p>\n x + y = 79<\/p>\n Considering students who took history as an elective we have,<\/p>\n 137 – y + x + 84 + 52 = 240<\/p>\n y – x = 33<\/p>\n Adding both the equation we get,<\/p>\n 2y = 112<\/p>\n y = 56<\/p>\n So, x = 23<\/p>\n So, number of students who took all the three electives = 23<\/p>\n Number of students who took only management = 56<\/p>\n Number of students who took only history = 137 – 56 = 81<\/p>\n So, the number of students who took only physics = 450 – 240 – 56 – 75 = 79<\/p>\n Thus, we get following venn diagram –<\/p>\n Number of students who selected both management and physics as elective = 75 + 23 = 98 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n To find the area of right angled triangle, we need to know the base and height or the length of any side if its a\u00a030-60-90 triangle<\/p>\n From statement I and III, two angles are 90\u00b0 and 60\u00b0, => Third angle = 30\u00b0<\/p>\n In a 30-60-90 triangle, the hypotenuse is always twice as long as the side opposite the 30\u00b0 angle and the side opposite the 60\u00b0 angle is\u00a0\u221a3 times as long as the side opposite the 30\u00b0 angle.<\/p>\n The ratio of sides opposite 30\u00b0, 60\u00b0 and 90\u00b0 angles =1\u00a0: $\\sqrt{3}$ : 2<\/p>\n Length\u00a0of the side opposite the 90\u00b0 angle (hypotenuse) = 5 cm<\/p>\n => Length of side opposite the 30\u00b0 angle = $\\frac{5}{2}=2.5$ cm<\/p>\n and length of side opposite to 60\u00b0 angle = $2.5 \\sqrt{3}$ cm<\/p>\n $\\therefore$ Area of triangle = $\\frac{1}{2} \\times 2.5 \\times 2.5 \\sqrt{3} = 3.125\\sqrt{3}$ $cm^2$<\/p>\n Thus, statement II is redundant.<\/p>\n => Ans – (B)<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Speed of train = 120 kmph<\/p>\n = $(120 \\times \\frac{5}{18})$ m\/s = $\\frac{100}{3}$ m\/s<\/p>\n Let length of the train = $l$ meters<\/p>\n Using, speed = distance\/time<\/p>\n => $\\frac{100}{3} = \\frac{l}{15}$<\/p>\n => $l=\\frac{100}{3} \\times 15$<\/p>\n => $l=100 \\times 5=500$ meters<\/p>\n => Ans – (C)<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given W:X=3:5 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $x^{2}+4x-221$=0 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Cost price of a fruit = 3\/12 = Rs. 0.25 Let us check the options one by one.<\/p>\n 9 fruits\/ 30 rupees => Cost of 1 fruit = 30\/9 = Rs.3.33 Therefore, option E is the right answer.<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n we have 100\/2 =50 numbers that are divisible by 2 in the first 100 numbers 9)\u00a0Answer\u00a0(E)<\/strong><\/p>\n here in this question given is the product of two consecutive prime numbers 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Using the formula, $M_1 D_1 = M_2 D_2$<\/p>\n => $36 \\times 14 = M_2 \\times 8$<\/p>\n => $M_2 = \\frac{36 \\times 14}{8} = 63$<\/p>\n $\\therefore$ Additional workers = 63 – 36 = 27<\/p>\n 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Time taken by A alone = 15 days<\/p>\n Efficiency of A\u00a0: B = 100\u00a0: 160<\/p>\n = 15\u00a0: 24<\/p>\n => Time taken by B = 24 days<\/p>\n $\\therefore$ (A + B)’s 1 day’s work<\/p>\n = $\\frac{1}{15} + \\frac{1}{24}$<\/p>\n = $\\frac{8 + 5}{120} = \\frac{13}{120}$<\/p>\n $\\therefore$ Required time = $\\frac{1}{\\frac{13}{120}}$<\/p>\n = $\\frac{120}{13} = 9\\frac{3}{13}$ days<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the total work to be done = 40 units<\/p>\n Rate at which B alone finishes the task = $\\frac{40}{10}$ = 4 units\/day<\/p>\n Rate at which\u00a0A & B together finishes the work = $\\frac{40}{8}$ = 5 units\/day<\/p>\n => Rate at which A alone finishes the work = 5 – 4 = 1 units\/day<\/p>\n $\\therefore$ Time taken by A to complete the task = $\\frac{40}{1}$ = 40 days<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 2a + 5b = 108 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let Q contribute Rs. x 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n We have $P((1+\\dfrac{50}{300})^{3}-1)=12700$<\/p>\n $P((1+\\dfrac{1}{6})^{3}-1)=12700$<\/p>\n $P(127\/216)=12700$ 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The lengths 140 cm, 280 cm and 420 cm should be the multiples of the length of the rod. The maximum possible length of the rod will be equal to the HCF of these 3 numbers.