790+ Mocks – Just Rs. 194. Use coupon: SBIDREAM70<\/a><\/p>\n Download SBI Clerk Puzzles Set-1 PDF<\/a><\/p>\n Question 1:\u00a0<\/b>In 8 years, Ramesh will be twice as old as Farheen was 8 years ago. If Ramesh is now 16 years older than Farheen, the present age of Farheen is :<\/p>\n a)\u00a032<\/p>\n b)\u00a040<\/p>\n c)\u00a036<\/p>\n d)\u00a044<\/p>\n e)\u00a024<\/p>\n Question 2:\u00a0<\/b>A was thrice as old as B when C was 2 years old. A will be twice as old as B when C is 10 years old. How old will B be when the difference between the ages of A and B will be equal to the age of C?<\/p>\n a)\u00a016 years<\/p>\n b)\u00a022 years<\/p>\n c)\u00a018 years<\/p>\n d)\u00a024 years<\/p>\n e)\u00a032 years<\/p>\n Question 3:\u00a0<\/b>Find the approximate value of $\\Large\\frac{169.87^{2}-13.94^{2}}{183.87}$ $\\large+$ $\\Large\\frac{3211.96}{44.002}$<\/p>\n a)\u00a0224<\/p>\n b)\u00a0236<\/p>\n c)\u00a0239<\/p>\n d)\u00a0229<\/p>\n e)\u00a0None of these<\/p>\n Download SBI Clerk Previous Papers PDF<\/a><\/p>\n Take a free mock test for SBI Clerk<\/a><\/p>\n Question 4:\u00a0<\/b>Find the approximate value of $17.33+142.895-76.795+235.008-4.779+38.102$<\/p>\n a)\u00a0321<\/p>\n b)\u00a0351<\/p>\n c)\u00a0281<\/p>\n d)\u00a0371<\/p>\n e)\u00a0None of these<\/p>\n Question 5:\u00a0<\/b>The average marks of a class increased by $\\large\\frac{1}{3}$ when a student\u2019s marks were wrongly entered as 64 instead of 46. The number of students in the class are<\/p>\n a)\u00a054<\/p>\n b)\u00a058<\/p>\n c)\u00a063<\/p>\n d)\u00a057<\/p>\n e)\u00a0None of these<\/p>\n Question 6:\u00a0<\/b>The average age of students in a class is 12. If the average age of boys is 18 and the average age of girls is 14 and If the number of boys are 22, then find the total number of students in the class<\/p>\n a)\u00a052<\/p>\n b)\u00a055<\/p>\n c)\u00a060<\/p>\n d)\u00a058<\/p>\n e)\u00a0None of these<\/p>\n Instructions<\/b><\/p>\n Select the correct option.<\/p>\n Question 7:\u00a0<\/b>Quantity 1:A number is increased by 85% and then it is decreased by 15% and result is approximately 357 a)\u00a0Quantity 1 $\\leq$Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 > Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Banking Study Material – 18000 Questions<\/a><\/p>\n Daily Free SBI Online Tests<\/a><\/p>\n Question 8:\u00a0<\/b>Quantity 1:If the cost price of total apples bought is Rs 288 and each apple is sold at Rs 14 each and the profit percent is 50\/3 then number of apples sold is a)\u00a0Quantity 1 $\\leq$Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 > Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Instructions<\/b><\/p>\n The following table shows the breakup of the marks scored by students of Class X of Sharda Mandir in Maths for 5 years from 2001-2005.<\/p>\n Note: A student can score a maximum of 100 marks.<\/p>\n Question 9:\u00a0<\/b>The students scoring below 40 are said to have failed the exam whereas the students scoring in between are 60-90 said to have passed the exam with distinction. a)\u00a0420:313<\/p>\n b)\u00a0417:305<\/p>\n c)\u00a0323:251<\/p>\n d)\u00a04:3<\/p>\n e)\u00a05:4<\/p>\n Question 10:\u00a0<\/b>The students scoring between 80 and 90 marks is said to have passed the exam with A grade. How many students got A grade during the given period?