<\/p>\n 140 = 2*70 As we can see, the HCF is 2*70 = 140 cm. 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n For a triangle to form the sum of two sides should be greater than the third side and the difference of two sides should be less than the third side 18)\u00a0Answer\u00a0(E)<\/strong><\/p>\n I : Let base of the triangle = $4x$ m<\/p>\n => Height of triangle = $\\frac{3}{4} \\times 4x = 3x$ m<\/p>\n There is no other info, so I alone is insufficient.<\/p>\n Similarly, II alone is also insufficient.<\/p>\n Combining both statements, and using pythagoras theorem, we get\u00a0:<\/p>\n => $(3x)^2 + (4x)^2 = (5^2)$<\/p>\n => $9x^2 + 16x^2 = 25$<\/p>\n => $x^2 = \\frac{25}{25} = 1$<\/p>\n => $x = \\sqrt{1} = 1$<\/p>\n => Base = 4 m and height = 3 m<\/p>\n $\\therefore$ Area of triangle = $\\frac{1}{2} \\times 3 \\times 4 = 6 m^2$<\/p>\n Thus, both statements together are sufficient.<\/strong><\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, average marks of B and D in Maths is 102.5 Marks obtained by A in Maths = 46.67% of 150 = $\\dfrac{7}{15}\\times150 = 70$<\/p>\n Marks obtained by C in Maths = 66.67% of 150 = $\\dfrac{2}{3}\\times150 = 100$<\/p>\n Marks obtained by E in Maths = 73.33% of 150 = $\\dfrac{11}{15}\\times150 = 110$<\/p>\n Therefore, Required average = $\\dfrac{70+100+110+205}{5} = \\dfrac{485}{5} = 97$<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Quantity I: Quantity II: Therefore, Quantity I < Quantity II<\/p>\n \u00a0<\/b><\/strong>
\n<\/a><\/p>\n
\n1) 75 students selected only management and physics.
\n2) 84 students selected only management and history.
\n3) 52 students selected only physics and history.
\n4) The number of students who selected only history is 137 less the number of students who selected only management.
\n5) In total 238 students selected management as an elective.
\n6) In total 240 students selected history as an elective.<\/p>\n
\nI. Length of the hypotenuse is 5 cm
\nII.perimeter of the triangle is four times its base
\nIII. one of the angles of the triangle is 60 degrees<\/p>\n
\nY:$y^{2}+36y+323$=0<\/p>\n
\nGive answer
\n(A) If the data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question
\n(B) If the data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question
\n(C) If the data either in statement I alone or in statement II alone are sufficient to answer the question
\n(D) If the data given in both statements I and II together are not sufficient to answer the question and
\n(E) If the data in both statements I and II together are necessary to answer the question.<\/p>\n
\nI.Height of right-angled triangle is \u00be th of its base
\nII.Length of the diagonal of the right angle triangle is 5 meters<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/DI18.JPG\")
<\/figure>\n
\nQuantity II: Find the units digit of $2567^{5242}$.<\/p>\n
\n$147^{45}$ has 7 in its units place and we have 7,9,3 and 1 as the power cycle of 7 and 45 when divided by 4 gives 1 as the remainder and so units digit will be $7^1 = 7$
\n$153^{52}$ has 3 in units place and power cycle of 3 has 3,9,7 and 1 and so when 52 is divided by 4 we have 0 as remainder and so units digit will be 1
\n$139^{32}$ has 9 in its units place and its power cycle has only 9 and 1 and so when 32 is divided by 2 we have 0 as remainder therefore units digit is 1
\nTherefore units digit of $131^{23}+147^{45}+153^{52}+139^{32}$ is 1+7+1+1=10.