<\/p>\n a)\u00a0925<\/p>\n b)\u00a0775<\/p>\n c)\u00a0825<\/p>\n d)\u00a0875<\/p>\n e)\u00a0900<\/p>\n 200+ Free GK Tests<\/a><\/p>\n Instructions<\/b><\/p>\n Each of the question given below is followed by three statements. You have to study the questions and all the three statements given to decide whether any information provided in the statement\/s is\/are redundant and can be dispensed with while answering the given questions.<\/p>\n Question 11:\u00a0<\/b>What is the maximum marks that can be obtained in the examination?<\/p>\n I: Ravi secured 35% of the maximum marks and failed by 7 marks. III: Soham secured 82 marks which was 5 marks more than the passing marks.<\/p>\n a)\u00a0Only I<\/p>\n b)\u00a0Both II and III<\/p>\n c)\u00a0Both I and II<\/p>\n d)\u00a0Only II<\/p>\n e)\u00a0None of these<\/p>\n Question 12:\u00a0<\/b>What is the share of B in the annual profits? a)\u00a0Only II<\/p>\n b)\u00a0Both I and III<\/p>\n c)\u00a0Only III<\/p>\n d)\u00a0Either II or III<\/p>\n e)\u00a0All the statements are required to answer the question.<\/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.<\/strong><\/p>\n Question 13:\u00a0<\/b>Quantity 1: Volume of a sphere of 20 cm radius. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 14:\u00a0<\/b>Quantity 1: Area of an equilateral triangle with side equals to 32 cm. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 15:\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 16:\u00a0<\/b>2 numbers are in the ratio 5:6. If the difference between the LCM and HCF of the 2 numbers is 1740, what is the sum of the 2 numbers?<\/p>\n a)\u00a0600<\/p>\n b)\u00a0300<\/p>\n c)\u00a0360<\/p>\n d)\u00a0660<\/p>\n e)\u00a01200<\/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.<\/strong><\/p>\n Question 17:\u00a0<\/b>Quantity 1: Simple interest charged by a bank on a sum of Rs. 1000 at the rate of 22% annum for 2 years. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 18:\u00a0<\/b>If the difference in compound interest and simple interest at 20% per annum in two years for certain sum is 80. Then find the difference in the compound and simple interest on the same sum at 10% annum after 3 years?<\/p>\n a)\u00a060<\/p>\n b)\u00a068<\/p>\n c)\u00a075<\/p>\n d)\u00a062<\/p>\n e)\u00a055<\/p>\n Question 19:\u00a0<\/b>5 bottles of apple juice and 9 bottles of mango juice together fill 71 glasses. Also, 7 bottles of apple juice and 11 bottles of mango juice together fill 93 glasses. How many glasses can be filled with 3 bottles of apple juice and 4 bottles of mango juice if each bottle of either of the juices fills an integer number of glasses?<\/p>\n a)\u00a037<\/p>\n b)\u00a041<\/p>\n c)\u00a034<\/p>\n d)\u00a045<\/p>\n e)\u00a032<\/p>\n Question 20:\u00a0<\/b>An ATM machine contains 100 notes. The notes are of Rs. 5, Rs. 10 and Rs. 20 denominations. The total value of the notes in the machine is Rs.1500. If the number of Rs.10 and Rs. 20 notes is exchanged, the net value goes down by Rs. 400. The number of 20 rupee notes in the machine is<\/p>\n a)\u00a010<\/p>\n b)\u00a020<\/p>\n c)\u00a040<\/p>\n d)\u00a060<\/p>\n e)\u00a080<\/p>\n General Knowledge Questions & Answers PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let us assume the present age of Farheen = x years Ramesh\u2019s age after 8 years will be = x+16+8 = x+24 years $\\Rightarrow$ x+24 = 2*(x-8) So the present age of Farheen will be 40 years.