\nUnits digit = 0.<\/p>\n
\nWe get following venn diagram –<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/venn_1.PNG\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/venn_2.PNG\")
<\/figure>\n
\nHence, option E is the correct choice.<\/p>\n
\nX:Y=7:8
\nW:X=21:35
\nX:Y=35:40
\nW:X:Y=21:35:40
\nY:Z=4:3
\nY:Z=40:30
\nW:X:Y:Z=21:35:40:30
\nGiven total share is Rs 2520
\nShare of X is $\\dfrac{35}{126}\\times2520$
\n= Rs.700<\/p>\n
\n$x^{2}-13x+17x-221$=0
\nx(x-13)+17(x-13)=0
\n(x+17)(x-13)=0
\nx=-17 and x=13
\n$y^{2}+36y+323$=0
\n$y^{2}+17y+19y+323$=0
\ny(y+17)+19(y+17)=0
\n(y+19)(y+17)=0
\ny=-19 and y=-17
\nSo X is either greater or equal to Y<\/p>\n
\nSelling price = 1.2*0.25 = Rs.0.3<\/p>\n
\n6 fruits\/15 rupees => Cost of 1 fruit = 15\/6 = Rs. 2.5
\n10 fruits\/ 6 rupees => Cost of 1 fruit = 6\/10 = Rs. 0.6
\n90 fruits\/30 rupees => Cost of 1 fruit = 30\/90 = Rs. 0.33
\n50 fruits\/ 15 rupees => Cost of 1 fruit = 15\/50 = Rs. 0.3<\/p>\n
\nSimilarly 100\/3 =33 numbers divisible by 3 in the first 100 numbers
\nBut both of them have some numbers and they are repeated twice so the LCM of 2 and 3 is 6
\n100\/6 =16 multiples of 6 are present in first 100 and so they are to be subtracted from them i.e
\n50+33-16=67
\nTherefore required probability = 67\/100 = 0.67.<\/p>\n
\ni.e 6=2*3
\n35=5*7
\n143=11*13
\n323=17*19
\n667=23*29<\/p>\n
\n108 – 5b = 2a
\nWe have to get an even number by subtracting a multiple of 5 from 108. So b must be an even number.
\nThere are 21 multiple of 5 below 108. Out of those 11 are odd, and 10 are even.
\nFor only 10 even multiples will get subtraction as an even number.
\nSo b = 2, 4, 6, \u2026, 20
\nHence, option C is the right answer.<\/p>\n
\nWe have $\\frac{55000*12}{x*8} = \\frac{5}{4}$
\nx = Rs. 66000
\nHence, option D is the right choice.<\/p>\n
\n$P((1+\\dfrac{r}{100})^{3}-1)=12700$<\/p>\n
\n$P=21600$
\nWe have P=Rs.21600 R=50\/3% t=3 years
\nSI=$\\dfrac{21600\\times50\\times3}{100\\times3}$
\n= Rs.10800<\/p>\n
\n280 = 4*70
\n420 = 6*70<\/p>\n
\nTherefore, option B is the right answer.<\/p>\n
\nTherefore (6+4)>s>(6-4)
\n10>s>2
\nSo all the values between 2 and 10 are possible as third side
\ni.e 3,4,5,6,7,8 and 9
\nTherefore 7 values are possible<\/p>\n
\nThen, Total marks obtained by B and D in Maths = 205.<\/p>\n
\nUnits digit of $9562^{2469}$ = Units digit of $2^{2469}$
\n$2^1 = 2$
\n$2^2 = 4$
\n$2^3 = 8$
\n$2^4 = 16$. Units digit = 6
\n$2^5 = 32$. Units digit = 2
\nHence, The cyclicity of 2 is 4.
\nDividing 2469 by 4, we get a remainder of 1.
\nHence, The units digit of $2^{2469}$ = Units digit of $2^1 = 2$<\/p>\n
\nUnits digit of $2567^{5242}$ = Units digit of $7^{5242}$
\n$7^1 = 7$
\n$7^2 = 49$. Units digit \u2192 9.
\n$7^3 = 343$. Units digit \u2192 3.
\n$7^4 = 2401$. Units digit \u2192 1.
\n$7^5 = 16807$. Units digit \u2192 7.
\nHence, The cyclicity of 7 is 4.
\nDividing 5242 by 4, we get a remainder of 2.
\nHence, The units digit of $7^{5242}$ = Units digit of $7^2 = 9$<\/p>\n