<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the age of A and B initially be \u2018a\u2019 and \u2018b\u2019 respectively.<\/p>\n When C = 2, a\/b = 3 When C is as old as the difference between the ages of A and B, i.e, when C is 24-8 = 16 years old, age of B will be 8+14 = 22 years. Therefore, option B is the right answer.<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\Large\\frac{169.87^{2}-13.94^{2}}{183.87}$ $\\large+$ $\\Large\\frac{3211.96}{44.002}\\simeq \\frac{170^{2}-14^{2}}{184}$ $\\large+$ $\\Large\\frac{3212}{44}$<\/p>\n $= \\Large\\frac{(170+14)(170-14)}{184}$ $+$ $73$<\/p>\n $= \\Large\\frac{184\\times156}{184}$ $+$ $73$<\/p>\n $= 156+73 = 229$<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $17.33+142.895-76.795+235.008-4.779+38.102 \\simeq 17+143-77+235-5+38$<\/p>\n Adding all positive terms: Adding all negative terms: $\\therefore$ $17+143-77+235-5+38 = 433-82 = 351$<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the number of students be ‘x’<\/p>\n Given that the average is increased by $\\Large\\frac{1}{3}$ due to the increase of $18$ marks ( $64$ $-$ $46$ $=$ $18$ marks )<\/p>\n Actual increase in marks $=$ $\\Large\\frac{1}{3}$ $\\times$ x<\/p>\n $\\Rightarrow$ $\\Large\\frac{1}{3}$ $\\times$ x $=$ $18$<\/p>\n $\\therefore$ Number of students(x) $=$ $54$<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the total age of boys be \u2018B\u2019<\/p>\n Average age of boys = 18<\/p>\n $\\Large\\frac{B}{22}$ $=$ $18$<\/p>\n $\\Rightarrow$ B $=$ $396$<\/p>\n Let the total age of Girls be ‘G’<\/p>\n Let the number of girls be ‘x’<\/p>\n Average age of girls $=$ $14$<\/p>\n $\\Large\\frac{G}{x}$ $=$ $14$<\/p>\n $\\Rightarrow$ G $=$ $14$x<\/p>\n Total average of students $=$ $12$<\/p>\n Total age of students $=$ Total age of boys $+$ Total age of girls $=$ $396$ $+$ $14$x<\/p>\n Total number of students $=$ Number of boys $+$ Number of girls $=$ $22$ $+$ x<\/p>\n Average age of class $=$ $12$<\/p>\n $\\Rightarrow$ $\\Large\\frac{396 + 14x}{22 + x}$ $=$ $12$<\/p>\n $\\Rightarrow$ $396$ $+$ $14$x $=$ $462$ $+$ $12$x<\/p>\n $\\Rightarrow$ $2$x $=$ $66$<\/p>\n $\\Rightarrow$ x $=$ $33$<\/p>\n $\\therefore$ Number of girls $=$ $33$<\/p>\n Total number of students $=$ $22$ $+$ $33$ $=$ $55$<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In Q1 we have x*1.85*0.85=357 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In Q1 we have Total CP=288 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Number of students scoring in between 0-20 = Total students appeared – Number of students scoring above 20 Thus, the required ratio = 2085:1525 = 417:305 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Number of students scoring in between 0-20 = Total students appeared – Number of students scoring above 20 Thus, 875 students got A grade during the given period. 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n let the maximum marks be \u2018x\u2019.<\/p>\n Ravi = 0.35x<\/p>\n passing marks = 0.35x + 7<\/p>\n Passing marks is also equal to 82 – 5 = 77 marks<\/p>\n Thus, 0.35x + 7 = 77<\/p>\n x = 200<\/p>\n Thus statements I and III both are required to answer the question.<\/p>\n Using statement II alone or with any other statement we cannot find the answer.<\/p>\n Hence, option D is right answer.<\/p>\n 12)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Let the intial investment amount be a, b and c by A, B and C respectively. b = 2(a+c) 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Volume of the sphere of 20 cm radius = $\\dfrac{4\\pi}{3}\\times 20^3$ = $\\dfrac{32000\\pi}{3}$ 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Area of equilateral triangle with side equals to 30cm = $\\dfrac{\\sqrt{3}}{4}\\times 32^2$ = 443.40 sq.cm 15)\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. 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the 2 numbers be 5x and 6x. 17)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Simple interest paid to bank = 1000*.22*2 = Rs. 440 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let us assume \u2018P\u2019 is the principal which is borrowed. Difference between compound and simple interest = 0.44P – 0.4P = 0.04P Simple interest accrued in 3 years at 10% per annum = $\\frac{2000 * 10 * 3}{100}$ = 600 So the difference in C.I. and S.I. at 10% annum after 3 years = 662-600 = 62<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let 1 bottle of apple juice fill \u2018a\u2019 glasses and 1 bottle of mango juice fill \u2018m\u2019 glasses. 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the number of 5 rupee notes be x, 10 rupee notes be y and 20 rupee notes be z. Multiplying equation (1) by 5 and substituting it in equations (2) and (3), we get, (4)*3 – (5) => 40z = 2400
\nQuantity 2:A number is increased by 82% and then it is decreased by 12% and the result is approximately 355.5<\/p>\n
\nQuantity 2:An article is marked up by 40% an a discount of 30% was given then 13 times the loss percent is<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/di-2409-1-1.PNG\")
<\/figure>\n
\nWhat is the ratio of the number of students that failed and the number of students that got distinction during the given period?<\/p>\n
\nII: The difference between marks scored by Ravi and Soham is 12 marks.<\/p>\n
\nI: Initial amount invested by B is double of the amount invested by A and C together.
\nII: The company generated 50% of intial investment as profit in the third year
\nIII: A, B and C invested a total amount of Rs. 57000 for three years.<\/p>\n
\nQuantity 2: Volume of a right circular cone whose height is 60cm and radius of the circular base is 25 cm.<\/p>\n
\nQuantity 2: Area of a square with side equals to 20 cm.<\/p>\n
\nQuantity 2: Compound interest charged by another bank on a sum of Rs. 1000 at the rate of 20% annum for 2 years compounded annually.<\/p>\n
\nSo present age of Ramesh will be = x + 16 years<\/p>\n
\nFarheen\u2019s age 8 years ago = x – 8 years<\/p>\n
\n$\\Rightarrow$ x= 40 years<\/p>\n
\n=> a = 3b
\nWhen C = 10, 8 years have passed.
\n=> (a+8)\/(b+8) = 2
\na+8 = 2b + 16
\na = 2b+8
\n=> b = 8 years and a = 24 years.<\/p>\n
\n$17+143+235+38 = 433$<\/p>\n
\n$77+5 = 82$<\/p>\n
\nx=357\/(1.85*0.85)
\nx=227
\nIn Q2 we have x*1.82*0.88=355.5
\nx=355.5\/(1.82*0.88)
\nx=220
\nTherefore Q1>Q2<\/p>\n
\nLet the number of apples be x
\nTotal SP=14x
\nTherefore ((14x-288)\/(288))*100=50\/3
\n14x-288=48
\n14x=288+48
\nx=336\/14
\nx=24
\nIn Q2 we have CP=100
\nMP=100*1.4=140
\nSP=MP-discount
\nSP=140-140*(30\/100)
\nSP=98
\nLoss percent=((100-98)\/(100))*100
\n=2%
\n13 times the loss percent=26
\nTherefore Q2 > Q1<\/p>\n
\nNumber of students scoring in between 20-40 = Number of students scoring above 20 – Number of students scoring above 40
\nNumber of students scoring in between 40-60 = Number of students scoring above 40 – Number of students scoring above 60
\nNumber of students scoring in between 60-80 = Number of students scoring above 60 – Number of students scoring above 80
\nNumber of students scoring in between 80-90 = Number of students scoring above 80 – Number of students scoring above 90
\nNumber of students scoring in between 90-100 = Number of students scoring above 90<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/di-2409-1-2.PNG\")
<\/figure>\n
\nHence, option B is the correct answer.<\/p>\n
\nNumber of students scoring in between 20-40 = Number of students scoring above 20 – Number of students scoring above 40
\nNumber of students scoring in between 40-60 = Number of students scoring above 40 – Number of students scoring above 60
\nNumber of students scoring in between 60-80 = Number of students scoring above 60 – Number of students scoring above 80
\nNumber of students scoring in between 80-90 = Number of students scoring above 80 – Number of students scoring above 90
\nNumber of students scoring in between 90-100 = Number of students scoring above 90<\/p>\n<\/figure>\n
\nHence, option D is the correct answer.<\/p>\n
\nFrom III we have, a + b + c = 570002<\/p>\n
\n3(a + c) = 57000
\na+c = 19000
\nb = 38000
\nThird year profit = 0.5 * 57000 = 28500
\nAmount received by B = 28500 * $\\frac{38000}{57000}$ = Rs. 19000
\nThus only if we use all the three statements we are able to answer the question.
\nhence, option E is the right choice.<\/p>\n
\nVolume of right-circular cone whose height is 20cm and radius of circular base is 15 cm.= $\\dfrac{\\pi}{3}\\times 25^2*60$ = $12500\\pi$
\nHence, we can say that Quantity 1 < Quantity 2. Option C is the correct answer.<\/p>\n
\nArea of square with side equals to 20cm = 20^2 = 400 sq. cm
\nHence, we can say that Quantity 1 > Quantity 2. Option A is the correct answer.<\/p>\n
\n280 = 4*70
\n420 = 6*70<\/p>\n
\nTherefore, option B is the right answer.<\/p>\n
\nHCF of the 2 numbers = x.
\nLCM of the 2 numbers = 30x
\nDifference between the LCM – HCF = 30x – x = 29x
\n29x = 1740
\n=> x = 60
\nSum of the 2 numbers = 5x + 6x = 11x = 660.
\nTherefore, option D is the right answer.<\/p>\n
\nCompound interest paid to bank = $1000(1 + \\dfrac{20}{100})^2 – 1000$ = Rs. 440
\nHence, we can say that Quantity 1 = Quantity 2. Option E is the correct answer.<\/p>\n
\nSimple interest accrued in 2 years = $\\frac{P * 20 * 2}{100}$ = 0.4P
\nCompound interest accrued in 2 years = P *$(1 + \\frac{20}{100} )^{2}$ – P = 0.44P<\/p>\n
\n$\\Rightarrow$ 80 = 0.04 P
\n$\\Rightarrow$ P = 2000<\/p>\n
\nCompound interest accrued in 3 years at 10% per annum = 2000 *$(1 + \\frac{10}{100} )^{3}$ – 2000 = 662<\/p>\n
\nWe have 5a + 9m = 71 and 7a + 11m = 93.
\nSolving we get a = 7 and m = 4.
\nSo 3 bottles of apple juice and 4 bottles of mango juice will fill = 7*3 + 4*4 = 37 glasses
\nHence, option A is the right choice.<\/p>\n
\nWe know that,
\nx + y + z = 100 ———————–(1)
\n5x + 10y + 20z = 1500————–(2)
\n5x + 20y + 10z = 1100————-(3)<\/p>\n
\n5y + 15z = 1000——-(4)
\n15y + 5z = 600 —-(5)<\/p>\n
\nz = 60
\nTherefore, the number of 20 rupee notes is 60.
\nTherefore, option D is the right answer.<\/p>\n